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145,770 |
One of the major motivations of Homotopy Type Theory is that it naturally builds in higher coherences from the beginning. One important setting where higher coherence requirements get annoying is higher category theory. It's easy to talk about $\infty$-groupoids in HoTT, they're just types and you build them as higher inductive types. What about the next step? How do you talk about $(\infty,1)$-categories? I looked around at the nlab and the relevant blogs, but didn't find anything. The natural setup is that you have a type of objects and a (dependent) type of morphisms. But composition seems to run into all the usual difficulties of coherence in higher category theory. Does the HoTT point of view simplify things at all here? Feel free to assume that I'm familiar with the discussion of ordinary categories in the HoTT book and the background in the HoTT book. On the other hand, also assume that I find all definitions of higher categories beyond dimension 2 at least somewhat confusing. My motivation is that I'm trying to understand what you would need to do in order to give a formal proof of the cobordism hypothesis in dimension 1.
|
This is an important open problem. There are several imaginable approaches, including but not limited to: Mimic some commonly used homotopical definition of $(\infty,1)$-category inside HoTT. The most likely candidate seems to be complete Segal spaces, since they have a space of objects rather than a set of objects. This would require a definition of "simplicial type" in HoTT, which is another important open problem (which is motivating some people to try modifying type theory). Use a definition in a more type-theoretic style. The $\infty$-groupoids of HoTT are naturally "algebraic" a la Grothendieck/Batanin, so maybe it would be more natural to use a similarly algebraic definition of $(\infty,1)$-category. One could, for instance, try to encode an operad of a suitable sort with an inductive definition. Invent a sort of "directed type theory" whose basic objects are $(\infty,1)$-categories, in the same way that the basic objects of HoTT are $\infty$-groupoids. Leverage the fact that HoTT admits models not just in $\infty$-groupoids but in other $(\infty,1)$-toposes, noting that complete segal spaces live inside the $(\infty,1)$-topos of simplicial spaces. I proposed this here ; Andre Joyal independently had the same idea. At this point I wouldn't presume to bet on which approach will prove the best, or whether it will be something entirely different.
|
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|
145,898 |
We know the following facts: (1) For all $1\leq n\leq 2$ the equation $x_{1}^{n}+x_{2}^{n}=x_{3}^{n}$ has a solution in $\mathbb{N}$. (2) For all $3\leq n$ the equation $x_{1}^{n}+x_{2}^{n}=x_{3}^{n}$ has no solution in $\mathbb{N}$. Question: Is the following generalization true? For all $2\leq m$ both of the following statements are true: (1) For all $1\leq n\leq m$ the equation $x_{1}^{n}+...+x_{m}^{n}=x_{m+1}^{n}$ has a solution in $\mathbb{N}$. (2) For all $m+1\leq n$ the equation $x_{1}^{n}+...+x_{m}^{n}=x_{m+1}^{n}$ has no solution in $\mathbb{N}$.
|
No this is not true. For $n=4$ one has
$$2682440^4 + 15365639^4 + 18796760^4 = 20615673^4$$
found by Elkies (as part of an infinite family of solutions).
Also earlier it was known, for example, for $n=5$,
$$27^5 + 84^5 + 110^5 + 133^5 = 144^5$$
by Lander and Parkin. But part 2 was a conjecture of Euler so you are in good company, and $n \ge 6$ is still open. See that page for further details.
|
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146,463 |
I first saw the term "entropy" in a chemistry course while studying thermodynamics.
During my graduate studies I encountered the term in many different areas of mathematics.
Can anyone explain why this term is used and what it means. What I am looking for is a few examples where the term "entropy" is used to describe some mathematical object/quantity and its meaning there.
|
Here is a simple story one can tell about the entropy $$H = -\sum_{i=1}^n p_i \log p_i$$ of a discrete probability distribution. Suppose you wanted to describe how surprised you are upon learning that some event $E$ happened. Call your surprise upon learning that $E$ happened $s(E)$, the "surprisal." Here are some plausible conditions that $s$ could satisfy: $s(E)$ is a decreasing function of the probability $\mathbb{P}(E)$. That is, the less likely something it is to happen, the more surprising it is that it ends up happening, and the likelihood of something happening is the only thing determining how surprising it is. For example, flipping $10$ heads in a row is more surprising than flipping $5$ heads in a row. If $E_1$ and $E_2$ are independent, then $s(E_1 \cap E_2) = s(E_1) + s(E_2)$. That is, your surprise at learning that two independent events happened should be the sum of your surprises at learning that each individual event happened. For example, flipping $10$ coins heads in a row is twice as surprising as flipping $5$ coins heads in a row. Exercise: These conditions imply that $s$ must be a positive scalar multiple of $- \log \mathbb{P}(E)$. Then the expected surprisal is a positive scalar multiple of the entropy. Note in particular that $H$ is minimized if some $p_i = 1$, which corresponds to the same thing always happening and which is not surprising at all.
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146,485 |
Let $f:J\rightarrow \mathbb{R}$ be an absolutely continuous.
Under what kind of extra condition for $f'$, (not $C$) holds the following relation?
$$
\Big | \frac{1}{|I_{1}|}\int_{I_{1}}f'(x)dx- \frac{1}{|I_{2}|}\int_{I_{2}}f'(x)dx\Big|\overset{|J|\rightarrow 0}{\longrightarrow} 0,
$$
for any $I_{1}\cap I_{2}=\emptyset$ and $I_{1}, I_{2}\subset J$ intervals.
|
Here is a simple story one can tell about the entropy $$H = -\sum_{i=1}^n p_i \log p_i$$ of a discrete probability distribution. Suppose you wanted to describe how surprised you are upon learning that some event $E$ happened. Call your surprise upon learning that $E$ happened $s(E)$, the "surprisal." Here are some plausible conditions that $s$ could satisfy: $s(E)$ is a decreasing function of the probability $\mathbb{P}(E)$. That is, the less likely something it is to happen, the more surprising it is that it ends up happening, and the likelihood of something happening is the only thing determining how surprising it is. For example, flipping $10$ heads in a row is more surprising than flipping $5$ heads in a row. If $E_1$ and $E_2$ are independent, then $s(E_1 \cap E_2) = s(E_1) + s(E_2)$. That is, your surprise at learning that two independent events happened should be the sum of your surprises at learning that each individual event happened. For example, flipping $10$ coins heads in a row is twice as surprising as flipping $5$ coins heads in a row. Exercise: These conditions imply that $s$ must be a positive scalar multiple of $- \log \mathbb{P}(E)$. Then the expected surprisal is a positive scalar multiple of the entropy. Note in particular that $H$ is minimized if some $p_i = 1$, which corresponds to the same thing always happening and which is not surprising at all.
|
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147,395 |
I'm curious about just how far the abstraction to a symplectic formalism can be justified by appeal to actual physical examples. There's good motivation, for example, for working over an arbitrary cotangent bundle of configuration space -- because there are natural problems where the configuration space is not trivial. But what motivation is there, from a physical standpoint, for passing further still into the realm of non-exact symplectic manifolds or those that can't be realized as a cotangent bundle? By an exact symplectic manifold, I mean one where the symplectic form is exact, the differential of a 1-form. (I also don't understand whether those two ideas -- non-exact symplectic manifolds and symplectic manifolds that can't be realized as a cotangent bundle -- are equivalent. The question was asked here , but I don't see a simple "yes" or "no." That could be because the answer isn't known, or it could be because I don't have a formal enough definition for 'sympletic manfiold that can be realized as a cotangent bundle.') José Figueroa-O'Farrill gave a partial answer to this question in his answer here . He writes: Not every space of states is a cotangent bundle, of course. One can
obtain examples by hamiltonian reduction from cotangent bundles by
symmetries which are induced from diffeomorphisms of the configuration
space, for instance. Or you could consider systems whose physical
trajectories satisfy an ODE of order higher than 2, in which case the
cotangent bundle is not the space of states, since you need to know
more than just the position and the velocity at a point in order to
determine the physical trajectory. I don't know much about symplectic reduction, but I don't see it as a very natural example, since there must have been a more fundamental problem that didn't demand a general symplectic manifold instead of a cotangent bundle. The example about ODEs of order greater than 2 is interesting. But I'm wondering if anyone can offer a fuller explanation about the role general symplectic manifolds play in physics rather than math. I suspect part of the story will be about quantization. EDIT : I just saw this question on whether symplectic reduction can be considered "interesting from a physical point of view." Figured it was appropriate to link to here, although I'm still interested in any bigger-picture insights that don't have to do with reduction.
|
Actually the first case in history of a symplectic manifold wasn't a cotangent space. It was the space of Keplerian motions of a planet, represented locally by its Keplerian elements. The Lagrange symplectic structure on this space is defined by the so-called Lagrange parenthesis he introduced at this time (three papers in 1808/09/10)(*). That manifold is actually even non Hausdorff, but its greatest hausdorff quotient is a still a manifold (this is known as the "regularization theorem"). This manifold is symplectic but not a cotangent (but however contractible to the sphere $S^3$ ). Extended with the repulsive motions, it is an algebraic manifold defined by the following equations [Sou]. $$
\left\{
\begin{array}{rcl}
||{A}||^2 -fx^2 & = & 1 \\
y^2 - f ||{B}||^2 & = & 1
\end{array}
\right.
\quad \& \quad
\left\{
\begin{array}{rcl}
A \cdot B - xy &=& 0 \\
xy - f\tau &=& 0,
\end{array}
\right.
$$ where $A,B \in {\bf R}^3$ and $x,y,f,\tau$ belong to $\bf R$ . Actually this manifold is the result of the gluing of $TS^3$ and $TH^3$ along $TS^2\times {\bf R}$ (where $H^3$ is the 3 dimensional pseudo-sphere). I made the following picture, for $A$ and $B$ in $\bf R$ , to get a visual idea of the manifold. The bottom represents the $TS^3$ part, the top represents $TH^3$ and it is glued along two lines representing $TS^2\times {\bf R} \simeq S^2 \times {\bf R}^3$ . Remark . There exists also the examples of compact symplectic manifolds representing internal degrees of symmetries, as mentioned in Tobias answer. In the same spirit there is the Grassmannian manifolds ${\rm Gr}(2,n+1)$ of $2$ -planes in ${\bf R}^{n+1}$ , representing the space of un-parametrized geodesics on the sphere $S^n$ . We can regard this space of geodesics as the space of light rays on the Euclidean sphere where the speed of light would be infinite. ---------- Edit March 28, 2017 On a conceptual point of view, I just finished to write a paper: Universal Structure Of Symplectic Manifolds http://math.huji.ac.il/~piz/documents/ESMIACO.pdf , ---------- Edit November 15, 2019 http://math.huji.ac.il/~piz/documents/ESMIALCO.pdf , This paper has been enhanced to make the symplectic manifold an orbit of the linear coadjoint action of a central extension of the group of Hamiltonian diffeomorphisms, independently of the group of periods. That is, even if the symplectic form is not integral. that proposes a way, based on diffeology, to understand the statement: "Every symplectic manifold is a coadjoint orbit". ---------- Notes (*) I published this paper on the origins of symplectic geometry, but in french, where Lagrange's construction is explained. ---------- Reference [Sou] Jean-Marie Souriau. Géométrie globale du problème à deux corps . In Modern Developments in Analytical Mechanics, pp. 369-418. Accademia della Scienza di Torino, 1983.
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148,731 |
Let $f(n)$ denote the number of (isomorphism classes of) groups of order $n$. A couple easy facts: If $n$ is not squarefree, then there are multiple abelian groups of order $n$. If $n \geq 4$ is even, then the dihedral group of order $n$ is non-cyclic. Thus, if $f(n) = 1$, then $n$ is a squarefree odd number (assuming $n \geq 3$). But the converse is false, since $f(21) = 2$. Is there a good characterization of $n$ such that $f(n) = 1$? Also, what's the asymptotic density of $\{n: f(n) = 1\}$?
|
$f(n)=1$ if and only if $\gcd(n,\phi(n))=1$, where $\phi$ is the Euler phi-function. These $n$ are tabulated at http://oeis.org/A003277 The result is found in Tibor Szele, Über die endichen Ordnungszahlen, zu denen nur eine Gruppe gehört, Comment. Math. Helv. 20 (1947) 265–267, MR0021934 (9,131b).
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148,925 |
Recently I gave a lecture to master's students about some nice properties of the group with two elements $\mathbb{Z}/2\mathbb{Z}$. Typically, I wanted to present simple, natural situations where the only group satisfying the given constraints is $\mathbb{Z}/2\mathbb{Z}$ (also $\mathbb{Z}/2\mathbb{Z}$ as a ring or as a field could qualify, but I'd prefer to stick to the group if possible). Here are some examples of theorems that I proved to the students : Let $G$ be a nontrivial group with trivial automorphism group. Then $G$ is isomorphic to $\mathbb{Z}/2\mathbb{Z}$. Let $G$ be a nontrivial quotient of the symmetric group on $n>4$ letters (nontrivial meaning here different from 1 and the symmetric group itself). Then $G$ is isomorphic to $\mathbb{Z}/2\mathbb{Z}$. Let $k$ be an algebraically closed field and let $k_0$ be a subfield such that $k/k_0$ is finite. Then $k/k_0$ is Galois and $G=\text{Gal}(k/k_0)$ is isomorphic to $\mathbb{Z}/2\mathbb{Z}$. (Moreover $k$ has characteristic $0$ and $k=k_0(i)$ where $i^2=-1$.) This is a theorem of Emil Artin and I actually did not prove it because my students did not have enough background in field theory. Let $k$ be a field with the following property: there exists a $k$-vector space $E$ of finite dimension $n>1$ and an isomorphism $E\simeq E^*$ between the space and its linear dual which does not depend on the choice of a basis, i.e. is invariant under $\text{GL}(E)$. Then $k=\mathbb{Z}/2\mathbb{Z}$, $n=2$ and the isomorphism $E\simeq E^*$ corresponds to the nondegenerate bilinear form given by the determinant. I am looking for some more fantastic apparitions of $\mathbb{Z}/2\mathbb{Z}$. Do you know some?
|
A nice theorem is: $\{\pm 1\}$ is the only group that can act freely on a sphere of even dimension. In contrast: There are infinitely many groups acting freely on every odd-dimensional sphere.
|
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148,937 |
Let $C \subset {\bf CP}^2$ be an irreducible algebraic smooth (projectively) planar curve over the complex numbers of degree $d$ (we allow finitely many points to be deleted from $C$ to make it smooth). Then a generic complex line in ${\bf CP}^2$ intersects $C$ in $d$ distinct points $z_1,\ldots,z_d$. Question 1: for any permutation $\sigma: \{1,\ldots,d\} \to \{1,\ldots,d\}$, is it possible to continuously deform the $d$ points $z_1,\ldots,z_d$ to $z_{\sigma(1)},\ldots,z_{\sigma(d)}$ while keeping all the points collinear, distinct, and on $C$ at all times? In other words, does there exist continuous maps $z_i: [0,1] \to C$ with $z_i(0) = z_i$, $z_i(1) = z_{\sigma(i)}$, and the $z_1(t),\ldots,z_d(t)$ distinct and collinear for all $t\in [0,1]$? I believe the answer to this question is yes, because one can transpose any two of the $z_i,z_j$ while keeping the other $z_k$ unchanged by bringing them close together near a generic point of $C$ (and making the collinear line close to the line of tangency of this point to $C$, which generically will not be tangent to anywhere else in $C$), performing the transposition, and then returning back to the original position. One can also phrase the question as follows: if we let $S \subset C \times Gr(2,1)$ be the set of incidences $(p,\ell)$ with $p \in C$ and $\ell \in Gr(2,1)$ a line through $p$, then (after passing to an open dense subset of $Gr(2,1)$), $S$ is a covering space over (most of) $Gr(2,1)$ whose fibre has $d$ points, and the question asserts that the fundamental group of the base acts completely transitively on the fibre (in that every permutation of the fibre shows up); equivalently, the question asserts that the $d^{th}$ exterior power of $S$ over $Gr(2,1)$ is irreducible. However, I'm having more trouble with proving the following generalisation of Question 1. Keep the same setup as before, but now suppose we also have another irreducible curve $C'$ covering $C$ with fibres of cardinality $d'$. Thus, above a generic $z_i$ in $C$ we have $d'$ points $w_{i,1},\ldots,w_{i,d'}$ in $C'$. Question 2: for any permutation $\sigma: \{1,\ldots,d\} \to \{1,\ldots,d\}$, and any $j_1,\ldots,j_d,k_1,\ldots,k_d \in \{1,\ldots,d'\}$ is it possible to continuously deform the $d$ points $w_{1,j_1},\ldots,w_{d,j_d}$ to $w_{\sigma(1),k_1},\ldots,w_{\sigma(d),k_d}$ while keeping all the points on $C'$, and their projections onto $C$ collinear and distinct? Like Question 1, one can phrase this question in terms of the transitivity properties of the action of the fundamental group of a suitable open dense subset of $Gr(2,1)$ on some fibre, but this formulation becomes rather complicated to state and I won't give it here. I believe I can resolve this problem in the affirmative if $C$ has a smooth closure, in which case I can use ad hoc arguments to move around each of the $j_i$ independently, but encountered a problem if $C$ has a singular point in its closure, and $C'$ ramifies above this point, as then the points $z_i$ appear to become entangled with each other near this point in a manner that I was not able to analyse by ad hoc topological arguments. [My motivation for this problem was to understand arcs in a finite plane ${\bf F}_p^2$, that is to say sets that do not contain any collinear triples, and to classify when these arcs can be constructed as the ${\bf F}_p$-points of a low-degree algebraic curve $C$ (or as the projection of the ${\bf F}_p$-points of a low-degree algebraic curve $C'$). Using a Lefschetz principle argument, one can transfer the problem (in the large $p$ limit) to a problem in the complex plane related to the questions above.]
|
The answer to question 1 is yes, and is known as Harris's Uniform Position Lemma. It was proved in Harris's 1980 paper Galois groups of enumerative problems .
You can find a nice exposition in Chapter 9 of Solving Polynomial Equations , by Bronstein, Dickenstein and Emiris. You should be warned that the analogous statement is not true in characteristic $p$, which you say is your motivation. For example, consider the curve $y=x^p$ in characteristic $p$. Any line $y=mx+b$ meets this curve at $p$ points, whose $x$ coordinates are the roots of $x^p=mx+b$. Note that the roots of this polynomial form an arithmetic progression: If $r$ and $s$ are roots, so is $r+k(s-r)$ for every $k \in \mathbb{Z}/p$. That means that the monodromy, as you move the line, must preserve this structure, so it is a subgroup of $\mathbb{Z}/p^{\ast} \ltimes \mathbb{Z}/p$. UPDATE I think question 2 is false! Let $C$ be a smooth cubic curve in $\mathbb{P}^2$, and choose a flex on $C$ to be the origin, so $C$ has a group structure. Take $C'$ to be equal to $C$, but with the map $C' \to C$ being the doubling map in the group law. So $d'=4$. If $L$ is a line in $\mathbb{P}^2$, and $L \cap C = \{ x,y,z \}$, then $x+y+z=0$ in the group law. If $2 x'=x$, $2y' =y$ and $2 z'=z$, then $x'+y'+z'$ is a $2$-torsion point of $C'$. If we move $(L, x',y',z')$ continuously, we can't change which $2$-torsion point $x'+y'+z'$ is. So there are (at least) four separate connected components in the space of $(x',y',z')$ triples.
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149,018 |
This might be a vague question, but I am troubled by the fact that fields do not admit a nifty categorical definition. An obvious attempt such a definition would be to say that fields are commutative groups in $\mathcal{Ab}$, using that rings are monoids. This fails because the tensor product isn't the categorical product and thus doesn't have diagonals. Can this be fixed using a more general definition of a group object [ 1 ]
or can the definition of a field be relaxed in a sensible way? Is this issue with definability related to why the category of fields is badly behaved? (eg. no limits / colimits, lack of a "free field") [ 2 ]
|
Fields are the simple objects in $\text{CRing}$. Edit: Some philosophical remarks. Elements having inverses is a property and not a structure, so in some sense it's not obviously a good idea to treat the inverse as extra structure. Talking about group objects instead of just monoid objects can really only be done in cartesian monoidal categories; in general you instead want to talk about monoid objects with some extra property. For example, Poisson-Lie groups are not group objects in the category of Poisson manifolds (which is not cartesian monoidal). Similarly, Hopf algebras are not group objects in the category of coalgebras (which is again not cartesian monoidal). For commutative rings "$x$ is invertible" should be thought of as "the ideal generated by $x$ is the unit ideal," and of course this holds for all nonzero $x$ if and only if there are no nontrivial quotients. This suggests that if we want to generalize the definition of a field to other categories similar to $\text{CRing}$ then we should try to generalize this condition. Example. For graded rings the natural generalization is "the homogeneous ideal generated by $x$ is the unit ideal." This is true precisely for graded rings such that every nonzero homogeneous element is invertible, such as the ring of Laurent polynomials over a field.
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|
149,219 |
Is there any characterization of continuous functions $f : \Bbb{R}\longrightarrow \Bbb{R}$ such that for any linearly independent set $A$ (over the rationals) $f(A)$ is also linearly independent ?
|
This may not be a complete answer to the question, but it is too long for a comment. Given such a function $f$, the restriction $f \restriction [0,\infty)$ is a power function $\alpha x^\beta$ where $\alpha \ne 0$ and $\beta \ge 0$, and similarly the restriction $f \restriction (-\infty,0]$ is a power function of $-x$. We will prove this for the restriction $f \restriction [0,\infty)$ because the other case is similar. Because $\sqrt{2}$ is irrational, for any $x > 0$ the set $\{x,\sqrt{2} x\}$ is linearly independent, so the set $\{f(x),f(\sqrt{2} x)\}$ is linearly independent, meaning that its elements are nonzero and either Their quotient $f(\sqrt{2} x)/f(x)$ is irrational, or They are the same element: $f(\sqrt{2} x) = f(x)$, so the quotient $f(\sqrt{2} x)/f(x)$ is 1. In either case the quotient $f(\sqrt{2} x)/f(x)$ must take a constant value on this interval because it is a continuous function of $x$ whose range is contained in the totally disconnected set $(\mathbb{R} \setminus \mathbb{Q}) \cup \{1\}$. Therefore the function $x \mapsto f(2 x)/f(x)$ must also take a constant value $K$ on the interval $(0,\infty)$, namely the square of the previous constant. Observe that for any positive integer $n$ we have $f(2^{1/n}) = K^{1/n}f(1)$, and in fact for any positive rational number $q$ we have $f(2^q) = K^q f(1)$.
This means that $f$ is a multiple of a power function on the set $\{2^q : q \in \mathbb{Q}\}$. By continuity this holds on the closure of this set, namely the interval $[0,\infty)$. Note that the constant factor $\alpha$ cannot be zero, and that the exponent $\beta$ must be non-negative for $f$ to be continuous at zero. Discussion of special cases: Constant case: If the exponent on the interval $[0,\infty)$ is zero, then
the function is a nonzero constant $\alpha$ on this interval.
Therefore $f(0) \ne 0$, which means that on the interval $(-\infty,0]$ the exponent must also be zero and the function must have the same constant value $\alpha$ by continuity. So in this case $f$ is a nonzero constant function. Another case is that the exponent $\beta$ is equal to $1$ on both the intervals $(-\infty,0]$ and $[0,\infty)$, so it is piecewise linear.
In this case the only restriction on the slopes is that they must be nonzero rational multiples of each other. I don't know if this is the only other case.
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149,492 |
Suppose I have a smooth manifold with a tangent bundle, and I have a connection. If this connection is curvature-free, is it guaranteed to be torsion-free? (I am not assuming a metric, just a finite-dimensional smooth manifold.) I know that in general curvature-free connections do not exist, and that in general torsion-free connections do. But is the existence of a curvature-free connection sufficient to prove the existence of a torsion-free connection?
|
Many manifolds have curvature-free (i.e., flat) connections on their tangent bundles. For example, any orientable $3$-manifold $M$ is parallelizable, i.e., its tangent bundle is trivial, so it carries a flat connection (in fact, many flat connections). However, nearly all of these connections will have torsion. In fact, they all will unless $M$ is very special; essentially, $M$ has to be what is known as a flat affine manifold. I don't understand your last question. For example, the $3$-sphere has a flat connection (just regard it as a Lie group and take the connection that makes the left-invariant vector fields parallel). However, the $3$-sphere certainly does not admit a torsion-free flat connection, since it is compact and simply-connected.
|
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|
149,656 |
I'd like to understand the structure of the free loop space of $S^n$ for small values of $n$. Here "understand" means roughly that I'd like to know a CW complex with the same homotopy type. I already "understand" the pointed loop space of $S^n$ using the James reduced product. (See Hatcher, Algebraic Topology , Section 4.J.) Is there something analogous for the free loop space? Alternatively, is there some simple relationship between the free loop space and the pointed loop space?
|
Stably the free loop space of the suspension of a connected space splits up, just as the based loop space does. Just as $\Omega\Sigma X$ is stably the wedge of the smash products $X^{\wedge n}$, $L\Sigma X$ is stably the wedge of $S^1_+\wedge_{C_n}X^{\wedge n}$. Here $C_n$ is a cyclic group of order $n$ acting freely on $S^1$ and permuting the factors in the smash product. This is even correct $S^1$-equivariantly in a weak sense. EDIT Here is an answer more directly relevant to the question: Suppose $X\sim BG$. Even if the group $G$ is not discrete, we still have the equivalence that Qiaochu mentioned in a comment, between $LX$ and the homotopy orbit space for the conjugation action of $G$ on $G$. And this in turn is equivalent to the two-sided bar construction for $G$ acting on itself on both sides, i.e. the cyclic bar construction $N^{cyc}G$. And if we have another topological monoid $M$ equivalent to $G$, for example the Moore loops on $X$, or the James construction $JY$ if $X=\Sigma Y$, then $N^{cyc}M$ is another model for $LX$. Furthermore, there is a nice way of thinking of (the realization of the simplicial space) $N^{cyc}M$, which informally goes like this: A point is given by a finite subset $T$ of $S^1$ together with a labeling: a function $T\to M$. The set $T$ and the labels can move. If $T$ moves in such a way that several points come together, the label of the new point is the product of the old labels. If a label becomes $1\in M$ then the point may be deleted from $T$. This is even equivariantly correct regarding the $S^1$-action, in some sense (not good for fixed-point spaces of the whole group, but OK for finite subgroups). When $X=S^{n}$ and $M=JS^{n-1}$, this gives a pretty good equivariant cell structure for $LS^n$.
|
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149,702 |
I learned of the following example in a recent seminar: if $j(\tau)$ denotes the usual $j$-invariant , and $\alpha = (-1+i\sqrt{163})/2$, then
\begin{align*}
\frac{j(i)}{1728} &= 1 \\
\frac{-j(\alpha)}{1728} &= 151931373056000 = 2^{12}5^323^329^3 \\
\frac{j(i)-j(\alpha)}{1728} &= 151931373056001 = 3^37^211^219^2127^2163.
\end{align*}
A computation revealts that $(a,b,c) = (1, 151931373056000, 151931373056001)$ is a reasonably high-quality abc triple : if $R$ denotes the radical of $abc$ (the product of the distinct primes dividing $abc$), then
$$
q(a,b,c) = \frac{\log c}{\log R} = 1.20362\dots.
$$ My question is: is this a freak coincidence (including the fact that $163$ divides $c$), or is this example part of a larger theory? Are there families of high-quality abc triples that come from differences of $j$-invariants?
|
There is a beautiful theory by Gross and Zagier (On singular moduli, J. Reine Angew. Math. 355 (1985), 191–220) that explains completely the factorizations of
$$
j(\tau_1)-j(\tau_2)
$$
for $\tau_1,\tau_2$ lying in (possibly two different) imaginary quadratic fields. There are recent extensions by Kristin Lauter and Bianca Viray , and there they state the result. So these numbers are always highly factorizable, this is not at all an isolated phenomenon! And the appearance of 163 here is predictable as well. However, generally $j(\tau)$ would lie in some abelian extension of an imaginary quadratic field, this is part of the CM theory for elliptic curves. These are almost never rational numbers, in fact there are exactly 13 imaginary quadratic irrationalities $\tau$, including $i$ and $\alpha$ from your example, for which $j(\tau)$ is in ${\mathbb Q}$ - Noam Elkies has a complete list in his answer. So you won't be able to construct many ABC examples like that, unfortunately.
|
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|
149,737 |
Suppose $m$ is a positive integer. A quantity of interest is $$
H_m = \liminf_{n\to\infty} \left(p_{n+m} - p_n \right)
$$ The twin prime conjecture, is, of course $H_1 = 2$, the the prime k-tuples conjecture of Hardy and Littlewood asserts that $H_2 = 6$, $H_3 = 8$ and so on. Goldston-Pintz-Yildirim showed that under the Elliot-Halberstam conjecture, $H_1 < \infty$, a result that Yitang Zhang has famously recently established unconditionally. It is my understanding that the methods of GPY were essentially restricted when dealing with $m>1$. One of their non-trivial results was that $g_n = p_{n+1} - p_n$ satisfies $g_n = o(\log p_n)$, or in other words that $\liminf \frac{g_n}{\log p_n} = 0$, which obviously trivially follows from Zhang's result. However, it is my understanding that even under the full Elliot-Halberstam conjecture, GPY were unable to prove even their weaker result for $m\geq 2$. Maynard and Tao (see http://arxiv.org/abs/1311.4600 ) have shown that in fact, that $H_m < \infty$. Apparently, this work only required the classical Bombieri-Vinogradov theorem, not even the variation of that Zhang's proof uses. So my question is this: what shortcoming did the GPY approach have that did not allow it to extend to $m>2$, and how has Maynard's approach solved this problem? PS: I apologize if this question is somehow inappropriate for Mathoverflow. I am an undergraduate trying to read up on these recent results by myself as much as I can, and I figured this would probably be the best place to ask my question.
|
The major difference is the choice of the Selberg Sieve weights. I strongly recommend reading Maynard's paper . It is well written, and the core ideas are nicely explained. In what follows, I'll give a brief explanation of what Selberg Sieve weights are, why they appear, and what was chosen differently in Maynard's paper compared to that of Goldston Pintz and Yildirim. Let $\chi_{\mathcal{P}}(n)$ denote the indicator function for the primes. Given an admissible $k$-tuple $\mathcal{H}=\{h_1<h_2<\cdots<h_k\}$ the goal is to choose a non-negative function $a(n)$ so that $$\sum_{x<n\leq 2x} \left(\sum_{i=1}^k \chi_{\mathcal{P}}(n)(n+h_i)\right)a(n)\geq \rho \sum_{x<n\leq 2x} a(n)\ \ \ \ \ \ \ \ \ \ (1)$$ for some constant $\rho>1$. If this is the case, then by the non-negativity of $a(n)$, we have that for some $n$ between $x$ and $2x$ $$\left(\sum_{i=1}^k\chi_{\mathcal{P}}(n)(n+h_i)\right)\geq \lceil\rho \rceil,$$ and so there are at least $\lceil\rho \rceil$ primes in an interval of length at most $h_k-h_1$. This would imply Zhangs theorem on bounded gaps between primes, and if we can take $\rho$ arbitrarily large, it would yield the Maynard-Tao theorem. However, choosing a function $a(n)$ for which equation $(1)$ this holds is very difficult. Of course we would like to take $$a(n)=\begin{cases}
1 & \text{when each of }n+h_{i}\text{ are prime}\\
0 & \text{otherwise}
\end{cases} ,$$
as this maximizes the ratio of the two sides, but then we cannot evaluate $\sum_{x<n\leq 2x} a(n)$ as that is equivalent to understanding the original problem. Goldston Pintz and Yildirim use what is known as Selberg sieve weights . They chose $a(n)$ so that it is $1$ when each of $n+h_i$ are prime, and non-negative elsewhere, in the following way: Let $\lambda_1=1$ and $\lambda_d=0$ when $d$ is large, say $d>R$ where $R<x$ will be chosen to depend on $x$. Then set $$a(n)=\left(\sum_{d|(n+h_1)\cdots(n+h_k)} \lambda_d\right)^2.$$ By choosing $a(n)$ to be the square of the sum, it is automatically non-negative. Next, note that if each of $n+h_i$ is prime, then since they are all greater than $x$, which is greater than $R$, the sum over $\lambda_d$ will consist only of $\lambda_1$. This means that $a(n)=1$ when each of the $n+h_i$ are prime, and hence $a(n)$ satisfies both of the desired properties. Of course, we do not want $a(n)$ to be large when none, or even very few of the $n+h_i$ are prime, so we are left with the difficult optimization problem of choosing $\lambda_d$. The optimization in the classical application of the Selberg sieve to bound from above the number of prime-tuples involves diagonalizing a quadratic form, leading to $$\lambda_d = \mu(d) \left(\frac{\log(R/d)}{\log R}\right)^k,$$ and so
$$a(n)=\left(\sum_{\begin{array}{c}
d|(n+h_{1})\cdots(n+h_{k})\\
d<R
\end{array}}\mu(d)\left(\frac{\log(R/d)}{\log R}\right)^{k}\right)^{2}.$$ This choice is not so surprising since the related function $\sum_{d|n}\mu(d) \left(\log(n/d)\right)^k$ is supported on the set of integers with $k$ prime factors or less. Goldston Pintz and Yildirim significantly modified this, and used a higher dimensional sieve to obtain their results. The critical difference in the approach of Maynard and Tao is choosing weights of the form $$a(n)=\left(\sum_{d_{i}|(n+h_{i})\ \forall i}\lambda_{d_{1}\cdots d_{k}}\right)^{2}.$$ This gives extra flexibility since each $\lambda_{d_1,\dots,d_k}$ is allowed depend on the divisors individually. Using this additional flexibility Maynard and Tao independently established equation $(1)$ for any $\rho>1$, with $k$ depending on $\rho$. Here are four great resources where you can read more: Maynard's paper . Tao's blog posts . Granville's article on the Maynard-Tao theorem and the work of Zhang. Soundararajan's exposition of Goldston Pintz and Yildirim's argument.
|
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|
149,745 |
The following problem is related to (and motivated by) the first open case of this MO question . It is difficult to believe that this is a hard problem; and yet, I do not have a solution. For two vectors $x,y\in {\mathbb R}^4$, let us say that $x$ is dominated by $y$ if there are (at least) two coordinates in which $x$ is strictly smaller than $y$. With this convention, the problem goes as follows: Does there exist a finite, non-empty set $S\subset {\mathbb R}^4$ with the property that for any two vectors $x,y\in S$, there exists yet another vector $z\in S$ such that the coordinate-wise maximum of $x$ and $y$ is dominated by $z$? Notice that it is easy to find a set of vectors with the property in question in ${\mathbb R}^5$ (see comments below), or to find a finite, non-empty set of vectors in ${\mathbb R}^4$ such that any vector is dominated by another one. On the other hand, there is no finite, non-empty set in ${\mathbb R}^4$ in which for any two vectors there is a third one exceeding their maximum in at least three coordinates. Post-factum: make sure you have not missed the extremely clever solution by zeb! It is not easy to follow, but I made an effort to trace it carefully and, as far as I can see, it is perfectly correct. Already the first step is very unusual: instead of looking at the smallest counterexample, zeb considers a counterexample which is by no means smallest, but instead has a comprehensible structure. Whether you have voted for the problem itself or not, consider voting for zeb's solution!
|
There is no such set $S$. Suppose for a contradiction that there was. By rescaling the coordinates, we can assume all coefficients of points in $S$ are positive integers. Now construct a set $S'$ as follows: for every point $(a,b,c,d)\in S$, put points $(a,b,0,0), (a,0,c,0), (a,0,0,d), (0,b,c,0), (0,b,0,d), (0,0,c,d)$ into $S'$. $S'$ will then be a solution to the problem if $S$ was, so replace $S$ by $S'$. By adding finitely many additional points to $S$, we may further assume that if we decrease a coefficient of a point in $S$ by $1$, then the resulting point is still in $S$ as long as the coefficient was greater than $1$ to start with. $S$ now has the following properties: Property 0: Every point in $S$ has exactly two positive coefficients, and the maximal $M$ such that $(M,1,0,0)\in S$ is the same as the maximal $M'$ such that $(M',0,1,0)\in S$. Property 1: If $(a,b,0,0)\in S$ and $(0,0,c,d)\in S$, then at least one of $(a+1,0,c+1,0), (a+1,0,0,d+1), (0,b+1,c+1,0), (0,b+1,0,d+1)$ is in $S$. Proof: Let $A$ be maximal such that $(A,b,0,0)\in S$ and let $D$ be maximal such that $(0,0,c,D)\in S$. Some element of $S$ dominates $(A,b,c,D)$, and by the choice of $A$ (respectively $D$) it can't have its last two (respectively first two) coefficients both equal to $0$. Property 2: If $(x,y,0,0)\in S$, then at least one of $(x+1,0,0,2), (0,y+1,0,2)$ is in $S$. Proof: Let $X$ be maximal such that $(X,y,0,0)\in S$, and let $M$ be maximal such that $(0,0,M,1)\in S$. Then some element $(a,b,c,d)$ of $S$ dominates $(X,y,M,1)$. By Property 0 we have $c\le M$, so $c = 0$. Since we can't have $(X+1,y+1,0,0)\in S$, $d$ must not be $0$, so $d \ge 2$ and either $a \ge X+1 \ge x+1$ or $b \ge y+1$. Now consider the following property, depending on a parameter $k$: Property $k$: If $(x,y,0,0)\in S$, then at least one of $(x+1,0,0,k), (0,y+1,0,k)$ is in $S$. If $S$ has Property $k$ for every integer $k$, then clearly $S$ must be infinite. We'll now prove that in fact Property $k$ implies Property $k+1$ for $k \ge 2$, and this will give our desired contradiction. Property $k$ implies Property $k+1$: Choose $A,B,C$ maximal such that $(A,0,0,k), (0,B,0,k), (0,0,C,k) \in S$. To see that such $A,B,C$ exist at all, we apply Property $k$ to the point $(1,1,0,0)$ to see that either $(2,0,0,k)$ or $(0,2,0,k)$ is in $S$, and then we apply Property $0$ to see that all of $(1,0,0,k), (0,1,0,k), (0,0,1,k)$ are in $S$. Now we construct a sequence of points $s_i \in S$, each with fourth coordinate equal to zero, as follows. Start with $s_0 = (x,y,0,0)$. For each $i$, split into several cases: If $s_i = (0,u,v,0)$, apply Property 1 to $(0,u,v,0)$ and $(A,0,0,k)$, and call the resulting dominating point $s_{i+1}$. If the fourth coordinate of $s_{i+1}$ is nonzero, stop here. If $s_i = (u,0,v,0)$, apply Property 1 to $(u,0,v,0)$ and $(0,B,0,k)$. Proceed similarly to the above. If $s_i = (u,v,0,0)$, apply Property 1 to $(u,v,0,0)$ and $(0,0,C,k)$. Proceed similarly to the above. The claim is that this process must stop, and the final $s_{i+1}$ produced will be the point we are looking for. To see this, first note that we can never end up in the same case twice in a row during this process. Furthermore, we can never pass through all three cases in the course of this process: for instance, if we were to start in case 3 at step $i-2$, then go through case 2 at step $i-1$, and then end up in case 1 at step $i$, then if we write $s_i = (0,u,v,0)$ we find that $u = B+1$ and $v = C+2$. Applying Property $k$ to $s_i$, we see that one of $(0,B+2,0,k), (0,0,C+3,k)$ is in $S$, contradicting the choice of $A,B,C$. Thus, the process alternates between case 3 and some other case, say case 2 for concreteness. At each step, the first coordinate of $s_i$ will then increase, and inductively we see that for $i\ge 0$, the first coordinate of $s_i$ is $x+i$. Since it can't increase without bound, the process must eventually end. If it ends after step $0$, then the dominating vector $s_1$ is either $(x+1,0,0,k+1)$ or $(0,y+1,0,k+1)$. If it ends after step $i$ for $i\ge 1$, then the dominating vector $s_{i+1}$ must be $(x+i+1,0,0,k+1)$, since otherwise it would be either $(0,B+2,0,k+1)$ or $(0,0,C+2,k+1)$ depending on whether $i$ was even or odd, contradicting the choice of $A,B,C$.
