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129,759 |
While there is tremendous progress happening in mathematics, most of it is just accessible to specialists. In many cases, the proofs of great results are both long and use difficult techniques. Even most research topologists would not be able to understand the proof of the virtually fibering conjecture or the Kervaire problem, to name just two recent breakthroughs in topology, without spending months on it. But there are some exceptions from this rule. As a topologist, I think here mostly about knot theory: The Jones and HOMFLY polynomials. While the Jones polynomial was first discussed from a more complicated context, a rather simple combinatorial description was found. These polynomials help to distinguish many knots. The recent proof by Pardon that knots can be arbitrarily distorted. This might be not as important as the Jones polynomial, but quite remarkably Pardon was still an undergrad then! Or an example from number theory are the 15- and 290-theorems: If a positive definite integer valued qudadratic form represents the first 290 natural numbers, it represents every natural number. If the matrix associated to the quadratic form has integral entries, even the first 15 natural numbers are enough. [Due to Conway, Schneeberger, Bhargava and Hanke.] [Edit: As mentioned by Henry Cohn, only the 15-theorem has a proof accessible to undergrads.] My question is now the following: What other major achievements in mathematics of the last 30 years are there which are accessible to undergraduates (including the proofs)?
|
Primes are in P . The proof is indeed accessible, see for example the article "Primes are in P: A breakthrough
for "Everyman", http://www.ams.org/notices/200305/fea-bornemann.pdf . The idea is really simple, based
on the observation, that, if the natural numbers
$a$ and $n$ are relatively prime, then $n$ is prime if and
only if
$$
(x − a)^n \equiv (x^n − a) \mod n
$$
in the ring of polynomials $\mathbb{Z}[x]$.
Of course, more precise results concerning complexity are not so easy.
|
{
"source": [
"https://mathoverflow.net/questions/129759",
"https://mathoverflow.net",
"https://mathoverflow.net/users/2039/"
]
}
|
129,762 |
I'm always looking for applications of homotopy theory to other fields, mostly as a way to make my talks more interesting or to motivate the field to non-specialists. It seems like most talks about Algebraic $K$-theory mention that we don't know $K(\mathbb{Z})$ and that somehow $K(\mathbb{Z})$ is worth computing because it contains lots of arithmetic information. I'd like to better understand what kinds of arithmetic information it contains. I've been unable to answer number theorists who've asked me this before. A related question is about what information is contained in $K(S)$ where $S$ is the sphere spectrum. I am aware of Vandiver's Conjecture and that it is equivalent to the statement that $K_n(\mathbb{Z})=0$ whenever $4 | n$. I also know there's some connection between $K$-theory and Motivic Homotopy Theory, but I don't understand this very well (and I don't know if $K(\mathbb{Z})$ helps). It seems difficult to search for this topic on google. Hence my question: Can you give me some examples of places where computations in $K(\mathbb{Z})$ or $K(S)$ would solve open problems in arithmetic or would recover known theorems with difficult proofs? I'm hoping someone who has experience motivating this field to number theorists will come on and give his/her usual spiel. Here are some potential answers I might give a number theorist if I understood them better...The wikipedia page for Algebraic K-theory mentions non-commutative Iwasawa Theory, L-functions (and maybe even Birch-Swinnerton-Dyer?), and Bass's conjecture. I don't know anything about this, not even whether knowing $K(\mathbb{Z})$ would help. Quillen-Lichtenbaum seems related to $K(\mathbb{Z})$, but it seems it would tell us things about $K(\mathbb{Z})$ not the other way around. Milnor's Conjecture (or should we call it Voevodsky's Theorem?) is definitely an important application of $K$-theory, but it's the $K$-theory of field of characteristic $p$, not $K(\mathbb{Z})$. There was a previous MO question about the big picture behind Algebraic K Theory but I couldn't see in those answers many applications to number theory. There's a survey written by Weibel on the history of the field, and that includes some problems it's solved (e.g. the congruence subgroup problem) but other than Quillen-Lichtenbaum I can't see anything which relies on $K(\mathbb{Z})$ as opposed to $K(R)$ for other rings. If $K(\mathbb{Z})$ could help compute $K(R)$ for general $R$ then that would be something I've love to hear about.