|
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|
150,335 |
I am learning differential geometry and have a few questions on curvature. -- Background: Gauss invented "Gauss curvature" to measure how surface curves. Riemann gives an ingenious generalization of Gauss curvature from surface to higher
dimensional manifolds using the "Riemannian curvature tensor" (sectional curvature
is exactly the Gauss curvature of the image of the "sectional" tangent 2-dimensional
subspace under the exponential map). In modern textbooks on differential geometry, people usually first define the notion
of connection, and then the Riemannian curvature tensor is expressed in terms of
connection. My questions are: Are there some other notions of "curvature", besides Gauss curvature and
Riemannian curvature tensor as its generalization, which people have invented to
measure how space curves? In history, who first introduced the notion of connection to describe
the Riemannian curvature tensor, and why is this idea natural? Thank you very much!
|
in addition to these excellent examples of non-local curvature quantities and their extensions to the non-smooth setting (which I am not sure the OP was anticipating), I might add the 'original' non-local curvature measure: the holonomy. Also, the OP was not precise about how to interpret the vague term 'space', so it is not at all clear how to answer the question about possible measurements of how 'space curves'. From the context of the question and the given description of the OP's background, I would guess that 'space' is intended to be interpreted as 'Riemannian manifold', in which case the OP is asking some fundamental questions that often aren't explicitly addressed in introductory expositions of Riemannian geometry: First, is the 'Riemann curvature tensor' the whole story in terms of understanding the local geometry of Riemannian manifolds (i.e., in describing how such 'spaces curve')? Second, how did the notion of a 'connection' arise historically, since Riemann didn't use it to define his measure of how 'space curves', and what was the motivation for introducing it? Both these questions frequently occur to beginners in the subject, so it's a bit surprising that they aren't treated a bit more explicitly in introductory books. For example, consider the following situation: Suppose that, instead of Riemannian geometry on smooth manifolds, we were considering Hermitian geometry on complex manifolds, i.e., our manifolds are complex and we only consider Riemannian metrics that are complex compatible, in the sense that $g(iv) = g(v)$ for all tangent vectors $v$. In this case, the Riemannian curvature tensor of $g$ is not the whole story in (complex) dimensions bigger than $1$; in addition to the Riemann curvature tensor, one must consider the exterior derivative of the canonical $2$-form $\omega_g$ associated to $g$. It is a nontrivial theorem that, in the case of Riemannian metrics, all of the 'pointwise' differential invariants of a metric $g$' (once these are properly defined) are generated by the 'Riemann curvature tensor' and its 'covariant derivatives'. (Note that this is stronger than what Riemann stated, which is that, if the Riemann curvature tensor vanishes, then the metric is locally Euclidean. It could have been, for example, that, when the Riemann curvature tensor vanishes identically, this overdetermined system of equations forces some other 'independent' invariants to vanish as well, due to integrability conditions of the overdetermined PDE system. As Weyl and Cartan proved, though, this does not happen in the case of Riemannian geometry.) [In the Hermitian case above, it turns out that the Riemann curvature tensor and $d\omega_g$, together with their 'covariant derivatives' suffice to generate all of the 'pointwise differential invariants of a Hermitian metric $g$ on a complex manifold'. This is also a nontrivial theorem.] As far as how the notion of connection arose and why it is used so much in Riemannian geometry (and other geometries), that has an interesting history, and I will only sketch it here: In an 1869 paper, Elwin Christoffel showed that one could compute what we now call the Riemann curvature tensor of a metric $g$ as a differential expression in the coefficients of $g$ in two stages. First, one computes certain expressions, the 'Christoffel symbols' $\Gamma^k_{ij}$, in terms of the coefficients $g_{ij}$ of $g= g_{ij} dx^i dx^j$ and their first partial derivatives (with respect to the chosen local coordinate system), and then one computes the Riemann curvature tensor $R^i_{jkl}$ from the $\Gamma^i_{jk}$ and their first partial derivatives (with respect to the chosen local coordinate system). This gave an efficient method of computing the Riemann curvature tensor and exhibited it explicitly as a second order partial differential expression in the coefficients of $g$ in a somewhat manageable form. Then, around 1890, Gregorio Ricci-Curbastro (for whom the Ricci tensor is named), realized that Christoffel's symbols could be used to define a notion of derivative for what we now call tensor fields in the presence of a background Riemannian metric $g$, one that would be independent of any choice of local coordinates, i.e., this derivative would be 'covariant', when one expresses both the tensor field and the metric in local coordinates. He and his former student, Tullio Levi-Civita, used this as the basis of their notion of 'absolute differential calculus'. This led to the idea of tensor fields 'parallel' (i.e., with vanishing absolute derivative) along curves and the notion of 'parallel transport', which gave a way of 'connecting' the tangent spaces along any (piecewise smooth) curve. This original notion of 'connection' (though I am not sure that Levi-Civita actually used this word as a noun) was then vastly generalized and applied to other geometries by Weyl, Schouten, and Cartan. The general idea of curvature (and, indeed, holonomy) as a measure of how parallel transport depends on the choice of curve joining two points then became a fundamental notion and, in fact, it turns out that the Riemann curvature tensor of a metric $g$ is precisely the curvature (in this sense of dependence of parallel transport on the path) of the 'connection' introduced by Levi-Civita. [From this point of view, Weyl's theorem that the Riemann curvature tensor and its covariant derivatives give all of the 'pointwise differential invariants' of $g$ is by no means obvious, and it takes some serious work (with the appropriate definitions) to prove it.]
|
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|
150,478 |
We know the followings :
$$\int_{0}^{\infty}\frac{{\sin}x}{x}dx=\int_{0}^{\infty}\frac{{\sin}^2x}{x^2}dx=\frac{\pi}{2},\int_{0}^{\infty}\frac{{\sin}^3x}{x^3}dx=\frac{3\pi}{8}.$$
Also, we can get
$$\int_{0}^{\infty}\frac{{\sin}^3x}{x^2}dx=\frac{3\log 3}{4},\int_{0}^{\infty}\frac{{\sin}^4x}{x^3}dx=\log 2.$$
Then, I got interested in their generalization. Question : Letting $p,q\in\mathbb N$, can we simplify the following?
$$\int_{0}^{\infty}\frac{{\sin}^px}{x^q}dx$$ I don't have any good idea. Could you show me how to simplify this? Remark : This question has been asked previously on math.SE without receiving any answers.
|
I'm posting an answer just to inform that the question has received an answer by Nick Strehlke on MSE . \begin{align*}
\int_0^\infty {\sin^p x\over x^q}\,dx & = \left\{\begin{array}{ll}
\displaystyle{(-1)^{(p+q)/2}\pi\over 2^{p+1}(q-1)!}\sum_{k = 0}^p(-1)^k{p\choose k} |p - 2k|^{q-1} & \text{$p,q$ even,} \\[2em]
\displaystyle {(-1)^{(p+q)/2-1}\pi\over 2^{p+1}(q-1)!}\sum_{k = 0}^p (-1)^k{p\choose k} \operatorname{sign}(p-2k) |p-2k|^{q-1} & \text{$p,q$ odd,} \\[2em]
\displaystyle {(-1)^{(p+q+1)/2} \over 2^p (q-1)!} \sum_{k = 0\atop k\not = p/2}^p (-1)^k {p\choose k} |p-2k|^{q-1}\log{|p - 2k|} & \text{$p$ even, $q$ odd,} \\[2em]
\displaystyle {(-1)^{(p+q-1)/2} \over 2^p (q-1)!} \sum_{k = 0\atop k\not = (p\pm1)/2}^p (-1)^k {p\choose k} \operatorname{sign}(p-2k) |p-2k|^{q-1}\log{|p - 2k|} & \text{$p$ odd, $q$ even,}
\end{array}\right.
\end{align*} By the way, I noticed that these formulas can be simplified a bit as the followings which might be easier to calculate :
$$\frac{(−1)^{(p+q+1)/2}}{2^{ p−q} (q−1)! } \sum_{k=0}^{ (p−4)/2} (−1)^k\binom pk\left(\frac p2−k\right)^{ q−1} \log\left(\frac p2−k\right) $$
for $p$ even and $q$ odd such that $3\le q\le p−1$ .
$$\frac{(−1)^{(p+q-1)/2}}{2^{ p−1} (q−1)! } \sum_{k=0}^{ (p−3)/2} (−1)^k\binom pk\left(p−2k\right)^{ q−1} \log\left(p−2k\right) $$
for $p$ odd and $q$ even such that $2\le q\le p−1$ .
|
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150,586 |
Let $f\in\mathbb{Z}[X]$ be an irreducible polynomial. Is there an integer $a\neq 0$ such that $f(X)+a$ is also irreducible in $\mathbb{Z}[X]$? Can this be also extended to $\mathbb{Q}[X]$?
|
Yes, and you don't need $f$ irreducible. The following irreducibility criterion suffices and shows that infinitely many $a$ work. Lemma: Let $g(x) = a_n x^n + ... + a_0 \in \mathbb{Z}[x]$ be such that $a_0$ is prime and $$|a_0| > |a_1| + ... + |a_n|.$$ Then $g(x)$ is irreducible. Proof. The condition on the coefficients ensures that all complex roots of $g$ have absolute value greater than $1$. But if $g$ is reducible then at least one of its irreducible factors has constant term $\pm 1$, hence at least one of its roots has absolute value less than or equal to $1$; contradiction. $\Box$
|
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|
150,835 |
My question is whether there are no nontrivial solutions in the ordinals of the equations arising in Fermat's last theorem $$x^n+y^n=z^n$$
where $n\gt 2$, and where we use the natural ordinal arithmetic , which is commutative. (Note: If we had used the usual ordinal arithmetic, there are some easy counterexamples, such as $1^3+\omega^3=\omega^3$.) The question spins off of my answer to Saint Georg's recent question, The Theory of Transfinite Diophantine Equations . Feldmann Denis pointed out in the comments there that for very small ordinals (below $\omega^\omega$), the question reduces to the corresponding question in polynomials, where it has an affirmative answer. Can we extend this observation to work for all ordinals?
|
There are no nontrivial solutions. This follows from Wiles’s proof, and the following observation. Proposition: If a set of Diophantine equations has a solution in (positive) ordinals using natural sum and product, then it has a solution in (positive) natural numbers. Proof: Every ordinal can be uniquely written in Cantor normal form
$$\tag{$*$}\alpha=\sum_{i<k}\omega^{\alpha_i}n_i,$$
where $\alpha_i>\alpha_{i+1}$ and $0<n_i<\omega$ for all $i<k$. Define a function $f\colon\mathrm{Ord}\to\omega$ by $f(\alpha)=\sum_jn_j$. Notice that $f(\alpha)=0$ only if $\alpha=0$, and $f(n)=n$ for all $n<\omega$. The proposition follows from Claim: $f$ is a homomorphism with respect to natural sum and product. This in turn follows from expression of the operations in terms of Cantor normal form. For sum, let $\alpha$ and
$$\beta=\sum_i\omega^{\alpha_i}m_i$$
be as in $(*)$, except that we allow $n_i$ and $m_i$ to be zero. Then their natural sum is
$$\def\ns{\mathbin\#}\alpha\ns\beta=\sum_i\omega^{\alpha_i}(n_i+m_i),$$
hence
$$f(\alpha\ns\beta)=\sum_i(n_i+m_i)=f(\alpha)+f(\beta).$$
As for product, the expression in $(*)$ is valid even if we interpret the sum and product there as the natural operations, and natural product is associative and distributive over natural sum, hence it suffices to consider only the case of $\alpha=\omega^{\alpha_0}$ and $\beta=\omega^{\beta_0}$. However, then their natural product is $\gamma=\omega^{\alpha_0\ns\beta_0}$, and $f(\gamma)=1=f(\alpha)f(\beta)$.
|
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|
150,949 |
This question concerns a statement in a short paper by S. P. Wang titled “A note on free subgroups in linear groups" from 1981. The main result of this paper is the following theorem. Theorem (Wang, 1981): For every field $k$ of characteristic 0 and subgroup $\Gamma$ of $GL(n, k)$, the group $\Gamma$ either has a nonabelian free subgroup or possesses a normal solvable subgroup of index < $\lambda(n)$. To prove this theorem, Wang needs the following statement: “It is easily shown that in any Lie group with finitely many connected components, there is a finite subgroup which meets every component.” Please give me a proof of this statement.
|
An immediate consequence of Theorem 3.1(ii) of Ch. XV of Hochschild's book "The structure of Lie groups" is that in such a Lie group, maximal compact subgroups meet every connected component (and are all conjugate to each other by part (iii)). So your question thereby reduces to the case of compact Lie groups. Consider a compact Lie group $K$, and let $T$ be a maximal torus in $K$. I claim that the normalizer $N_K(T)$ (whose identity component is $T$) maps onto $\pi_0(K)$, so we could then replace $K$ with $N_K(T)$ to reduce to the case when $K^0$ is a torus. To prove the claim, for any $\gamma \in \pi_0(K)$ and $k \in K$ lifting $\gamma$, the conjugate $kTk^{-1}$ is a maximal torus in $K^0$, so it is $K^0$-conjugate to $T$. That enables us to change $k$ by left multiplication against an element of $K^0$ so that $kTk^{-1}=T$, thereby verifying that $N_K(T) \rightarrow \pi_0(K)$ is surjective. Now we may assume $K^0$ is a torus $T$, so we seek to show that if $G$ is an extension of a finite group $\Gamma$ by a torus $T$ then $G$ contains a finite subgroup mapping onto $\Gamma$. Since $T$ is commutative, it is a $\Gamma$-module via $G$-conjugation on $T$ (since $G/T = \Gamma$). This enables us to define ${\rm{H}}^2(\Gamma, T)$, and the isomorphism class of $G$ as an extension of $\Gamma$ by $T$ is classified by a class in this cohomology group (with $T$ encoding the analytic structure on $G$). Since this cohomology group is killed by the size $n$ of $\Gamma$, we see that ${\rm{H}}^2(\Gamma, T[n]) \rightarrow {\rm{H}}^2(\Gamma, T)$ is surjective (consider the $\Gamma$-cohomology sequence attached to the exact sequence $1 \rightarrow T[n] \rightarrow T \stackrel{n}{\rightarrow} T \rightarrow 1$). Hence, we get a finite group $E$ that is an extension of $\Gamma$ by $T[n]$ such that its pushout along the inclusion $T[n] \rightarrow T$ is $G$. That identifies $E$ as a finite subgroup of $G$ mapping onto $\Gamma = \pi_0(G)$. QED EDIT : I should also note that the analogue for linear algebraic groups is much more elementary to prove insofar as it avoids the hard work on maximal compact subgroups in the presence of disconnectedness, and perhaps this suffices for the motivating application (which was not explained much in the question posted). Namely, if $G$ is a linear algebraic group over a field $F$ of characteristic 0 then we claim there is a finite $F$-subgroup in $G$ that meets all connected components of $G_{\overline{F}}$. Indeed, if $T$ is a maximal $F$-torus in $G$ then $N_G(T)$ maps onto the finite etale $F$-group $G/G^0$ by a calculation with $\overline{F}$-points exactly as in the compact case above, so we reduce to the case when $G^0$ is a torus. If $F$ is algebraically closed then we can carry out the degree-2 group cohomology argument as in the compact case above to conclude. For general $F$ we can proceed similarly by working with the abelian category of $\Gamma$-modules for the etale topology on the category of finite type $F$-schemes, with $\Gamma$ a finite $F$-group, and considering the sheafified "standard cochain" complex for such $\Gamma$.
|
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|
150,956 |
I am a graduate student. Occasionally for some reason I am asked to give a talk on my research at a conference whose stated purpose is almost completely unrelated to my research. To preserve my anonymity, I won't say what I actually do, but as an example: Imagine I do research in some particular aspect of exterior differential systems, and I am asked to speak at a special session on some particular aspect of numerical analysis. In the past when I have done this, it has not been a pleasant experience. I sit through a bunch of talks I don't understand, try to socialize with a group of people who already know each other and who I'll likely never see again, and then give my talk, which I feel is seen as irrelevant. But I am still encouraged (by my advisor and others) to give such talks, because it is good for my career. In particular, I've been told that it's good to advertise my work to a broad audience, and that people in my audience may someday referee my papers. This seems very unlikely to me, but I am uncertain. Is giving such a talk really a worthwhile endeavor?
|
Rather than an unqualified "go anyway" or "stay home", I would suggest exploring why you are getting invited to these conferences. Unless the conference is a total sham (such things do exist; does it have a hefty registration fee?), the organizers are noted experts in (following your example) numerical analysis, and they have some reason to think your talk would be interesting to the audience. If it isn't clear to you why that is, you should try to find out. Perhaps there are connections between numerical analysis and exterior differential systems that you're not aware of. Talk to your advisor, and also try to find a numerical analyst to talk to (your advisor may know someone if you don't). Ask them, "What aspects of my work would be most interesting to numerical analysts?" If you can learn about what these connections are, you can use them as a guide for what you should highlight in your talk. You could also ask the organizers who invited you. "Thanks for the invitation. I've recently been working on topic X, do you think that would fit well with the theme of your session?" They might offer a suggestion as to aspects that might interest their intended audience (especially the other invitees), and you can plan your talk accordingly. Or they can say "We were actually hoping you'd speak on your recent work on Y", which gives you another hint. Of course, there's the remote possibility that you've been invited by mistake. Do you have a similar name to a famous numerical analyst? If this should be the case, contacting the organizers as above will help them figure it out and gently fix the error without embarrassing anyone. (If they say "We were hoping you would speak on Y", and Y is something you've never heard of, then this would suggest a mistake has been made.) Like the other answers, I don't think you should stay away just because you're not 100% comfortable; one can often get a lot out of conferences that seemed unlikely. But if you really have absolutely no idea why you're there, and therefore your talk is completely irrelevant, I agree with you that it doesn't seem like it will give you positive exposure. So my answer is neither "accept regardless" or "stay away", but "find out why you're there" and "give a more relevant talk"!
|
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|
151,159 |
It is a long-standing conjecture (probably just as old as the twin prime conjecture, which has gotten a lot of attention as of late since Zhang and Maynard's breakthrough results this year) that there exist infinitely many primes of the form $x^2 + 1$. The first few primes of this shape are $5, 17, 37, 101, \cdots$. Of course, no one has been able to prove this theorem, but perhaps if one relaxed the problem one would be able to obtain positive results. So my question is: what is the least value of $n$ such that one can prove that infinitely many elements of $P_n$ (where $P_n$ is the set of numbers with at most $n$ prime divisors) are of the form $x^2 + 1$? A similar result to this is a result of Selberg who proved that the polynomial $x(x+2)$ assumes values which have at most 7 prime factors, in particular there exist infinitely many elements of $P_7$ which are of the shape $x(x+2)$. The parity problem in sieve theory would likely prevent any progress on obtain a lower bound for primes instead of almost primes. Any insight would be appreciated.
|
Yes, Iwaniec proved in "Almost primes represented by quadratic polynomials"
(Invent. math. 47(1978) 171–188), that if $f$ is a quadratic polynomial with $f(0)$ an odd integer, then $f$ contains infinitely many elements of $P_2$. For an arbitrary polynomial $f$ that is irreducible and doesn't have a fixed prime divisor, one can say that $f$ represents infinitely many elements of $P_n$, where $n=1+\deg f$. This was proven by Buhstab in "A combinatorial strengthening of the Eratosthenian sieve method", Usp. Mat. Nauk 22, no. 3, 199–226.
|
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|
151,488 |
Obviously, this question is not a research level mathematics question at all. But, I've just met an extremely mathematically talented $11$ years old student and I don't know how I can help him. For years I was working at a special school for young gifted and talented students. But, I had never met such talent at such a young age in my life. I talked to a mathematician friend of mine (who himself was an IMO gold medalist at a very young age), but he had no idea unfortunately. Basically, there is no one around with any idea. That is why I came to MO. Just in case that this question gets closed please e-mail me at asghari.amir at gmail if you have any idea.
|
Hmm. I was that kid. I was/am very strong in math. I was in a very strong gifted program from gradeschool to high school. I went through various Academic Olympiads. I have a trophy where I was the second best HS kid in Chicago one year... Why second? Because I didn't try. I didn't like trying. I flunked several classes in HS and college. I have severe anxiety at times, though getting better - a very supportive wife makes me wonder what I could have done if I dealt with anxiety sooner. My parents both had anxiety and depression. There were other home issues that made me more anxious and made "success" not that important. I read everything I could find, but the things I was supposed to read. No risk of judgement there. My point being, all the books you could give me wouldn't have helped. My issues were mental, and some of safety. If you want this kid (or any kid for that matter) to reach their full potential, issues of mental state, safety, even food and shelter may be more important than any books you could give them. I'm not saying this person has my rollup of stuff, just to be aware that there is a lot more out there than access to contests. With the Internet, access to learning materials is 1000x easier than when I was that age, so your help will be even more needed on the mental side. One big thing I forgot to add is - struggle, and how to deal with "failure". Though learning how to push through is hard for everyone, a smart kid has a lot of expectations on them. Help them deal with not getting it right the first time. The confusion of "this is normally so easy". Help them with patience to work it through, to not get frustrated or quit because something is hard for once. That's probably the lesson I most wish I learned earlier.
|
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|
151,706 |
Consider the integer lattice points in the positive quadrant $Q$ of $\mathbb{Z}^2$.
Say that a point $(x,y)$ of $Q$ is visible from the origin if the
segment from $(0,0)$ to $(x,y) \in Q$ passes through no other point of $Q$.
So points block visibility, and the only points visible from the origin are those with
a clear line of sight: Let $\nu(n)$ be the ratio of the number of points visible within the
square with corner $(n,n)$ to $n^2$. For example, for $n=8$, $43$ of the $64$ points
are visible, so $\nu(n) = 43/64 \approx 0.67$. Q : What is $\lim_{n \to \infty} \nu(n)$? It appears to be approaching $\approx 0.614$: The question bears some similarity to Polya's Orchard Problem
(T.T. Allen, "Polya's orchard problem," The American Mathematical Monthly 93(2): 98-104 (1986). Jstor link ), but I cannot see my question
is answered in that literature. One could ask the same question for $\mathbb{Z}^d$. Answered by Pete Clark: the limit is $\frac{1}{\zeta(d)}$.
Thus in higher
dimensions, almost all the lattice points are visible (because $\zeta(d)$ approaches $1$).
|
The limiting ratio is $\frac{1}{\zeta(2)} = \frac{6}{\pi^2}= 0.6079271018540266286632767792\ldots$ The question is easily seen to be equivalent to "What is the probability that two integers are relatively prime?" This old chestnut of elementary number theory has been addressed before on this site: see here . The idea is incredibly simple and appealing: we consider the events " $a$ and $b$ are both divisible by $p$ " -- as $p$ runs over all primes -- as independent and use the Euler product for $\zeta(2)$ . It is not completely obvious how to make this reasoning rigorous, but it can be and has been done in any number of ways. The linked to answer contains one of them. Since the OP is highly involved in discrete geometry I wanted to mention a more general result: if $n \geq 2$ and $\Omega \subset \mathbb{R}^n$ is a bounded, Jordan measurable region, then the number of primitive (= visible from the origin) lattice points in the dilate $r\Omega$ of $\Omega$ is asymptotic to $\left(\frac{\operatorname{Vol} \Omega}{\zeta(n)} \right) r^n$ . For $n =2$ this is proved for instance in $\S$ 24.10 (of the sixth edition, but I don't think that matters) of Hardy and Wright's An Introduction to the Theory of Numbers . It is stated in $\S$ 7.3 of these notes and placed there in a more general context: namely it is related to the celebrated Minkowski-Hlawka Theorem . (The proof was left to two students in the graduate seminar I was then running. They presented a proof, but I didn't get around to incorporating their arguments into the writeup. Anyway the transition from the two-dimensional case is straightforward.)
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|
151,773 |
Sometimes (often?) a structure depending on several parameters turns out to be symmetric w.r.t. interchanging two of the parameters, even though the definition gives a priori no clue of that symmetry. As an example, I'm thinking of the Littlewood–Richardson coefficients: If defined by the skew Schur function $s_{\lambda/\mu}=\sum_\nu c^\lambda_{\mu\nu}s_\nu$, where the sum is over all partitions $\nu$ such that $|\mu|+|\nu|=|\lambda|$ and $s_{\lambda/\mu}$ itself is defined e.g. by $
s_{\lambda/\mu}= \det(h _{\lambda_i-\mu_j-i+j}) _{1\le i,j\le n}$, it is not at all straightforward to see from that definition that $c^\lambda_{\mu\nu} =c^\lambda_{\nu\mu} $. Granted that this way of looking at it may seem a bit artificial, as I guess that in many of such cases, it is possible to come up with a "higher level" definition that shows the symmetry right away (e.g. in the above example, the usual (?) definition of $c_{\lambda\mu}^\nu$ via $s_\lambda s_\mu =\sum c_{\lambda\mu}^\nu s_\nu$), but showing the equivalence of both definitions may be more or less involved. So I am aware that it might just be a matter of "choosing the right definition". Therefore, maybe it would be better to think of the question as asking especially for cases where historically, the symmetry of a certain structure has been only stated 'later', after defining or obtaining it in a different way first. Another example that would fit here: the Perfect graph theorem , featuring a 'conceptual' symmetry between a graph and its complement. What are other examples of "unexpected" or at least surprising symmetries? (NB. The 'combinatorics' tag seemed the most obvious to me, but I won't be surprised if there are upcoming examples far away from combinatorics.)
|
If $a$ and $b$ are positive integers, and you make the definition
$$ a \cdot b = \underbrace{a + \cdots + a}_{b \text{ times} }$$
then it's a slightly surprising fact that $a \cdot b$ is actually equal to $b \cdot a$.
|
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|
151,916 |
Imagine you have a finite number of "sites" $S$ in the positive quadrant
of the integer lattice $\mathbb{Z}^2$,
and from each site $s \in S$, one connects $s$ to every lattice point to which it
has a clear line of sight, in the sense used
in my earlier question : No other lattice point lies along that line-of-sight.
This creates a (highly) nonplanar graph;
here $S=\{(0,0),(5,2),(3,7),(11,6)\}$: Now, for every pair of edges that properly cross in this graph, delete the longer edge,
retaining the shorter edge.
In the case of ties, give preference to the earlier site, in an initial sorting of
the sites.
The result is a planar graph, because all edge crossings have been removed: Q1 . Is this graph $4$-colorable? Some nodes of this graph (at least those on the convex hull)
have a (countably) infinite degree. More generally, Q2 . Is every infinite planar graph $4$-colorable?
Which types of "infinite planar graphs" are $4$-colorable? The context here is that I am considering a type of "lattice visibility Voronoi diagram."
One can ask many specific questions of this structure, but I'll confine myself
to the $4$-coloring question, which may have broader interest.
|
The answer to both questions is "yes", by the De Bruijn–Erdős theorem .
|
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152,352 |
Apparently Euclid died about 2,300 years ago (actually 2,288 to be more precise), but the title of the question refers to the rallying cry of Dieudonné, "A bas Euclide! Mort aux triangles!" (see King of Infinite Space: Donald Coxeter, the Man Who Saved Geometry by Siobhan Roberts, p. 157), often associated in the popular mind with Bourbaki's general stance on rigorous, formalized mathematics (eschewing pictorial representations, etc.). See Dieudonné's address at the Royaumont seminar for his own articulated stance. In brief, the suggestion was to replace Euclidean Geometry (EG) in the secondary school curriculum with more modern mathematical areas, as for example Set Theory, Abstract Algebra and (soft) Analysis. These ideas were influential, and Euclidean Geometry was gradually demoted in French secondary school education. Not totally abolished though: it is still a part of the syllabus, but without the difficult and interesting proofs and the axiomatic foundation. Analogous demotion/abolition of EG took place in most European countries during the 70s and 80s, especially in the Western European ones. (An exception is Russia!) And together with EG there was a gradual disappearance of mathematical proofs from the high school syllabus, in most European countries; the trouble being (as I understand it) that most of the proofs and notions of modern mathematical areas which replaced EG either required maturity or were not sufficiently interesting to students, and gradually most of such proofs were abandoned. About ten years later, there were general calls that geometry return, as the introduction of the alternative mathematical areas did not produce the desired results. Thus EG came back, but not in its original form. I teach in a University (not a high school), and we keep introducing new introductory courses, for math majors, as our new students do not know what a proof is. [Cf. the rise of university courses in the US that come under the heading "Introduction to Mathematical Proofs" and the like.] I am interested in hearing arguments both for and against the return of EG to high school curricula. Some related questions: is it necessary for high-school students to be exposed to proofs? If so, is there a more efficient mathematical subject in comparison to EG, for high school students, in order to learn what is a theorem, an axiom and a proof? Full disclosure : currently I am leading a campaign for the return of EG to the syllabus of the high schools of my country (Cyprus). However, I am genuinely interested in hearing arguments both pro and con.
|
When I was in high school (in the early 1960's), Euclidean geometry was the only course in the standard curriculum that required us to write proofs. These proofs, however, were in a very rigid format, with statements on the left side of the page and a reason for every statement on the right side. So I fear that many students got an inaccurate idea of what proofs are really like. They also got the idea that proofs are only for geometry; subsequent courses (in the regular curriculum, not honors courses) didn't involve proofs. The textbook that we used also had some defects concerning proofs. For example, Theorem 1 was word-for-word identical with Postulate 19; Theorem 1 was given a proof that didn't involve Postulate 19, so, in effect, we were shown that Postulate 19 is redundant, but the redundancy was never mentioned, and I still don't know why a redundant postulate was included in the first place. Another defect of the standard courses in geometry was that, because of the need to gently teach how to find and write proofs (in that rigid format), very little interesting geometry was taught; the class was mostly proving trivialities. I was fortunate to be in an honors class, with an excellent instructor who showed us some really interesting things (like the theorems of Ceva and Menelaus), but most students at my school had no such advantage. I conjecture that Euclidean geometry can be used for a good introduction to mathematical proof, but, as the preceding paragraph shows, there are many things that can go wrong. (There are other things that can go wrong too. I mentioned that I had an excellent teacher. But my school also had math teachers who knew very little about proofs or about geometry beyond what was in the textbook.) So my advice is, if you want to develop a course such as you described in the question, proceed, but be very careful. Incidentally, many years ago, I recommended to my university department that we use a course on projective geometry as an "introduction to proof" course. The idea was that there are fairly easy proofs, and the results are not as obvious, intuitively, as equally easy results of Euclidean geometry. My suggestion was not adopted. Qiaochu Yuan's suggestion of discrete math instead of geometry might have similar advantages as my projective geometry proposal, but it will still be subject to many of the pitfalls that I indicated above, plus one more: Most high school math teachers know less about discrete math than they do about geometry.
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152,353 |
I am trying to prove the following claim (may be it has been proven). Claim: Consider a set of points $\phi=\{x_1,x_2,...,x_i,...\}$ generated by a homogeneous PPP with rate $\lambda$ in the 2-D plane $\mathbb R^2$ . Then we generate the Voronoi cells with the $k$ nearest points ( $k$ order Voronoi cell WiKi , Demo ). Can we claim that the expected sum area of typical cells, in which a point $x_i\in \phi$ , takes part is the same for any point $\{x_1,x_2,...,x_i,...\}$ ? PS: Defining expectation for a point seems tricky to me. Because the points will change with each trial. Please help to formulate the expected number of cells a point $x_i$ takes part in, as a integral. I have a similar question on the number of cells but I posted it separately in order not to put too much in one question https://math.stackexchange.com/questions/612562/poisson-point-process-ppp-and-voronoi-cells Suggestions or a references are most welcome, thanks.
|
When I was in high school (in the early 1960's), Euclidean geometry was the only course in the standard curriculum that required us to write proofs. These proofs, however, were in a very rigid format, with statements on the left side of the page and a reason for every statement on the right side. So I fear that many students got an inaccurate idea of what proofs are really like. They also got the idea that proofs are only for geometry; subsequent courses (in the regular curriculum, not honors courses) didn't involve proofs. The textbook that we used also had some defects concerning proofs. For example, Theorem 1 was word-for-word identical with Postulate 19; Theorem 1 was given a proof that didn't involve Postulate 19, so, in effect, we were shown that Postulate 19 is redundant, but the redundancy was never mentioned, and I still don't know why a redundant postulate was included in the first place. Another defect of the standard courses in geometry was that, because of the need to gently teach how to find and write proofs (in that rigid format), very little interesting geometry was taught; the class was mostly proving trivialities. I was fortunate to be in an honors class, with an excellent instructor who showed us some really interesting things (like the theorems of Ceva and Menelaus), but most students at my school had no such advantage. I conjecture that Euclidean geometry can be used for a good introduction to mathematical proof, but, as the preceding paragraph shows, there are many things that can go wrong. (There are other things that can go wrong too. I mentioned that I had an excellent teacher. But my school also had math teachers who knew very little about proofs or about geometry beyond what was in the textbook.) So my advice is, if you want to develop a course such as you described in the question, proceed, but be very careful. Incidentally, many years ago, I recommended to my university department that we use a course on projective geometry as an "introduction to proof" course. The idea was that there are fairly easy proofs, and the results are not as obvious, intuitively, as equally easy results of Euclidean geometry. My suggestion was not adopted. Qiaochu Yuan's suggestion of discrete math instead of geometry might have similar advantages as my projective geometry proposal, but it will still be subject to many of the pitfalls that I indicated above, plus one more: Most high school math teachers know less about discrete math than they do about geometry.
|
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|
152,405 |
This question complement a previous MO question: Examples of theorems with proofs that have dramatically improved over time . I am looking for a list of Major theorems in mathematics whose proofs are very hard but was not dramatically improved over the years. (So a new dramatically simpler proof may represent a much hoped for breakthrough.) Cases where the original proof was very hard, dramatical improvments were found, but the proof remained very hard may also be included. To limit the scope of the question: a) Let us consider only theorems proved at least 25 years ago. (So if you have a good example from 1995 you may add a comment but for an answer wait please to 2020.) b) Major results only. c) Results with very hard proofs. As usual, one example (or a few related ones) per post. A similar question was asked before Still Difficult After All These Years . (That question referred to 100-years old or so theorems.) Answers (Updated Oct 3 '15) 1) Uniformization theorem for Riemann surfaces ( Koebe and Poincare, 19th century) 2) Thue-Siegel-Roth theorem (Thue 1909; Siegel 1921; Roth 1955) 3) Feit-Thompson theorem (1963); 4) Kolmogorov-Arnold-Moser (or KAM) theorem (1954, Kolgomorov; 1962, Moser; 1963) 5) The construction of the $\Phi^4_3$ quantum field theory model.
This was done in the early seventies by Glimm, Jaffe, Feldman, Osterwalder, Magnen and Seneor. (NEW) 6) Weil conjectures (1974, Deligne) 7) The four color theorem (1976, Appel and Haken); 8) The decomposition theorem for intersection homology (1982, Beilinson-Bernstein-Deligne-Gabber); ( Update: A new simpler proof by de Cataldo and Migliorini is now available) 9) Poincare conjecture for dimension four , 1982 Freedman (NEW) 10) The Smale conjecture 1983, Hatcher; 11) The classification of finite simple groups (1983, with some completions later) 12) The graph-minor theorem (1984, Robertson and Seymour) 13) Gross-Zagier formula (1986) 14) Restricted Burnside conjecture , Zelmanov, 1990. (NEW) 15) The Benedicks-Carleson theorem (1991) 16) Sphere packing problem in R 3 , a.k.a. the Kepler Conjecture (1999, Hales) For the following answers some issues were raised in the comments. The Selberg Trace Formula- general case (Reference to a 1983 proof by Hejhal) Oppenheim conjecture (1986, Margulis) Quillen equivalence (late 60s) Carleson's theorem (1966) (Major simplification: 2000 proof by Lacey and Thiele.) Szemerédi’s theorem (1974) (Major simplifications: ergodic theoretic proofs; modern proofs based on hypergraph regularity, and polymath1 proof for density Hales Jewett.) Additional answer: Answer about fully formalized proofs for 4CT and FT theorem .
|
The Four Colour Theorem might perhaps be a canonical example of a very hard proof of a major result which has improved, but is still very hard- no human-comprehensible proof exists, as far as I know, and all known proofs require computer computations.
|
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152,823 |
My question is the following: Let $f\in C^\infty(a,b)$, such that $f^{(n)}(x)\ne 0$, for every $n\in\mathbb N$, and every $x\in (a,b)$. Does that imply that $f$ is real analytic? EDIT. According to a theorem of Serge N. Bernstein ( Sur les fonctions absolument monotones , Acta Mathematica, 52 (1928) pp. 1–66) if $f\in C^\infty(a,b)$ and $f^{(n)}(x)\ge 0$, for all $x\in(a,b)$, then $f$ extends analytically in a ball centered at $(a,0)\in\mathbb C$ and radius $b-a$!
|
If $f$ is $C^{\infty}$ every derivative is continuous, so the hypothesis on $f$ implies that each derivative $f^{(n)}$ has constant sign. Such functions were studied by S. Bernstein and called regularly monotonic . In particular he proved in 1926 that a regularly monotonic function is real analytic. This 1971 AMM article by R.P. Boas provides a proof, more history, and further results along these lines. See also this 1975 PAMS article of J. McHugh .
|
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|
153,069 |
The article Friedrich Wille: Galerkins Lösungsnäherungen bei monotonen Abbildungen ,
Math. Z. 127 (1972), no. 1, 10-16 is written in the form of a lengthy poem, in a style similar to that
of the works of Wilhelm Busch . Are there any other examples of original mathematical research published in a similar form?
|
A famous example is Tartaglia's solution of the equation of degree 3, which he gave to Cardano (after much discussion) in the following form : Quando chel cubo con le cose appresso se agguaglia qualche numero
discreto trovan dui altri
differenti in esso. Dapoi terrai questo per consueto Che'l lor produtto sempre sia eguale Al terzo cubo delle cose neto, El residuo poi suo generale Delli lor lati cubi ben sottratti Varra la tua cosa principale...
|
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|
153,092 |
The undecidability of the halting problem states that there is no general procedure for deciding whether an arbitrary sufficiently complex computer program will halt or not. Are there some large $n$ and interesting computer languages (say C++, Scheme) for which we have a constructive halting algorithms for programs of length up to $n$? Note that for any fixed bound, there is some finite lists of zero's and one's that stores whether all the programs less than that bound halt or not, so non-constructively for any $n$ there is always a decision procedure for the bounded problem. But I would like to know whether we have some $n$ for which we can constructively write a decision procedure for programs of length less than that bound. By stipulating large $n$, I'd like to avoid $n$ and languages which are trivial in some sense, for instance, decision procedure for languages where the first number of programs all halt and we can easily see that they halt. Note that the usual proof of the undecidability of the halting program prima facia does not apply, since the decision procedure for programs bounded by $n$ might have length much larger than $n$, so our decision procedure need not correctly apply to our diagonlization program. What kind of work has been done on this? Are there any good references out there? Thank you!
|
As you noticed in your question, for any particular value of $n$, there is a constructive algorithm that solves the halting problem for instances of size at most $n$. Since for a particular value of $n$, there are only finitely many instances, one may simply hard-code the finitely many answers into the program itself. So in this sense, yes, for any particular $n$, we have a constructive solution of the halting problem for instances of size at most $n$. Meanwhile, there can be no uniform method of describing these algorithms, since from any uniform method of describing the programs, we could extract a computable solution of the halting problem. If there were a uniform solution, then given any instance of the halting problem, we could produce the program that is suitable for that instance, and then run it, thereby solving the halting problem. So there can be no general method to describe these particular instances. This issue of uniformity was also an issue in some other MO questions, particularly Gordon Royle's question Can a problem be simultaneously polynomial time and undecidable? . But worse, what I claim is that for any particular background theory, such as PA or ETCS or ZFC or ZFC plus large cardinals, there is a particular value of $n$ (that we can compute from the theory), such that there is a program of size at most $n$ such that the theory simply does not settle the question of whether it halts or not. This $n$ is simply the size of the program that searches for a proof of a contradiction in the background theory, halting only when one is found. We can easily design such a program for any particular theory, such as ZFC or whatever (c.e.) theory you like (as long as it was consistent), and experts could compute the particular value of $n$, although this particular value would depend on the details of the computability formalism. For this $n$, there can be no constructive algorithm that you can prove in your background theory is a computable solution of the halting problem, since if such an algorithm truly and provably solved the halting problem for the program that searched for a contradiction, then it would also
settle the consistency question of the background theory, which is impossible. So the conclusion is that there is a comparatively small value of $n$ for which no specific constructive algorithm can provably and truly settle the halting problem for instances of size at most $n$. This can be taken as a negative answer to your question, since there can be no truly large values of $n$ for which we can provably settle the halting problem by a specific algorithm.
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153,272 |
Let $T_0$ be $\mathsf{ZFC}$ and, for $n\in\omega$ , set $T_{n +1}=T_{n}+\mathrm{Con}(T_{n})$ . Question 1: Is there a natural number $n$ such that $T_{n}$ is equiconsistent with $\mathsf{ZFC}+$ some large cardinal axiom? Question 2: Is the consistency strength of the $T_{n}$ bounded by some large cardinal axiom? More precisely, is there a large cardinal axiom $A$ such that $$\mathrm{Con}(\mathsf{ZFC}+A)\Longrightarrow \forall n\,\mathrm{Con}(T_{n})?$$ Continue the iteration transfinitely, so for each ordinal $\alpha$ , $T_{\alpha +1}=T_{\alpha}+\mathrm{Con}(T_{\alpha})$ and, if $\alpha$ is limit, then $T_{\alpha}=\bigcup_{\beta\in \alpha} T_{\beta}$ . Finally, let $T_{\infty}=\bigcup_\alpha T_{\alpha}$ . Question 3: For which large cardinal axiom $A$ and ordinal $\alpha$ we have that $T_{\alpha}$ is equiconsistent with $\mathsf{ZFC}+ A$ ? Question 4: Is the consistency strength of all $T_{\alpha}$ s bounded by some large cardinal axiom? Precisely, is there a large cardinal axiom $A$ such that $$\mathrm{Con}(\mathsf{ZFC}+A)\Longrightarrow \forall \alpha\,\mathrm{Con}(T_{\alpha})?$$ In the other words what is the consistency strength of $T_{\infty}$ in the large cardinal hierarchy?
|
The procedure you suggest really cannot get too far. Here is an abstract result explaining what I mean (see Aspects of incompleteness by Per Lindström): Given an r.e. sequence of r.e. theories that interpret arithmetic and whose union is consistent, there is a $\Pi^0_1$ formula that is not provable by any of them. That a sequence is r.e. means here that we have a way of listing all pairs $(\phi,n)$ with $\phi$ an axiom of the $n$-th theory. (The connection comes by realizing that $\mathsf{Con}(T)$ is $\Pi^0_1$ for r.e. $T$.) Arguing specifically within your setting: There is a natural way of iterating the consistency operator, to define a sequence $T_\alpha$, $\alpha<\omega_1^{CK}$. Although each $T_\alpha$ is strictly stronger than its predecessors, their union is still pretty weak: The theory $U_1=\mathsf{ZFC}+$"$\mathsf{ZFC}$ has an $\omega$-model" is stronger than all of them. This theory can also be iterated $\omega_1^{CK}$ times ($U_2$ would be $U_1+$"$U_1$ has an $\omega$-model", etc). All these theories are strictly weaker than $S_1=\mathsf{ZFC}+$"$\mathsf{ZFC}$ has a transitive set model", which, again, we can iterate $\omega_1^{CK}$ times. All these theories are strictly weaker than asserting that some $V_\alpha$ is a model of $\mathsf{ZFC}$ ($\alpha$ is what Joel Hamkins calls a worldly cardinal ). Worldliness can also be iterated, and the resulting theories are strictly weaker than asserting that there is a weakly inaccessible cardinal. (Stopping at $\omega_1^{CK}$ is just an artifact of my wanting to keep all the theories recursive. I don't quite see how to make sense of iterating these theories beyond the recursive ordinals. What do we mean by $\mathsf{Con}(T)$ in such a case, since $T$ is no longer r.e.? Of course, we can make sense of this semantically, and just require the existence of models (after some pains formalizing this, as the relevant language in which the theories are formulated would evolve with the ordinal), but even then it seems to me we need to argue that the models are sufficiently correct to see the relevant ordinals, and this seems a distraction from the main goal.) Above, I mentioned that we iterate the consistency operator "naturally". By contrast, Solomon Feferman and others have investigated ways of iterating "consistency" operators, so that the claims above fail, with the "$\omega$-th step" being considerably stronger than just $\mathsf{ZFC}+\mathrm{Con}(\mathsf{ZFC})+\mathrm{Con}(\mathrm{Con}(\mathsf{ZFC}))+\dots$ There are also other alternatives, where we "add" to $\mathsf{ZFC}$ all true $\Pi^0_1$ statements, then to this "add" all true $\Pi^0_2$-statements, etc. This is typically formalized via reflection sequences , see here .
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153,277 |
There seems to be a lot of recent activity concerning topological modular forms (TMF), which I gather is an extraordinary cohomology theory constructed from the classical theory of modular forms on the moduli space of elliptic curves. I gather that the homotopy theorists regard it as a major achievement. My question is as follows : what sorts of more classical topological problems is it useful for solving? It is clear why non-algebraic-topologists should care about the more classical extraordinary cohomology theories (e.g. K-theory and cobordism theory); they allow elegant solutions to problems that are obviously of classical interest. Does TMF also allow solutions to such problems, or is it only of technical interest within algebraic topology? My background : my research interests are in geometry and topology, but I am not an algebraic topologist (though I have used quite a bit of algebraic topology in my research).
|
One of the closer connections to geometric topology is likely from invariants of manifolds. The motivating reason for the development of topological modular forms was the Witten genus . The original version of the Witten genus associates power series invariants in $\mathbb{C}[[q]]$ to oriented manifolds, and it was argued that what it calculates on M is an $S^1$-equivariant index of a Dirac operator on the free loop space $Map(S^1,M)$. It is also an elliptic genus, which Ochanine describes much better than I could here . This is supposed to have especially interesting behavior on certain manifolds. An orientation of a manifold is a lift of the structure of its tangent bundle from the orthogonal group $O(n)$ to the special orthogonal group $SO(n)$, which can be regarded as choosing data that exhibits triviality of the first Stiefel-Whitney class $w_1(M)$. A Spin manifold has its structure group further lifted to $Spin(n)$, trivializing $w_2(M)$. For Spin manifolds, the first Pontrjagin class $p_1(M)$ is canonically twice another class, which we sometimes call "$p_1(M)/2$"; a String manifold has a lift to the String group trivializing this class. Just as the $\hat A$-genus is supposed to take integer values on manifolds with a spin structure, it was argued by Witten that the Witten genus of a String manifold should take values in a certain subring: namely, power series in $\Bbb{Z}[[q]]$ which are modular forms. This is a very particular subring $MF_*$ isomorphic to $\Bbb{Z}[c_4,c_6,\Delta]/(c_4^3 - c_6^2 - 1728\Delta)$. The development of the universal elliptic cohomology theory ${\cal Ell}$, its refinement at the primes $2$ and $3$ to topoogical modular forms $tmf$, and the so-called sigma orientation were initiated by the desire to prove these results. They produced a factorization of the Witten genus $MString_* \to \Bbb{C}[[q]]$ as follows:
$$
MString_* \to \pi_* tmf \to MF_* \subset \Bbb{C}[[q]]
$$
Moreover, the map $\pi_* tmf \to MF_*$ can be viewed as an edge morphism in a spectral sequence. There are also multiplicative structures in this story: the genus $MString_* \to \pi_* tmf$ preserves something a little stronger than the multiplicative structure, such as certain secondary products of String manifolds and geometric "power" constructions. What does this refinement give us, purely from the point of view of manifold invariants? The map $\pi_* tmf \to MF_*$ is a rational isomorphism, but not a surjection. As a result, there are certain values that the Witten genus does not take, just as the $\hat A$--genus of a Spin manifold of dimension congruent to 4 mod 8 must be an even integer (which implies Rokhlin's theorem). Some examples: $c_6$ is not in the image but $2c_6$ is, which forces the Witten genus of 12-dimensional String manifolds to have even integers in their power series expansion; similarly $\Delta$ is not in the image, but $24\Delta$ and $\Delta^{24}$ both are. (The full image takes more work to describe.) The map $\pi_* tmf \to MF_*$ is also not an injection; there are many torsion classes and classes in odd degrees which are annihilated. These actually provide bordism invariants of String manifolds that aren't actually detected by the Witten genus, but are morally connected in some sense because they can be described cohomologically via universal congruences of elliptic genera. For example, the framed manifolds $S^1$ and $S^3$ are detected, and Mike Hopkins' ICM address that Drew linked to describes how a really surprising range of framed manifolds is detected perfectly by $\pi_* tmf$. These results could be regarded as "the next version" of the same story for the relationship between the $\hat A$-genus and the Atiyah-Bott-Shapiro orientation for Spin manifolds. They suggest further stages. And the existence, the tools for construction, and the perspective they bring into the subject have been highly influential within homotopy theory, for entirely different reasons. Hope this provides at least a little motivation.
|
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153,302 |
What stories, puzzles, games, paradoxes, toys, etc from everyday life are better understood after learning homology theory? To be more precise, I am teaching a short course on homology, from chapter two of Hatcher's book. Before diving into the details of Delta-complexes, good pairs, long exact sequences, degrees, and so on, I would like to present a collection of real-life phenomena that are greatly illuminated by actually knowing homology theory. Ideally, I would refer back to these examples as the course progressed and explain them with the new tools the students learn. Here is what I have thought of so far. Tavern puzzles: before trying to solve a tavern puzzle, it is always a good idea to check that the two pieces are topologically unlinked. You can approximate this by computing the linking number, via the degree of the map from the torus to the two-sphere. Cowlicks: after getting a very short haircut, there is a place on your scalp (typically near the crown) where the hair is standing up. This is "explained" by the hairy ball theorem: a tangent vector field on the two-sphere must vanish somewhere. "Explained" is in quotes because your hair is a vector field on a disk, not on a sphere. When you knead dough, and then push it back into its original shape, there is always a bit of dough that has returned to its original position. This is the Brouwer fixed-point theorem. This example is a bit tricky, because it is impossible to see the fixed bit of dough. (Now, when you are hiking, you can clearly see that there is a point of the map lying directly above the point it represents. However, this is better explained by the contraction mapping principle.) There are no draws in the board game Hex. This is equivalent to the Brouwer fixed-point theorem. This isn't a perfect example, because most people don't know the game. Noticeably missing are puzzles, etc that rely on homological algebra (diagram chasing, long exact sequences, etc). $\newcommand{\SO}{\operatorname{SO}} \newcommand{\ZZ}{\mathbb{Z}}$
EDIT: Let me also give some non-examples, to clarify what I am asking. Linking number also arises in discussions of DNA replication; see discussions of topoisomerase. However DNA is not an everyday object, so doesn't work as an example. The plate (or belt) trick; this is a fancy move that a waiter can make with your plate, but it is more likely to appear in a juggling show. It is explained by knowing $\pi_1(\SO(3)) = \ZZ/2\ZZ$. However, this is a fact about the fundamental group, not about homology. Impossible objects such as the Penrose tribar that exist locally, but not globally. These can be explained via non-trivial cohomology classes. But, cohomology is in the next class, so if/when I teach that...
|
Here is an example of economic interpretation of cohomology from some Russian popular lecture that I read (unfortunately, I do not remember the source; I only remember it was in Russian). Consider a geographic map with some countries in it. By some
stretch of imagination, assume that the countries make an open covering:-)
Now each country has a currency, and on the "common boundary" (the intersection of two open countries), an exchange rate between the two currencies exists. The question is whether a universal
currency can be established, so that each country's currency can have an exchange rate with respect to this universal currency, so that the existing exchange rates between pairs of currencies are compatible with it. EDIT. I found a book which has this example (this is not the Russian book I mentioned in the first paragraph, but another book, which is full of such examples): Robert Ghrist, Elementary applied topology . This example in in Chapter 6, section 6.
The book is freely readable online.
|
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153,484 |
Let $M$ be a compact manifold and $\varphi : M \longrightarrow M$ be a diffeomorphism which is isotopic to the identity. Does there exist a vector field $ X $ on $M$ such that $\varphi$ is the flow at time $1$ of $X$? If that is not always the case, where does the obstruction for such a $\varphi$ to be a flow lives in ?
|
The answer is no. In fact, there are diffeomorphisms arbitrarily close to the identity which are not contained in flows (which are also often called $1$-parameter subgroups; here one is thinking of the set of vector fields on $M$ as a sort of "Lie algebra" for the diffeomorphism group). Here's an example which I learned about from Warning 1.6 of Milnor's paper "Remarks on infinite-dimensional Lie groups" (which I highly recommend reading if you are interested in things like this). Regard $S^1$ as $\mathbb{R}/2\pi$. Fix some integer $n$ and some $\epsilon$ such that $0<\epsilon<1/n$. Define
$$f : S^1 \longrightarrow S^1$$
$$f(x) = x + \pi/n + \epsilon \sin^2(n x).$$
Choosing $n$ large enough and $\epsilon$ small enough, we can make $f$ arbitrarly close to the identity in the $C^{\infty}$
topology. It is a fun exercise to show that $f$ does not lie in any $1$-parameter subgroup of $\text{Diff}(S^1)$. It is not hard to generalize this kind of behavior to any manifold. I don't think there is any easy description of the obstruction here; it seems to be a delicate problem in dynamics. However, it is the case that every element of $\text{Diff}^0(M)$ can be written as a product of finitely many elements that are contained in $1$-parameter subgroups. Indeed, the set of elements of $\text{Diff}^0(M)$ that are contained in $1$-parameter subgroups generate a subgroup $\Gamma$ that is easily seen to be normal. But Thurston proved that $\text{Diff}^0(M)$ is simple for any compact manifold $M$, so we must have $\Gamma = \text{Diff}^0(M)$. For a discussion and proof of Thurston's theorem, I recommend Banyaga's book "The Structure of Classical Diffeomorphism Groups".
|
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|
153,542 |
For $f(x)=x$ , the half-derivative of $f$ is $$\frac{d^{\frac{1}{2}}}{dx^{\frac{1}{2}}} x = 2 \sqrt{\frac{x}{\pi}} \;.$$ Is there some geometric interpretation of ( Q1 ) this specific derivative, and, ( Q2 ) of the half-derivative more generally? I have read that fractional derivatives are nonlocal,
but it seems strange to me that integral derivatives
can be described in terms of local
geometry only, while fractional derivatives cannot be so described.
This would suggest an odd discontinuity between, say, $d^{1}$ and $d^{1.01}$ .
This seems especially at odds with the many applications of fractional derivatives ,
which (superficially) suggests continuity should reign. I'd appreciate someone clearing up my elementary confusions—Thanks in advance! Addendum ( 5Jan14 ).
@AlexR. found this geometric interpretation of
the fractional integral in Richard Herrmann's book, Fractional Calculus: An Introduction for Physicists ,World Scientific, 2011:
|
A possible mechanical interpretation of the half-derivative can be given in terms of Abel's solution to a classical problem from the calculus of variations ( the tautochrone problem ). Let there be a heavy particle which is constrained to slide without friction along
the curve $y=y(t)$ in uniform gravity to its
lowest point. Then, given a function $T(y)$ that specifies the total
time of descent for a given starting height what is an equation of the
curve that yields this result? The principle of conservation of energy implies that the distance $S=S(t)$ travelled by the particle along the curve from the initial height $y_0$ satisfies the equation $$\left(\frac{dS}{dt}\right)^2=2g(y_0-y).$$ This is equivalent to the integral equation $$T(y_0)=\frac{1}{\sqrt{2g}}\int_0^{y_0}\frac{1}{(y_0-y)^{1/2}}\frac{dS}{dy}dy.$$ The r.h.s. of the latter equation is nothing else but the Riemann–Liouville fractional integral of $f=\pi^{1/2}(2g)^{-1/2}dS/dy$ , i.e. $$D^{-\alpha}f(x)=\frac{1}{\Gamma(\alpha)}\int_0^x (x-y)^{\alpha-1}f(y)dy$$ of the order $\alpha=1/2$ . The solution to Abel's integral equation $dS/dy$ can be now interpreted (up to a constant factor) as the half-derivative of $T=T(y_0)$ .
|
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|
153,558 |
This 1895 painting of Nikolai Bogdanov-Belsky shows mental calculations in the public school of Sergei Rachinsky. Boys in a Russian village school try to calculate $(10^2+11^2+12^2+13^2+14^2)/365$ in their heads. One of the methods of solution is based on the equality $10^2+11^2+12^2=13^2+14^2$. Now this Rachinsky equality can be considered as a generalization of the well-known Pythagorean triple (3,4,5), $3^2+4^2=5^2$, and in analogy with the Pythagorean triples one can define Rachinsky quintets as a set of five positive integers $(a,b,c,d,e)$ such that $a^2+b^2+c^2=d^2+e^2$. It is known that all primitive Pythagorean triples $(a,b,c)$ such that $a^2+b^2=c^2$ are generated by Euclid's formula $a=m^2-n^2$, $b=2mn$, $c=m^2+n^2$, where $m$ and $n$ are positive integers such that $m>n$, $m$ and $n$ are coprime, and $m \not\equiv n \bmod 2$. Can one establish an analogous result for Rachinsky quintets?
|
The equation $a^2+b^2+c^2=d^2+e^2$ defines a quadric $Q\subset\mathbb{P}^4$, with a rational point $p=(1,0,0,0,1)$. Therefore it is rational : projecting from $p$, say on the hyperplane $e=0$, defines a birational map $Q --> \mathbb{P}^3$. The inverse of that map, namely
$$ (x,y,z,t)\mapsto (x-\lambda ,y,z,t,\lambda )\quad \mbox{with }\lambda :=(x^2+y^2+z^2-t^2)/2x$$
gives a parametrization of all rational points in $Q$ with $x\neq 0$; to get integral points just multiply all coordinates by $2x$. To get the remaining points replace $p$ by $p'=(0,1,0,0,1)$, etc.
|
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|
153,604 |
I have read (but I cannot now find where) that V. I. Arnold & J.-P. Serre had a public debate on the value of Bourbaki. Does anyone have more details, or remember or know what was said?
|
I was there. Arnol'd is one of my big mathematical heros, but I found the whole thing really sad. It was in French, but my French is decent. Arnol'd began his part with a phrase I've heard him say before: “In Russia it is impolite to talk ill of the dead, so I will not talk about Bourbaki” and then he proceeded to lambast Bourbaki in as nasty a way as you've ever read in any of his writings.
I just felt like hanging my head. It was embarrassing watching him insult French Mathematics in front of 500 French men and saying things that seemed silly. It went on from there, Serre with a kind of sad understated dignity, not fighting, Arnol'd wanting a fight, hurling insults. The two barely even addressed each other.
And yes, he did mention Toth, and if memory serves, he stated that Toth was probably Thales and had most likely come up with Newton's inverse square law. For me, the whole event was sad, embarrassing, and myth-crushing. Well, us mathematicians, we are all humans.
|
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153,740 |
Why we need to study representations of matrix groups? For example, the group $\operatorname{SL}_2(\mathbb F_q)$ , where $\mathbb F_q$ is the field with $q$ elements, is studied by Drinfeld. I think that these groups are already given by matrices. The representation theory is to represent elements in an algebra or group (or other algebraic structure) by matrices. Why we still need to study representations of matrix groups? Thank you very much.
|
I will answer a more general question. If $X$ is an important mathematical object, why should we study the representation theory of $Aut(X)$? Maybe algebraic geometers care about $X$, or maybe mathematical physicists, or maybe homotopy theorists, etc. Each discipline has techniques that can build a new object $X'$ out of $X$. These new objects are usually more inspiring (if less fundamental) than $X$ itself. Maybe $X$ is projective $n$-space and algebraic geometers take $X'$ to be a Hilbert scheme. Or maybe $X$ is a manifold and mathematical physicists build the cotangent bundle of the cotangent bundle of $X$. Or maybe $X$ is a genus 3 surface and homotopy theorists build the loop space of the suspension of the space of maps from a genus 5 surface, or whatever. The point is, these constructions preserve the $Aut(X)$ symmetry. For some reason, every branch of mathematics seems to have a bunch of nice functors to the category of vector spaces over $\mathbb{C}$. Let $H$ be one such functor. Now $HX'$ will be a $\mathbb{C}$-vector space carrying information about the complicated object $X'$. But it will also be a representation of the group $Aut(X)$. If the representation theory or $Aut(X)$ is already known, then we gain access to a powerful set of tools with which to study $HX'$. If it's not already known, then we should try to decompose $HX'$ anyway--this is the way to find irreducibles in the first place. Knowing the representation theory of $Aut(X)$ lets us prepare for any possible construction of $X'$ and any possible functor $H$.
|
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153,745 |
What is the "strongest" core model to this day? In particular, how far are we from a core model for supercompact cardinals? There are rumors of some notes from a workshop in 2004: http://www.math.cmu.edu/~eschimme/AIM/LongDescription.html But I couldn't find any more details with respect to a new core model. Also, could someone recommend the clearest and most rigorous exposition of the "strongest" core model so far?
|
${}$Hi Ioanna, I. The answer probably depends on how we define core model. At the level of "there are no Woodin cardinals in any inner model", we can finally show that core models exist, provably in $\mathsf{ZFC}$. The result was known before, of course, but we needed extra assumptions (such as: There is a measurable cardinal $\kappa$) that allowed us to build the model, but only locally (say, in $V_\kappa$). That $\mathsf{ZFC}$ suffices has recently been established by Jensen and Steel, in $K$ without the measurable . This paper also contains an axiomatization of what we mean by a core model (see their Theorem 1.1). Past one Woodin cardinal, the theory turns complicated for a variety of reasons, the first of which being that iterability is no longer an absolute notion, so additional assumptions on the universe of sets are needed to establish the appropriate results. The idea is simple, but making it rigorous takes work: We can relativize the construction of $K$ to $K_x$ where, instead of considering premice, we look at $x$-premice, that are just like premice but have $x$ as an additional object "added at the bottom". If $x$ is itself a mouse, what we are doing is building mice that extend $x$, but now we only look at iterations that do not require us going back to the extenders of $x$. This way, we can then build appropriate versions of the core model for any finite number of Woodin cardinals, inductively (we start with $x$ having one Woodin cardinal, and build $K$ over $x$, establish somehow that weak covering is violated so the construction actually reaches a Woodin cardinal, so now we have models with two Woodin cardinals, we can use these models $y$ to build $K$ on top, etc. Making this precise requires that we be able to "patch" together local structures into global ones, which is typically why we need additional assumptions (such as, $V$ is closed under sharps). The precise nature of these assumptions obviously ends up depending on the setting we work with, but we can axiomatize the whole process. This leads to the core model induction which, finally, we can truly describe as an induction, The specific requirement on $V$ then becomes closure under appropriate "mice operators". An excellent description of this approach, and how far it can get, can be seen in the Schindler-Steel monograph The core model induction . As explained in the book, how far the induction can reach (that is, how strong can we obtain a core model) is intimately tied up with how much determinacy we can prove (and so, measuring the strength of the core models becomes a problem of the strength of determinacy assumptions). This is because to obtain closure under the appropriate operators requires that we prove different instances of "mice capturing", for which determinacy appears essential. Nowadays, we understand that the relevant descriptive set theory and inner model theory are so intimately related, that we talk of Descriptive inner model theory . We can reach far beyond what is described in the Schindler-Steel book, and the theory keeps advancing rapidly, so a precise statement of how far we are at is hard to locate. The strongest results are due to Sargsyan, and the strongest written accounts are due to him. See On the strength of $\mathsf{PFA}$. I and the page on hod mice on his website . Part of what Sargsyan did was to identify what seems to be the right hierarchy of mice within models of determinacy. From earlier work of Woodin we knew that the relevant $\mathsf{ZFC}$ models to study where the $\mathsf{HOD}$ of the models of determinacy. The appropriate hierarchy of hybrid mice we now call hod mice . Their "hybrid" nature
means that they are not pure mice, but add fragments of the relevant iteration strategies to them. Their strength is then measured via the Solovay sequence of the determinacy models we consider. In turn, we use these mice to prove determinacy of stronger models, so the process is truly inductive. Here, the Solovay sequence is the sequence of "local" $\theta$ ordinals: $\theta_0$ is the smallest non-zero ordinal not the surjective image of $\mathbb R$ via an ordinal definable map. We can then define $\theta_1$ as the smallest non-zero ordinal not the surjective image of $\mathbb R$ via maps that are ordinal definable in $A$, where $A$ is any set of reals of Wadge degree $\theta_0$, etc. The hierarchy stops once we reach (true) $\Theta$, the first non-zero ordinal not the surjective image of $\mathbb R$. The expectation Sargsyan has is that the Solovay hierarchy should at least reach as far as the large cardinal hierarchy. The core models we build this way would have strength measured in terms of the Solovay sequence, and to identify their strength in traditional large cardinal terms would then require a further translation. This is described in some detail in the specific case of his results on the strength of failures of square, see his $\mathsf{PFA}$ paper. As you see there, the final translation (in this case, from "there is an inner model $M$ containing $\mathbb R$ and satisfying "$\mathsf{AD}_{\mathbb R}+\Theta$ is regular" to something past "there is an inner model of $\mathsf{ZFC}$ where there is a proper class of Woodin cardinals and a proper class of strong cardinals") requires a non-trivial amount of work. It usually builds on results of Neeman (on models with a Woodin limit of Woodin cardinals) and Jensen-Schimmerling-Schindler-Steel on Stacking mice . From the above, you probably see that we can currently produce core models essentially in the neighborhood of "There is a Woodin limit of Woodin cardinals". There are natural benchmarks to go from here, say: Models with a measurable Woodin cardinal, or with a subcompact cardinal. This is still significantly below supercompact cardinals, and there are many technical obstacles to overcome before reaching that far. II. Independently of the line of developments outlined above, Woodin has a program that attempts to identify what should be the ultimate core model, for all large cardinal assumptions. It is this programs and his current results that are described on his papers on Ultimate $L$ and what he calls suitable extender models . Woodin's high level program has two parts: First, we identify the coarse features of appropriate models of strong large cardinals, in particular, we identify what their extender sequences should look like, and what the appropriate comparison process and iterability assumptions should be. (This is akin to the isolation of Martin-Steel models for Woodin cardinals, that predated the development of their fine-structure by Mitchell-Steel). This part has had reasonable success. Of course, we do not have proofs yet of the appropriate iterability assumptions, but have instead identified what seem to be the relevant conjectures to pursue in this area. The success of the second part seems to be still far away (even assuming the truth of the relevant conjectures). Woodin has started working on the fine structure of suitable extender models, but I do not know the current state of his results. (You may want to ask him for a draft of his paper.)
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|
154,431 |
Recently, I have proved that Kazhdan's property (T) is theoretically provable
by computers ( arXiv:1312.5431 ,
explained below), but I'm quite lame with computers and have
no idea what they actually can do. So, my question is how feasible is it to
prove property (T) of a given group, say $\mathrm{Out}(F_{r>3})$ (a famous
open problem), by solving the equation below by a computer? Even the case of $\mathrm{SL}_{r>2}({\mathbb Z})$ where property (T) is known is unclear. A group $\Gamma$ , generated by a finite subset $S$ and with its non-normalized Laplacian denoted by $$\Delta=\sum_{x\in S} (1-x)^*(1-x)=\sum_{x\in S} (2-x-x^{-1})\in{\mathbb Z}[\Gamma],$$ has property (T) iff the equation in ${\mathbb Z}[\Gamma]$ , $$ m \Delta^2 = n \Delta + \sum_{i=1}^k l_i \xi_i^*\xi_i $$ has a solution in $k,m,n,l_i\in{\mathbb Z}_{>0}$ and $\xi_i\in{\mathbb Z}[\Gamma]$ .
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Using the $\Delta^2- \epsilon \Delta$ approach, Tim Netzer and I have verified Kazhdan's property (T) for ${\rm SL}(3,\mathbb Z)$ . For the standard generators $e_{ij}$ ( $i\neq j$ ) we can show a spectral gap of the normalized Laplace operator of $1/120$ . There is a lot of room for further improvement. To my knowledge, the best previously known lower bound was about $1/3500000$ (and the best upper bound $1/3$ ). The approach uses a positive semi-definite programming package in MatLab, which we use to guess a large positive semi-definite matrix and Mathematica to verify symbolically (computing with fractions etc.) that this indeed yields a sum-of-squares decomposition + some easy theoretical argument that deals with the error terms. The final argument is purely symbolic and does not involve any numeric computation that could involve errors because of rounding etc. We are planning to write a short note and make the computation available in the internet. Now, we attempt to see what we get for ${\rm Aut}(F_4)$ . Edit on November 11, 2014: We have now uploaded the preprint with an attached Mathematica notebook to the arXiv, https://arxiv.org/abs/1411.2488 .
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|
154,450 |
Sorry for asking a basic question but this did not get answered on M.SE. Let $\Omega \subset \mathbb{R}^n$ be a Lipschitz domain. How do I show rigorously that
$$\int_{\partial\Omega} \frac{1}{|y|^{n-2}} dS(y) < \infty$$
where $dS$ is the surface measure. My problem is that there is no easy way (for me) to convert this integral via chart maps etc because it's a Lipschitz domain (and not a graph which would be easy). If the integral were over an open subset of $\mathbb{R}^d$ (i.e. a normal integral) then it's easy to show using polar coordinates.
|
Using the $\Delta^2- \epsilon \Delta$ approach, Tim Netzer and I have verified Kazhdan's property (T) for ${\rm SL}(3,\mathbb Z)$ . For the standard generators $e_{ij}$ ( $i\neq j$ ) we can show a spectral gap of the normalized Laplace operator of $1/120$ . There is a lot of room for further improvement. To my knowledge, the best previously known lower bound was about $1/3500000$ (and the best upper bound $1/3$ ). The approach uses a positive semi-definite programming package in MatLab, which we use to guess a large positive semi-definite matrix and Mathematica to verify symbolically (computing with fractions etc.) that this indeed yields a sum-of-squares decomposition + some easy theoretical argument that deals with the error terms. The final argument is purely symbolic and does not involve any numeric computation that could involve errors because of rounding etc. We are planning to write a short note and make the computation available in the internet. Now, we attempt to see what we get for ${\rm Aut}(F_4)$ . Edit on November 11, 2014: We have now uploaded the preprint with an attached Mathematica notebook to the arXiv, https://arxiv.org/abs/1411.2488 .
|
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|
154,498 |
The formula $V-E+F=2$ is so simple that I can't believe that it was really Euler (or perhaps Descartes) who first observed it (I mean the formula itself in some generality, not necessarily a valid proof). To have a concrete question: Is there any reference to this formula in ancient mathematics?
|
there is no doubt the answer to your question is "no"; for a wonderful and scholarly recent book on the whole story, see Euler's Gem: The Polyhedron Formula and the Birth of Topology by David Richeson. They all missed it. The ancient Greeks -- mathematical luminaries
such as Phythagoras, Theaetetus, Plato, Euclid, and
Archimedes, who where infatuated with polyhedra -- missed it.
Johannes Kepler, the great astronomer, so in awe of the beauty of polyhedra that he based an early model of the solar system on them, missed it.