|
$\newcommand\Z{\mathbf{Z}}$
$\newcommand\Q{\mathbf{Q}}$ I'm a number theorist who already thinks of the algebraic $K$-theory of $\Z$ as part of number theory anyway, but let me make some general remarks. A narrow answer : Since (following work of Voevodsky, Rost, and many others) the $K$-groups of $\Z$ may be identified with Galois cohomology groups (with controlled ramification) of certain Tate twists $\Z_p(n)$, the answer is literally "the information contained in the $K$-groups is the same as the information contained in the appropriate Galois cohomology groups."
To make this more specific, one can look at the rank and the torsion part of these groups. The ranks (of the odd $K$-groups) are related to $H^1(\Q,\Q_p(n))$ (the Galois groups will be modified by local conditions which I will suppress), which is related to the group of extensions of (the Galois modules) $\Q_p$ by $\Q_p(n)$. A formula of Tate computes the Euler characteristic of $\Q_p(n)$, but the cohomological dimension of $\Q$ is $2$, so there is also an $H^2$ term. The computation of the rational $K$-groups by Borel, together with the construction of surjective Chern classes by Soulé allows one to compute these groups explicitly for positive integers $n$. There is no other proof of this result, as far as I know (of course it is trivial in the case when $p$ is regular). The (interesting) torsion classes in $K$-groups are directly related to the class groups of cyclotomic extensions. For example, let $\chi: \mathrm{Gal}(\overline{\Q}/\Q) \rightarrow \mathbf{F}^{\times}_p$ be the mod-$p$ cyclotomic character. Then one can ask whether there exist extensions of Galois modules: $$0 \rightarrow \mathbf{F}_p(\chi^{2n}) \rightarrow V \rightarrow \mathbf{F}_p \rightarrow 0$$ which are unramified everywhere. Such classes (warning: possible sign error/indexing disaster alert) are the same as giving $p$-torsion classes in $K_{4n}(\Z)$. The non-existence of such classes for all $n$ and $p$ is Vandiver's conjecture. Now we see that:
The finiteness of $K$-groups implies that, for any fixed $n$, there are only finitely many $p$ such that an extension exists. An, for example, an explicit computation of $K_8(\Z)$ will determine explicitly all such primes (namely, the primes dividing the order of $K_8(\Z)$). As a number theorist, I think that Vandiver's conjecture is a little silly --- its natural generalization is false and there's no compelling reason for it to be true. The "true" statement which is always correct is that $K_{2n}(\mathcal{O}_F)$ is finite. Regulators . Also important is that $K_*(\Z)$ admits natural maps to real vector spaces whose image is (in many cases) a lattice whose volume can be given in terms of zeta functions (Borel). So $K$-theory is directly related to problems concerning zeta values, which are surely of interest to number theorists. The natural generalization of this conjecture is one of the fundamental problems of number theory (and includes as special cases the Birch--Swinnerton-Dyer conjecture, etc.). There are also $p$-adic versions of these constructions which also immediately lead to open problems, even for $K_1$ (specifically, Leopoldt's conjecture and its generalizations.) A broader answer : A lot of number theorists are interested in the Langlands programme, and in particular with automorphic representations for $\mathrm{GL}(n)/\Q$. There is a special subclass of such representations (regular, algebraic, and cuspidal) which on the one hand give rise to regular $n$-dimensional geometric Galois representations (which should be irreducible and motivic), and on the other hand correspond to rational cohomology classes in the symmetric space for $\mathrm{GL}(n)/\Q$, which (as it is essentially a $K(\pi,1)$) is the same as the rational cohomology of congruence subgroups of $\mathrm{GL}_n(\Z)$. Recent experience suggests that in order to prove reciprocity conjectures it will also be necessary to understand the integral cohomology of these groups. Now the cohomology classes corresponding to these cuspidal forms are unstable classes, but one can imagine a square with four corners as follows: stable cohomology over $\mathbf{R}$ : the trivial representation. unstable cohomology over $\mathbf{R}$ : regular algebraic automorphic forms for $\mathrm{GL}(n)/\Q$. stable cohomology over $\mathbf{Z}$: algebraic $K$-theory. unstable cohomology over $\mathbf{Z}$: ?"torsion automorphic forms"? , or at the very least, something interesting and important but not well understood. From this optic, algebraic $K$-theory of (say) rings of integers of number fields is very naturally part of the Langlands programme, broadly construed. Final Remark: algebraic K-theory is a (beautiful) language invented by Quillen to explain certain phenomena; I think it is a little dangerous to think of it as being an application of "homotopy theory". Progress in the problems above required harmonic analysis and representation theory (understanding automorphic forms), Galois cohomology, as well as homotopy theory and many other ingredients. Progress in open questions (such as Leopoldt's conjecture) will also presumably require completely new methods.