In his investigation of polyhedra the mathematician and philosopher René
Descartes was but a few logical steps away from discovering it, yet he
too missed it. These mathematicians, and so many others, missed a
relationship that is so simple that it can be explained to any
schoolchild, yet is so fundamental that it is part of the fabric of
modern mathematics. The great Swiss mathematician Leonhard Euler did not miss it. On
November 14, 1750, in a letter to his friend, the number theorist
Christian Goldbach, Euler wrote, "It astonishes me that these general
properties of stereometry have not, as far as I know, been noticed by
anyone else". Centuries later, we remain astonished.
|
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|
155,367 |
One example of a subset of a group $G$ which has to be closed in any topology on $G$ compatible with the group operations is a centraliser. Are there any other interesting examples?
|
Subsets of a group that are closed with respect to any Hausdorff group topology
are called unconditionally closed . Clearly, all algebraic sets are unconditionally closed, where a subset of a group $G$ is called algebraic if it is an intersection of finite unions of the sets of solutions to some equations with coefficient from $G$. A.A.Markov proved that for countable groups the converse is also true:
$$
\hbox{unconditionally closed = algebraic}.
$$
For uncountable groups, this is not always the case as follows from works of S. Shelah (under CH) and G. Hesse.
|
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|
155,731 |
The "Fundamental Theorem of Space Curves"
( Wikipedia link ; MathWorld link )
states that there is a unique (up to congruence)
curve in space that simultaneously realizes
given continuous curvature $\kappa(s)$ and torsion $\tau(s)$ functions. (Image from this link .) Q . What is the closest equivalent generalization for surfaces
in $\mathbb{R}^3$, and more generally for Riemannian manifolds
in $\mathbb{R}^d$?
|
I've added a few sentences to my answer to clarify something that some readers may be wondering about, which is why there isn't as simple an answer for surfaces in $3$-space as there is for curves in $3$-space. First, a word of caution about the so-called 'Fundamental Theorem of Space Curves': You need to assume that the curvature $\kappa(s)$ is nowhere vanishing in order to get $\tau$ well-defined. The theorem is really about space curves endowed with a Frenet frame, and a Frenet frame for a given smooth curve may not be unique (or even be continuous) if $\kappa$ is allowed to vanish. Second (still about curves), if one looks carefully at the FTSC, one sees that one actually has to specify three pieces of data along an interval $I\subset \mathbb{R}$ in order to determine a curve $c:I\to\mathbb{R}^3$ up to rigid motion: The functions $\kappa>0$, $\tau$, and the element of arc , $\mathrm{d}s$, which is a $1$-form. When $\kappa' := \mathrm{d}\kappa/\mathrm{d}s$ is nonzero, one could, alternatively, specify the triple of functions $(\kappa,\tau,\kappa')$ along $I$, for, then, one could recover $\mathrm{d}s$ as $(\mathrm{d}\kappa)/\kappa'$. This is satisfying because one has exactly three pieces of data along the domain of the curve to specify the three components of the curve up to rigid motion. Now, for surfaces, one could hope that it would be possible to determine an immersed surface $f:S\to\mathbb{R}^3$ (which depends on a choice of three 'arbitrary' functions on $S$) uniquely up to rigid motion by specifying some three functions that are constructed out of $f$ by differentiation operations and algebra and that are unchanged when $f$ is replaced by $Rf$, where $R$ is a rigid motion of $\mathbb{R}^3$, i.e, what are often called 'Euclidean differential invariants' in the differential geometry literature. There certainly are plenty of such 'Euclidean differential invariants; for example, the Gauss curvature $K$ and the mean curvature $H$ (or, if one is worried about orientation effects, say $H^2$) are second order invariants, and higher order invariants are easily defined (such as, for example, $|\mathrm{d}H|^2$, etc.). However, it is a (nontrivial) theorem that no such triple of differential invariants exists that will uniquely characterize each immersion up to rigid motion. (Even stronger, it turns out that there is no differential invariant 'tensor' of rank $3$ that can do this, which includes data such as the first fundamental form, etc.) It turns out that the best one can do is specify 'excess data' to get the 'unique up to rigid motion characterization' property and then analyze what conditions one needs to put on the 'excess' in order to determine when the specified data actually does come from an immersion $f$. Bonnet's Theorem (discussed below) is one such example, and probably the most well-known (because we teach it in every beginning differential geometry course). Now, for surfaces in $\mathbb{R}^3$, there is a different (and more traditional) generalization from the result that Anton mentions that may be what you want. It is usually known as Bonnet's Theorem, and it goes like this: For any immersion of an oriented surface $f:S\to\mathbb{R}^3$, one defines two quadratic forms: The first fundamental form , $\mathrm{I}_f = \mathrm{d} f\cdot \mathrm{d} f>0$, and the second fundamental form , $\mathrm{I\!I}_f = - \mathrm{d} f\cdot \mathrm{d} n$, where $n:S\to \mathbb{R}^3$ is the oriented unit normal to the immersion. Then the uniqueness theorem says that $f$ is uniquely determined up to oriented isometry by the pair $(\mathrm{I}_f,\mathrm{I\!I}_f)$, and the existence theorem says that, if $(\mathrm{I},\mathrm{I\!I})$ are given quadratic forms on a connected, oriented, simply connected surface $S$ that satisfy the Gauss equation and the Codazzi equations (see below), then there exists an immersion $f:S\to\mathbb{R}^3$ such that $(\mathrm{I},\mathrm{I\!I}) = (\mathrm{I}_f,\mathrm{I\!I}_f)$. The Gauss equation is the condition $\det\mathrm{I\!I} = K(\mathrm{I})\ \det\mathrm{I}$, where $K(\mathrm{I})$ is the Gauss curvature of the (positive definite) quadratic form $\mathrm{I}$ and $\det$ means the operation on quadratic forms that, in local coordinates, is given by
$$
\det( e\, dx^2 + 2f\, dx\ dy + g\, dy^2) = (eg-f^2) (dx\wedge dy)^{\otimes 2}
$$
The Gauss equation is a single second order partial differential equation on the pair $(\mathrm{I},\mathrm{I\!I})$. The Codazzi equations are $\delta_{\mathrm{I}}(\mathrm{I\!I})=0$, where $\delta_{\mathrm{I}}:C^\infty\bigl(S^2(T^*\!S)\bigr)\to C^\infty\bigl(T^*\!S\bigr)$ is a linear, first-order differential operator that is the $\mathrm{I}$-covariant derivative $\nabla_\mathrm{I}:C^\infty\bigl(S^2(T^*\!S)\bigr)\to C^\infty\bigl(T^*\!S\otimes S^2(T^*\!S)\bigr)$ followed by the appropriate contraction operation $c:T^*\!S\otimes S^2(T^*\!S)\to T^*\!S$. They constitute $2$ first-order partial differential equations on the pair $(\mathrm{I},\mathrm{I\!I})$. Note that this uniqueness and existence theorem makes sense: A choice of $f:S\to\mathbb{R}^3$ is a choice of $3$ 'arbitrary' functions on the surface $S$ (subject only to the 'open' condition that they define an immersion of $S$ into $\mathbb{R}^3$) while a choice of a pair $(\mathrm{I},\mathrm{I\!I})$ represents a choice of $6$ 'arbitrary' functions on $S$ (the coefficients of the two quadratic forms in local coordinates, subject only to the condition that $\mathrm{I}$ be positive definite). Subjecting those $6$ functions to $3$ partial differential equations (i.e., the Gauss and Codazzi equations) restores the 'balance of arbitrariness' to $3$ arbitrary functions on the surface. In higher dimensions, for submanifolds of $\mathbb{R}^d$, there is a similar theorem, but now $\mathrm{I\!I}$ gets replaced by a quadratic form taking values in a 'normal bundle' $N$, which is a vector bundle endowed with an Euclidean connection. The Gauss equation and the Codazzi equations generalize in a straightforward way, but when the 'normal bundle' has dimension greater than $1$, one also has a further set of equations relating the curvature of the connection on the normal bundle with the quadratic form $\mathrm{I\!I}$. The exact statement can be found in any good book on submanifold theory. If the ambient space is not $\mathbb{R}^d$ or, more generally, a space of constant sectional curvature, then the situation is more complicated, and I refer you to a book on submanifold theory for the ideas and details.
|
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|
156,008 |
What is the maximum number $k$ of unit radius, infinitely long cylinders with mutually disjoint interiors that can touch a unit ball? By a cylinder I mean a set congruent to the Cartesian product of a line and a circular disk. The illustrations, copied from 2 , show various configurations of six cylinders, perhaps all possible - up to isometry. The question is about 25 years old; the answer is conjectured to be 6, but the conjecture is still unconfirmed. Heppes and Szabó 1 proved in 1991 that $k\le 8$ ; nine years later Brass and Wenk 2 improved this result to $k\le 7$ . Also, one would like to know: What does the configuration space of six unit cylinders touching a unit ball look like? In particular: Is the configuration space connected? Even more specifically, referring to the configurations shown below: Within the configuration space, is a continuous transition possible from the configuration in Figures 1 and 2 to the configuration in Figure 3? The last question may be possible to verify by a natural candidate, but the necessary computations seem too tedious to me... 1 Heppes, Aladar and Szabó, Laszló. "On the number of cylinders touching a ball." Geom. Dedicata 40 (1991), no. 1, 111–116; MR1130481. 2 Brass, Peter and Wenk, Carola. "On the number of cylinders touching a ball." Geom. Dedicata 81 (2000), no. 1-3, 281–284; MR1772209.
|
Here is an idea. Consider the following parameterization, which is supposed to cover the configuration space in question. $$\mathcal{C}_7:=\left\{\pmatrix{x_k\\y_x\\z_k},\pmatrix{a_k\\b_k\\c_k}_{1\leq k\leq 7}\in{\mathbb{R}^{3\times 2}}^7\,\middle |\, \text{such that conditions 1.-4. are satisfied} \right\} $$ Conditions: $x_k^2+y_k^2+z_k^2=1$ $\left\langle\pmatrix{x_k\\y_k\\z_k},\pmatrix{a_k\\b_k\\c_k} \right\rangle=0$ $a_k^2+b_k^2+c_k^2=1$ $d(l_i,l_j)\geq 2$ for $1\leq i<j\leq 7,$ where we define the line
$$l_k:=\left\{2\pmatrix{x_k\\y_k\\z_k}+\alpha\pmatrix{a_k\\b_k\\c_k}\,\middle|\,\alpha\in\mathbb{R} \right\}$$ and denote with $d(\cdot,\cdot)$ the distance between two lines. Note that condition 4. can be rewritten as polynomial inequalities. Hence $\mathcal{C}_7$ is a semi-algebraic set in $\mathbb{R}^{42}$. The $(x,z,y)$ are the points, where the unit cylinder is tangent to the unit sphere. The corresponding $(a,b,c)$ gives the direction in the tangent space and the lines $l$ are the cores of the cylinders. (Note that $(-a,-b,-c)$ gives the same cylinder.) The question " Is $\mathcal{C}_7$ empty? " should be decidable. Maybe an algorithmic approach could help from here. For the other questions the study of an analogues defined $\mathcal{C}_6$, which we know to be non-empty might be worthwhile. I wrote a little program that tries to find points in the described semi-algebraic sets. Here's what it found for $\mathcal{C}_6$ (click here for an animation). Let's take a slightly different point of view. Fix the radius of the ball to be $1$, but let the radii of the $k$ cylinders vary while making sure all cylinders have the same radius. We can then ask: What is the largest radius $r_k$, so that we can find $k$ non-overlapping cylinders of radius $r_k$, that
touch the unit ball? Hence the question is: $r_7\geq 1?$ An obvious lower bound on $r_k$ comes from the packing that allows a equatorial section which is a circle packing, as for $k=6$ in figure 1 and figure 2 in the question post. We therefore have:
$$r_k\geq \frac{\operatorname{sin}(\frac{\pi}{k})}{1-\operatorname{sin}(\frac{\pi}{k})}$$
Here's a list of decimal approximations for small $k$s:
$$\begin{array}{c|cccccc}k&3&4&5&6&7&8\\\hline
\frac{\operatorname{sin}(\frac{\pi}{k})}{1-\operatorname{sin}(\frac{\pi}{k})}
&6.464101& 2.414213& 1.425919& 1&
0.766421& 0.619914\end{array}$$
A perhaps surprising result of my calculations is the fact that $r_6>1$, indeed
$$r_6> 1.04965$$
So in other words there is configuration of $6$ cylinders where the cylinders have radius larger than $1.04965$. Here is a picture of the configuration
(again click here for an animation): I also drew cylinders of radius $1$ with the same tangent points, so one can see the difference. The configuration space can be viewed as subset of the the $6$th power of the unit tangent bundle of the sphere $(T^1(S^2))^6$ (see conditions 1.-4. and Henrik Rüping's comment). The upshot of finding a configuration with larger radius is: the configuration space contains an open subset of $(T^1(S^2))^6$ and hence is $18$-dimensional locally . Edit: Here is list of lower bounds on $r_k$ for small $k$: For $k=3$ and $k=4$ I conjecture the trivial bound for $r_k$ given above to be sharp. For $k=5$ one can find a configuration that shows:
$r_5>1.45289>1.425919$ For $k=6$ we have $r_6>1.04965 >1$ as mentioned above. For $k=7$ I found a configuration that shows $r_7>0.846934>0.766421$. Here is a picture of this configuration (again click here for an animation):
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156,238 |
Yesterday I was shocked to discover that function extensionality (the statement that if two functions $f$ and $g$ on the same domain satisfy $f\left(x\right) = g\left(x\right)$ for all $x$ in the domain, then $f = g$) is not an axiom in the standard constructive logic of Coq. Of course, one can add it as an axiom in one's files, but it is not obviously available in any pre-defined tactics. I am left wondering... 1. What is the thinking behind considering function extensionality as foreign to the calculus of constructions? I thought that from a computational perspective, a function really is there to be applied to things, and cannot carry any more information than what it does to them (and its type). For a moment I suspected that Coq avoids function extensionality in order to allow applying results to models like arbitrary enriched categories whose internal homomorphisms carry some more information than plain morphisms. But this is not the case: Coq (since version 8.4) has a implementation of eta-expansion (saying that any function $f$ equals the function sending every $x$ in its domain to $f\left(x\right)$, provided the types are right). In an enriched category, this would pour any additional structure of an internal Hom down the drain. (I must say the eta-expansion in Coq feels rather weird, too -- it is triggered by the reflexivity tactic. I expected it to be a tactic on its own...) Having eta-expansion but no extensionality is seriously confusing: one can have $f\left(x\right) = g\left(x\right)$ for all $x$, and yet one cannot rewrite the $f\left(x\right)$ in "the function sending every $x$ to $f\left(x\right)$" as a $g\left(x\right)$. And there I thought the bound variable in a lambda term would be like the bound variable in a forall quantification? 2. On a more practical note (and more on-topic in MathOverflow), how much do the axiom of function extensionality and the (weaker) axiom of eta-expansion contribute to the strength of the logic? (At this point I have to admit that I don't really know the definition of the logic involved, so I'll just say I'm talking about the logic of Coq with no additional axioms assumed; it has so far been agreeing with my intuitive understanding of constructivism, until extensionality came along.) If I can define two terms $a$ and $b$ of type $\mathrm{nat}$ (natural numbers) and use function extensionality (resp. eta-reduction, both ways) to prove their equality, can I also do it without? If I can prove a (more complicated) statement using function extensionality, is there a way to transform it into a (possibly clumsier) statement which can be proven without function extensionality and which can be transformed into the former statement using some straightforward applications of function extensionality? I'm sorry for logical naivety. The way I am posing the question, I fear it would qualify as soft; nevertheless I am pretty sure that there is some precise statements to be made here (or maybe just a reference to a textbook to be given).
|
I am going to draw heavily from Github discussion on HoTT book pull request 617 . There are different kinds of equality. Let us say that equality is "intensional" if it distinguishes objects based on how they are defined, and "extensional" if it distinguishes objects based on their "extension" or "observable behavior". In Frege's terminology, intensional equality compares sense and extensional equality compares reference. To use Russell's example, intensionally the morning star and the evening star are clearly not the same (because their definitions are different), but they are extensionally the same because they both denote the same object (planet Venus). A more mathematical example is comparison of $x + y$ and $y + x$. These are extensionally equal, but intensionally differ because (the usual) definition of $+$ treats its arguments asymmetrically. It should be clear that two functions may be extensionally equal (have same behavior) even though they differ intensionally (have different definitions). It is possible for two kinds of equality to coexist. Thus in type theory there are two equalities. The intensional one is called "judgmental" or "definitional equality" $\equiv$ and the extensional one is known as "propositional equality" $=$. Mathematicians are aware of $=$ as a "real thing" while they think of $\equiv$ as "formal manipulation of symbolic expressions" or some such. We may control the two kinds of equality and the relationship between them with additional axioms. For instance, the reflection rule collapses $\equiv$ and $=$ by allowing us to conclude $a \equiv b$ from $a = b$ (the other direction is automatic). There are also varying degrees of extensionality of $=$. Without any extra axioms, $=$ is already somewhat extensional. For instance, we can prove commutativity of $+$ on natural numbers by induction in the form $x + y = y + x$, but we cannot prove $x + y \equiv y + x$. Function extensionality is an axiom which describes what constitutes an "observation" on functions: by saying that two functions are equal when they give equal values we are in essence saying that only values matter (but not for example the "running time" or some other aspect of evaluation). Another axiom which makes $=$ "more extensional" is the Univalence axiom. It is hard to do mathematics without function extensionality, but type theorists have their reasons for not including it as an axiom by default. But before I explain the reason, let me mention that there is a standard workaround. We may introduce user-defined equalities on types by equipping types with equivalence relations. This is what Bishop did in his constructive mathematics, and this is what we do in Coq when we use setoids . With such user-defined equalities we of course recover function extensionality by construction. However, setoids are often annoying to work with, and they drag in technical problems which we would prefer to avoid. Incidentally, the setoid model shows that function extensionality does not increase the power of type theory (it is a model validating function extensionality built in type theory without function extensionality). So why don't type theorist adopt function extensionality? If we want to have a type theory with nice properties, and a useful proof assistant, then $\equiv$ should be "easy" to deal with. Technically speaking, we would like a strongly normalizing $\equiv$. By assuming function extensionality we throw into type theory a new constant funext without explaining how it interacts with the process of strong normalization, and things break. Type theorists would say that we failed to explain the computational meaning of funext . Consequently, Coq does not adopt function extensionality because that would lead to a lot of problems. Coq would not be able to handle $\equiv$ automagically anymore, and the whole system would just have worse behavior. Type theorists of course recognize that having a good computational explanation of function extensionality, and more generally of the univalence problem, would be extremely desirable. This is why the HoTTest open problem is to give a computational interpretation of the univalence axiom. Once this is accomplished, we ought to have at our disposal type systems and proof assistants which are much more natural from a mathematician's point of view. Until then, you can always assume funext as an axiom and work around the resulting complications. To see how this can be done, have a loot at the Funext axiom in the HoTT library. [This P.S. is outdated after the question was edited.] P.S. The title of your question points to a common leap of reasoning from "not accepting function extensionality" to "denying function extensionality". While there are models in which function extensionality has counter-examples, one should be aware of the difference between "not accept" and "deny". (I am complaining because this sort of leap is often made about the law of excluded middle, and there it leads to absurdity.)
|
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|
156,246 |
The famous Higman embedding theorem says that every recursively presented group embeds in a finitely presented group. This is a convenient tool to construct finitely presented groups with bizarre properties from recursively presented ones, which are usually easier to construct. One cannot hope for an exact analogue of Higman's theorem in the setting of residually finite groups because finitely presented residually finite groups have solvable word problem and hence their finitely generated subgroups do as well. But Meskin constructed finitely generated recursively presented groups with undecidable word problem. I know of no other obstruction to embedding a finitely generated residually finite group into a finitely presented one, so I ask the following question. Does there exist a finitely generated residually finite group with decidable word problem that cannot be embedded in a finitely presented residually finite group?
|
I am going to draw heavily from Github discussion on HoTT book pull request 617 . There are different kinds of equality. Let us say that equality is "intensional" if it distinguishes objects based on how they are defined, and "extensional" if it distinguishes objects based on their "extension" or "observable behavior". In Frege's terminology, intensional equality compares sense and extensional equality compares reference. To use Russell's example, intensionally the morning star and the evening star are clearly not the same (because their definitions are different), but they are extensionally the same because they both denote the same object (planet Venus). A more mathematical example is comparison of $x + y$ and $y + x$. These are extensionally equal, but intensionally differ because (the usual) definition of $+$ treats its arguments asymmetrically. It should be clear that two functions may be extensionally equal (have same behavior) even though they differ intensionally (have different definitions). It is possible for two kinds of equality to coexist. Thus in type theory there are two equalities. The intensional one is called "judgmental" or "definitional equality" $\equiv$ and the extensional one is known as "propositional equality" $=$. Mathematicians are aware of $=$ as a "real thing" while they think of $\equiv$ as "formal manipulation of symbolic expressions" or some such. We may control the two kinds of equality and the relationship between them with additional axioms. For instance, the reflection rule collapses $\equiv$ and $=$ by allowing us to conclude $a \equiv b$ from $a = b$ (the other direction is automatic). There are also varying degrees of extensionality of $=$. Without any extra axioms, $=$ is already somewhat extensional. For instance, we can prove commutativity of $+$ on natural numbers by induction in the form $x + y = y + x$, but we cannot prove $x + y \equiv y + x$. Function extensionality is an axiom which describes what constitutes an "observation" on functions: by saying that two functions are equal when they give equal values we are in essence saying that only values matter (but not for example the "running time" or some other aspect of evaluation). Another axiom which makes $=$ "more extensional" is the Univalence axiom. It is hard to do mathematics without function extensionality, but type theorists have their reasons for not including it as an axiom by default. But before I explain the reason, let me mention that there is a standard workaround. We may introduce user-defined equalities on types by equipping types with equivalence relations. This is what Bishop did in his constructive mathematics, and this is what we do in Coq when we use setoids . With such user-defined equalities we of course recover function extensionality by construction. However, setoids are often annoying to work with, and they drag in technical problems which we would prefer to avoid. Incidentally, the setoid model shows that function extensionality does not increase the power of type theory (it is a model validating function extensionality built in type theory without function extensionality). So why don't type theorist adopt function extensionality? If we want to have a type theory with nice properties, and a useful proof assistant, then $\equiv$ should be "easy" to deal with. Technically speaking, we would like a strongly normalizing $\equiv$. By assuming function extensionality we throw into type theory a new constant funext without explaining how it interacts with the process of strong normalization, and things break. Type theorists would say that we failed to explain the computational meaning of funext . Consequently, Coq does not adopt function extensionality because that would lead to a lot of problems. Coq would not be able to handle $\equiv$ automagically anymore, and the whole system would just have worse behavior. Type theorists of course recognize that having a good computational explanation of function extensionality, and more generally of the univalence problem, would be extremely desirable. This is why the HoTTest open problem is to give a computational interpretation of the univalence axiom. Once this is accomplished, we ought to have at our disposal type systems and proof assistants which are much more natural from a mathematician's point of view. Until then, you can always assume funext as an axiom and work around the resulting complications. To see how this can be done, have a loot at the Funext axiom in the HoTT library. [This P.S. is outdated after the question was edited.] P.S. The title of your question points to a common leap of reasoning from "not accepting function extensionality" to "denying function extensionality". While there are models in which function extensionality has counter-examples, one should be aware of the difference between "not accept" and "deny". (I am complaining because this sort of leap is often made about the law of excluded middle, and there it leads to absurdity.)
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156,263 |
Imagine the beginning of a game of pool, you have 16 balls, 15 of them in a triangle <| and 1 of them being the cue ball off to the left of that triangle. Imagine that the rack (the 15 balls in a triangle) has every ball equally spaced apart and all balls touching all other appropriate balls. All balls are perfectly round. Now, imagine that the cue ball was hit along a friction free surface on the center axis for this triangle O-------<| and hits the far left ball of the rack dead center on this axis. How would the rack react? I would imagine this would be an extension of newtons cradle and only the 5 balls on the far end would move at all. But in what way would they move? Thanks
|
This question was cross-posted on Math Stack Exchange. Here is a copy of my answer for it there . This is it. The perfectly centered billiards break. Behold. Setup This break was computed in Mathematica using a numerical differential equations model. Here are a few details of the model: All balls are assumed to be perfectly elastic and almost perfectly rigid. Each ball has a mass of 1 unit and a radius of 1 unit. The cue ball has a initial speed of 10 units/sec. The force between two balls is given by the formula
$$
F \;=\; \begin{cases}0 & \text{if }d \geq 2, \\ 10^{11}(2-d)^{3/2} & \text{if }d<2,\end{cases}
$$
where $d$ is the distance between the centers of the balls. Note that the balls overlap if and only if $d < 2$. The power of $3/2$ was suggested by Yoav Kallus in the comments, because it follows Hertz's theory of non-adhesive elastic contact . The initial speed of the cue ball is immaterial -- slowing down the cue ball is the same as slowing down time. The force constant $10^{11}$ has no real effect as long as it's large enough, although it does change the speed at which the initial collision takes place. The Collision For this model, the entire collision takes place in the first 0.2 milliseconds, and none of the balls overlap by more than 0.025% of their radius during the collision. (These figures are model dependent -- real billiard balls may collide faster or slower than this.) The following animation shows the forces between the balls during the collision, with the force proportional to the area of each yellow circle. Note that the balls themselves hardly move at all during the collision, although they do accelerate quite a bit. The Trajectories The following picture shows the trajectories of the billiard balls after the collision. After the collision, some of the balls are travelling considerably faster than others. The following table shows the magnitude and direction of the velocity of each ball, where $0^\circ$ indicates straight up. $$
\begin{array}{|c|c|c|c|c|c|c|c|c|c|c|}
\hline
\text{ball} & \text{cue} & 1 & 2,3 & 4,6 & 5 & 7,10 & 8,9 & 11,15 & 12,14 & 13 \\
\hline
\text{angle} & 0^\circ & 0^\circ & 40.1^\circ & 43.9^\circ & 0^\circ & 82.1^\circ & 161.8^\circ & 150^\circ & 178.2^\circ & 180^\circ \\
\hline
\text{speed} & 1.79 & 1.20 & 1.57 & 1.42 & 0.12 & 1.31 & 0.25 & 5.60 & 2.57 & 2.63 \\
\hline
\end{array}
$$ For comparison, remember that the initial speed of the cue ball was 10 units/sec. Thus, balls 11 and 15 (the back corner balls) shoot out at more than half the speed of the original cue ball, whereas ball 5 slowly rolls upwards at less than 2% of the speed of the original cue ball. By the way, if you add up the sum of the squares of the speeds of the balls, you get 100, since kinetic energy is conserved. Linear and Quadratic Responses The results of this model are dependent on the power of $3/2$ in the force law -- other force laws give other breaks. For example, we could try making the force a linear function of the overlap distance (in analogy with springs and Hooke's law ), or we could try making the force proportional to the square of the overlap distance. The results are noticeably different Stiff Response Glenn the Udderboat points out that "stiff" balls might be best approximated by a force response involving a higher power of the distance (although this isn't the usual definition of "stiffness"). Unfortunately, the calculation time in Mathematica becomes longer when the power is increased, presumably because it needs to use a smaller time step to be sufficiently accurate. Here is a simulation involving a reasonably "stiff" force law
$$
F \;=\; \begin{cases}0 & \text{if }d \geq 2, \\ 10^{54}(2-d)^{10} & \text{if }d<2.\end{cases}
$$ As you can see, the result is very similar to my initial answer on Math Stack Exchange. This seems like good evidence that the behavior discussed in my initial answer is indeed the limiting behavior in the case where this notion of "stiffness" goes to infinity. As you might expect, most of the energy in this case is transferred very quickly at the beginning of the collision. Almost all of the energy has moves to the back corner balls in the first 0.02 milliseconds. Here is an animation of the forces: After that, the corner balls and the cue ball shoot out, and the remaining balls continue to collide gently for the next millisecond or so. While the simplicity of this behavior is appealing, I would guess that "real" billard balls do not have such a force response. Of the models listed here, the intial Hertz-based model is probably the most accurate. Qualitatively, it certainly seems the closest to an "actual" break. Note: I have now posted the Mathematica code on my web page .
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157,938 |
This question was asked by a student (in a slightly different form), and I was unable to answer it properly. I think it's quite interesting. The problem is to produce an example of the following situation: find a short exact sequence
$$ 0 \to X_1 \to X_2 \to X_3 \to 0$$
(in some category of your choice), and a second exact sequence
$$ 0 \to Y_1 \to Y_2 \to Y_3 \to 0$$
such that there are isomorphisms $X_n \cong Y_n$ for all $n$, BUT in such a way that there is no commutative diagram whatsoever between the two sequences, with the vertical maps being isomorphisms. It is impossible to find such an example in the category of vector spaces, or of finitely-generated abelian groups. I don't know about the general case, though. I would be grateful for any example, but would be disappointed if the chosen category were constructed specifically to answer the problem. EDIT / COMMENT: in the first version of this question I was asking for sequences which could potentially be infinite. Some great examples came in the comments straightaway. I'm very thankful for them, but I recall only now that the student's original question was about short exact sequences, so I've edited accordingly. (I'm sorry for the confusion, the student asked me this question several months ago, and I was posting only now, for some reason... and got it wrong. I'm very happy to know about the examples involving infinite sequences, though.)
|
$\def\ZZ{\mathbb{Z}}$
This can happen in finitely generated abelian groups. Let $p$ be prime, set $G = \ZZ/p^2 \oplus \ZZ/p$ and set $H = \ZZ/p^3 \oplus \ZZ/p^2 \oplus \ZZ/p$. Then there are two non-isomorphic short exact sequences $0 \to G \to H \to G \to 0$. The first one is the sum of the extensions:
$$\begin{matrix}
0 & \to & \ZZ/p & \to & \ZZ/p^3 & \to & \ZZ/p^2 &\to& 0 \\
0 & \to & \ZZ/p^2 & \to & \ZZ/p^2 & \to & 0 &\to& 0 \\
0 & \to & 0 & \to & \ZZ/p & \to & \ZZ/p & \to & 0 \\
\end{matrix}$$
and the second is the sum of
$$\begin{matrix}
0 & \to & \ZZ/p^2 & \to & \ZZ/p^3 & \to & \ZZ/p &\to& 0 \\
0 & \to & 0 & \to & \ZZ/p^2 & \to & \ZZ/p^2 &\to& 0 \\
0 & \to & \ZZ/p & \to & \ZZ/p & \to & 0 & \to & 0 \\
\end{matrix}$$ Let's see that these are not isomorphic. Write $\alpha$ for the map $G \to H$. In the first extension, $\alpha(p G) \cap p^2 H = (0)$; in the second extension, $\alpha(pG) = p^2 H$. In any reasonable category, isomorphism classes of extensions $0 \to X \to ?? \to Z \to 0$ are classified by the orbits of $\mathrm{Aut}(X) \times \mathrm{Aut}(Z)$ on $\mathrm{Ext}^1(Z,X)$. You are asking for cases where there is more than one orbit which makes the center term isomorphic as an abstract element of the category. There is no reason this shouldn't happen, so I would expect it to happen basically any time there are nontrivial extension groups available. To add some more high level context, the analogous example works in $k[t]$-modules. (That is to say, extending $k[t]/t^2 \oplus k$ by itself to get $k[t]/t^3 \oplus k[t]/t^2 \oplus k$.) In general, for any partition $\lambda$, let $M(\lambda)$ be the $k[t]$ module $\bigoplus_i k[t]/t^{\lambda_i}$. Then $\mathrm{Ext}^1(M(\mu), M(\lambda))$ is stratified into locally closed pieces according to the isomorphism type of the extension, and the number of components where the extension is isomorphic to $M(\nu)$ is the Littlewood-Richardson number $c_{\lambda \mu}^{\nu}$. This example is $c_{(21) (21)}^{(321)}=2$. Since $Aut(M(\lambda))$ is always a connected group (it is a unipotent extension of $\prod GL(\lambda_i)$), it can't mix the components. You can start to get context for this from the first two chapters of Schiffmann's Lectures on Hall Alegbras . Unfortunately, Schiffmann always takes $k$ to be a finite field so that he can count points, which makes it harder to talk about positive dimensional subvarieties of $\mathrm{Ext}^1$.
|
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|
157,967 |
In $\textit{Set Theory}$ by Jech 1978 edition, in the proof of Lemma 32.5 which you can hopefully see at the Google book link . In the course of the proof using the tree property, he produces from any weakly compact cardinal $\kappa$ a non principal $L_\alpha$-ultrafilter $U$ on $\kappa$, which is $L_\alpha$-$\kappa$ complete, and moreover the intersection of $\kappa$ many elements of $U$ (taken in $V$) is nonempty. The latter implies that countable intersection (taken in $V$) is nonempty. This fact implies iterability for example by 19.13 of Kanamori. So $L_\alpha$ with this ultrafilter can be iterated of length $\text{Ord}$ However, $0^\sharp$ follows from the existence of a mouse (iterable premouse). But $0^\sharp$ has stronger consistency strength than a weakly compact. Is this $(L_\alpha, U)$ not a pre-mouse? I have found varying definition of pre-mouse. One definition has the addition condition that $\kappa$ should be the largest cardinal of $L_\alpha$. This seems like it may not hold since at the beginning of the proof, the proof chooses an $\alpha$ such that $L_\alpha \models ZF^-$. I would like to have the $\text{Ord}$ length iteration for what I am trying to do; however, I am troubled by whether or not this would give an iterable premouse and hence imply sharps. Thanks for any clarification you can provide.
|
$\def\ZZ{\mathbb{Z}}$
This can happen in finitely generated abelian groups. Let $p$ be prime, set $G = \ZZ/p^2 \oplus \ZZ/p$ and set $H = \ZZ/p^3 \oplus \ZZ/p^2 \oplus \ZZ/p$. Then there are two non-isomorphic short exact sequences $0 \to G \to H \to G \to 0$. The first one is the sum of the extensions:
$$\begin{matrix}
0 & \to & \ZZ/p & \to & \ZZ/p^3 & \to & \ZZ/p^2 &\to& 0 \\
0 & \to & \ZZ/p^2 & \to & \ZZ/p^2 & \to & 0 &\to& 0 \\
0 & \to & 0 & \to & \ZZ/p & \to & \ZZ/p & \to & 0 \\
\end{matrix}$$
and the second is the sum of
$$\begin{matrix}
0 & \to & \ZZ/p^2 & \to & \ZZ/p^3 & \to & \ZZ/p &\to& 0 \\
0 & \to & 0 & \to & \ZZ/p^2 & \to & \ZZ/p^2 &\to& 0 \\
0 & \to & \ZZ/p & \to & \ZZ/p & \to & 0 & \to & 0 \\
\end{matrix}$$ Let's see that these are not isomorphic. Write $\alpha$ for the map $G \to H$. In the first extension, $\alpha(p G) \cap p^2 H = (0)$; in the second extension, $\alpha(pG) = p^2 H$. In any reasonable category, isomorphism classes of extensions $0 \to X \to ?? \to Z \to 0$ are classified by the orbits of $\mathrm{Aut}(X) \times \mathrm{Aut}(Z)$ on $\mathrm{Ext}^1(Z,X)$. You are asking for cases where there is more than one orbit which makes the center term isomorphic as an abstract element of the category. There is no reason this shouldn't happen, so I would expect it to happen basically any time there are nontrivial extension groups available. To add some more high level context, the analogous example works in $k[t]$-modules. (That is to say, extending $k[t]/t^2 \oplus k$ by itself to get $k[t]/t^3 \oplus k[t]/t^2 \oplus k$.) In general, for any partition $\lambda$, let $M(\lambda)$ be the $k[t]$ module $\bigoplus_i k[t]/t^{\lambda_i}$. Then $\mathrm{Ext}^1(M(\mu), M(\lambda))$ is stratified into locally closed pieces according to the isomorphism type of the extension, and the number of components where the extension is isomorphic to $M(\nu)$ is the Littlewood-Richardson number $c_{\lambda \mu}^{\nu}$. This example is $c_{(21) (21)}^{(321)}=2$. Since $Aut(M(\lambda))$ is always a connected group (it is a unipotent extension of $\prod GL(\lambda_i)$), it can't mix the components. You can start to get context for this from the first two chapters of Schiffmann's Lectures on Hall Alegbras . Unfortunately, Schiffmann always takes $k$ to be a finite field so that he can count points, which makes it harder to talk about positive dimensional subvarieties of $\mathrm{Ext}^1$.
|
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|
158,302 |
It seems that when $p>3$ is a prime, then each group of order $p(p^2+1)/2$ is abelian as I checked by Gap for small $p$. Is it true for each $p$?
Thanks for your answers
|
The smallest counterexample I could find is for $p=53$; then $(p^2+1)/2=5\times281$ and the cyclic group of order 281 has an automorphism of order 5, so the corresponding semidirect product will not be abelian.
|
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158,378 |
Let $C$ be the standard Cantor middle-third set. As a consequence of the Baire category theorem, there are numbers $r$ such that $C+r$ consists solely of irrational numbers, see here . What would be an explicit example of a number $r$ with this property? Short of an explicit example, are there any references addressing this question? A natural approach would be to see that all irrationals in $C$ are transcendental, so it would suffice to take $r=\sqrt2$. But this is open, see here . Many thanks for the answers. (It would be interesting to know whether $\sqrt2$ works.)
|
One way to obtain explicit examples, which combines the ideas of (weak forms of) randomness and base 3 expansions, is to use the fact that normality in a given base is preserved under rational addition, which was proved by D. D. Wall in his 1949 Berkeley PhD Dissertation. (I'm relying on D. Doty, J. H. Lutz, and S.Nandakumar [ Finite-state dimension and real arithmetic , Information and Computation 205(11):1640-1651, 2007] for this reference.) Here a number $r$ is normal in base $b$ if for any finite nonempty string $\sigma$ drawn from the alphabet $\{0,...,b-1\}$, the limiting frequency of the appearances of $\sigma$ as a substring of the base $b$ expansion of $r$ is $b^{-|\sigma|}$. Since elements of $C$ are not normal in base $3$, any number $r$ that is normal in base $3$ has the desired property. Examples of such numbers can be found at http://en.wikipedia.org/wiki/Normal_number , for instance. In fact, normality is overkill. Let $r$ be disjunctive in base $3$, i.e., every finite ternary string appears as a substring of the ternary expansion of $r$ (which is both a comeager and a conull property). We need to show is that if $q$ is a positive rational then $r+q \notin C$. (Here addition is mod $1$.) If the ternary expansion of $q$ has infinitely many $1$'s then the fact that the ternary expansion of $r$ contains $0^n$ for all $n$ means that $q+r \notin C$. Otherwise, the fact that this expansion contains $0^m10^n$ for all $m,n$ does the trick.
|
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158,505 |
Let $G$ and $H$ be two topological groups. Assume that $\phi:\pi_{1}(G) \to \pi_{1}(H)$ is a group homomorphism. Is there a continuous function $f:G\to H$ such that $f_{*}=\phi$?
|
No. Take $G=SO(5)$ and $H=SO(3)$, both of which have fundamental group $\mathbb{Z}/2$. I claim that there is no continuous map $f: G\to H$ which induces the identity homomorphism. If there were, then $f$ would induce a nontrivial homomorphism $f_\ast: H_1(G;\mathbb{Z}/2)\to H_1(H;\mathbb{Z}/2)$, and a graded ring homomorphism $f^\ast: H^\ast(H;\mathbb{Z}/2)\to H^\ast(G;\mathbb{Z}/2)$ which is nontrivial on $H^1$. To see that no such ring homomorphism exists, recall (see Hatcher's Algebraic Topology, section 3.D) that $$H^\ast(H;\mathbb{Z}/2)\cong \mathbb{Z}/2[\alpha_1]/(\alpha_1^4)$$ and $$H^\ast(G;\mathbb{Z}/2)\cong \mathbb{Z}/2[\beta_1]/(\beta_1^8)\otimes\mathbb{Z}/2[\beta_3]/(\beta_3^2),$$ where each $\alpha_i,\beta_i$ is in degree $i$.
|
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|
158,509 |
Consider a set of $N$ individuals and let their distance be given by $R$, a $N\times N$ matrix. In that, $R(1,2)$ is the distance between individual 1 and 2. Now lets say that I want to separate the individuals into $k$ subsets (or clusters) such that the distance between individuals in each subset is the minimum possible. For instance, lets assume that $N=5$, $k=3$ and $R$ is 0.00000 0.56570 0.93140 0.98990 0.98990 0.56570 0.00000 0.84850 0.64850 0.98990 0.93140 0.84850 0.00000 0.78990 1.13140 0.98990 0.64850 0.78990 0.00000 0.93140 0.98990 0.98990 1.13140 0.93140 0.00000 One possible solution is $S_1=\{ \{1,2,3\}, \{4\}, \{5\} \}$ where subset 1 is $\{1,2,3\}$, subset 2 is $\{4\}$ and subset 3 is $\{5\}$. In this case the total distance is given by
\begin{equation}
R(1,2)+R(1,3)+R(2,3)=2.3456.
\end{equation}
In the above, consider that the distance for subsets $\{4\}$ and $\{5\}$ is zero since there is only one individual in each of these subsets. Although $S_1$ is a feasible solution, it is not the optimal one. The optimal solution in this case is $S_2=\{ \{1,2,4\}, \{3\}, \{5\} \}$ since the total distance is
\begin{equation}
R(1,2)+R(1,4)+R(2,4)=2.2041,
\end{equation}
which is lower than that given by $S_1$. Is it possible to formulate this problem as a clustering problem where k is the number of clusters? Could I apply k-means? If so, what should the centroid be? Alternatively, I have been trying to formulate this problem as integer programming problem. In that, there is $N$ integer variables bounded by $k$. A solution can then be represented by $X=[X_1\ldots X_N]$ and the variables can only take values as give by $X_i \in \{1,...,k\}$, for $i=1,...N$. For the above problem, the optimal solution is given by $X=[1~1~2~1~3]$, which means that individual 1 corresponds to subset 1, individual 2 corresponds to subset 1, individual 3 corresponds to subset 2, individual 4 corresponds to subset 1, and individual 5 corresponds to subset 3. The objective function can be formulated as minimize the sum of all distances between individuals that belong to the same subset, for all subsets. One constraint for this problem is that all subsets (or clusters) need to have at least one individual. For instance, for $k=3$ the solution $X=[1~1~1~1~1]$ is unfeasible because subset 2 and subset 3 have no individuals. I think this is an NP-hard problem and it will take long for an optimization method to find the optimal solution if $N$ increases beyond some threshold. Therefore, I was thinking that formulating this problem as a clustering problem should be more appropriate. Any ideas on this?
|
No. Take $G=SO(5)$ and $H=SO(3)$, both of which have fundamental group $\mathbb{Z}/2$. I claim that there is no continuous map $f: G\to H$ which induces the identity homomorphism. If there were, then $f$ would induce a nontrivial homomorphism $f_\ast: H_1(G;\mathbb{Z}/2)\to H_1(H;\mathbb{Z}/2)$, and a graded ring homomorphism $f^\ast: H^\ast(H;\mathbb{Z}/2)\to H^\ast(G;\mathbb{Z}/2)$ which is nontrivial on $H^1$. To see that no such ring homomorphism exists, recall (see Hatcher's Algebraic Topology, section 3.D) that $$H^\ast(H;\mathbb{Z}/2)\cong \mathbb{Z}/2[\alpha_1]/(\alpha_1^4)$$ and $$H^\ast(G;\mathbb{Z}/2)\cong \mathbb{Z}/2[\beta_1]/(\beta_1^8)\otimes\mathbb{Z}/2[\beta_3]/(\beta_3^2),$$ where each $\alpha_i,\beta_i$ is in degree $i$.
|
{
"source": [
"https://mathoverflow.net/questions/158509",
"https://mathoverflow.net",
"https://mathoverflow.net/users/47383/"
]
}
|
158,769 |
To three elements $a_1$, $a_2$, $a_3$ in the finite field $\mathbb F_q$
of $q$ elements we associate the number $N(a_1,a_2,a_3)$
of elements $a_0\in \mathbb F_q$ such that the polynomial
$x^4+a_3x^3+a_2x^2+a_1x+a_0$ splits into four distinct linear factors over
$\mathbb F_q$. The number
$$\sum_{(a_1,a_2,a_3)\in\mathbb F_q^3}{N(a_1,a_2,a_3)\choose 2}$$
counts then the number of minimal pairs in the Craig lattice $C_{q-1,3}$
(for $q$ a prime number). Experimentally (for all primes up to 2000)
this number seems to be given by
$$\frac{1}{1152}q(q-1)(q^3-21q^2+171q-c_q)$$
(the factors $q$ and $(q-1)$ are easy to explain through the action of the affine group) where
$$c_q=\left\lbrace\begin{array}{ll}
455\qquad&q\equiv 5\pmod{24}\\
511&q\equiv 7\pmod{24}\\
583&q\equiv 13\pmod{24}\\
383&q\equiv 23\pmod{24}\end{array}\right.$$
and no nice formula seems to exist for the remaining cases (which after exclusion of powers of $2$ and $3$ form a subgroup of the multiplicative group
$(\mathbb Z/24\mathbb Z)^\ast$). Is there an explanation for these identities? Remark: One can of course define similarly $N(a_1,a_2)$ or $N(a_1,a_2,a_3,a_4)$.
Nice formulae for $\sum{N(\ast)\choose 2}$ exist for all primes in the first case (and are easy to prove using quadratic reciprocity). I could see nothing in the second case (computations become however quite heavy and I could not get very far).
|
For prime $q \geq 5$ write the count as
$$
\frac1{1152} q (q-1) (q^3 - 21q^2 + 171 q - c_q)
$$
where
$$
c_q = 483 + 36 \left(\frac{-1}{q}\right) + 64 \left(\frac{-3}{q}\right)
+ \delta_q.
$$
Then for $(\frac{-2}{q}) = -1$ Ronald Bacher 's calculations
indicate $\delta_q=0$. If $(\frac{-2}{q}) = +1$ then
$q$ can be written as $m^2 + 2n^2$, uniquely up to changing
$(m,n)$ to $(\pm m, \pm n)$, and we have
$$
\delta_q = 24(m^2 - 2n^2) + 192 + 72 \left(\frac{-1}{q}\right).
$$ The explanation is as follows. Start as did Will Sawin by considering the variety of $(s_1,s_2,s_3,s_4,t_1,t_2,t_3,t_4)$
such that for $i=1,2,3$ the $i$-th elementary symmetric function
of the $s$'s equals the $i$-th elem.sym.fn. of the $t$'s.
We may apply any $aX+b$ transformation to all $8$ variables,
which explains the $q(q-1)$ factor. (The factor $1152 = 2 \cdot 4!^2$
is from coordinate permutations that respect the partition of the
$8$ variables into two sets of $4$.) In odd characteristic, there's
a unique representative with $\sum_{i=1}^4 s_i = \sum_{i=1}^4 t_i = 0$;
this takes care of the translations, and then we mod out by scalars
by going to projective space. We end up with the complete intersection
of a quadric and a sextic in ${\bf P}^5$. This threefold, call it
${\cal M}$, turns out to be rational. (This has probably been known
for some time, because ${\cal M}$ classifies perfect multigrades of order $4$,
and such things have been studied since the mid-19th century, see the Prouhet-Tarry-Escott problem ; I outline a proof below.)
However, by requiring that all coordinates be distinct
we're removing some divisor ${\cal D}$ on this threefold,
so the final count decreases by the outcome of an inclusion-exclusion formula
whose terms are point counts over some subvarieties of ${\cal M}$.
Most of these sub varieties are rational curves, or points that may be defined
over ${\bf Q}(i)$ or ${\bf Q}(\sqrt{-3})$, the latter explaining
the appearance of Legendre symbols $(\frac{-1}{p})$, $(\frac{-3}{p})$
in the counting formula. But the two-dimensional components of ${\cal D}$
are isomorphic K3 surfaces, arising as a complete intersection of
a quadric and a cubic in ${\bf P}^4$; and those components make a more complicated
contribution. Fortunately these K3 surfaces are "singular" (i.e. their
Picard number attains the maximum of $20$ for a K3 surface in
characteristic zero) $-$ I computed that they're birational with
the universal elliptic curve over $X_1(8)$ $-$ and it is known that
the point-count of this singular K3 surface can be given by a formula
that involves $m^2-2n^2$ when $(\frac{-2}{q}) = +1$. To show that $\cal M$ is rational, it is convenient to apply a linear
change of variables from the "$A_3$" coordinates $s_i,t_i$ to
"$D_3$" coordinates, say $a,b,c$ and $d,e,f$, with
$$
s_i = a+b+c, \phantom+ a-b-c, \phantom+ -a+b-c, \phantom+ -a-b+c
$$
and likewise $t_i = d+e+f, \phantom. d-e-f, \phantom. -d+e-f, \phantom. -d-e+f$.
Then $\sum_{i=1}^4 s_i = \sum_{i=1}^4 t_i = 0$ holds automatically,
and the quadric and cubic become simply
$$
a^2+b^2+c^2 = d^2+e^2+f^2,
\phantom\infty
abc = def.
$$
Let $d=pa$ and $e=qb$. Then $f=(pq)^{-1}c$, and the quadric becomes
a conic in the $(a:b:c)$ plane with coefficients depending on $p,q$:
$$
(p^2-1)a^2 + (q^2-1)b^2 + ((pq)^{-2}-1) c^2 = 0.
$$
So $\cal M$ is birational to a conic bundle over the $(p,q)$ plane,
and this conic bundle has a section $(a:b:c:d:e:f) = (1:p:pq:p:pq:1)$
which lets us birationally identify $\cal M$ with the product of the
$(p,q)$ plane with ${\bf P}^1$. This is a rational threefold, QED .
|
{
"source": [
"https://mathoverflow.net/questions/158769",
"https://mathoverflow.net",
"https://mathoverflow.net/users/4556/"
]
}
|
158,781 |
A pencil of quadrics in $\mathbb{P}^n$ is a line in $\mathbb{P}^N$, where $N=\frac{n(n+3)}{2}$. So the space of pencil of quadrics is the Grassmannian $Gr(2,N+1)$. The group $SL_{n+1}(\mathbb{C})$ acts on $Gr(2,N+1)$ by $T\circ (a_1A+a_2B)=a_1T^tAT+a_2T^tBT$ for any $T\in SL_{n+1}(\mathbb{C})$ and pencil $a_1A+a_2B$.(Here, $A,B$: symmetric $(n+1)\times (n+1)$ matrices corresponding to quadrics.) Thinking $Gr(2,N+1)$ as embedded into projective space via $Pl\ddot{u}cker$ mapping we have the notion of stability of GIT for pencils of quadrics. And $D(a_1,a_2):=det(a_1A+a_2B)$ is a polynomial of degree $n+1$ in the two variables. $D(a_1,a_2)$ has roots in $\mathbb{P}^1$. There is natural $SL_{2}(\mathbb{C})$ action. From this paper , "stability of pencils of quadrics in $\mathbb{P}^n$"="stability of $n+1$ points in $\mathbb{P}^1$." I want to know I can tell same(or similar) thing about "stability of linear systems(net, web or any linear system) of quadrics". Can You give any results or references?
|
For prime $q \geq 5$ write the count as
$$
\frac1{1152} q (q-1) (q^3 - 21q^2 + 171 q - c_q)
$$
where
$$
c_q = 483 + 36 \left(\frac{-1}{q}\right) + 64 \left(\frac{-3}{q}\right)
+ \delta_q.
$$
Then for $(\frac{-2}{q}) = -1$ Ronald Bacher 's calculations
indicate $\delta_q=0$. If $(\frac{-2}{q}) = +1$ then
$q$ can be written as $m^2 + 2n^2$, uniquely up to changing
$(m,n)$ to $(\pm m, \pm n)$, and we have
$$
\delta_q = 24(m^2 - 2n^2) + 192 + 72 \left(\frac{-1}{q}\right).
$$ The explanation is as follows. Start as did Will Sawin by considering the variety of $(s_1,s_2,s_3,s_4,t_1,t_2,t_3,t_4)$
such that for $i=1,2,3$ the $i$-th elementary symmetric function
of the $s$'s equals the $i$-th elem.sym.fn. of the $t$'s.
We may apply any $aX+b$ transformation to all $8$ variables,
which explains the $q(q-1)$ factor. (The factor $1152 = 2 \cdot 4!^2$
is from coordinate permutations that respect the partition of the
$8$ variables into two sets of $4$.) In odd characteristic, there's
a unique representative with $\sum_{i=1}^4 s_i = \sum_{i=1}^4 t_i = 0$;
this takes care of the translations, and then we mod out by scalars
by going to projective space. We end up with the complete intersection
of a quadric and a sextic in ${\bf P}^5$. This threefold, call it
${\cal M}$, turns out to be rational. (This has probably been known
for some time, because ${\cal M}$ classifies perfect multigrades of order $4$,
and such things have been studied since the mid-19th century, see the Prouhet-Tarry-Escott problem ; I outline a proof below.)
However, by requiring that all coordinates be distinct
we're removing some divisor ${\cal D}$ on this threefold,
so the final count decreases by the outcome of an inclusion-exclusion formula
whose terms are point counts over some subvarieties of ${\cal M}$.
Most of these sub varieties are rational curves, or points that may be defined
over ${\bf Q}(i)$ or ${\bf Q}(\sqrt{-3})$, the latter explaining
the appearance of Legendre symbols $(\frac{-1}{p})$, $(\frac{-3}{p})$
in the counting formula. But the two-dimensional components of ${\cal D}$
are isomorphic K3 surfaces, arising as a complete intersection of
a quadric and a cubic in ${\bf P}^4$; and those components make a more complicated
contribution. Fortunately these K3 surfaces are "singular" (i.e. their
Picard number attains the maximum of $20$ for a K3 surface in
characteristic zero) $-$ I computed that they're birational with
the universal elliptic curve over $X_1(8)$ $-$ and it is known that
the point-count of this singular K3 surface can be given by a formula
that involves $m^2-2n^2$ when $(\frac{-2}{q}) = +1$. To show that $\cal M$ is rational, it is convenient to apply a linear
change of variables from the "$A_3$" coordinates $s_i,t_i$ to
"$D_3$" coordinates, say $a,b,c$ and $d,e,f$, with
$$
s_i = a+b+c, \phantom+ a-b-c, \phantom+ -a+b-c, \phantom+ -a-b+c
$$
and likewise $t_i = d+e+f, \phantom. d-e-f, \phantom. -d+e-f, \phantom. -d-e+f$.
Then $\sum_{i=1}^4 s_i = \sum_{i=1}^4 t_i = 0$ holds automatically,
and the quadric and cubic become simply
$$
a^2+b^2+c^2 = d^2+e^2+f^2,
\phantom\infty
abc = def.
$$
Let $d=pa$ and $e=qb$. Then $f=(pq)^{-1}c$, and the quadric becomes
a conic in the $(a:b:c)$ plane with coefficients depending on $p,q$:
$$
(p^2-1)a^2 + (q^2-1)b^2 + ((pq)^{-2}-1) c^2 = 0.
$$
So $\cal M$ is birational to a conic bundle over the $(p,q)$ plane,
and this conic bundle has a section $(a:b:c:d:e:f) = (1:p:pq:p:pq:1)$
which lets us birationally identify $\cal M$ with the product of the
$(p,q)$ plane with ${\bf P}^1$. This is a rational threefold, QED .
|
{
"source": [
"https://mathoverflow.net/questions/158781",
"https://mathoverflow.net",
"https://mathoverflow.net/users/47518/"
]
}
|
158,881 |
I was wondering if the Hölder's inequality was true for matrix induced norms, i.e. if
$$\|AB\|_1 \leq \|A\|_p\|B\|_q, \quad\forall p,q \in [1,\infty] \text{ s.t. } \tfrac{1}{p}+\tfrac{1}{q} = 1.$$
But it seems that this does not hold in general, in fact
$$A = \begin{bmatrix}1 & 2\\ 0 & 0 \end{bmatrix}, \; B = \begin{bmatrix}1 & 0\\ 2 & 0 \end{bmatrix}, \; p = 1, \; q = \infty $$
is a simple counterexample, and it is not hard to find similar ones for other choices of $p$ and $q$. Question Does a Hölder-like inequality hold for matrix induced norms?
|
There are (at least two) "generalizations" of Hölder inequality to the non-commutative case. One is the so called tracial matrix Hölder inequality: $$
|\langle A, B \rangle_{HS} |= |\mathrm{Tr} (A^\dagger B) | \le \| A\|_p \,\, \| B\|_q
$$ where $\| A\|_p$ is the Schatten $p$ -norm and $1/p+1/q=1$ . You can find a proof in Bernhard Baumgartner, An Inequality for the trace of matrix products, using absolute values . Another generalization is very similar to what you wrote and reads $$ \parallel|AB|\parallel \, \le\, \parallel |A|^p\parallel^{1/p} \,\, \parallel|B|^q \parallel^{1/q}
$$ where $|M|:=(M^\dagger M)^\frac12$ and it holds whenever $ \parallel \cdot \parallel$ is a unitarily invariant norm. You can find a proof in the book of Bhatia Matrix Analysis .
|
{
"source": [
"https://mathoverflow.net/questions/158881",
"https://mathoverflow.net",
"https://mathoverflow.net/users/41123/"
]
}
|
159,268 |
So I walked into this very innocent-looking combinatorics problem,
and quite soon I ended up with the problem to prove that any doubly stochastic $n \times n$ matrix has a non-zero permanent. Now clearly, this follows from the Van der Waerden conjecture (which is now a theorem), which give a lower (positive) bound for the permanent.. However, in my case, it feels like overkill to reference this theorem, so I wonder if there is some elementary argument that shows that the permanent of a doubly stochastic matrix is positive. (Although, the lower bound mentioned above converges to 0, as the matrix size grows, so it must be non-trivial...). Or, is proving that the permanent is non-zero "as hard as" proving the lower bound?
|
It's known (again using Hall's theorem, or using convex analysis) that the doubly stochastic matrices are a convex combination of the permutation matrices (these are the extreme points of the collection of doubly stochastic matrices). Accordingly each doubly stochastic matrix is a finite positive linear combination of permutation matrices. Then that the permanent is non-zero is immediate.
|
{
"source": [
"https://mathoverflow.net/questions/159268",
"https://mathoverflow.net",
"https://mathoverflow.net/users/1056/"
]
}
|
159,350 |
Let $\gamma$ be a piecewise smooth curve in $\mathbb{R}^n$. Recall that the centroid of $\gamma$ is the point $(\overline{x}, \overline{y})$ where $\overline{x}$ is the average value of $x$ on $\gamma$ and $\overline{y}$ is the average value of $y$ on $\gamma$: $$\overline{x} = \frac{1}{\text{Length}(\gamma)} \int_\gamma x\, d\gamma, \hspace{1cm} \overline{y} = \frac{1}{\text{Length}(\gamma)} \int_\gamma y\, d\gamma$$ My question is: if $\gamma_n$ is a sequence of piecewise smooth curves which converge uniformly to a piecewise smooth curve $\gamma$, is it true that $(\overline{x_n}, \overline{y_n}) \to (\overline{x}, \overline{y})$? If it is more convenient to replace "piecewise smooth" with "rectifiable" or something else, I don't mind. A hint that this might not be completely trivial is the observation that $\text{Length}(\gamma_n)$ need not converge to $\text{Length}(\gamma)$: the standard example is a sequence of finer and finer staircase curves converging uniformly to a diagonal line. However, the sequence of centroids does converge to the right limit in this example.
|
The answer is "no". In the unit square, take the "half-finished staircase" converging to left half of the diagonal line with its right half running straight along the diagonal (see drawing below). The sequence of these "combined" curves converges uniformly to the diagonal, but the centroids do not converge to the midpoint of the diagonal. Added by request: The length of the left-hand half is $\sqrt2$ times the length of the right-hand half, and the centroid of each of the two pieces is in its middle. Therefore the centroid of every curve in the sequence is the weighted average $\frac{\sqrt2(1/4,\ 1/4) + (3/4,\ 3/4)}{1+\sqrt2}\neq(1/2,1/2)$. ${\qquad\qquad\qquad}$
|
{
"source": [
"https://mathoverflow.net/questions/159350",
"https://mathoverflow.net",
"https://mathoverflow.net/users/4362/"
]
}
|
159,989 |
I am looking for a closed statement (i.e. not depending on any parameter objects) which is true in the internal logic of the topos of simplicial sets, but is not an intuitionistic tautology. Ideally, I would like it to be a simple universal statement of propositional logic (e.g. "for all propositions P, Q, and R, blah blah", like LEM or de Morgan's law). (To clarify: this has nothing to do with higher topoi or homotopy type theory; it's purely a question about ordinary 1-categorical 1-topos theory.)
|
Short answer: the Kreisel-Putnam axiom $(\lnot p \to (q \lor r)) \to ((\lnot p \to q) \lor (\lnot p \to r))$ is not an intuitionistic tautology but it is valid for any subobjects of an object in the topos of simplicial sets. The longer answer relies on an interesting characterization of the subobject classifier $\Omega$ of the topos of simplicial sets. (Thanks to Zhen Lin for helping me explain this characterization.) Sieves on $[n]$ in $\Delta$ can be identified with (possibly empty) abstract simplicial complexes $A$ with vertices drawn from $[n] = \{0,\ldots,n\}$. The sieve corresponding to $A$ consists of all order preserving maps $f:[m]\to[n]$ such that $\{f(0),\dots,f(m)\} \in A$. So, for each $n$, $\Omega(n)$ can be identified with the set of all such abstract simplicial complexes and for $g:[m]\to[n]$, $\Omega(g):\Omega(n)\to\Omega(m)$ takes each $A \in \Omega(n)$ to $B = \{x \subseteq [m] : \{g(i) \mid i \in x\} \in A\}$. Interestingly, the lattice of abstract simplicial complexes on $[n]$ (with intersection and union) is the free distributive lattice $D_{n+1}$ on $n+1$ generators with a free top element (but no free bottom element though it still has a bottom element). Every finite bounded distributive lattice is a Heyting algebra by defining implication via $$(p \to q) \equiv \bigvee \{r \mid p \land r \leq q\},$$ where the big join makes sense since there are only finitely many possibilities for $r$. The logical operators $\land,\lor,\to:\Omega\times\Omega\to\Omega$ can be computed pointwise using the Heyting algebra structure of $D_{n+1}$. In other words, ${\land}(n):\Omega(n)\times\Omega(n)\to\Omega(n)$ takes each pair of abstract simplicial complexes $A_n,B_n$ on $[n]$ to $A_n \cap B_n$, and similarly for $\lor$ and $\to$. Thus any propositional formula which is true in $D_{n+1}$ for every $n$ will be valid for generalized truth values in the topos of simplicial sets. I don't know a simple characterization of this logic but it at least satisfies the Kreisel-Putnam axiom. To verify the Kreisel-Putnam axiom in $D_{n+1}$, suppose $P$, $Q$, $R$ are abstract simplicial complexes over $[n]$. Note that $\lnot P$ consists of all nonempty $x \subseteq [n]$ that are disjoint from every element of $P$. If nonempty, which is the interesting case, this simplicial complex has a maximal element that I will denote $z$. Now $\lnot P \to (Q \lor R)$ consists of all $x \subseteq [n]$ such that $x \cap z \in Q \cup R$. Since $\lnot P \to Q$ (resp. $\lnot P \to R$) similarly consist of all $x \subseteq [n]$ such that $x \cap z \in Q$ (resp. $x \cap z \in R$), we see that $\lnot P \to (Q \lor R)$ and $(\lnot P \to Q) \lor (\lnot P \to R)$ correspond to the same abstract simplicial complexes.
|
{
"source": [
"https://mathoverflow.net/questions/159989",
"https://mathoverflow.net",
"https://mathoverflow.net/users/49/"
]
}
|
160,161 |
Is there an unlabeled locally-finite graph which is a Cayley graph of an infinitely many non-isomorphic groups with respect to suitably chosen generating sets?
|
Here I answer your additional question about finitely presented groups. The answer is then no. Affirmation. Only finitely many finitely presentable groups may have the same given Cayley graph. This works as follows. Let $X_1$ be the Cayley graph of some f.p. group. By finite presentability, there exists $n_0$ such that gluing $k$-gons each time you have a $k$-cycle in $X_1$ with $k\le n_0$, the resulting polygonal complex $X$ is simply connected. Now consider the locally compact group $G=\mathrm{Aut}(X_1)=\mathrm{Aut}(X)$. A finitely generated group with Cayley graph $X_1$ is the same as a group with a simply transitive action on $X_1$, hence letting $K$ be the stabilizer in $G$ of some vertex $x_0$, it is the same as a subgroup of $G$ whose intersection with each coset $gK$ is a singleton; let $\mathcal{W}$ be the set of subgroups of $G$ with this property. Using the action on $X$, a standard argument shows the following: let $S_G$ be the set of elements of $G$ mapping $x_0$ to a neighbor of $x_0$; then for every $\Gamma\in\mathcal{W}$, the group $\Gamma$ admits a presentation using $\Gamma\cap S_G$ as set of generators and relators of length $\le\max(2,n_0)$ (2 is necessary because of the possible generators of order 2, unless these are represented by double edges). Since there are only finitely many such presentations, we are done. Note that the proof even shows that there are finitely many marked groups $(\Gamma,S)$ with a given Cayley graph, in the finitely presentable case.
|
{
"source": [
"https://mathoverflow.net/questions/160161",
"https://mathoverflow.net",
"https://mathoverflow.net/users/10443/"
]
}
|
160,183 |
I have seen exact sequences appearing a lot in algebraic texts with different purposes. But I've never seen names of the people associated with it. Also I don't understand what's so good about showing a certain sequence is exact. Does anyone know who first invented the exact sequences or how they came to be? What were they invented for originally and what's so useful about them?
|
From Lefschetz's obituary of Hurewicz : “At a later date (1941) and in a very short abstract of this Bulletin Hurewicz introduced the concept of exact sequence whose mushroom like expansion in recent topology is well known.” Here is the abstract transcribed in its entirety: 329. Witold Hurewicz: On duality theorems . Let $A$ be a locally compact space, $B$ a closed subset of $A$ , and $H^n(A)$ , $H^n(B)$ , $H^n(A-B)$ the $n$ -dimensional cohomology groups of the sets $A$ , $B$ , and $A-B$ (with integers as coefficients). Consider “natural homomorphisms” $H^n(A)\rightarrow H^n(B)\rightarrow H^{n+1}(A-B)\rightarrow$ $H^{n+1}(A)\rightarrow H^{n+1}(A-B)$ . It can be shown that the kernel of each of these homomorphisms is the image of the preceding homomorphism. This statement contains Kolmogoroff’s generalization of Alexander’s duality theorem and has many applications. Using the preceding theorem one can prove: If $A$ and $B$ are compact spaces of dimension $n$ and $m$ respectively, the necessary and sufficient condition that the topological product $A\times B$ be of dimension $n+m$ is the existence of an open set $U \subset A$ and an open set $V \subset B$ such that $H^n(U)$ and $H^m(V)$ contain elements $\alpha$ and $\beta$ satisfying the following conditions: If the integer $d$ is a factor of the order of $\alpha$ , then $\beta \not\equiv 0\ \textrm{modulo}\ d$ (that is, there is no element $\gamma$ of $H^m(V)$ satisfying $\beta=d\gamma$ ); if the integer $e$ is a factor of the order of $\beta$ , then $\alpha \not\equiv 0\ \textrm{modulo}\ e$ . (Received May 3, 1941.) Raster images of the abstract are below.
|
{
"source": [
"https://mathoverflow.net/questions/160183",
"https://mathoverflow.net",
"https://mathoverflow.net/users/44220/"
]
}
|
160,265 |
Recently I wrote a blog post entitled "The Scientific Case for P≠NP" . The argument I tried to articulate there is that there seems to be an "invisible electric fence" separating the problems in P from the NP-complete ones, and that this phenomenon should increase our confidence that P≠NP. So for example, the NP-complete Set Cover problem is approximable in polynomial time to within a factor of ln(n), but is NP-hard to approximate to within an even slightly smaller factor (say, 0.999ln(n)). Notice that, if either the approximation algorithm or the hardness result had been just slightly better than it was, then P=NP would've followed immediately. And there are dozens of other examples like that. So, how do our algorithms and our NP-hardness results always manage to "just avoid" crossing each other, even when the avoidance requires that they "both know about" some special numerical parameter? To me, this seems much easier to explain on the P≠NP hypothesis than on the P=NP one. Anyway, one of the questions that emerged from the discussion of that post was sufficiently interesting (at least to me) that I wanted to share it on MO. The question is this: When, in the history of mathematics, have problems "like P vs. NP" arisen and then been solved? In those cases, what were the resolutions? I'd better clarify what I mean by a problem being "like P vs. NP"! I mean the following: Mathematicians managed to classify a large number of objects of interest to them into two huge classes. (Ideally, these would be two equivalence classes, like P and the NP-complete problems, which they conjectured to be disjoint. But I'd settle for two classes one of which clearly contains the other, like P and NP, as long as many of the objects conjectured to be in $\operatorname{Class}_2 \setminus \operatorname{Class}_1$ were connected to each other by a complex web of reductions, like the NP-complete problems are---so that putting one of these objects in $\operatorname{Class}_1$ would also do so to many others.) Mathematicians conjectured that the two classes were unequal, but were unable to prove or disprove that for a long time, even as examples of objects in the two classes proliferated. Eventually, the conjecture was either proved or disproved. Prior to the eventual solution, the two classes appeared to be separated by an "invisible fence," in the same sense that P and the NP-complete problems are. In other words: there were many results that, had they been slightly different (say, in some arbitrary-looking parameter), would have collapsed the two classes, but those results always stopped short of doing so. To give a sense of what I have in mind, here are the best examples we've come up with so far (I'd say that they satisfy some of the conditions above, but probably not all of them): David Speyer gave the example of Diophantine sets of integers versus recursively enumerable sets. The former was once believed to be a proper subset of the latter, but now we know they're the same. Sam Hopkins gave the example of symplectic manifolds versus Kähler manifolds. The former contains the latter, but the containment was only proved to be strict by Thurston in the 1970s. I gave the example of independence results in set theory. Until Cohen, there were many proven statements about transfinite sets, and then a whole class of other statements---V=L, GCH, CH, AC, Zorn's Lemma, well-orderability...---that were known to be interrelated by a web of implications (or equivalent, as in the last three cases), but had resisted all proof attempts. Only with forcing were the two classes "separated," an outcome that some (like Gödel) had correctly anticipated.
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This isn't an exact analogue to P != NP, in which two large classes exist and it is undecided whether they are equal or not; instead, two large "universes" exist, of which only one is the truth, with one of them strongly believed to not exist, but for which all attempts to disprove this parallel universe have been defeated by an invisible fence. (Perhaps a complexity theory analogue would be a scenario in which we knew that of Impagliazzo's five worlds , only Algorithmica or Cryptomania were possible, but we could not determine which, with both worlds showing an equal propensity to "want to exist".) Anyway, the situation is in analytic number theory, where there are two worlds (which, very roughly , would correspond to "Algorithmica" and "Cryptomania" in Impagliazzo's list): Siegel zero : The primes conspire (i.e. show extremely anomalous correlation) with some multiplicative function, such as a Dirichlet character $\chi$; roughly speaking, this means that there is some modulus q such that there is a huge bias amongst the primes to be quadratic nonresidues mod q rather than quadratic residues. (Dirichlet's theorem tells us that the bias will die down eventually - for primes exponentially larger than q - but this is not useful in many applications). The most common way to describe this scenario is through a "Siegel zero" - a zero of an L-function that is really, really far away from the critical line (and really close to 1). Weirdly, such a conspiracy actually makes many number theory problems about the primes easier than harder , because one gets to "pretend" that the Mobius function is essentially a character. For instance, there is a cute result of Heath-Brown that if there are an infinite family of Siegel zeroes, then the twin prime conjecture is true. (Basically, the principle is that at most one conspiracy in number theory can be in force for any given universe; a Siegel zero conspiracy sucks up all the "conspiracy oxygen" for a twin prime conspiracy to also hold.) It does lead to some other weird behaviour though; for instance, the existence of a Siegel zero forces many of the zeroes of the Riemann zeta function to lie on the critical line and be almost in arithmetic progression. Standard model: this is the universe which is believed to exist, in which the primes do not exhibit any special correlation with any other standard multiplicative function. In this world, GRH is believed to be true (and the zeroes should be distributed according to GUE, rather than in arithmetic progressions (this latter hypothesis has occasionally been called the " Alternative hypothesis ")). [This is an oversimplification; much as how complexity theorists have not ruled out the intermediate worlds between Algorithmica and Cryptomania, we don't have as strong of a separation between these two number-theoretic worlds as we would like. For instance there could conceivably be intermediate worlds where there are no Siegel zeroes, but GRH or GUE still fails (somewhat analogous to Impagliazzo's "Pessiland"). So in practice we have to weaken one or the other of these worlds, for instance by replacing GRH with a much weaker zero-free region. I'm glossing over these technical details though for this discussion. My feeling is that we have some chance with current technology of eliminating some more of these intermediate worlds, but we're quite stuck on eliminating either of the extreme worlds.] In many different problems in analytic number theory, one has to split into two cases depending on which of the above universes is actually the one we live in, and use different arguments for each (which is the major source of the notorious ineffectivity phenomenon in analytic number theory, that many of out estimates involve completely ineffective constants, as they could depend on the conductor of a Siegel zero, for which we have no upper bound); finding a way around this dichotomy in even just one of these problems would be a huge breakthrough and would likely lead soon to the elimination of one of these universes from consideration. But there is an invisible fence that seems to block us from doing so; both universes exhibit a surprising amount of "self-consistency", suggesting that one could modify our mathematical universe very slightly one way or the other to "force" one of the two scenarios to be in effect (a bit like how one can force P to equal or not equal NP by relativising). The " parity barrier " in sieve theory is one big (and relatively visible) section of this fence, but this fence seems to be much longer and more substantial than just this parity barrier. I discuss these things a bit more in this blog post . See also this survey of Conrey that was also linked to above.
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|
160,811 |
Admittedly this question is vague. But I hope to convey my point. Feel free to downvote this. Permit me to define prime number the following way: A number $n>1$ is a prime if all integers $d$ with $1< d \leq \sqrt{n}$ give non-zero remainders while dividing $n$. This will expose the primes 2 and 3: they qualify as primes as this condition is vacuous (there are simply no integers in that interval to check divisibility). Lot of exceptions to some theory (quadratic forms, elliptic curves, representation theory) happen at primes 2 and 3. Is it because 2 and 3 entered The Prime Club by this backdoor? It is my wild guess. Perhaps experts might be able to say if I am wrong. Or possibly there are more conceptual reasons. Kindly
enlighten me.
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I think that in different theories, there is often a "primitive" fact (which is hard to explain further) that lies at the heart of the complication you mention. Let me give examples. As for the "2 is the oddest prime" credo in number theory , often it boils down to the fact that $\mathbb{Q}$ contains exactly the second roots of unity. Or equivalently, the unit group of $\mathbb{Z}$ is $2$-torsion. I do not know if this can be embedded in a conceptual explanation; maybe it's a fact one has to live with, with ever-occuring consequences. In the theory of algebraic groups and Lie algebras , e.g. in Chevalley bases and related stuff, the coefficients will be (or have as prime factors) only $2$ or $3$. A consequence is that many integral structures are $p$-integral only for primes $\ge 5$, and this pops up again and again in the theory. See Dietrich Burde's answer for more. I think here an ultimate explanation for this occurrence of $2$ and $3$ is that they appear in the basic combinatorics of root systems. That is the "primitive" fact. As for the characteristic $2$ exception for quadratic forms , it is the non-equivalence of quadratic and symmetric bilinear forms that causes trouble. This in turn seems to be "primitive", just try to show equivalence and see that you have to invert $2$. And of course one should expect that for something quadratic, the number $2$ plays a special role. I guess if we were more interested in some tri-linear stuff, or more in things that can be given as $7$-tuples than in pairs, the cases of characteristic $3$ or $7$ would need more attention. So this translates the question into why bilinear things, and pairs, are often natural. (Remark that such a basic thing as multiplication , including Lie brackets and other non-associative stuff, is a bilinear map and thus will have a tendency to need special treatment in characteristic $2$. Same for any duality, pairings etc.) As for $2$ and $3$ as bad primes for elliptic curves , the story seems to be a little different. The answers by jmc and Joe Silverman suggest the following view: there is a family of objects (abelian varieties) which can be parametrised roughly by certain numbers (dimension), and exceptional patterns are related to this parameter; and because elliptic curves are the ones where the parameter is small, there are small numbers that behave irregular. Now one would think that this is just a high-brow version of Alex Degtyarev's comment. But there is an interesting subtlety: It is not that in the general theory there are numbers different from $2, 3$ that misbehave (while these become nice), but there are more than just them. In other words: Granted that for every single number you might find some monstrosity somewhere in the general theory. But one might find it surprising that there are some numbers that always need care, even in the most specialised, well-behaved cases. For this, I have no better explanation than: The strong law of small numbers : Small numbers (not necessarily primes) give exceptional patterns. Because naturally, there are so few of them, and they "have to satisfy too much at once". Maybe this is as far as one gets if one seeks after a common pattern between the "primitive" explanations above.