|
{
"source": [
"https://mathoverflow.net/questions/129762",
"https://mathoverflow.net",
"https://mathoverflow.net/users/11540/"
]
}
|
129,955 |
I asked the same question on math.se but got no answer there. Since it pertains to my current research, I decided to ask here: Let $n\in 2\mathbb{N}$ be an even number. I want to evaluate
$$I_n
:=
\int_0^1\mathrm{d} u_1 \cdots \int_0^1 \mathrm{d} u_n \frac{\delta(1-u_1-\cdots-u_n)}{(u_1+u_2)(u_2+u_3)\cdots(u_{n-1}+u_n)(u_n+u_1)}.
$$
For small $n$, this is computable by simply parameterizing the $\delta$ function, and
$I_2 = 1$, $I_3 = \pi^2/4$, $I_4 = 2\pi^2/3$. The values of $I_5$ and $I_6$ are numerically $18.2642 \approx 3\pi^4/16$ and $51.9325\approx 8\pi^4/15$. I strongly suspect that
$$
I_{2n+2} \stackrel{?}{=} (2\pi)^{2n} \frac{(n!)^2}{(2n+1)!} =
\frac{(2\pi)^{2n}}{\binom{2n+1}{n}(n+1)}
= (2\pi)^{2n}\mathrm{B}(n+1,n+1),
$$
where $\mathrm{B}$ is the Beta function. Dividing by $(2\pi)^{2n}$, this is Sloane's A002457 . For $I_6$, this conjecture is equivalent to
$$\int_0^1\mathrm{d}x \Bigl(\mathrm{Li}_2(\frac{x-1}{x})\Bigr)^2 \stackrel{?}{=} \frac{17}{180}\pi^4$$
(with $\mathrm{Li}_2$ the dilogarithm), which seems to be true numerically, but I could neither prove it nor find it in the literature. As a last remark, it is possible to get rid of the $\delta$ function by using the identity
$$I_n = \int_{(0,\infty)^n}\mathrm{d}u \frac{f(\lvert u\rvert_1)}{(u_1+u_2)\cdots(u_n+u_1)} \Bigm/\int_0^\infty\mathrm{d}t \frac{f(t)}{t}$$
for any $f:(0,\infty)\to\mathbb{R}$ that makes both integrals finite. Using $f(t) = t 1_{[0,1]}(t)$ where $1_{[0,1]}$ is the characteristic function of the interval $[0,1]$, one can write $I_n$ as an integral over an $n$-dimensional simplex.
|
We can think of $I_{n}$ as being a classical partition function for $n$ beads on a circle which cannot pass through each other, with logarithmic interaction potential between each bead and its next-to-nearest neighbors on either side. For $I_{2n}$ the beads fall into two ``colors" which do not have logarithmic interactions with each other; while for $I_{2n+1}$ the beads do not fall into two independent groups. We make two changes of variable. First, we can label the coordinates of the $k^{th}$ bead as $y_k$, where $y_1=0$ is fixed (exploiting the translation invariance of the problem) and we define $y_{2n+k} = 1+y_k$ (because of the periodic nature of the circle):
$$
u_i = y_{i+1}-y_i\ ,\qquad y_1=0\ ,\qquad y_{2n+i}\equiv 1+y_i \ .