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"https://mathoverflow.net/questions/160811",
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|
160,986 |
For any commutative ring $R$ let $R[x]$ denote the ring of polynomials with coefficients in $R$. Any polynomial $p \in R[x]$ naturally induces a function $\hat{p} :R \rightarrow R$. In some cases, a nonzero polynomial will induce the zero function. For example, with $R=\mathbb{Z}_6$, the polynomials $x^5 +3x^2+2x$, $3x^4 + 5x^3 + 4x$, and $x^4 - x^3 - x^2 + x$ all induce the zero function. Some things that I already know: If $R$ is finite, $R = \{r_1, r_2, ... r_n\}$, then $(x-r_1)(x-r_2)...(x-r_n)$ induces the zero function. If $R=\mathbb{Z}/p\mathbb{Z}$ then the ideal $I$ of polynomials inducing the zero function is generated by $x^p - x$ (and in fact this coincides with the polynomial given above). If $R$ is an infinite integral domain, then no nonzero polynomial induces the zero function. If $R$ is infinite but not an integral domain, then there may or may not be nonzero polynomials that induce the zero function. (I have examples of each.) It's this last point that brings me here. I would like to know if there are necessary and sufficient conditions on $R$ that ensure that $\hat{p}=0$ only for $p=0$. Such conditions would have to be stronger than "infinite" but weaker than "infinite integral domain". Does anybody know what those conditions are?
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$R$ has a nonzero polynomial that induces the zero function if and only if there are ideals $I$, $J$ such that $I$ is nontrivial, $IJ=0$, and $R/J$ is a ring satisfying the following condition: There exists $n$ such that, for any $n$ elements $x_1, \dots x_n \in R/J$, the discriminant $\prod_{i<j} (x_i-x_j)=0$. So $R$ is glued together out of an arbitrary ring and a ring that satisfies a specific polynomial identity. Proof of only if: Let $I$ be the ideal generated by the coefficients of the nonzero polynomial and let $J$ be the ideal of zero divisors of $I$. We will show that for all $n$ elements $x_1,\dots x_n \in R$, $I\prod_{i<j} (x_i-x_j)=0$, so that $R/J$ satisfies this identity. To see this, take any $n$ elements $x_1,\dots,x_n\in R$ and take the Vandermonde matrix. Multiplying this by the vector of coefficients of the nonzero polynomial gives $0$. So multiplying on the other side by the adjugate matrix we still get $0$, but a matrix times its adjugate is just the identity times the determinant, so the determinant time each coefficient of the polynomial is $0$. The determinant of the Vandermonde matrix is just the discriminant $\prod_{i<j} (x_i-x_j)$. Proof of if: Let $a$ be a nonzero element of $I$, let $n$ be the smallest $n$ such that $a\prod_{i<j} (x_i-x_j)=0$ for all $x_1,\dots,x_n \in R$, so that we have some $y_1,\dots y_{n-1}\in R$ that satisfy $a\prod_{i<j} (y_i-y_j)\neq0$ . Then all $x$ satisfy: $$a\prod_{i<j} (y_i-y_j) \prod_{i=1}^{n-1} (x-y_i)=0$$ and the leading coefficient of that polynomial is nonzero. Since $R/I$ is arbitrary, I think we should study rings satisfying $\prod_{i<j} (x_i-x_j)=0$. Such rings have all residue fields bounded in size by $n$, so all prime ideals are maximal and their spectra are totally disconnected topological spaces. Since the geometry of this kind of space is not very easy to classify, we might want to restrict our attention to the local rings: If $R$ is a local ring with maximal ideal $m$, there is some $n$ such that, for all $x_1,\dots x_n \in R$, $\prod_{i<j} (x_i-x_j)=0$ if and only if $R/m$ finite and there is some $N$ such that every element $x\in m$ satisfies $x^N=0$. Proof of only if: If $R/m$ is infinite, take $x_1,\dots x_n$ to be lifts of distinct elements in $R/m$. Set $N=\frac{n^3-n}{6}$. For all $x \in m$ we can take $x_i= x^i$ for $1\leq i \leq n$. Then $$0=\prod_{i<j} (x_i-x_j) = \prod_{i<j} x^i (1-x^{j-i})$$ which is $x^{\frac{n^3-n}{6}}$, times a unit, so $x^N=0$. Proof of if: If $a_1,\dots,a_m$ are lifts of all the elements of the residue field, $\prod_{i=1}^m (x-a_i)^N$ is monic and vanishes for all $x$, so the bottom row of the Vandermonde matrix of any $x_1,\dots, x_{mN}$ is a weighted sum of the previous rows, and so the determinant of the Vandermonde vanishes. I don't think one can improve the classification much beyond this. Consider the examples $\mathbb Z[a,b]/(a^2,ab,2a)$, which is just slightly off from a nice infinite integral domain, but every element satisfies the identity $a x(x-1)=0$, and $\mathbb F_2[a_1,a_2,\dots]/(a_1^2,a_2^2,\dots)$, which is an extension of Manny Reyes's example where every element satisfies the identity $x^2(x-1)^2=0$. But I would be excited to see any further insights.
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"https://mathoverflow.net/questions/160986",
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|
161,768 |
The mapping class group of a manifold is the group $\pi_0 Diff(M)$ of components of the diffeomorphism group. There are several variations: oriented manifolds and orientation preserving diffeomorphisms, etc. I am interested in finding out if we know any sorts of general properties of these groups. Are they finitely presented? finitely generated? ... ? The mapping class groups of spheres can be identified, via a clutching construction, with the group of exotic spheres one dimension higher. Via Kervaire-Milnor, which relates these these groups to the stable homotopy groups of spheres, and the fact that the stable homotopy groups of spheres are finite, we know that these mapping class groups are finite. However this uses a lot of heavy machinery from surgery. The general case seems far from obvious (though maybe there is an obvious counter example out there?). I would also be interested in statements about the higher homotopy groups as well.
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These things can be pretty wild. For instance, for $n \geq 5$ the mapping class group of the $n$ -torus is not even finitely generated : it is a split extension of $\text{GL}_n(\mathbb{Z})$ by an infinite rank abelian group. See Theorem 4.1 of A. E. Hatcher, Concordance spaces, higher simple-homotopy theory, and applications. Proc. Sympos. Pure Math., XXXII, Part 1, pp. 3-21, Amer. Math. Soc., Providence, R.I., 1978. and Example 4 of Hsiang, W. C.; Sharpe, R. W. Parametrized surgery and isotopy .
Pacific J. Math. 67 (1976), no. 2, 401–459. I also recommend Hatcher's survey about the homotopy type of the diffeomorphism groups of manifolds here ; it discusses a lot of things that are known about $\pi_0$ . It should be remarked that the above trouble happens because we are considering manifolds with infinite fundamental groups, which often causes terrible things to happen in high-dimensional topology. If $M$ is a compact simply-connected manifold of dimension at least $5$ , then Sullivan proved that the mapping class group of $M$ is an arithmetic group, and hence has every possible finiteness property. This was extended to manifolds with finite fundamental groups by Triantafillou. For all of this, I recommend the survey Triantafillou, Georgia
The arithmeticity of groups of automorphisms of spaces. Tel Aviv Topology Conference: Rothenberg Festschrift (1998), 283–306,
Contemp. Math., 231, Amer. Math. Soc., Providence, RI, 1999.
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"https://mathoverflow.net/questions/161768",
"https://mathoverflow.net",
"https://mathoverflow.net/users/184/"
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|
162,030 |
This was an interesting question posed to me by a friend who is very interested in commutative algebra. It also has some nice geometric motivation. The question is in two parts. The first, as stated in the title, asks whether every Noetherian commutative ring a quotient of a Noetherian Domain? Geometrically, this question asks if every Noetherian affine scheme can be embedded as a closed subscheme of an integral scheme. The second part of the question is that if the first part is answered in the affirmative, then is every (regular) Noetherian ring a quotient of a regular Noetherian domain? Geometrically, this asks whether every affine Noetherian scheme can be embedded as a closed subscheme of a smooth integral scheme. We managed to make some progress on the first part of the question. We looked at finite products of Noetherian domains and showed that if the Noetherian domains A and B contain a common Noetherian subring C such that A and B are essentially of finite type over C, then $A \times B$ is a quotient of a Noetherian domain. But we weren't able to remove the essentially finite condition and the simplest example which we were unable to work out was $\mathbb{Q} \times \mathbb{C}$.
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No for cardinality reasons. Let $F$ a finite field and $G$ a field with cardinality strictly greater than the continuum. Then $F\times G$ is not the homomorphic image of a noetherian integral domain by lemma 2.1 in http://spot.colorado.edu/~kearnes/Papers/residue_final.pdf Lemma 2.1. Let $R$ be a Noetherian integral domain that is not a finite field and let $I$ be a proper ideal of $R$. If $|R| = \rho$ and $|R/I| = \kappa$, then $ \kappa + \aleph_0 \leq \rho \leq \kappa^{\aleph_0}$. [note: I am not an expert and have not checked.] [Edit by Joël: for convenience, I add the proof of $\rho \leq \kappa^{\aleph_0}$ taken from the cited article. Since $I$ is finitely generated, $I^n/I^{n+1}$ is a finite $R/I$-module, hence has cardinality at most $\kappa$ (resp. is finite if $\kappa$ is finite). Since $R/I^{n+1}$ is a successive extension of $I^k/I^{k+1}$ for $k=0,1,\dots,n$, the cardinality of $R/I^{n+1}$ is also at most $\kappa$ (resp. finite if $\kappa$ is finite). By Krull's lemma, $\cap_n I^n = 0$, so $R$ injects into $\prod_n R/I^n$ which has cardinality at most $\kappa^{\aleph_0}$, QED.]
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"source": [
"https://mathoverflow.net/questions/162030",
"https://mathoverflow.net",
"https://mathoverflow.net/users/48273/"
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|
162,031 |
Let $X = (X_1, X_2, \ldots , X_n)$ be an $n$-dimensional random variable, where each $X_i$ is a random variable on finite discrete set $S$. In addition, $X_i$ are independent of each other (but not identically distributed). That is, if we denote $X \sim \mathcal{D}$ and $X_i \sim \mathcal{D}_i$, then $\mathcal{D}$ is just the product of $\{ \mathcal{D}_i \}$. Similarly, let $Y = (Y_1, Y_2, \ldots , Y_n) \sim \mathcal{D}'$ and $Y_i \sim \mathcal{D}'_i$ independently on $S$ (the same $S$ as $X_i$'s). Then $\mathcal{D}'$ is the product of $\{ \mathcal{D}'_i \}$. $X$ and $Y$ are independent. Now I don't know $\mathcal{D}$ and $\mathcal{D}'$ precisely but I can sample from them (so that I can sample from all the marginal distributions $\mathcal{D}_i$ and $\mathcal{D}'_i$ as well). I would like to estimate the total variation between $\mathcal{D}$ and $\mathcal{D}'$. Let $d = \|\mathcal{D} - \mathcal{D}'\|_{TV}$ and $\hat{d}$ be our estimator for $d$. Let $N$ be the number of samples we need. I hope to get an error-confidence bound of the form
$$\Pr[ |\hat{d} - d| \ge \epsilon] \le \delta$$
where $\delta$ is at least polynomially small w.r.t. $n, |S|, N$ and $\epsilon$. I just found a similar question . However, the sample space of $X$ and $Y$ here is $S^n$, which is exponentially large, making the bound in that post not applicable to this question. Besides, here we have $\mathcal{D}$ and $\mathcal{D}'$ being product distributions, which could probably make things easier. Thank you.
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No for cardinality reasons. Let $F$ a finite field and $G$ a field with cardinality strictly greater than the continuum. Then $F\times G$ is not the homomorphic image of a noetherian integral domain by lemma 2.1 in http://spot.colorado.edu/~kearnes/Papers/residue_final.pdf Lemma 2.1. Let $R$ be a Noetherian integral domain that is not a finite field and let $I$ be a proper ideal of $R$. If $|R| = \rho$ and $|R/I| = \kappa$, then $ \kappa + \aleph_0 \leq \rho \leq \kappa^{\aleph_0}$. [note: I am not an expert and have not checked.] [Edit by Joël: for convenience, I add the proof of $\rho \leq \kappa^{\aleph_0}$ taken from the cited article. Since $I$ is finitely generated, $I^n/I^{n+1}$ is a finite $R/I$-module, hence has cardinality at most $\kappa$ (resp. is finite if $\kappa$ is finite). Since $R/I^{n+1}$ is a successive extension of $I^k/I^{k+1}$ for $k=0,1,\dots,n$, the cardinality of $R/I^{n+1}$ is also at most $\kappa$ (resp. finite if $\kappa$ is finite). By Krull's lemma, $\cap_n I^n = 0$, so $R$ injects into $\prod_n R/I^n$ which has cardinality at most $\kappa^{\aleph_0}$, QED.]
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"https://mathoverflow.net/questions/162031",
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|
162,076 |
Mathematica knows that: $$\log(n) = \lim\limits_{s \rightarrow 1} \zeta(s)\left(1 - \frac{1}{n^{(s - 1)}}\right)\;\;\;\;\;\;\;\;\;\;\;\; (1)$$ The von Mangoldt function should then be: $$\Lambda(n)=\lim\limits_{s \rightarrow 1} \zeta(s)\sum\limits_{d|n} \frac{\mu(d)}{d^{(s-1)}} \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; (2)$$ The question is if the following Riemann zeta function product is equal to the Fourier transform of the von Mangoldt function? $$\text{Fourier Transform of } \Lambda(n) \sim \sum\limits_{n=1}^{n=\infty} \frac{1}{n} \zeta(1/2+i \cdot t)\sum\limits_{d|n}\frac{\mu(d)}{d^{(1/2+i \cdot t-1)}} \;\;\;\;\;\;\; (3)$$ In the Fourier transform the von Mangoldt function has this form: $$\log (\text{scale}) ,\log (2),\log (3),\log (2),\log (5),0,\log (7),\log (2),\log (3),0,\log (11),0,...,\Lambda(\text{scale})$$ $$scale = 1,2,3,4,5,6,7,8,9,10,...k$$ Or as latex: $$\Lambda(n) = \begin{cases} \log q & \text{if }n=1, \\\log p & \text{if }n=p^k \text{ for some prime } p \text{ and integer } k \ge 1, \\ 0 & \text{otherwise.} \end{cases}$$ $$n=1,2,3,4,5,...q$$ TableForm[Table[Table[If[n == 1, Log[q], MangoldtLambda[n]], {n, 1, q}], {q, 1, 12}]] Put in pictures: Is this Dirichlet series: equal to this Fourier transform? The pictures are respectively:
Spectral Riemann zeta or Riemann zeta function product found below as Image 5.
Fourier transform of von Mangoldt function found below as Image 6. Edit 4 4 2014:
Matrix $T_2$ defined below as the Dirichlet inverse of the Euler totient function as a function of the Greatest Common Divisor (GCD) of row index $n$ and column index $k$ ; $$T_2(n,k) = a(GCD(n,k))$$ where: $$a(n) = \lim\limits_{s \rightarrow 1} \zeta(s)\sum\limits_{d|n} \mu(d)(e^{d})^{(s-1)}$$ starts: $$\displaystyle T_2 = \begin{bmatrix} +1&+1&+1&+1&+1&+1&+1&\cdots \\ +1&-1&+1&-1&+1&-1&+1 \\ +1&+1&-2&+1&+1&-2&+1 \\ +1&-1&+1&-1&+1&-1&+1 \\ +1&+1&+1&+1&-4&+1&+1 \\ +1&-1&-2&-1&+1&+2&+1 \\ +1&+1&+1&+1&+1&+1&-6 \\ \vdots&&&&&&&\ddots \end{bmatrix}
$$ By the answer at Mathematics Stackexchange given by joriki, we know that the Dirichlet series in the rows and columns tend to the von Mangoldt function $\Lambda(n)$ and $\Lambda(k)$ : $$\displaystyle \begin{bmatrix}
\frac{T_2(1,1)}{1 \cdot 1}&+\frac{T_2(1,2)}{1 \cdot 2}&+\frac{T_2(1,3)}{1 \cdot 3}+&\cdots&+\frac{T_2(1,k)}{1 \cdot k} \\
\frac{T_2(2,1)}{2 \cdot 1}&+\frac{T_2(2,2)}{2 \cdot 2}&+\frac{T_2(2,3)}{2 \cdot 3}+&\cdots&+\frac{T_2(2,k)}{2 \cdot k} \\
\frac{T_2(3,1)}{3 \cdot 1}&+\frac{T_2(3,2)}{3 \cdot 2}&+\frac{T_2(3,3)}{3 \cdot 3}+&\cdots&+\frac{T_2(3,k)}{3 \cdot k} \\
\vdots&\vdots&\vdots&\ddots&\vdots \\
\frac{T_2(n,1)}{n \cdot 1}&+\frac{T_2(n,2)}{n \cdot 2}&+\frac{T_2(n,3)}{n \cdot 3}+&\cdots&+\frac{T_2(n,k)}{n \cdot k} \end{bmatrix} = \begin{bmatrix} \frac{\infty}{1} \\ +\frac{\Lambda(2)}{2} \\ +\frac{\Lambda(3)}{3} \\ \vdots \\ +\frac{\Lambda(n)}{n} \end{bmatrix}$$ $$=\;\;\;\;\;\;\;\;\;\;\;$$ $$\displaystyle \begin{bmatrix} \frac{\infty}{1}&+\frac{\Lambda(2)}{2}&+\frac{\Lambda(3)}{3}+&\cdots&+\frac{\Lambda(k)}{k} \end{bmatrix}\;\;\;\;\;\;\;\;\;\;\;$$ Also by the answer below by GH from MO we know that the von Mangoldt function can written as a Dirichlet generating function when the $s$ goes towards the value $1$ . Numerically for non complex values of $s$ the following seems to be true: $$\displaystyle \begin{bmatrix}
\frac{T_2(1,1)}{1 \cdot 1^s}&+\frac{T_2(1,2)}{1 \cdot 2^s}&+\frac{T_2(1,3)}{1 \cdot 3^s}+&\cdots&+\frac{T_2(1,k)}{1 \cdot k^s} \\
\frac{T_2(2,1)}{2 \cdot 1^s}&+\frac{T_2(2,2)}{2 \cdot 2^s}&+\frac{T_2(2,3)}{2 \cdot 3^s}+&\cdots&+\frac{T_2(2,k)}{2 \cdot k^s} \\
\frac{T_2(3,1)}{3 \cdot 1^s}&+\frac{T_2(3,2)}{3 \cdot 2^s}&+\frac{T_2(3,3)}{3 \cdot 3^s}+&\cdots&+\frac{T_2(3,k)}{3 \cdot k^s} \\
\vdots&\vdots&\vdots&\ddots&\vdots \\
\frac{T_2(n,1)}{n \cdot 1^s}&+\frac{T_2(n,2)}{n \cdot 2^s}&+\frac{T_2(n,3)}{n \cdot 3^s}+&\cdots&+\frac{T_2(n,k)}{n \cdot k^s} \end{bmatrix} = \begin{bmatrix} \frac{\zeta(s)}{1} \\ +\frac{\zeta(s)\sum\limits_{d|2} \frac{\mu(d)}{d^{(s-1)}}}{2} \\ +\frac{\zeta(s)\sum\limits_{d|3} \frac{\mu(d)}{d^{(s-1)}}}{3} \\ \vdots \\ +\frac{\zeta(s)\sum\limits_{d|n} \frac{\mu(d)}{d^{(s-1)}}}{n} \end{bmatrix}$$ $$=\;\;\;\;\;\;\;\;\;\;\;$$ $$\displaystyle \begin{bmatrix} \frac{\infty}{1^s}&+\frac{\Lambda(2)}{2^s}&+\frac{\Lambda(3)}{3^s}+&\cdots&+\frac{\Lambda(k)}{k^s} \end{bmatrix}\;\;\;\;\;\;\;\;\;\;\;$$ but I don't know for sure. In order for the sum of columns sums to converge at all, the first term can not be allowed to be $\frac{\infty}{1}$ . Therefore I set the terms in the first column equal to zero, and the matrix becomes: $$\displaystyle \begin{bmatrix}
\frac{0}{1 \cdot 1^s}&+\frac{T_2(1,2)}{1 \cdot 2^s}&+\frac{T_2(1,3)}{1 \cdot 3^s}+&\cdots&+\frac{T_2(1,k)}{1 \cdot k^s} \\
\frac{0}{2 \cdot 1^s}&+\frac{T_2(2,2)}{2 \cdot 2^s}&+\frac{T_2(2,3)}{2 \cdot 3^s}+&\cdots&+\frac{T_2(2,k)}{2 \cdot k^s} \\
\frac{0}{3 \cdot 1^s}&+\frac{T_2(3,2)}{3 \cdot 2^s}&+\frac{T_2(3,3)}{3 \cdot 3^s}+&\cdots&+\frac{T_2(3,k)}{3 \cdot k^s} \\
\vdots&\vdots&\vdots&\ddots&\vdots \\
\frac{0}{n \cdot 1^s}&+\frac{T_2(n,2)}{n \cdot 2^s}&+\frac{T_2(n,3)}{n \cdot 3^s}+&\cdots&+\frac{T_2(n,k)}{n \cdot k^s} \end{bmatrix} = \begin{bmatrix} \frac{\zeta(s)}{1}-\frac{1}{1} \\ +\frac{\zeta(s)\sum\limits_{d|2} \frac{\mu(d)}{d^{(s-1)}}}{2} -\frac{1}{2} \\ +\frac{\zeta(s)\sum\limits_{d|3} \frac{\mu(d)}{d^{(s-1)}}}{3} -\frac{1}{3} \\ \vdots \\ +\frac{\zeta(s)\sum\limits_{d|n} \frac{\mu(d)}{d^{(s-1)}}}{n} -\frac{1}{n} \end{bmatrix}$$ $$=\;\;\;\;\;\;\;\;\;\;\;$$ $$\displaystyle \begin{bmatrix} \frac{\Lambda(1)}{1^s}&+\frac{\Lambda(2)}{2^s}&+\frac{\Lambda(3)}{3^s}+&\cdots&+\frac{\Lambda(k)}{k^s} \end{bmatrix}\;\;\;\;\;\;\;\;\;\;\;$$ So if this is correct then the sum of row sums and the sum of column sums would be: $$-\frac{\zeta'(s)}{\zeta(s)}=\sum\limits_{k=1}^{\infty} \frac{\Lambda(k)}{k^s} = \sum\limits_{n=1}^{\infty} \left(\frac{1}{n}\zeta(s)\sum\limits_{d|n} \frac{\mu(d)}{d^{(s-1)}}-\frac{1}{n}\right)\;\;\;\;\;\;\;\;\;\;\;\;(4)
$$ as in wikipedia about the von Mangoldt function. This leads to the question: Derivative of Riemann zeta, is this inequality true? Answered by Raymond Manzoni. Edit Apr 2 2014 after GH from MO answer below: The algorithm originating from Michael Rubinstein -> Brian Conrey -> Chris King -> Heike,
with the modification that we let: $$\Lambda(1) = H_{\text{scale}}$$ $$f(n) = H_{\text{scale}} + \sum _{n=2}^{n=scale} \Lambda (n)$$ Then $$\text{Fourier Transform of } \Lambda(n) = g(k)$$ $$ g(k) = \frac{\Im(\sum _{x=0}^{\log (\text{scale})} \left(e^x\right)^{(i k+\frac{1}{2}-1)} \left(f \left\lfloor e^x\right\rfloor -e^x\right))}{\text{length of vector x}}$$ which simplified is: $$ g(k) = \Im(\sum _{x=0}^{\log (\text{scale})} \left(e^x\right)^{(i k+\frac{1}{2}-1)} \left(f \left\lfloor e^x\right\rfloor -e^x\right))$$ according to the algorithm, where "scale" is some number that ideally should go to infinity, but for computation is set to some chosen large number. Again the input (f[[Floor[x]]] - x) to the Fourier transform In the algorithm; Clear[f, x]
scale = 1000;
f = ConstantArray[0, scale];
f[[1]] = N@HarmonicNumber[scale];
Monitor[Do[
f[[i]] = N@MangoldtLambda[i] + f[[i - 1]], {i, 2, scale}], i]
xres = .002;
x = Exp[Range[0, Log[scale], xres]];
tmax = 60;
tres = .015;
Monitor[errList =
Table[(f[[Floor[x]]] - x), {k, Range[0, 60, tres]}][[1 ;; 20]];, k]
ListLinePlot[errList, DataRange -> {0, 60}, PlotRange -> {-10, 15},
PlotStyle -> Blue, ImageSize -> Large] looks like this: From this one can conjecture that the Fourier transform of an
exponential sawtooth wave will give the Riemann zeta function. Then going back from the Riemann zeta function to the original
input domain with the following two Mathematica lines: x = Exp[Range[0, Log[scale], xres]];
a = -FourierDCT[Log[x]*FourierDST[(SawtoothWave[x] - 1)*(x)^(-1/2)]]; as part of a longer code snippet, where FourierDCT is the Fourier discrete cosine transform and FourierDST is the Fourier discrete sine transform, one gets the following plot: Mimicing this plot I have written: (*Mathematica 8*)
Clear[x, xx]
scale = 200;
xres = .001;
x = Exp[Range[0, Log[scale], xres]];
xx = Flatten[{0, Differences[Floor[Exp[Range[0, Log[scale], xres]]]]}];
ListLinePlot[xx*x^(-1/2), PlotRange -> {-0.1, 0.8},
ImageSize -> Large] Although this kind of numerical mimicing does not anymore give accurate plots of the Riemann zeta function, I have based on it tried to imagine what the von Mangoldt function to be Fourier transformed, might look like with the following program: (*Mathematica 8*)
Clear[x, xx]
scale = 200;
xres = .001;
x = Exp[Range[0, Log[scale], xres]];
xx = Flatten[{0, Differences[Floor[Exp[Range[0, Log[scale], xres]]]]}];
yy = Accumulate[xx]*xx;
zz = Table[
If[yy[[i]] == 0, 0, MangoldtLambda[yy[[i]]]], {i, 1, Length[yy]}];
ListLinePlot[zz*x^(-1/2), PlotRange -> {-0.1, 0.8},
ImageSize -> Large] which gives this plot: But this probably far from the correct input signal to the Fourier transform
to get the similarity with the Riemann zeta function product involving
the Möbius function. To find the correct logarithmic Dirac comb like input one should probably look at the exponential sawtooth for each row in the von Mangoldt function Matrix $T_2$ below, see what sawtooth wave each accumulated row is, and Fourier transform it twice. Then possibly one could match it to a variant of the von Mangoldt function. A hint of what it might be is given in this plot: Mobius function -> Dirichlet series -> Riemann zeta function product -> Fourier transform -> von Mangoldt function: for which the program is found lower down. In the question I ask: Is it equal to? What I mean in the third equation $(3)$ with the $\sim$ symbol is if the left hand side and the right hand side are equal to each other if you multiply one side with a scale factor $a$ like this $$\Im(\sum _{x=0}^{\log (\text{scale})} \left(e^x\right)^{(i k+\frac{1}{2}-1)} \left(f \left\lfloor e^x\right\rfloor -e^x\right)) = a \cdot \sum\limits_{n=1}^{n=\infty} \frac{1}{n} \zeta(1/2+i \cdot t)\sum\limits_{d|n} \frac{\mu(d)}{d^{(1/2+i \cdot t-1)}} \;\;\;\;\;\;\; (3)$$ Introduction: In the case of the logarithm of $2$ , log(2), this is equal to a table (red and green) called $\zeta(s)$ : $k \mid n : 1$ else $0$ . matrix multiplied by a second table (red, green and yellow) called $(1 - 2^{1-s})$ : $n=k: 1$ else if $n=2 \cdot k: -2$ else $0$ Picture of tables $\zeta(s)$ and $(1 - 2^{1-s})$ [Image 1]: Taking the matrix product of the two tables above we get [Image 2]: The first column then has the numerators of the alternating series. This is as said: $$\log(2) = \lim\limits_{s \rightarrow 1} \zeta(s)\left(1 - 2^{(1 - s)}\right)$$ which with a sign change in the exponent is equal to: $$\log(2) = \lim\limits_{s \rightarrow 1} \zeta(s)\left(1 - \frac{1}{2^{(s - 1)}}\right)$$ Before I get to the actual question I show the Dirichlet series for logarithm of n. We had already: $$1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}... =\log(2)$$ which is the second column in: Logarithms of $n$ : $$T_1 = \begin{bmatrix} 0&0&0&0&0&0&0 \\ 1&-1&1&-1&1&-1&1 \\ 1&1&-2&1&1&-2&1 \\ 1&1&1&-3&1&1&1 \\ 1&1&1&1&-4&1&1 \\ 1&1&1&1&1&-5&1 \\ 1&1&1&1&1&1&-6 \end{bmatrix}$$ It turns out that: $$1+\frac{1}{2}-\frac{2}{3}+\frac{1}{4}+\frac{1}{5}-\frac{2}{6}... =\log(3)$$ and: $$\displaystyle \log(n)=\sum\limits_{k=1}^{\infty}\frac{T(n,k)}{k}$$ in general. That is same as the Mathematica known formula above. The recurrence describing matrix $T_1$ is: $$\displaystyle T_1(1,n)=0; n>1: T_1(n,k) = \sum\limits_{i=1}^{n-1} T_1(n,k-i)$$ It is natural to ask what this recurrence might do when applying it symmetrically,
like this: $$\displaystyle T_2(n,1)=1, T_2(1,k)=1, n>=k: -\sum\limits_{i=1}^{k-1} T_2(n-i,k), n<k: -\sum\limits_{i=1}^{n-1} T_2(k-i,n)$$ von Mangoldt function matrix: $$\displaystyle T_2 = \begin{bmatrix} +1&+1&+1&+1&+1&+1&+1&\cdots \\ +1&-1&+1&-1&+1&-1&+1 \\ +1&+1&-2&+1&+1&-2&+1 \\ +1&-1&+1&-1&+1&-1&+1 \\ +1&+1&+1&+1&-4&+1&+1 \\ +1&-1&-2&-1&+1&+2&+1 \\ +1&+1&+1&+1&+1&+1&-6 \\ \vdots&&&&&&&\ddots \end{bmatrix}
$$ Here we then have $\log(2)$ in the second column and second row. Like wise $\log(3)$ in third row and third column. By doing a Möbius inversion on each row or column one can conjecture that the von Mangoldt function is found as: $$\Lambda(n)=\lim\limits_{s \rightarrow 1} \zeta(s)\sum\limits_{d|n} \frac{\mu(d)}{d^{(s-1)}}$$ Alternatively by inspecting the numbers in the decimal digits one also finds conjecturally that: $$\Lambda(k)=\sum\limits_{n=1}^{\infty} \frac{T_2(n,k)}{n}$$ Also, matrix $T_2$ is a matrix with the Greatest Common Divisor as lookup index: $$T_2(n,k) = a(GCD(n,k))$$ where conjecturally: $$a(n) = \lim\limits_{s \rightarrow 1} \zeta(s)\sum\limits_{d|n} \mu(d)(e^{d})^{(s-1)}$$ which is also known as the Dirichlet inverse of the Euler Totient starting: $$1,-1,-2,-1,-4,+2,-6,...$$ And as a side note we can by a result of Wolfgang Schramm calculate the Möbius function from this matrix as: $$\displaystyle \mu(n) = \frac{1}{n} \sum\limits_{k=1}^{k=n} T_2(n,k) \cdot e^{i 2 \pi \frac{k}{n}}$$ So much for the elementary introduction. Next we will look at the function: $$\Lambda(n)=\lim\limits_{s \rightarrow 1} \zeta(s)\sum\limits_{d|n} \frac{\mu(d)}{d^{(s-1)}}$$ in the complex plane where $$s=a+ib$$ is a complex number, and thereby get closer to the question, The plot of this I call the Riemann zeta function product. With a modification of Jeffrey Stopple's code; Show[Graphics[
RasterArray[
Table[Hue[
Mod[3 Pi/2 +
Arg[Sum[Zeta[sigma + I t]*
Total[1/Divisors[n]^(sigma + I t - 1)*
MoebiusMu[Divisors[n]]]/n, {n, 1, 30}]],
2 Pi]/(2 Pi)], {t, -30, 30, .1}, {sigma, -30, 30, .1}]]],
AspectRatio -> Automatic] it looks like this: The Riemann zeta function product (30-th partial sum) [Image 3]: Again as a sidenote we can compare with the usual Riemann zeta function: Normal or usual zeta [Image 4]: which is plotted in the same (complex) region as the Riemann zeta function product (-30 to +30 and -30i to +30i). The critical strip of the Riemann zeta function product; scale = 50; (*scale = 5000 gives the plot below*)
Print["Counting to 60"]
Monitor[g1 =
ListLinePlot[
Table[Re[
Zeta[1/2 + I*k]*
Total[Table[
Total[MoebiusMu[Divisors[n]]/Divisors[n]^(1/2 + I*k - 1)]/(n*
k), {n, 1, scale}]]], {k, 0 + 1/1000, 60, N[1/6]}],
DataRange -> {0, 60}, PlotRange -> {-0.15, 1.5}], Floor[k]] looks like this: The Riemann zeta function product [Image 5]: What a coincidence then that if we take the Fourier transform of the von Mangoldt function with Heike 's algorithm; Clear[f]
scale = 100000;
f = ConstantArray[0, scale];
f[[1]] = N@HarmonicNumber[scale];
Monitor[Do[
f[[i]] = N@MangoldtLambda[i] + f[[i - 1]], {i, 2, scale}], i]
xres = .002;
xlist = Exp[Range[0, Log[scale], xres]];
tmax = 60;
tres = .015;
Monitor[errList =
Table[(xlist^(1/2 + I k - 1).(f[[Floor[xlist]]] - xlist)), {k,
Range[0, 60, tres]}];, k]
ListLinePlot[Im[errList]/Length[xlist], DataRange -> {0, 60},
PlotRange -> {-.01, .15}] where we have set the first term in the von Mangoldt function sequence: Table[Limit[Zeta[s]*Total[MoebiusMu[Divisors[n]]/Divisors[n]^(s - 1)], s -> 1], {n, 1, 32}] $$\infty ,\log (2),\log (3),\log (2),\log (5),0,\log (7),\log (2),\log (3),0,\log (11),0,$$ equal to a Harmonic number: $$H_{\text{scale}} ,\log (2),\log (3),\log (2),\log (5),0,\log (7),\log (2),\log (3),0,\log (11),0,...,\Lambda(\text{scale})$$ and where scale is the scale is the scale of the Fourier transform matrix and the number of terms used in the von Mangoldt function sequence, we get the same plot: Fourier transform of von Mangoldt function [Image 6]: This is unclear to me. But the parameter scale in the program anyways, is the value of the Harmonic Number $H_{\text{scale}}$ . Put simpler I have taken a matrix $A$ : $$ A = \left(
\begin{array}{cccccccccc}
1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
2^{\frac{1}{2}-i t} & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
3^{\frac{1}{2}-i t} & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
4^{\frac{1}{2}-i t} & 2^{\frac{1}{2}-i t} & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\
5^{\frac{1}{2}-i t} & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\
6^{\frac{1}{2}-i t} & 3^{\frac{1}{2}-i t} & 2^{\frac{1}{2}-i t} & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\ 7^{\frac{1}{2}-i t} & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\
8^{\frac{1}{2}-i t} & 4^{\frac{1}{2}-i t} & 0 & 2^{\frac{1}{2}-i t} & 0 & 0 & 0 & 1 & 0 & 0 \\ 9^{\frac{1}{2}-i t} & 0 & 3^{\frac{1}{2}-i t} & 0 & 0 & 0 & 0 & 0 & 1 & 0 \\
10^{\frac{1}{2}-i t} & 5^{\frac{1}{2}-i t} & 0 & 0 & 2^{\frac{1}{2}-i t} & 0 & 0 & 0 & 0 & 1 \end{array}\right)$$ calculated it's matrix inverse and multiplied by the Riemann zeta function: $$B=\zeta(\frac{1}{2}+i t)\left(
\begin{array}{cccccccccc}
1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
-2^{\frac{1}{2}-i t} & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
-3^{\frac{1}{2}-i t} & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & -2^{\frac{1}{2}-i t} & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\
-5^{\frac{1}{2}-i t} & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\
6^{\frac{1}{2}-i t} & -3^{\frac{1}{2}-i t} & -2^{\frac{1}{2}-i t} & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\ -7^{\frac{1}{2}-i t} & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & -2^{\frac{1}{2}-i t} & 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & -3^{\frac{1}{2}-i t} & 0 & 0 & 0 & 0 & 0 & 1 & 0 \\ 10^{\frac{1}{2}-i t} & -5^{\frac{1}{2}-i t} & 0 & 0 & -2^{\frac{1}{2}-i t} & 0 & 0 & 0 & 0 & 1\end{array}\right)$$ and then plotted the function that is matrix $B$ . What I seem to get is the Riemann zeta zero spectrum as a Riemann zeta function product resembling the Fourier transform of the von Mangoldt function where $\Lambda(1) = H_{scale}$ So the question is if image 5 and image 6 are equivalent, differing only by a scale factor? Or as in the title: Is the Riemann zeta function product $f(t)$ (Spectral Riemann zeta): $$f(t)=\sum\limits_{n=1}^{n=\infty} \frac{1}{n} \zeta(1/2+i \cdot t)\sum\limits_{d|n} \frac{\mu(d)}{d^{(1/2+i \cdot t-1)}}$$ similar or equal to the Fourier transform of: $$\Lambda(n)=\lim\limits_{s \rightarrow 1} \zeta(s)\sum\limits_{d|n} \frac{\mu(d)}{d^{(s-1)}}$$ times a scale factor? In support of this claim is that if we take the Fourier transform of the Riemann zeta function product at the critical line (real part equal 1/2); Clear[n, k, t, A, nn, B]
nn = 60
A = Table[
Table[If[Mod[n, k] == 0, 1/(n/k)^(1/2 + I*t - 1), 0], {k, 1,
nn}], {n, 1, nn}]; MatrixForm[A];
B = FourierDCT[
Table[Total[
1/Table[n, {n, 1, nn}]*(Total[
Transpose[Re[Inverse[A]*Zeta[1/2 + I*t]]]] - 1)], {t, 1/1000,
600, N[1/6]}]];
g1 = ListLinePlot[B[[1 ;; 700]], DataRange -> {0, 60},
PlotRange -> {-7, 25}];
mm = 11.35/Log[2];
g2 = Graphics[
Table[Style[Text[n, {mm*Log[n], 5 - (-1)^n}],
FontFamily -> "Times New Roman", FontSize -> 14], {n, 1, 16}]];
Show[g1, g2, ImageSize -> Large] we get: Mobius function -> Dirichlet series -> The Riemann zeta function product -> Fourier transform -> von Mangoldt function: which looks a lot like the von Mangoldt function in Heikes algorithm, in the way it is input there. Notice that the line in the last program above B[[1 ;; 700]]*Table[Sqrt[n], {n, 1, 700}] is not analytically correct. A larger image of the same plot: http://i.stack.imgur.com/02A1p.jpg The partial sums of the Riemann zeta function product: 12 first curves together or partial sums: and a larger plot: http://i.stack.imgur.com/5LirM.jpg This whole question has previously been asked here: https://math.stackexchange.com/questions/424530/is-this-similarity-to-the-fourier-transform-of-the-von-mangoldt-function-real Clear[n, k, t, A, nn, h]
nn = 60;
h = 2; (*h=2 gives log 2 operator, h=3 gives log 3 operator and so on*)
A = Table[
Table[If[Mod[n, k] == 0,
If[Mod[n/k, h] == 0, 1 - h, 1]/(n/k)^(1/2 + I*t - 1), 0], {k, 1,
nn}], {n, 1, nn}];
MatrixForm[A];
g1 = ListLinePlot[
Table[Total[
1/Table[n*t, {n, 1, nn}]*
Total[Transpose[Re[Inverse[A]*Zeta[1/2 + I*t]]]]], {t, 1/1000,
nn, N[1/6]}], DataRange -> {0, nn}, PlotRange -> {-3, 7}];
mm = N[2*Pi/Log[h], 12]
g2 = Graphics[
Table[Style[Text[n*2*Pi/Log[h], {mm*n, 1}],
FontFamily -> "Times New Roman", FontSize -> 14], {n, 1, 32}]];
Show[g1, g2, ImageSize -> Large] Matrix inverse of Riemann zeta times log 2 operator: Clear[n, k, t, A, nn, dd]
dd = 220;
Print["Counting to ", dd]
nn = 20;
A = Table[
Table[If[Mod[n, k] == 0, 1/(n/k)^(1/2 + I*t - 1), 0], {k, 1,
nn}], {n, 1, nn}];
Monitor[g1 =
ListLinePlot[
Table[Total[
1/Table[n*t, {n, 1, nn}]*
Total[Transpose[
Re[Inverse[
IdentityMatrix[nn] + (Inverse[A] - IdentityMatrix[nn])*
Zeta[1/2 + I*t]]]]]], {t, 1/1000, dd, N[1/100]}],
DataRange -> {0, dd}, PlotRange -> {-7, 7}];, Floor[t]]
mm = N[2*Pi/Log[2], 20]
g2 = Graphics[
Table[Style[Text[n, {mm*n, 1}], FontFamily -> "Times New Roman",
FontSize -> 14], {n, 1, 32}]];
Show[g1, g2, ImageSize -> Large] Matrix Inverse of (matrix inverse times zeta function (on critical line)): The following is a relationship: Let $\mu(n)$ be the Möbius function, then: $$a(n) = \sum\limits_{d|n} d \cdot \mu(d)$$ $$T(n,k)=a(GCD(n,k))$$ $$T = \left( \begin{array}{ccccccc} +1&+1&+1&+1&+1&+1&+1&\cdots \\ +1&-1&+1&-1&+1&-1&+1 \\ +1&+1&-2&+1&+1&-2&+1 \\ +1&-1&+1&-1&+1&-1&+1 \\ +1&+1&+1&+1&-4&+1&+1 \\ +1&-1&-2&-1&+1&+2&+1 \\ +1&+1&+1&+1&+1&+1&-6 \\ \vdots&&&&&&&\ddots \end{array} \right)$$ $$\sum\limits_{k=1}^{\infty}\sum\limits_{n=1}^{\infty} \frac{T(n,k)}{n^c \cdot k^s} = \sum\limits_{n=1}^{\infty} \frac{\lim\limits_{z \rightarrow s} \zeta(z)\sum\limits_{d|n} \frac{\mu(d)}{d^{(z-1)}}}{n^c} = \frac{\zeta(s) \cdot \zeta(c)}{\zeta(c + s - 1)}$$ which is part of the limit: $$\frac{\zeta '(s)}{\zeta (s)}=\lim_{c\to 1} \, \left(\zeta (c)-\frac{\zeta (c) \zeta (s)}{\zeta (c+s-1)}\right)$$ $$\int \frac{\zeta '(s)}{\zeta (s)} ds = \log(\zeta(s))$$ $$\exp(\log(\zeta(s)))$$ $$=\zeta(s)$$ Plot of the Dirichlet generating function for the matrix on the critical line: Clear[s, c, t]
c = 1 + 1/100;
Plot[Re[Zeta[1/2 + I*t]*Zeta[c]/Zeta[1/2 + I*t + c - 1]], {t, 0, 60},
PlotRange -> {-5, 105}]
|
Your first two relations follow quickly from the asymptotic behavior of the functions $s\mapsto\zeta(s)$ and $s\mapsto d^{1-s}$ as $s\to 1$. Let me demonstrate this for the second relation for $n>1$ (note that it fails for $n=1$). For $|s-1|<1$ we have
$$\zeta(s)=1/(s-1)+O(1)\quad\text{and}\quad d^{1-s}=1-(s-1)\log d+O_d((s-1)^2),$$ hence
$$ \sum_{d\mid n}\frac{\mu(d)}{d^{s-1}} =\sum_{d\mid n}\mu(d)-(s-1)\sum_{d\mid n}\mu(d)\log d+O_n((s-1)^2).$$
On the right hand side the first sum is zero, and the second sum is $-\Lambda(n)$, because the convolution of the Möbius function with $1$ and $\log$ at $n>1$ equals $0$ and $\Lambda(n)$, respectively. It follows that
$$ \sum_{d\mid n}\frac{\mu(d)}{d^{s-1}} \sim (s-1)\Lambda(n)\quad\text{as}\quad s\to 1, $$
whence
$$ \lim_{s\to 1}\zeta(s)\sum_{d\mid n}\frac{\mu(d)}{d^{s-1}} =\Lambda(n).$$ Now I don't understand your main question. What do you mean by the Fourier transform of $\Lambda(n)$? The Fourier transform is initially defined for sequences in $L^1(\mathbb{Z})$, and then by a limit procedure for sequences in $L^2(\mathbb{Z})$. However, $\Lambda(n)$ is not even bounded. Please define precisely the left hand side of your third display, and explain what you mean by the symbol $\sim$ there. P.S. Similarly, I don't understand what you mean by $f(t)$. You define it by an infinite series, but I don't see that the series converges.
|
{
"source": [
"https://mathoverflow.net/questions/162076",
"https://mathoverflow.net",
"https://mathoverflow.net/users/25104/"
]
}
|
162,097 |
Most books on commutative algebra explain Grobner bases in the non graded case and minimal free resolutions in the local case. I like projective geometry and want to compute the minimal free resolution of a coherent sheaf on projective space. What follows is my understanding of how this should work, along with some points at which I am confused: We are given a coherent sheaf $ \mathscr{F}$ on $ \mathbb{P}^n$ via generators and relations. That is, we have an exact sequence
$$ \oplus \mathscr{O}(-b^2_{p}) \to \oplus \mathscr{O}(-b^1_{p}) \to \mathscr{F} \to 0 $$
with $ b^2_{p} \geq b^1_{q} $ for each $ p,q$. The first map in this sequence is given by a matrix $ (F_{ij})$ of homogeneous polynomials such that $ F_{ij}$ has degree $ b^2_j - b^1_i$. For the resolution which we are going to compute to be minimal we want $b_p^2 > b^1_q$ for each $p,q$. This brings us to our first question: Question 1: How can we modify the presentation so that $ b^2_p > b_q^1$? Lets continue assuming that $ b_p^2 > b_q^1 $. Write $ S$ for the homogeneous coordinate ring of $ \mathbb{P}^n$. Then we have a map $ (F_{ij}) : \oplus S \to \oplus S $. Suppose that we run Schreyer's Algorithm as explained on page 338 of Eisenbud's book. This gives us a matrix $(G_{ij}) : \oplus S \to \oplus S $ such that
$$ \oplus S \xrightarrow{(G_{ij})} \oplus S \xrightarrow{(F_{ij})} \oplus S $$
Is exact. Question 2: If we run the version of Schreyer's Algorithm from Eisenbud's book, then will it be true that the $ G_{ij}$ are homogeneous? Also, is there any way to modify the algorithm to ensure that the $G_{ij}$ all have positive degree? Suppose that now we have produced a free resolution $ \oplus S \to \oplus S \to \dots \to \oplus S $ where the matrix entries are all homogeneous with positive degree. By modifying the gradings on all the copies of $S$, we do not destroy exactness, so we have an exact sequence of graded maps
$$ \oplus S (-b^d_p) \to \dots \to \oplus S(-b^2_p) \to \oplus S(-b^1_p) $$
where $b_p^i > b_q^{i-1}$ for all $p,q$. Applying the graded $ \tilde{} $ functor gives us a free resolution of $ \mathscr{F}$. Question 3: Is the overall picture correct? One of the main things I am worried about if forgetting the grading when we actually do the Grobner bases computation, but if we try and carry it along then things get really messy (at least for a human). OK, I guess that is it. Thanks for reading all that!
|
Your first two relations follow quickly from the asymptotic behavior of the functions $s\mapsto\zeta(s)$ and $s\mapsto d^{1-s}$ as $s\to 1$. Let me demonstrate this for the second relation for $n>1$ (note that it fails for $n=1$). For $|s-1|<1$ we have
$$\zeta(s)=1/(s-1)+O(1)\quad\text{and}\quad d^{1-s}=1-(s-1)\log d+O_d((s-1)^2),$$ hence
$$ \sum_{d\mid n}\frac{\mu(d)}{d^{s-1}} =\sum_{d\mid n}\mu(d)-(s-1)\sum_{d\mid n}\mu(d)\log d+O_n((s-1)^2).$$
On the right hand side the first sum is zero, and the second sum is $-\Lambda(n)$, because the convolution of the Möbius function with $1$ and $\log$ at $n>1$ equals $0$ and $\Lambda(n)$, respectively. It follows that
$$ \sum_{d\mid n}\frac{\mu(d)}{d^{s-1}} \sim (s-1)\Lambda(n)\quad\text{as}\quad s\to 1, $$
whence
$$ \lim_{s\to 1}\zeta(s)\sum_{d\mid n}\frac{\mu(d)}{d^{s-1}} =\Lambda(n).$$ Now I don't understand your main question. What do you mean by the Fourier transform of $\Lambda(n)$? The Fourier transform is initially defined for sequences in $L^1(\mathbb{Z})$, and then by a limit procedure for sequences in $L^2(\mathbb{Z})$. However, $\Lambda(n)$ is not even bounded. Please define precisely the left hand side of your third display, and explain what you mean by the symbol $\sim$ there. P.S. Similarly, I don't understand what you mean by $f(t)$. You define it by an infinite series, but I don't see that the series converges.
|
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|
162,125 |
I have a student who is looking to start a PhD in pure mathematics. She is talented and motivated, and will do quite well. She is still in a phase of her development where she is still open to the possibility of working in a broad range of research areas. In order to offset the risk of not finding an academic job post-PhD she would like to write a dissertation that will give her increased likelihood of finding work in industry. She wants to do research in pure mathematics however, i.e. prefers proving theorems to writing code or testing models. Question: What areas of pure mathematics research are best for a post-PhD transition to industry? Please be as specific as possible. Answers to the questions here and here are certainly relevant, but these questions are obviously distinct from my question. I think this question is useful to the pure mathematics community in that it addresses the fact that there are so many qualified academic job applicants in recent years. (For this reason, I hope the community gives the question a chance.)
|
Probability. A strong background in probability will permit to qualify for jobs in
statistics and financial math. See AMS Notices where a lot of statistics about the
employment of recent PhD is published yearly. And a salary survey. EDIT. My second guess was combinatorics/coding theory or PDE.
But my friend explained me that PURE combinatorics is not so hot in the (industrial)
job market,
coding theory is not pure math, and pure PDE is very different from numerical PDE,
the last thing is of course in great demand. EDIT2. Reply on Peter Shor's comment. The distinction between "pure" and "applied" math is not sharp. On my opinion, if a problem arises from a "real world application", it is
applied math (like math physics, control theory, coding theory). If a problem arises
from the inner logic of development of math then it is pure math (like Fermat's last theorem). But of course, the difference is fuzzy, and one can trace almost all math problems to the "real world". Frequently it happens like this: a problem arises in the
real world, mathematicians like it, and start working on it, and then work and work,
forgetting its real world origin. (Example: constructions with compass and ruler, etc.)
Probability was also applied math at its origin.
And it becomes pure math. Similar things we see in coding theory, math physics and computer science.
A lot of "computer science" is pure math.
|
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|
162,583 |
99-Graph: Is there a graph with 99 vertices in which every edge (i.e. pair of joined vertices) belong to a unique triangle and every nonedge (pair of unjoined vertices) to a unique quadrilateral?
|
First we will prove the graph is regular. Let $x,y$ be two non-adjacent vertices, and let $a,b$ be their common neighbours. Define $X$ to be the neighbourhood of $x$ other than $a,b$, and $Y$ to be the neighbourhood of $y$ other than $a,b$. Considering the edge $ax$, there is a unique vertex $u\in X$ adjacent to both of them. Considering the non-edge $yu$, there must be exactly one edge from $u$ to $Y$. Similarly for the common neighbourbour of $b$ and $x$. For a vertex in $v\in X$ not adjacent to $a$ or $b$, the two common neighbours of $v$ and $y$ must lie in $Y$. Consider the bipartite graph with parts $X,Y$ and the edges between them. We have proved that each part has 2 vertices of degree 1 and the others of degree 2. This is only possible if $|X|=|Y|$, which proves that $x$ and $y$ have the same degree. This proves the graph is regular. A simple count shows the degree must be 14. Now you are looking for a strongly-regular graph of order 99, degree 14, $\lambda=1$ and $\mu=2$. According to Andries Brouwer's table , the existence is unknown.
|
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|
162,836 |
I'm trying to understand why John Nash's 1950 2-page paper that was published in PNAS was such a big deal. Unless I'm mistaken, the 1928 paper by John von Neumann demonstrated that all n-player non-cooperative and zero-sum games possess an equilibrium solution in terms of pure or mixed strategies. From what I understand, Nash used fixed point iteration to prove that non-zero-sum games would also have the analogous result. Why was this such a big deal in light of the earlier work by von Neumann? There are two references I provide that are good: One is this discussion on simple proofs of Nash's theorem and this one is a very well done (readable and accurate) survey of the history in PNAS.
|
I think von Neumann dealt with the case $n=2$, and it was by no means obvious how to extend the concept of equilibrium for the general case and prove that it always exists. More precisely, $n$ players before Nash were reduced to the $n=2$ case by partioning the players into two groups in all possible ways. Once you regard several players as a single player, they are meant to cooperate as they must act like a single player. Nash is very clear about this in his 1951 Annals paper: Von Neumann and Morgenstern have developed a very fruitful theory of
two-person zero-sum games in their book Theory of Games and Economic Behavior.
This book also contains a theory of $n$-person games of a type which
we would call cooperative. This theory is based on an analysis of the interrelationships
of the various coalitions which can be formed by the players of the
game. Our theory, in contradistinction, is based on the absence of coalitions in that
it is assumed that each participant acts independently, without collaboration or
communication with any of the others. The notion of an equilibrium point is the basic ingredient in our theory. This
notion yields a generalization of the concept of the solution of a two-person zero-sum game.
It turns out that the set of equilibrium points of a two-person zero-sum game is
simply the set of all pairs of opposing "good strategies."
In the immediately following sections we shall define equilibrium points and
prove that a finite non-cooperative game always has at least one equilibrium
point. We shall also introduce the notions of solvability and strong solvability
of a non-cooperative game and prove a theorem on the geometrical structure of
the set of equilibrium points of a solvable game.
|
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|
163,329 |
Let me start by acknowledging the existence of this thread: Mathematics and cancer research It is well-known that mathematical modeling and computational biology are effective tools in cancer research. When I started college and declared the math major, this was the direction I envisioned myself pursuing. However, I quickly fell in love with algebra, number theory, and "pure" math. Are there any ways for an algebraist to contribute to cancer research? I've recently learned that algebraic geometry has been useful in studying phylogenetic trees in evolutionary biology, so while I cannot even imagine what an affirmative answer to my question might look like, I am hopeful that one exists.
|
You might like to look at this paper: Monica Nicolau, Arnold J. Levine, and Gunnar Carlsson, Topology based data analysis identifies a subgroup of breast cancers with a unique mutational profile and excellent survival , Proceedings of the National Academy of Sciences, February 2011. There is also a lot of other stuff by the same group at http://comptop.stanford.edu Only a small proportion involves biology, but that might be enough for you. You might also like to look at the work of Maria-Grazia Ascenzi: http://ortho.ucla.edu/body.cfm?id=218&ref=39 Her PhD is in mathematics but her current work is in biomedical science. I have heard that she is interested in applying algebraic geometry but I do not know the details.
|
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|
163,711 |
Does Peano's theorem apply to spaces with infinite dimension? Or is there a counterexample? Here, Peano's theorem is: Let $E$ be a space with finite dimension. Consider a point $(t_0,x_0) \in \Re \times E$, constants $ a, b > $ 0 and a continuous function $$F: [t_0 - a, t_0 + a] \times B_b[x_0] \longrightarrow E$$
Then for every $ M> $ 0 satisfying
$$\sup \{||F(t,x)||:(t,x) \in [t_0 - a, t_0 + a] \times B_b[x_0]\} < M$$
the Cauchy problem
$$x'(t)=F(t,x(t));\ \ x(t_0)=x_0$$
admits at least one solution in the interval:
$$\big[t_0 - \min(a,\frac{b}{M}),t_0 + \min(a,\frac{b}{M})\big] $$
An infinite-dimensional counterexample would be of great help. Thank you very much.
|
No, Peano's existence theorem fails completely in infinite-dimensional spaces: there are counterexamples in every infinite-dimensional Banach space. This is a theorem of Godunov (A. N. Godunov, Peano's theorem in Banach spaces , Functional Analysis and its Applications 9 (1975), 53-55, http://dx.doi.org/10.1007/BF01078180 ), while the first counterexample in some Banach space was due to Dieudonné (J. Dieudonné, Deux exemples singuliers d'équations différentielles , Acta Sci. Math. Szeged. 12:B (1950), 38-40; see http://acta.fyx.hu/ ). It's not hard to see that the finite-dimensional proof fails (the weak point is generally when Arzelà–Ascoli is applied), but I still find it surprising that the theorem fails as well, since it naively sounds like it should obviously be true.
|
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|
163,847 |
The fundamental group of the Hawaiian earring is very complicated, but since it's "1-dimensional" one might guess that the higher homotopy groups vanish. Do they? Since the Hawaiian earring does not have a universal cover, the standard approach to showing that higher homotopy groups of graphs vanish does not apply.
|
Despite the non-existence of a universal covering space, there is still an object that acts like a universal covering space for the earring space. It is actually quite useful for understanding the structure of the fundamental group as a subgroup of the inverse limit $\varprojlim F_n$ of free groups. The "generalized universal covering" still provides all the same unique path lifting properties for maps out of locally path connected spaces that one might want. The study of these particular coverings for wild spaces is mainly due to Hanspeter Fischer and Andreas Zastrow. I think you should be able to use this to mimic the argument for graphs. Let $\mathbb{E}$ be the earring space, which is a shrinking wedge of a sequence of circles. Let $\widetilde{\mathbb{E}}$ be the set of homotopy classes of paths starting at the basepoint. Take $p:\widetilde{\mathbb{E}}\to \mathbb{E}$ , $p([\alpha])=\alpha(1)$ to be the endpoint projection and give $\widetilde{\mathbb{E}}$ the usual "whisker" topology that you do to construct universal covers: the basic neighborhoods are $B([\alpha],U)=\{[\alpha\cdot\epsilon]|\epsilon([0,1])\subset U\}$ where $U$ is open in $\mathbb{E}$ . It turns out that $p:\widetilde{\mathbb{E}}\to \mathbb{E}$ is an open map which has unique lifting of all paths and homotopies of paths. Moreover, if $Y$ is path connected, locally path connected, and simply connected, then any based map $f:Y\to \mathbb{E}$ has a unique lift $\tilde{f}:Y\to \widetilde{\mathbb{E}}$ such that $p\tilde{f}=f$ . Consequently, $p$ induces isomorphisms on higher homotopy groups. It also turns out that $\widetilde{\mathbb{E}}$ has the structure of an $\mathbb{R}$ -tree (uniquely arcwise connected metric space where each arc is isometric to a sub-arc of the reals), and these are known to be contractible (this is due to J. Morgan I believe). Now you should have all the usual ingredients to mimic the usual argument for graphs. In fact, the earring is not special here. This should work out for all 1-dimensional Peano continua.
|
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|
164,092 |
We see that $$\frac{2}{5}=\frac{36}{90}=\frac{6^2}{90}=\frac{\zeta(4)}{\zeta(2)^2}=\prod_p\frac{(1-\frac{1}{p^2})^2}{(1-\frac{1}{p^4})}=\prod_p \left(\frac{(p^2-1)^2}{(p^2+1)(p^2-1)}\right)=\prod_p\left(\frac{p^2-1}{p^2+1}\right)$$ $$\implies \prod_p \left(\frac{p^2-1}{p^2+1}\right)=\frac{2}{5},$$ But is this the only way to compute this infinite product over primes? It seems like such a simple product, one that could be calculated without the zeta function. Note that $\prod_p(\frac{p^2-1}{p^2+1})$ also admits the factorization $\prod_p(\frac{p-1}{p-i})\prod_p(\frac{p+1}{p+i})$. Also notice that numerically it is quite obvious that the product is convergent to $\frac{2}{5}$: $\prod_p(\frac{p^2-1}{p^2+1})=\frac{3}{5} \cdot \frac{8}{10} \cdot \frac{24}{26} \cdot \frac{48}{50} \cdots$.
|
I found a proof of
$$5 \sum_{m=1}^{\infty} \frac{1}{m^4} = 2 \left( \sum_{n=1}^{\infty} \frac{1}{n^2} \right)^2$$
by rearranging sums and wrote it up . The argument is just 1.5 pages, the other 4.5 are explanations and context. Here is a summary using divergent sums; see the write up for a correct version. Set
$$h(m,n) = \begin{cases}
\frac{1}{m^3 (n-m)} & m \neq n,\ m \neq 0, \\
0 & m=n \ \text{or} \ m=0.
\end{cases}$$
Then we should have
$$\sum_{(m,n) \in \mathbb{Z}^2} h(m,n) - h(n,2n-m) =0$$
as every value occurs twice with opposite signs. So $\sum_{(m,n) \in \mathbb{Z}^2} g(m,n)=0$ where
$$g(m,n) := h(m,n) - h(n,2n-m) = \begin{cases}
\frac{m^2+mn+n^2}{m^3 n^3} & m \neq n,\ m, n \neq 0 \\
- \frac{1}{m^4} & n=0,\ m \neq 0 \\
- \frac{1}{n^4} & m=0,\ n \neq 0 \\
0 & m=n \end{cases}$$
Group together the terms where $(|m|, |n|)$ have a common value; we get $\sum_{(m,n) \in \mathbb{Z}_{\geq 0 }^2} f(m,n) =0$ where
$$f(m,n) = \begin{cases}
\frac{4}{m^2 n^2} & m \neq n,\ m,n >0, \\
- \frac{2}{m^4} & m>n=0, \\
- \frac{2}{n^4} & n>m=0, \\
- \frac{2}{m^4} & m=n>0, \\
0 & m=n=0.
\end{cases} $$
Writing this out, $4\zeta(2)^2 - 6 \zeta(4) - 4 \zeta(4)=0$, as desired. Has anyone seen this? If this is new, I'm thinking of sending it to the Monthly .
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|
164,148 |
In comments on Quora (see, for example, here , here , here ), Ron Maimon has repeatedly expressed the strong opinion that Hilbert's program was not killed by Gödel's results in the way typically claimed. More precisely, he argues that the consistency of axioms sufficient for all meaningful mathematics should be provable from "intuitively self-evident" statements about various computable ordinals being well-defined. Here, of course, he points to Gentzen's 1936 consistency proof , which proves Con(PA) from primitive recursive arithmetic plus induction up to the ordinal ε 0 ; as well as more recent results from the field of ordinal analysis , which show that the consistency of various weaker set theories than ZF (for example, Aczel's constructive ZF and Kripke-Platek set theory) can also be reduced to the well-definedness of various computable ordinals. Maimon then goes on to say that Con(ZF) should similarly be reducible to the well-definedness of some "combinatorially-specified," computable ordinal, although the details haven't been worked out yet. (And indeed, the Wikipedia page on ordinal analysis says that it hasn't been done "as of 2008.") This sounds amazing, especially since I'd never heard anything about this problem before! So, here are my questions: Is there a general theorem showing that Con(ZF) must be reducible to the well-definedness of some computable ordinal, i.e. something below the Church-Kleene ordinal (even if we don't know how to specify such an ordinal "explicitly")? Is finding an "explicit description" of a computable ordinal whose well-definedness implies Con(ZF) a recognized open problem in set theory? Do people work on this problem? Or is there some reason why it's generally believed to be impossible, or possible but uninteresting? Or does it just come down to vagueness in what would count as an "explicit description"? Addendum: There's a connection here to a previous MO question of mine , about the existence of Π 1 -statements independent of ZF with lots of iterated consistency axioms. In particular, using an observation from Turing's 1938 PhD thesis, I now see that it's indeed possible to define a "computable ordinal" whose well-definedness is equivalent to Con(ZF), but only because of a "cheap encoding trick." Namely, one can define the ordinal ω via a Turing machine which lists the positive integers one by one, but which simultaneously searches for a proof of 0=1 in ZF, and which halts and outputs nonsense if it ever finds one. What I suppose I'm asking for, then, is a computable ordinal α such that Con(ZF) can be reduced to the statement that there's some Turing machine that correctly defines α.
|
Rom Maimon is describing the program of proof-theoretic ordinal analysis. First, as you observed in your addendum, it isn't interesting to find some encoding of an ordinal whose well-foundedness implies Con(ZFC), but rather an ordinal such that the well-foundedness of any representation implies Con(ZFC). One hopes that it suffices to consider natural representations of the ordinal, which has been true in practice, but is unproven (and probably unprovable, given the difficulty of making precise what counts as a natural representation). It's possible to prove that the smallest ordinal which ZFC fails to prove well-founded is computable by noticing that the computable notations for ordinals provably well-founded in ZFC are a $\Sigma_1$ subset of the computable notations for ordinals, so certainly $\Sigma^1_1$, and by a result of Spector, any $\Sigma_1^1$ subset of the computable notations for ordinals is bounded. As pointed out in the answer Timothy Chow links to above, it's typically true that this notion of proof-theoretic ordinal ends up being the same as ordinals with other nice properties (like implying Con(ZFC)), but there's no proof that that will always happen (and can't be, since there are defective examples that show it's not always true), nor a proof that covers ZFC. However it's generally believed that for "natural" theories, including ZFC, the different notions of proof-theoretic ordinal will line up. Finding an explicit description of the ordinal for ZFC is an active problem in proof theory, but progress has been very slow. The best known results are by Rathjen and Arai (separately) at the level of $\Pi^1_2$-comprehension (a subtheory of second order arithmetic, so much, much weaker than ZFC), and after nearly 20 years, those remain unpublished. The results in the area got extremely difficult and technical, and didn't seem to provide insight into the theories the way the smaller ordinals had, so it's not nearly as active an area as it once was. Wolfram Pohlers and his students still seem to working in the area, and some other people seem to be thinking about other approaches rather than directly attacking it (Tim Carlson and Andreas Weiermann and their students come to mind).
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164,473 |
I have hit upon major (for me —relative to my trivial accomplishments)
insights in my research
in various sleep-deprived altered states of consciousness,
e.g., long solo car-drives extending through the night into the morning.
But I have never actually solved a problem in my sleep.
I have awakened thinking That's it! , but never was it actually it . Q . Can anyone report an actual significant advance in their research
that occurred during and emerged from their sleep? Of course this is entirely subjective, but you would know it if it happened to you. Poincaré's famous step onto the bus in 1908
("At the moment when I put my foot on the step the idea came to me...")
indicates significant unconscious processing,
and his insomnia account (quoted below) adds further credence to
such "background" processing.
But I am not aware of first-hand reports of
significant and accurate reasoning
occurring during sleep. One evening, contrary to my custom, I drank black coffee and could not sleep. Ideas rose in crowds; I felt them collide until pairs interlocked, so to speak, making a stable combination. ... ( Link )
|
On several occasions it has happened that I have made a key insight while sleeping or drifting in and out of sleep. For example, one of the critical ideas in my paper Joel David Hamkins , Gap forcing , Israel J. Math. 125 (2001), 237--252, came to me this way, and waking up with the mathematical idea, I tore myself out of bed to work it out more fully on paper. It was totally right and formed the basis of later work. I remember sitting in my night attire in the bare moonlight at the table in my apartment, looking out at the empty sidewalk at Wall Street and Williams, where I lived at the time, pondering the approximation property applied to ultrafilters. Because this has now happened several times, I now quite regularly try to prime myself, by intentionally focusing on a particular mathematical issue just as I am going to sleep. My mind floods with mathematical ideas just as I drift off. On welcome rare occasions, the problem is solved in the hypnagogic state, and having awoken I lay in bed pondering it, trying to check it, and wondering if it really is right (sometimes, of course, what seems right is later found to be mistaken). More often, though, when there is welcome news it consists not of a full solution but rather of a new perspective, which later forms the framework of a solution. That is, the result of the unconscious thought is a new way of thinking about the problem, rather than a complete logical proof. At times, naturally, it is an interesting (or obsessive) MathOverflow question that I set myself to thinking about as I lay myself down. But let me say categorically that it has never been the case ( ahem, cough, cough ) that an hour or two after going to bed, I would wake with an answer and crawl out to my computer to type up an MO answer in the dark, while the rest of the household is sleeping, only to realize at that point, right before clicking "Post Your Answer" that the solution was totally flawed or wrong. What a downer that would be, to be sitting in the dark in the middle of the night, tired, with nothing to show for it but a wrong mathematical idea. That has NEVER happened... :-)
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164,588 |
I asked this question two months ago on MSE , where it earned the rare Tumbleweed badge for garnering zero votes, zero answers, and 25 views over 61 days.
Perhaps justifiably so! Here I repeat it with slight improvements. Let $P$ be a polyhedron, all of whose vertices are at points of $\mathbb{Z}^3$,
all of whose edges are parallel to an axis, with every face simply connected, and the surface topologically a sphere.
Let $A(P)$ be the area sequence , the sorted list of areas of $P$'s
faces. For example: Using regular expression notation, this sequence can be written
as $1^4 2^2 3^2$. In analogy with golygons , I wondered if there is a $P$ with
$A(P)= 1^1 2^1 3^1 4^1 5^1 \cdots$. I don't think so, i.e.,
I conjecture there are no golyhedra . Q1 . Can anyone prove or disprove
this? Easier is to achieve $A(P)= 1^+ 2^+ 3^+ \cdots$, where $a^+$ means one or more
$a$'s.
For example, this polyhedron achieves $1^+ 2^+ 3^+ 4^+ 5^+ 6^+$: Q2 . But can $A(P)= 1^n 2^n 3^n \cdots$ be achieved, for some $n$?
The above example is in some sense close, with $A(P) = \cdots 4^4 5^4 6^4 \cdots$,
but end effects destroy the regularity. The broadest question is: Q3 . Which sequences $A(P)$ are achievable?
Can they be characterized? Or at least constrained? Update ( 30Apr14 ). Q1 and Q2 are answered by Adam Goucher's
brilliant example that achieves
$1^1 2^1 3^1 \cdots 32^1$. In light of this advance,
a more specific version of Q3 may be in order: Q3a : Identify some sequence that is not realized by any $A(P)$. Update ( 9Jun14 ): Alexey Nigin has constructed a 15-face golyhedron,
described on Adam Goucher's blog .
And later a 12-face golyhedron .
|
I found a 32-face example with face areas $\{ 1, 2, \dots, 32 \}$: It took a reasonable amount of experimentation to stop it from self-intersecting.
|
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164,624 |
A few weeks ago, I asked on math.stackexchange.com how many quadruples of non-negative integers $(a,b,c,d)$ satisfy the following equation: $$2^a + 3^b = 2^c + 3^d \quad (a \neq c)$$ I found 5 quadruples: $5 = 2^2 + 3^0 = 2^1 + 3^1$, $11 = 2^3 + 3^1 = 2^1 + 3^2$, $17 = 2^4 + 3^0 = 2^3 + 3^2$, $35 = 2^5 + 3^1 = 2^3 + 3^3$, $259 = 2^8 + 3^1 = 2^4 + 3^5$ I didn't get an answer, but only a link to an OEIS sequence (no more quadruples below $10^{4000}$), so I'm asking the same question here. Is there a way to prove that they are [not] infinite? And, more generally, are there known tuples for which the following equation:
$$p_{i_1}^{a_1} + p_{i_2}^{a_2} + ... + p_{i_n}^{a_n}=p_{i_1}^{b_1} + p_{i_2}^{b_2} + ... + p_{i_n}^{b_n}$$
holds for infinitely many (or holds only for finitely many) $a_i,b_i$?
|
That these are all the solutions to $2^a+3^b=2^c+3^d$ was conjectured by Pillai in the 1930s and proved by Stroeker and Tijdeman in 1982, using lower bounds for linear forms in logarithms. Finiteness was proved by Pillai. Oh, and I should have added that the answer is "no" to your second question via a result of Evertse, provided you have no "vanishing subsums".
|
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164,959 |
I probably don't have the appropriate background to even ask this question. I know next to nothing about formal or computer-aided proof, and very little even about group theory. And this question is more "tech support" than math. But: after reading that Georges Gonthier and collaborators had formalized a proof of the Feit-Thompson theorem in Coq , I thought it would be fun to try to verify it on my own computer. So I installed Coq (8.4pl2), OCaml, etc, on my Ubuntu box, and then went here and downloaded feit_thompson.tar.gz (v2.0). I unpacked it and ran make . It ran various invocations of coqc for about two hours, and then finished with no error reported. A bunch of .vo files were produced, including one for theories/stripped_odd_order_theorem.v , which even the casual observer can see contains what appears to be a statement of the Feit-Thompson theorem. Is that it? Does that mean that Coq agrees with the correctness of the proof? Or are there more steps? I was particularly curious because I saw that Makefile.coq has a validate target (which isn't run by just doing make ). So I ran make -f Makefile.coq validate . This started a coqchk command which ran for about five minutes and then failed with: User error: Signature components for label fun_of_fin do not match
make: *** [validate] Error 1 Was I not supposed to do that? Or did I just find a flaw in their proof? :-) Again, I freely admit that I do not really know what I'm doing. In particular, it's not clear to me whether compilation of a Coq file actually verifies the theorems and proofs within it, or merely prepares it for later validation. If an expert answers "Go away until you understand Coq" I'll meekly accept it. But it does seem like it might be helpful for even a newbie to be able to understand how to properly run Coq on a proof created by someone else, and I couldn't find documentation that clearly explained it.
|
The error you get is a real one, but is not in the proof of the odd order theorem. It is in Coq. Let me be more clear: a bug in the kernel of Coq
was making the .vo files (the files coqchk checks) incomplete. Some universe
constraints coming from module sub typing were forgotten. coqchk correctly spots it, and actually revealed the bug I fixed a while ago. A patchlevel release of Coq including the fix is already on its way. The bugreport: https://coq.inria.fr/bugs/show_bug.cgi?id=3243 The commit that fixed the bug: https://github.com/coq/coq/commit/a07deb4eac1d5f886159784ef5d8d006892be547
|
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165,116 |
We say that a subset A in a topological space X is anti-compact if every covering of A by closed sets has a finite subcover. Clearly if X is Hausdorff then all anti-compact subsets of X are finite. What is the structure of anti-compact sets in non-Hausdorff spaces? Is there any reference?
|
When looking for a dual concept we should be careful not to be tricked by a shallow symmetry. I will not comment on your definition of anti-compactness. Instead I would like to explain what the "true" dual to compactness is. The notion dual to compactness is overtness . The concept has appeared in different forms in various approaches to topology. It is not easy to grasp for a classical topologist, and therefore some resistance and ignoring is to be expected. If you think not, consider this: Theorem: All spaces are overt. Even without knowing what "overt" means, the theorem has convinced you that it must be a useless concept. Anyhow, let me define the concept. Given a space $X$ let $\mathcal{O}(X)$ be its topology, equipped with the Scott topology . Homeomorphically, for a nice enough $X$ , the space $\mathcal{O}(X)$ corresponds to the space of continuous maps $\mathcal{C}(X, \Sigma)$ , equipped with the compact-open topology. Here $\Sigma = \{\bot, \top\}$ is the Sierpinski space . Define the map $A_X : \mathcal{O}(X) \to \Sigma$ by $$A_X(U) = \begin{cases}
\top & \text{if $U = X$}\\
\bot & \text{else}
\end{cases}$$ In essence, $A_X$ is the universal quantifier , for it tells us whether every point of $x \in X$ is in $U$ . It would be logical to actually write $$A_X(U) = (\forall x \in X \,.\, x \in U).$$ Now we have: Theorem: A space $X$ is compact if, and only if, $A_X$ is continuous. Let us dualize. Define the existential quantifier $E_X : \mathcal{O} \to \Sigma$ by $$E_X(U) = \begin{cases}
\bot & \text{if $U = \emptyset$}\\
\top & \text{else}
\end{cases}$$ or in logical notation $$E_X(U) = (\exists x \in X \,.\, x \in U).$$ Is this just shallow symmetry? Well, certainly we have a symmetry between universal and existential quantifier, and it is cool that there is a direct connection between compactness and logic. Definition: A space $X$ is overt when $E_X$ is continuous. As I already stated, all spaces are overt in classical topology . That's why it is very hard to discover overtness (and this is an example of a monoculture being unable to make progress for a long time). But overtness has been independently discovered in point-free topology , in computable topology , and in constructive topology . Because there it is not a vacuous notion. For instance, speaking somewhat vaguely, in computable topology a subspace $S \subseteq X$ is computably overt when it is semidecidable whether $U \in \mathcal{O}(X)$ intersects $S$ . Dually, a subspace $S$ is computably compact when it is semidecidable whether $U \in \mathcal{O}(X)$ covers $S$ . It is not hard to come up with subspaces which are not computably overt, and subspaces which are compact but not computably compact: take the closed interval $[-\alpha, \alpha] \subseteq \mathbb{R}$ where $\alpha$ is a non-computable real. Here is a further category-theoretic observation that makes the duality more convincing. I am going to skip over technicalities and side conditions, you can read about Paul Taylor's Abstract Stone Duality to get them all and more. The assignment $X \mapsto \mathcal{O}(X)$ which takes a space to its topology is a contravariant functor from spaces to frames . A continuous map $f : X \to Y$ is mapped to the inverse image map $\mathcal{O}(f) : \mathcal{O}(Y) \to \mathcal{O}(X)$ , defined by $\mathcal{O}(f)(U) = f^{-1}(U)$ . There is a unique map $t_X : X \to 1$ to the singleton space. This map is taken to a map $$\mathcal{O}(t_X) : \mathcal{O}(1) \to \mathcal{O}(X)$$ where we observe that $\mathcal{O}(1)$ is just $\Sigma$ , so we have $$\mathcal{O}(t_X) : \Sigma \to \mathcal{O}(X)$$ in the category of frames. Since frames are posets, we can ask for left and right adjoint of $\mathcal{O}(t_X)$ , i.e., Galois connections. And it turns out that: the right adjoint exists (and is $A_X$ ) if, and only if, $X$ is compact the left adjoint exists (and is $E_X$ ) if, and only if, $X$ is overt [NB: in the category of frames morphisms are supposed to preserve all joins and finite meets, which gives a punch to existence of adjoints.] The characterization of quantifiers in terms of adjoints works very generally, and is in fact the basis for category-theoretic treatment of predicate calculus. Thus not only we have a duality between compactness and overtness, but also (again) a connection with logic. A couple of theorems about compactness which dualize using overtness, where we note that the dual of Hausdorff (closed diagonal) is discrete (open diagonal): Theorem: A closed subspace of a compact space is compact. An open subspace of an overt space is overt. Theorem: (assuming the exponential $Y^X$ exists) If $X$ is compact and $Y$ is discrete then $Y^X$ is discrete. If $X$ is overt and $Y$ is Hausdorff then $Y^X$ is Hausdorff.