$$
Then the integral can be written as a path ordered expression without the delta function constraint as
$$
I_{n}= \int_0^1 dy_{n} \int_0^{y_{n}} dy_{n-1}\cdots\int_0^{y_3} dy_2\, \prod_{k=1}^{2n}\frac{1}{y_{k+2}-y_k}\ .
$$
The second change of variables to $\{y_2,\ldots,y_n\}\to \{s_2,\ldots,s_n\}$ in order change the integration domain to a unit hypercube:
$$
y_{k} =\prod_{j=k}^{n} s_{j}\ ,
$$
with Jacobian
$$
J_n = \prod_{j=3}^{n} s_j^{j-2}\ .
$$
With this change of variables, $I_{2n}$ becomes (for $n\ge 2$)
$$
I_{2n} = \int_0^1 d^{2n-1}{\bf s}\, \prod_{j=2}^{2n-1} \,
\frac{1}{1-s_j s_{j+1}} \frac{1}{1-s_{2n}{\bf S}_{2n+1}}\equiv \int_0^1d^{2n-1}{\bf s}\, {\cal F}_{2n}({\bf s})\ $$
where $d^{2n-1}{\bf s}=ds_2\cdots ds_{2n}$, and the integral sign indicates that each of the $s$ variables is being integrated from zero to one, and we have defined
$$
{\bf S}_{k}\equiv 1+(-1)^k\prod_{j=2}^{k-2} s_j\ ,\qquad
{\cal F}_{2n}({\bf s}) = \prod_{j=2}^{2n-1} \,
\frac{1}{1-s_j s_{j+1}} \frac{1}{1-s_{2n}{\bf S}_{2n+1}}\ ,
$$
with
$$
S_3=0\ ,\qquad {\cal F}_{2}({\bf s}) =1\ .
$$
Note that for odd $k$, ${\bf S}_k<1$, while for even $k$, ${\bf S}_k>1$.
This object ${\bf S}_k$ has the property for any $k$
$$
{\bf S}_{k+1} -s_{k-1} = 1-s_{k-1} {\bf S}_k\ .
$$ The strategy is to consider developing a recursion relation when integrating over $ds_{2n}$ and $ds_{2n-1}$, relating $I_{2n}$ to $I_{2n-2}$. To that end it is useful to define the following functions of $x$, $y$ in the domain $ 0<x<1,\ 0<y<1$:
$$
{\cal P}_k(x,y) = \frac{1}{(2k)!}
\prod_{i=1}^k
\left(\pi^2 (2k-1)^2 + \ln^2\left[\frac{1-x}{x(1-y)}\right]\right)\ ,\qquad {\cal P}_0(x,y)\equiv 1\ ,
$$
and
$$
{\cal G}(\alpha,x,y) = \sum_{n=0}^\infty \left(\frac{\alpha}{\pi}\right)^{2n} {\cal P}_n(x,y) =
\frac{1}{2 \sqrt{1-\alpha ^2}}\left[\left(\frac{1-x}{x(1-y)}\right)^{c}+\left(\frac{1-x}{x(1-y) }\right)^{-c}\,\right]\ ,
$$
$$
c\equiv \frac{\sin ^{-1}(\alpha )}{\pi }\ .
$$
We generalize the problem to considering the integral
$$
{\cal I}_{2n}(\alpha) = \int_0^1d^{2n-1}{\bf s}\, {\cal F}_{2n}({\bf s})\,{\cal G}(\alpha,s_{2n},{\bf S}_{2n+1})\ .