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165,303 |
Given an alphabet with $n$ characters, what is the shortest sequence that contains all $n!$ permutations as subsequences? A subsequence can be obtained from a sequence by deleting any characters, thus it's different from a substring, whose elements have to be contiguous in the original sequence. I say this because the similar problem of finding the shortest sequence having all permutations as substrings seems to be more studied, but it's different from what I'm asking here. Some examples of shortest supersequences: $n=2\quad-\quad121$ (length 3) $n=3\quad-\quad1213121$ (length 7) $n=4\quad-\quad1234123142134$ (length 13 - not proven to be shortest). It's easy to see that $n^2$ is an upper bound, since a sequence
$$12\ldots n\, 12\ldots n\, \ldots\, 12\ldots n $$
($n$ times) contains all permutations. A simple lower bound is $n(n+1)/2$, basically because
$$12\ldots n\; 12\ldots (n-1)\; 12\ldots (n-2)\;\ldots $$
is too short (this can be proven rigorously). Is anything more known about this problem? The question was asked on stack exchange , but the answer there is far from satisfactory since it gives only a broken link and no reference.
|
This seems to be an open problem. It is listed in the OEIS as sequence A062714 , as noted by Ilya. Summarising the most important results: Let $m$ be the length of such a sequence. Then Newey (amongst others) describes a simple algorithm to generate them with
$$m = n^2 -2n +4.$$
This is not, however, the best one. Radomirovic proves that the shortest sequences obey the tighter upper bound of
$$m \le \left\lceil n^2 - \frac73 n + \frac{19}3\right\rceil.$$
A lower bound better than the trivial one shown by OP is given by Kleitman and Kwiatkowski :
$$m \ge n^2 - C_\epsilon n^{7/4+\epsilon},$$
for any $\epsilon > 0$.
|
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165,359 |
"Adleman refers to integers which factor completely into small primes as “smooth” numbers." (ME Hellman, JM Reyneri. Advances in Cryptology , 1983: citation link .) Does anyone know what is the intuitive reason behind the choice of the adjective
"smooth" for this concept?
|
I asked Ron Rivest, and he replied: Yes, I coined the term "smooth number" to refer to a number that
has only small prime factors. I don't recall now much about the
thinking process, except that smooth is rather the opposite of
"lumpy"...
|
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165,364 |
Recently I read Dale Rolfsen's paper –A surgical view of Alexander’s polynomial. This is a good paper. But there is a lemma which I don’t know how to prove.The lemma is following: Lemma:In the cover$R^{1}\times R^{2}\to S^{3}$- {trivial knot},let $\alpha,\beta$ be disjoint closed oriented curves downstairs which lift to closed oriented curves $\tilde{\alpha},\tilde{\beta}$ in $R^{1}\times R^{2}$. Then the following formula holds: $lk(\alpha,\beta)=\sum_{k}lk(\tilde{\alpha},t^{k}\tilde{\beta})$. assuming the covering map preserves orientation. Thanks a lot!
|
I asked Ron Rivest, and he replied: Yes, I coined the term "smooth number" to refer to a number that
has only small prime factors. I don't recall now much about the
thinking process, except that smooth is rather the opposite of
"lumpy"...
|
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165,405 |
To put it short: In which active research areas of (pure) mathematics no (or only minimal) knowledge in category theory is required ? To put it long: I know almost nothing about category theory - but I know that I do not like solving problems and providing arguments by diagram chasing and very "high-level" arguments. I'm currently getting near the end of my time as a student and want to start a PhD. And while I'm not lacking a skill in algebra (which has been taught to us in an entirely category-theory-free manner), and therefor assume I could manage with some amount of category theory, I'm just not a fan of "high-level" arguments. So it's more a thing of personal taste. Therefore I would prefer to specialize in an area which does not exhibit too much category-theory (ideally: none).
Now the thing is, when reading around in this forum it seems to me that also in areas, in which I wouldn't expect category theory to come up, like analysis, it actually does come up ("Lipschitz categories"). Thus I would like to know, which sub(sub)fields of big fields like PDE, or number theory can be dealt with only a very small amount of category theory. Especially: Are there any subfields of geometry that are category free ? I'm currently taking a course in differential geometry and to me it seems that category theory is (in disguise) almost everywhere, since our professor constantly explains how some diagram, that proves some assertion about manifolds is actually some category-theoretic notion. When browsing for example through the work of Terence Tao suprisingly few diagrams and mentionings of "categories" come up, so it seems to me that certain areas of PDEs and number theory may fit the bill. But this perception may be due to that fact, that currently I understand almost nothing about what I read and therefore may have missed some categorical arguments. Side question: Are there conversely any areas of applied mathematics that heavily use category theory ? Do there exist, for example, some applications of category theory on numerical mathematics ?
|
As a (slowly) recovering category-phobe, allow me to suggest that you change the way you think of category theory. Specifically, don't think of category theory as a "theory". A theory in mathematics generally consists of three components: a collection of related definitions, a collection of nontrivial theorems about the objects defined, and a collection of interesting examples to which the theorems apply. To learn a theory is to understand the proofs of the main theorems and how to apply them to the examples. Category theory is different: there is an incredibly rich supply of definitions of examples, but very few theorems compared to other established "theories" like group theory or algebraic topology. Moreover, the proofs of the theorems are almost trivial (the Yoneda lemma is one of the most important theorems in category theory and it is not even called a "theorem"). A consequence of this is that you don't have to sit around reading a category theory book before you make contact with the language of categories: the very act of understanding how people express results from "ordinary" mathematics in the language of categories and functors is learning category theory . So now I'll try to answer your question. It is possible to work in nearly any area of pure mathematics without much category theory, and most areas have people everywhere on the category theory spectrum (with the possible exception of algebraic geometry, wherein the language of derived functors is basically built into the foundations). Analysis in particular seems to be somewhat resistant (but not immune) to categorification, and if you are really committed to avoiding categories then you might consider exploring the more analytic aspects of what interests you (e.g. geometric PDE's or analytic number theory). But before you make that commitment, try to find some examples of categorical language in action in what you already understand. The sheer ingenuity of it all might change your mind.
|
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165,464 |
"Without loss of generality" is a standard in the mathematical lexicon, and I am writing to ask if anyone knows where the expression was popularized. (The idea has been around since antiquity, I'm sure, but the expression itself might not be that old.)
|
I think one reason JSTOR doesn't have “loss of generality” before 1831 is that fewer scientists wrote in English. But one finds (with minor variants merged, translations *starred, and year first published in [brackets]): French (1674–1831): sauf l'universalité ( Leibniz 1674 [1903], 1679 [1761]) sans faire tort à la généralité ( Leibniz 1679 [1761]) sans affaiblir la généralité ( Euler 1757 ) sans nuire à la généralité ( Condorcet 1775 ; Lagrange 1785 ; Legendre 1798 ; Abel 1827 ; Dirichlet 1828 ; Quetelet 1828 ) sans déroger à la généralité ( Lagrange 1783 , 1788 ) sans diminuer la généralité ( Cousin 1787 ; Legendre 1798 , 1825 ; Lacroix 1799 , 1803 ; Brisson 1808 ; Poisson 1823 ; Abel 1823 , 1829 , 1829 ) sans rien ôter à la généralité ( Lagrange 1788 , 1811 ; Poncelet 1822 ) sans altérer la généralité ( Monge 1795 ; Legendre 1797 , 1802 ; Garnier 1808 ; Fourier 1829 ; Liouville 1831 ) sans détruire la généralité (* Euler 1796 ) sans perdre de sa généralité ( Lacroix 1797 ; Legendre 1798 ; Biot 1829 ) sans restreindre la généralité ( Legendre 1798 ; Lagrange 1815 ; Poisson 1829 , 1831 ) sauf la généralité ( Pagani 1826 ). Latin (1676–1842): salva generalitate ( Leibniz 1676 [1858], 1676 [1993]; Euler 1770 ; Jacobi 1842 ) salva universalitate ( Leibniz 1679 [1875]; Christmann 1815 ) non impedit generalitatem ( Leibniz 1691 [2009]) sine detrimento universalitatis ( Euler 1744 , 1748 , 1764 ) non restringitur amplitudo ( Euler 1747 , 1750 , 1764 , 1770 ) non limitatur generalitas ( Euler 1761 ) sine detrimento amplitudinis ( Euler 1764 , 1766 , 1769 ) sine detrimento generalitatis ( Euler 1769 ; Gauss 1818 ) nihil de universalitate perire ( Euler 1782 ) nihil de amplitudine amittere ( Euler 1783 ) non restringitur generalitas ( Euler 1794 ). Pig latin (1695, 1735): a generalitate hindert aber nichts ( Leibniz 1695 [2004: III, 6]) pro praesenti negotio general genug ( Bernoulli 1735 [1843]). German (1779–1830): ohne der Allgemeinheit Abbruch zu tun ( Lagrange 1779 ; Plücker 1828 , 1828 , 1829 ) unbeschadet der Allgemeinheit (* de Bicquilley 1788 ; * Lagrange 1791 , 1797 ; Fischer 1792 ; Bolzano 1804 ; Littrow 1823 ; Möbius 1827 ; Plücker 1828 ; Naumann 1830 ; * Euler 1830 ) ohne dass die Allgemeinheit leidet (* Euler 1791 ; Unger 1827 ) ohne der Allgemeinheit zu schaden ( Lorenz 1792 ; Umpfenbach 1823 ; Littrow 1823 , 1827 ; Grunert 1824 ; von Ettingshausen 1827 ; Jacobi 1828 ) ohne Allgemeinheit zu verlieren (* Lagrange 1797 ; * Lacroix 1800 , 1822 ; von Münchow 1826 ) ohne der Allgemeinheit Eintrag zu tun (* Lacroix 1822 ; * Lagrange 1823 ; Umpfenbach 1823 ; Littrow 1823 , 1827 ; Jacobi 1828 ; * Euler 1829 ) ohne die Allgemeinheit zu schmälern (* Lacroix 1822 ) ohne die Allgemeinheit zu beschränken ( Littrow 1827 ; Fischer 1829 ) ohne Nachtheil für die Allgemeinheit ( Gauss 1828 ) ohne Beeinträchtigung der Allgemeinheit (* Euler 1829 ; Moth 1829 ). Italian (1792-1824): senza nulla togliere alla generalità ( Paoli 1792 , 1799 , 1803 , 1804 ; Brunacci 1808 ; Poletti 1824 ) senza limitare la generalità ( Pezzi 1792 ; Forni 1811 ; Frullani 1816 ) salva la generalità ( Paoli 1799 , 1803 , 1804 ; Forni 1811 ) senza alterare la generalità ( Brunacci 1804 , 1807 ; Magistrini 1806 ). English (1809-1830): without losing its generality (* Legendre 1809 ; * Lacroix 1816 ; Wilder 1830 ; Morton 1830 ) without affecting the generality (* Lacroix 1816 ; Ryan 1828 ) without diminishing the generality ( Morton 1830 ) without detracting from the generality ( Morton 1830 ). Summing up: Clearly Euler had a significant role in popularizing the expression after 1740. Yet there remains, for now, the puzzle of the missing link between Euler and anything published by Leibniz (or any precursor or successor of Leibniz). End Note: The French also often expressed the same idea by simply writing “ce qui est permis” . Such rhetorical turns of phrase are briefly discussed and called “indifferent hypotheses” in François Rostand, Sur la clarté des démonstrations mathématiques (Vrin, Paris, 1962, pp. 79–80) .
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165,868 |
I originally asked this on math.stackexchange, where I asked if there could exist a closed manifold that could be given different geometric structures of constant curvature (not at the same time, of course). It was pointed out that the Chern-Gauss-Bonnet theorem shows that no such manifold exists in even dimensions. Also, it seems that by looking at the universal cover we see that no such manifold can be given both a spherical and hyperbolic, or both a spherical and euclidean structure. So, the only case that remains is showing that, in odd dimensions, no manifold can be given both a Euclidean and a hyperbolic structure. Is there an example of a such a manifold, or a proof that no such manifold exists?
|
Here is another approach to impossibility. Some decades after Bieberbach, Milnor showed that the ball of radius $R$ in the universal cover of a compact manifold is basically a bunch of copies of fundamental domain, copies indexed by the ball of a similar radius in the fundamental group under the word metric. Thus a metric of negative curvature requires a fundamental group with exponential growth, while a flat metric requires a fundamental group with polynomial growth. These are not compatible, so no compact manifold can admit both metrics.
|
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166,297 |
Weil's bound for Kloosterman sums states that for $(a,b)\not=(0,0)$,
$$
|K(a,b;q)|:=\left|\sum_{x\in\mathbb{F}_q^*}\chi(ax+bx^{-1})\right|\leq 2\sqrt{q},
$$
where $\chi$ is a non-trivial additive character on $\mathbb{F}_q$ (the field with $q$ elements). My question is, is it known to be false that $\sqrt{q}$ can be replaced by $\sqrt{q-1}$? Here's what is known (to me): Weil's bound follows from the fact that $K(a,b;q)=\alpha+\beta$ where $\alpha\beta =q$ and $|\alpha|=|\beta|=\sqrt q$. Thus there is a unique angle $\theta(a,b;q)$ in $[0,\pi]$ such that
$$
\frac{K(a,b;q)}{2\sqrt q}=\cos\theta(a,b;q)
$$ My question then asks, is there $a,b,q$ such that
$$
|\cos\theta(a,b;q)|>\sqrt{1-\frac 1q}?\qquad (*)
$$ "Vertical" equidistribution of Kloosterman angles implies that as $q\to\infty$
$$
\frac 1{q-1}\sum_{\lambda\in F_q^*}f(\theta(1,\lambda;q))\to\frac 2\pi\int_0^\pi f(\theta)\sin^2\theta\,d\theta
$$ Thus for any fixed $\delta>0$, as $q\to\infty$ the proportion of angles $\theta(a,b;q)\in [0,\delta]$ approaches $\frac 1\pi (\delta-\frac 12\sin(2\delta))\approx \frac{2\delta^3}{3\pi}$. $(*)$ is roughly equivalent to $|\theta(a,b;q)|<q^{-1/2}$, so by equidistribution the expected number of such angles is $\approx 2(q-1)\frac{2}{3\pi} q^{-3/2}\approx \frac{4}{3\pi} q^{-1/2}$, which is (much) less than 1. So one might ask how good is the concentration around the expected number of angles? And how good is this approximation of the expectation to begin with? Probably the most reasonable approach is to just search by computer. For $q=p$ prime and $p\leq 61$ there are no counterexamples, but this isn't very convincing.
|
Going a bit beyond $61$, I find that the first counterexample to $|K(a,b;q)| < 2 \sqrt{q-1}$ with prime $q$ has $(q,ab) = (139,38)$, when $K(a,b;q) = -23.51308393\ldots = -2 \sqrt{138.216\ldots}\,$, and there are no further prime counterexamples up to $10^3$. [ added later ] Extending the search overnight reached a bit beyond $10^4$
and found five more cases, at $q=1747$, $3121$, $3593$, $3853$, and $10973$.
The smallest $\delta$ for $|K(a,b;q)| = 2\sqrt{q-\delta}$ is about $0.2892$
for $(q, ab) = (1747, 461)$. The other $\delta$'s are about $0.653$,
$0.830$, $0.833$, and $0.2999$, the last for $(q,ab) = (10973, 8093)$. The gp code I ran is about an order of magnitude faster than
yesterday's, mostly thanks to storing a table of cosines instead of
computing each $\chi(ax+bx^{-1})$ as it arises. But it still takes
about $q^2$ time per $q$, and thus about $x^3$ to try all $q \leq x$.
There's a factor of about $q$ (and thus of about $x$) to be saved
by setting up the computation as a fast Fourier transform over
either ${\bf F}_q$ or ${\bf F}_q^*$, but I'll leave that to
somebody else to implement (or has it been done already?).
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167,323 |
It is well known that there are functions $f \colon \mathbb{R} \to \mathbb{R}$ that are everywhere continuous but nowhere monotonic (i.e. the restriction of $f$ to any non-trivial interval $[a,b]$ is not monotonic), for example the Weierstrass function . It’s easy to prove that there are no such functions if we add the condition that $f$ is continuously differentiable, so it is natural to ask the same question, with $f$ only differentiable. This seems to me a non trivial question, since, at least a priori, the derivative $f'$ could change sign on any non trivial interval, so we cannot use the standard results to prove the monotonicity of $f$. Question: Does it exist a function $f \colon \mathbb{R} \to \mathbb{R}$ that is everywhere differentiable but nowhere monotonic?
|
Everywhere differentiable but nowhere monotonic real functions do exist. It seems that the first correct examples were found by A. Denjoy in this paper . A short existence proof, based on Baire's category theorem, was given by C. E. Weil in this paper .
|
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167,349 |
I asked the following question on math.stackexchange several months ago: Let $n,m,p>1$ be such that $S_n \times S_m \hookrightarrow S_p$. Does it imply that $p \geq n+m$? Derek Holt gave a positive answer. However, because the problem is easy to solve for small values of $p$, I am really curious to know if there exists an "elementary" solution.
|
I think I can do this with an elementary argument, but I have to rush off somewhere soon, so I will answer quickly and hope I get it roughly right! Assume $p < n+m$, so we can assume also that $n>p/2$. $S_n$ must have a faithful orbit in its action on $p$ points, and if that orbit has size $k$ then $n \le k \le p < 2n$. Let $H$ be the point stabilizer of the action of $S_n$ on this orbit of length $k$. Then since $|S_n:H| < 2n$ and any proper subgroup of $A_n$ has index at least $n$ (I am assuming $n \ge 5$, since you can do small cases separately), we cannot have $H < A_n$, so $H$ must be a maximal subgroup of $S_n$. But then the action of $S_n$ on this orbit is primitive, and primitive groups have trivial centralizer. (That's because the point stabilizer in a primitive group fixes a unique point, since otherwise its fixed points would be a block, and hence the centralizer fixes that point and hence all points.) So the other factor, $S_m$ must fix all points in that orbit, but that's impossible, because $m > p-k$.
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167,577 |
In this question of mine in a comment to the accepted answer, someone remarked: There are ways to express even basic things in analysis, such as the
spectral theorem or the Lebesgue integral, using the language of
categories. But many of the hard theorems in analysis boil down to
subtle estimates which (so far!) have not been simplified by clever
categorical arguments. I replied to this requesting a reference where the Lebesgue integral is expressed using categories but unfortunately I didn't get an answer. So I'm posting it as a question (I wouldn't mind for a reference concerning the spectral theorem either) with the additional kind request to also provide an explanation why it is difficult to describe estimates using category theory.
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Sorry to refer to my own work, but I think this answers your question directly: http://www.maths.ed.ac.uk/~tl/glasgowpssl/ That link is to a very short note, but I might as well repeat the result here. Let's agree that a "map" of Banach spaces is a map of norm $\leq 1$, and let's also agree that when $X$ and $Y$ are Banach spaces, we equip $X \oplus Y$ with the norm $\| (x, y) \| = (\|x\| + \|y\|)/2$. Now let $\mathcal{C}$ be the category of triples $(X, \xi, u)$ where $X$ is a Banach space, $\xi$ is a map $X \oplus X \to X$, and $u$ is an element of the closed unit ball of $X$ such that $\xi(u, u) = u$. Theorem : The initial object of $\mathcal{C}$ is $(L^1[0, 1], \gamma, 1)$ where $1$ is the constant function $1$ and $\gamma$ concatenates two functions then scales the domain by a factor of $1/2$. Another object of $\mathcal{C}$ is $(\mathbb{R}, \text{mean}, 1)$. The unique map in $\mathcal{C}$ from the initial object to this object is Lebesgue integration, $\int_0^1: L^1[0, 1] \to \mathbb{R}$. While I'm at it, I'll add another result that isn't in that note (or written up anywhere yet). This characterizes Lebesgue integrability and integration on arbitrary finite measure spaces. Let $\mathbf{Meas}$ be the category of finite measure spaces and "embeddings" (by which I mean maps that are isomorphisms to their images). Let $\mathbf{Ban}$ be the category of Banach spaces (with maps as above). Let $\mathcal{D}$ be the category of pairs $(F, u)$, where $F$ is a functor $\mathbf{Meas} \to \mathbf{Ban}$ and $u$ assigns to each measure space $X = (X, \mu)$ an element $u_X \in F(X, \mu)$, subject to two laws: first, $\|u_X\| \leq \mu(X)$, and second, whenever
$$
Y \stackrel{i}{\longrightarrow} X \stackrel{j}{\longleftarrow} Z
$$
in $\mathbf{Meas}$ with $X = iY \sqcup jZ$ (disjoint union) then $(Fi)u_Y + (Fj)u_Z = u_X$. Theorem The initial object of $\mathcal{D}$ is $(L^1, I)$, where $I_X \in L^1(X)$ is the constant function $1$. (In the case of this initial object, the equation "$(Fi)u_Y + (Fj)u_Z = u_X$" says that when $X$ is partitioned into subsets $Y$ and $Z$, the indicator function of $Y$ plus the indicator function of $Z$ is the constant function $1$.) Another object of $\mathcal{D}$ is $(K, t)$, where $K$ has constant value $\mathbb{R}$ (or $\mathbb{C}$, depending on our choice of ground field) and $t_X = \mu(X)$ for a measure space $X = (X, \mu)$. The unique map in $\mathcal{D}$ from the initial object to this object is integration. To spell that out a bit more: the maps in $\mathcal{D}$ are natural transformations satisfying the obvious condition, and in this case, the $X$-component of the unique map $(L^1, I) \to (K, t)$ is $\int_X: L^1(X) \to \mathbb{R}$.
|
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|
167,613 |
It seems that for any prime number $p$ and for any non-zero element $a$ in the finite field $\mathbb F_p$, the polynomial $x^p-x+a$ is irreducible over $\mathbb F_p$. (It is of course obvious that there are no linear factors.) Are there any general irreducibility criteria which can help to prove such a result? (More generally, it seems that all irreducible factors over $\mathbb F_p$
of $$a+\sum_{j=0}^n{n\choose j}(-1)^jx^{p^j}$$
(where $a$ is of course a non-zero element of $\mathbb F_p$)
have a common degree given by a non-trivial power of $p$.)
|
This is true. Pass to an extension field where the polynomial has a root $r$ , notice that the other roots are of the form $r+1$ , $r+2$ , ..., $r+p-1$ . Suppose that $x^p - x +1 = f(x) g(x)$ , with $f, g \in \mathbb{F}_p\left[x\right]$ and $\deg f = d$ . Then $f(x) = (x-r-c_1) (x-r-c_2) \cdots (x-r-c_d)$ for some subset $\{ c_1, c_2, \ldots, c_d \}$ of $\mathbb{F}_p$ . So the coefficient of $x^{d-1}$ in $f$ is $-\sum (r+c_i) = - (dr + \sum c_i)$ ; we deduce that $dr \in \mathbb{F}_p$ . If $d \neq 0 \bmod p$ , then $r \in \mathbb{F}_p$ which is plainly false, so $d=0$ or $p$ and the factorization is trivial. If I were to try to turn this proof into a general technique, I suppose I would frame it as "prove that the Galois group is contained in a cyclic group of order $p$ , and observe that it can't be the trivial group, so it must be the whole group." I can also prove the generalization. Define a $\mathbb{F}_p$ -module endomorphism $T$ of any commutative $\mathbb{F}_p$ -algebra by $T(u) = u^p-u$ . Set $F(n) = \mathbb{F}_{p^{p^n}}$ . Observe that $$T^r(x) = \sum_{k=0}^r (-1)^{r-k} x^{p^k} \binom{r}{k}.$$ Lemma $T$ is an $\mathbb{F}_p$ -linear endomorphism of $F(n)$ whose Jordan-normal form is a single nilpotent block (of size $p^n$ ). Proof Obviously, $T$ is $\mathbb{F}_p$ -linear. Observe that $$T^{p^n}(x) = \sum_{k=0}^{p^n} (-1)^{p^n-k} x^{p^k} \binom{p^n}{k} = x^{p^{p^n}} - x$$ so $T^{p^n}$ is zero on $F(n)$ and we know that $T$ is nilpotent. Finally, $T(x) = x^p-x$ so the kernel of $T$ is one dimensional, and we see that there is only one Jordan block. $\square$ Now, let $p^{n-1} \leq r < p^n$ . Roland's polynomial is $T^r(x) = a$ for $a$ a nonzero element of $\mathbb{F}_p$ . Using the Lemma, the image of $T^r: F(n) \to F(n)$ is the same as the kernel of $T^{p^n-r}$ . In particular, since $a \in \mathrm{Ker}(T)$ , we see that $a$ is in the image of $T^r: F(n) \to F(n)$ . Using the Lemma again, all nonzero fibers of $T^r$ are of size $p^r$ , so there are $p^r$ roots of $T^r(x)=a$ in $F(n)$ . Since Roland's polynomial only has degree $p^r$ , we see that all roots of $T^r(x)=a$ are in $F(n)$ . All proper subfields of $F(n)$ are contained in $F(n-1)$ . But, since $r \geq p^{n-1}$ , the Lemma shows that $T^r(x)=0$ on $F(n-1)$ . So none of the roots of $T^r(x)=a$ are in $F(n-1)$ . We conclude that all the factors of Roland's polynomial are of degree $p^n$ .
|
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|
167,685 |
I asked this question on stackexchange, but despite much effort on my part have been unsuccesful in finding a solution. Does the inequality
$$2(|a|+|b|+|c|) \leq |a+b+c|+|a+b-c|+|a+c-b|+|b+c-a|$$
hold for all complex numbers $a,b,c$ ?
For real values a case analysis will verify the inequality.
What is desired is a proof using the triangle inequality or a counterexample.
Thanks in advance.
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It seems that your inequality is just an incarnation of Hlawka's inequality which says that for any vectors $x, y, z$ in an inner product space $V$ we have \begin{equation*}
\|x+y\| + \|y+z\|+\|z+x\| \le \|x\|+\|y\| + \|z\| + \|x+y+z\|.
\end{equation*} Using $x=a+b-c$, $y=a+c-b$, and $z=b+c-a$ we obtain the inequality in the OP. Additional remarks: To add some more context, please see the paper linked here , which provides quite a nice summary of work related to Hlawka's inequality, which apparently stems back to a 1942 paper of Hornich (also cited by Zurab below). The paper linked to above explores the interesting generalization:
\begin{equation*}
f(x+y) + f(y+z) + f(z+x) \le f(x+y+z) + f(x)+f(y)+f(z),
\end{equation*}
where $x,y,z$ may come from an Abelian group, or a linear space, or the real line---each with its own set of conditions on the mapping $f$. The functional form of Hlawka's inequality is credited to a 1978 paper of Witsenhausen.
|
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|
167,701 |
Suppose $A$ and $B$ are finitely generated Abelian groups. Are all exact sequences of the form $0 \rightarrow A \rightarrow A \oplus B \rightarrow B \rightarrow 0$ split? If not, is there an example?
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This is true more generally for finitely generated modules over a noetherian ring. Your question is equivalent to asking whether the sequence $$0\rightarrow \operatorname{Hom}(B,A)\rightarrow \operatorname{Hom}(A\oplus B,A)\rightarrow \operatorname{Hom}(A,A)$$ is surjective on the right. To prove this, it suffices to localize and then complete at an arbitrary prime $P$ , so we can assume we're working over a complete local ring where $P$ is the maximal ideal. Now it suffices to check surjectivity after modding out an arbitrary power $P^n$ , which allows us to assume that all the modules are of finite length. Surjectivity follows because the lengths of the left-hand and right-hand modules add up to the length of the module in the middle. Edited to add: The comments above (which I read after I posted this) remind me that I've posted this same argument before. If people think this instance should be deleted, I'm fine with that.
|
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|
167,951 |
I read in an information theory textbook the Brunn-Minkowski inequality follows from the Entropy Power inequality. The first one says that if $A,B$ are convex polygons in $\mathbb{R}^d$, then $$ m(A+B)^{1/d} \geq m(A)^{1/d} + m(B)^{1/d} $$ where $A+B = \{ a+b: a \in A, b \in B\}$ is the Minkowski sum . The entropy power inequality states that if $X$ and $Y$ are independent random variables, and $H(X) = \int f(x) \ln f(x) \, dx$ then $$ e^{2H(X+Y)/d} \geq e^{2H(X)/d}+ e^{2H(Y)/d}$$ If you plug in Gaussian random variables with respective covariance matrices
$X = N(0,C_1)$ and $Y = N(0,C_2)$ then you get the version for boxes: $$ \det(C_1+C_2)^{1/d} \geq \det(C_1)^{1/d} + \det(C_2)^{1/d}$$ I would like to understand better why Convex geometry should have anything to do with information theory, and that the natural probability distributions are associated to a convex set. The article by Gardner has the "ancestor" of the Brunn-Minkowski inequality as the Brascamp-Lieb inequality, while the Entropy power inequality follows from Young's inequality. Brunn-Minkowski Inequality Brascamp-Lieb inequality Young inequality In that case, my question is what role to convex polygons have there? To me seen seems a "light bulb" these difficult inequalities from functional analysis are based on intuition from convex geometry - especially since nobody draws pictures in a functional analysis textbook.
|
The similarity between the entropy power inequality and the Brunn-Minkowski inequality is not directly related to convexity - after all, Brunn-Minkowski can be generalised to bounded open sets that are not necessarily convex. (However, many of the proofs of both inequalities use ideas from convexity theory, of course.) Taking the microstate (i.e. Boltzmann) interpretation of entropy, one can heuristically view the entropy power inequality as a high-dimensional "99%" analogue of Brunn-Minkowski, in which one considers the "99%-sumset" $A \stackrel{99\%}{+} B$ of two high-dimensional sets A and B (this is a somewhat vaguely defined concept, but roughly speaking it is the bulk of the support of two random variables distributed on A and B respectively) rather than the "100%-sumset" A+B. However, thanks to the concentration of measure phenomenon, the 99%-sumset is considerably smaller than the 100%-sumset in high dimensions, leading to the additional exponent of 2 in the entropy power inequality. For instance, consider in high dimensions two balls $B(0,R)$, $B(0,r)$ of radii $R,r$ respectively. Their 100%-sumset is $B(0,R+r)$ of course. But if one takes a random element of $B(0,R)$ and adds it to a random element of $B(0,r)$, the sum concentrates in a much smaller ball - asymptotically, $B(0,\sqrt{R^2+r^2})$ (in fact it concentrates to the boundary of this ball). This boils down to the basic fact that pairs of vectors in high dimensions are typically almost orthogonal to each other. So one morally has $$ B(0,R) \stackrel{99\%}{+} B(0,r) = B(0,\sqrt{R^2+r^2}) \qquad (1)$$ in the high-dimensional limit. Anyway, the EPI can be thought of as a rigorous formulation of a ``99% Brunn-Minkowski inequality'' $$ |A \stackrel{99\%}{+} B|^{1/N} \geq \sqrt{ (|A|^{1/N})^2 + (|B|^{1/N})^2 } \qquad (2)$$ in the high-dimensional limit $N \to \infty$ (with $A,B \subset \mathbb{R}^N$ varying appropriately with $N$), which is of course consistent with (1). To see this interpretation, one observes from the microstate interpretation of entropy that if $X$ is a continuous random variable on $\mathbb{R}^n$ with a nice distribution function (e.g. $C^\infty_c$), and $M$ is large, then taking $M$ independent copies $X_1,\dots,X_M$ of $X$ gives a random vector $X^{\otimes M} := (X_1,\dots,X_M)$ in $\mathbb{R}^{N}$ for $N := Mn$ which (by the asymptotic equipartition property ) is concentrated in a subset of $\mathbb{R}^N$ of measure $e^{M (H(X)+o(1))}$ in the limit $M \to \infty$ (this is a nice calculation using Stirling's formula; it may help to first work out the case when the probability distribution of $X$ is a simple function rather than a test function). Similarly, if $Y$ is another random variable independent of $X$, then $Y^{\otimes M}$ will be concentrated in a set of measure $e^{M(H(Y)+o(1))}$, while $(X+Y)^{\otimes M}$ is concentrated in a set of measure $e^{M(H(X+Y)+o(1))}$. EPI is then morally a consequence of (2) in the limit $M \to \infty$. Despite the significant differences between the $99\%$-sumset and $100\%$-sumset in high dimensions, it is still good to think of these concepts as being closely analogous, so that almost any sumset inequality should have an entropy counterpart and vice versa (although in most cases we do not have a direct logical implication between the sumset inequality and the entropy inequality; instead, the inequalities typically have analogous, but not completely identical, proofs). See e.g. this recent paper of Kontoyiannis and Madiman (and the references therein) for some instances of this. EDIT: Of course, by bounding the $99\%$-sumset by the $100\%$-sumset one can get some connection between the two types of inequalities, but usually one gets an inferior estimate when one uses this approach (it completely ignores the effect of concentration of measure), so this is not the "right" way to relate sumset inequalities with their entropy counterparts. For instance, by directly applying the EPI to uniform random variables on $A,B \subset \mathbb{R}^n$ and then using Jensen's inequality, one only gets a weak form
$$ |A+B|^{1/n} \geq \sqrt{ (|A|^{1/n})^2 + (|B|^{1/n})^2 }$$
of the Brunn-Minkowski inequality (compare with (2)). The problem here, of course, is that the sum of two uniformly distributed independent random variables is almost never uniformly distributed.
|
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|
168,586 |
Expander graphs ("sparse graphs that have strong connectivity properties")
burst onto the mathematical scene around the millennium, but I have not
been successful in tracing the origin of
(a) the concept, and
(b) the name expander .
Does anyone know? And can provide a citation? Paley graphs (connecting pairs of elements that differ in a quadratic residue ) are expanders.
|
The concept (but not the name) was introduced by Barzdin and Kolmogorov in A. N. Kolmogorov and Y. M. Barzdin, “On the realization of networks in
three-dimensional space” in Selected Works of Kolmogorov, vol. 3, Kluwer,
Dordrecht, 1993, 194–202. which was published in 1967. They proved that they exist via a probabilistic argument. They were then rediscovered and named expanders by Pinsker in his paper M. S. Pinsker, "On the complexity of a concentrator'', Proceedings of the Seventh International Teletraffic Congress (Stockholm, 1973), pp. 318/1–318/4, Paper No. 318. available here (see the appendix). He also proves they exist via a probabilistic argument. The first explicit examples were found by Margulis in his paper G. Margulis, Explicit constructions of concentrators, Problemy Peredachi Informatsii, 9(4) (1973), pp. 71-80; Problems Inform. Transmission, 10 (1975), pp. 325-332. and by Gabber-Galil in their paper O. Gabber and Z. Galil, Explicit constructions of linear size superconcentrators, Proc. 20th Annual Symposium on the Foundations of Computer Science, 1979, pp. 364-370. By the way, I learned the above history from the following lovely paper: M. Gromov and L. Guth,
Generalizations of the Kolmogorov-Barzdin embedding estimates.
Duke Math. J. 161 (2012), no. 13, 2549–2603.
|
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