$$
We can perform the $s_{2n}$ and $s_{2n-1}$ integrals in ${\cal I}_{2n}$ using the results (using the properties of ${\bf S_k}$ above) For $0<s_{2n-1}<1$ and $0<{\bf S}_{2n+1}<1$:
$$
\frac{1}{2} \int_0^1 ds_{2n} \frac{1}{(1-s_{2n-1} s_{2n})(1-s_{2n}{\bf S}_{2n+1})}\left[ \left(\frac{1-s_{2n}}{s_{2n}(1-{\bf S}_{2n+1})}\right)^c+ \left(\frac{1-s_{2n}}{s_{2n}(1-{\bf S}_{2n+1})}\right)^{-c}\right]
= \frac{\pi \csc (\pi c) \left(\left(\frac{1-s_{2n-1}}{1-{\bf S}_{2n+1}}\right)^{-c}-\left(\frac{1-s_{2n-1}}{1-{\bf S}_{2n+1}}\right)^c\right)}{2(s_{2n-1}-{\bf S}_{2n+1})}
=\frac{\pi \csc (\pi c)\left(\left(\frac{1-s_{2n-1}}{s_{2n-1}({\bf S}_{2n}-1)}\right)^{-c}-\left(\frac{1-s_{2n-1}}{s_{2n-1}({\bf S}_{2n}-1)}\right)^c\right)}{2(1-s_{2n-1}{\bf S}_{2n})}\
$$ For $0<s_{2n-2}<1$ and $1<{\bf S}_{2n}$:
$$
\frac{1}{2} \int_0^1 ds_{2n-1} \frac{1}{(1-s_{2n-2}s_{2n-1})(1-{\bf S}_{2n}s_{2n-1})}
\left[ \left(\frac{1-s_{2n-1}}{s_{2n-1}({\bf S}_{2n}-1)}\right)^c- \left(\frac{1-s_{2n-1}}{s_{2n-1}({\bf S}_{2n}-1)}\right)^{-c}\right]
= -\frac{\pi \csc (\pi c)
\left(\left(\frac{1-s_{2n-2}}{{\bf S}_{2n}-1}\right)^{-c}+\left(\frac{1-s_{2n-2}}{{\bf S}_{2n}-1}\right)^c-2 \cos
(\pi c)\right)}{2 (s_{2n-2}-{\bf S}_{2n})}
= -\frac{\pi \csc (\pi c)
\left(\left(\frac{1-s_{2n-2}}{1-s_{2n-2}{\bf S}_{2n-1}}\right)^{-c}+\left(\frac{1-s_{2n-2}}{1-s_{2n-2}{\bf S}_{2n-1}}\right)^c-2 \cos
(\pi c)\right)}{2 (1-s_{2n-2}{\bf S}_{2n-1})}
$$ With these integrals we can perform the integrations over $s_{2n}$ and $s_{2n-1}$ in our generalized integral ${\cal I}_{2n}(\alpha)$, obtaining $$
{\cal I}_{2n}(\alpha)
=\int d^{2n-3}{\bf s} \, {\cal F}_{2n-2}({\bf s})\, \left[\pi^2\csc^2(c\pi)\left({\cal G}(\alpha,s_{2n-2},{\bf S}_{2n-1}) -\frac{ \cos c\pi}{ \sqrt{1-\alpha^2}}\right) \right]
\,
=\int d^{2n-3}{\bf s}{\cal F}_{2n-2}({\bf s})\, \left[\frac{\pi^2}{\alpha^2}\left({\cal G}(\alpha,s_{2n-2},{\bf S}_{2n-1})-1\right) \right]
$$
where to get the second line we just plugged in $\pi c=\sin^{-1}\alpha$. Referring to the definition of ${\cal G}$ in eq.(\ref{gdef}), we can equate powers of $\alpha$ on both sides of the above equation with the result that for every $k\ge 0$,
$$
\int_0^1d^{2n-1}{\bf s}\, {\cal F}_{2n}({\bf s})\, {\cal P}_k(s_{2n},{\bf S}_{2n+1})
= \int_0^1d^{2n-3}{\bf s}\, {\cal F}_{2n-2}({\bf s})\, {\cal P}_{k+1}(s_{2n-2},{\bf S}_{2n-1})
$$
which is a pretty result. The above result allows us to write for the desired $2n$-dimensional integrals as one-dimensional integrals
$$
I_{2n}=\int_0^1d^{2n-1}{\bf s}\, {\cal F}_{2n}({\bf s})\, {\cal P}_0(s_{2n},{\bf S}_{2n+1})
=\int_0^1ds_2 {\cal P}_{n-1}(s_{2},0)\ .
$$
The above results then imply that
$$
\sum_{n=0}^\infty \left(\frac{\alpha}{\pi}\right)^{2n} I_{2n+2} = \int_0^1dx\,\sum_{n=0}^\infty \left(\frac{\alpha}{\pi}\right)^{2n} {\cal P}_n(x,0)
= \int_0^1dx\, {\cal G}(\alpha,x,0)
=\int_0^1dx\frac{1}{2 \sqrt{1-\alpha ^2}}\left[\left(\frac{1-x}{x}\right)^{c}+\left(\frac{1-x}{x }\right)^{-c}\right]
= \frac{\sin^{-1}\alpha}{\alpha\sqrt{1-\alpha^2}}
= \sum_{n=0}^\infty (2\alpha)^{2n} B(n+1,n+1)\ .
$$
Equating powers of $\alpha$ between the first and last expressions answers the posted question. This solution was found in collaboration with E. Mereghetti (we're physicists, so the language might look odd).
|
{
"source": [
"https://mathoverflow.net/questions/129955",
"https://mathoverflow.net",
"https://mathoverflow.net/users/21850/"
]
}
|
129,970 |
Consider three 2-torus ($S^1*S^1$) living in four space. Can I have links of these objects, which is generalization of links of circles in 3D? If so, how can I judge whether three 2-torus are linked or not?
|
We can think of $I_{n}$ as being a classical partition function for $n$ beads on a circle which cannot pass through each other, with logarithmic interaction potential between each bead and its next-to-nearest neighbors on either side. For $I_{2n}$ the beads fall into two ``colors" which do not have logarithmic interactions with each other; while for $I_{2n+1}$ the beads do not fall into two independent groups. We make two changes of variable. First, we can label the coordinates of the $k^{th}$ bead as $y_k$, where $y_1=0$ is fixed (exploiting the translation invariance of the problem) and we define $y_{2n+k} = 1+y_k$ (because of the periodic nature of the circle):
$$
u_i = y_{i+1}-y_i\ ,\qquad y_1=0\ ,\qquad y_{2n+i}\equiv 1+y_i \ .
$$
Then the integral can be written as a path ordered expression without the delta function constraint as
$$
I_{n}= \int_0^1 dy_{n} \int_0^{y_{n}} dy_{n-1}\cdots\int_0^{y_3} dy_2\, \prod_{k=1}^{2n}\frac{1}{y_{k+2}-y_k}\ .
$$
The second change of variables to $\{y_2,\ldots,y_n\}\to \{s_2,\ldots,s_n\}$ in order change the integration domain to a unit hypercube:
$$
y_{k} =\prod_{j=k}^{n} s_{j}\ ,
$$
with Jacobian
$$
J_n = \prod_{j=3}^{n} s_j^{j-2}\ .
$$
With this change of variables, $I_{2n}$ becomes (for $n\ge 2$)
$$
I_{2n} = \int_0^1 d^{2n-1}{\bf s}\, \prod_{j=2}^{2n-1} \,
\frac{1}{1-s_j s_{j+1}} \frac{1}{1-s_{2n}{\bf S}_{2n+1}}\equiv \int_0^1d^{2n-1}{\bf s}\, {\cal F}_{2n}({\bf s})\ $$
where $d^{2n-1}{\bf s}=ds_2\cdots ds_{2n}$, and the integral sign indicates that each of the $s$ variables is being integrated from zero to one, and we have defined
$$
{\bf S}_{k}\equiv 1+(-1)^k\prod_{j=2}^{k-2} s_j\ ,\qquad
{\cal F}_{2n}({\bf s}) = \prod_{j=2}^{2n-1} \,
\frac{1}{1-s_j s_{j+1}} \frac{1}{1-s_{2n}{\bf S}_{2n+1}}\ ,
$$
with
$$
S_3=0\ ,\qquad {\cal F}_{2}({\bf s}) =1\ .
$$
Note that for odd $k$, ${\bf S}_k<1$, while for even $k$, ${\bf S}_k>1$.
This object ${\bf S}_k$ has the property for any $k$
$$
{\bf S}_{k+1} -s_{k-1} = 1-s_{k-1} {\bf S}_k\ .
$$ The strategy is to consider developing a recursion relation when integrating over $ds_{2n}$ and $ds_{2n-1}$, relating $I_{2n}$ to $I_{2n-2}$. To that end it is useful to define the following functions of $x$, $y$ in the domain $ 0<x<1,\ 0<y<1$:
$$
{\cal P}_k(x,y) = \frac{1}{(2k)!}
\prod_{i=1}^k
\left(\pi^2 (2k-1)^2 + \ln^2\left[\frac{1-x}{x(1-y)}\right]\right)\ ,\qquad {\cal P}_0(x,y)\equiv 1\ ,
$$
and
$$
{\cal G}(\alpha,x,y) = \sum_{n=0}^\infty \left(\frac{\alpha}{\pi}\right)^{2n} {\cal P}_n(x,y) =
\frac{1}{2 \sqrt{1-\alpha ^2}}\left[\left(\frac{1-x}{x(1-y)}\right)^{c}+\left(\frac{1-x}{x(1-y) }\right)^{-c}\,\right]\ ,
$$
$$
c\equiv \frac{\sin ^{-1}(\alpha )}{\pi }\ .
$$
We generalize the problem to considering the integral
$$
{\cal I}_{2n}(\alpha) = \int_0^1d^{2n-1}{\bf s}\, {\cal F}_{2n}({\bf s})\,{\cal G}(\alpha,s_{2n},{\bf S}_{2n+1})\ .
$$
We can perform the $s_{2n}$ and $s_{2n-1}$ integrals in ${\cal I}_{2n}$ using the results (using the properties of ${\bf S_k}$ above) For $0<s_{2n-1}<1$ and $0<{\bf S}_{2n+1}<1$:
$$
\frac{1}{2} \int_0^1 ds_{2n} \frac{1}{(1-s_{2n-1} s_{2n})(1-s_{2n}{\bf S}_{2n+1})}\left[ \left(\frac{1-s_{2n}}{s_{2n}(1-{\bf S}_{2n+1})}\right)^c+ \left(\frac{1-s_{2n}}{s_{2n}(1-{\bf S}_{2n+1})}\right)^{-c}\right]
= \frac{\pi \csc (\pi c) \left(\left(\frac{1-s_{2n-1}}{1-{\bf S}_{2n+1}}\right)^{-c}-\left(\frac{1-s_{2n-1}}{1-{\bf S}_{2n+1}}\right)^c\right)}{2(s_{2n-1}-{\bf S}_{2n+1})}
=\frac{\pi \csc (\pi c)\left(\left(\frac{1-s_{2n-1}}{s_{2n-1}({\bf S}_{2n}-1)}\right)^{-c}-\left(\frac{1-s_{2n-1}}{s_{2n-1}({\bf S}_{2n}-1)}\right)^c\right)}{2(1-s_{2n-1}{\bf S}_{2n})}\
$$ For $0<s_{2n-2}<1$ and $1<{\bf S}_{2n}$:
$$
\frac{1}{2} \int_0^1 ds_{2n-1} \frac{1}{(1-s_{2n-2}s_{2n-1})(1-{\bf S}_{2n}s_{2n-1})}
\left[ \left(\frac{1-s_{2n-1}}{s_{2n-1}({\bf S}_{2n}-1)}\right)^c- \left(\frac{1-s_{2n-1}}{s_{2n-1}({\bf S}_{2n}-1)}\right)^{-c}\right]
= -\frac{\pi \csc (\pi c)
\left(\left(\frac{1-s_{2n-2}}{{\bf S}_{2n}-1}\right)^{-c}+\left(\frac{1-s_{2n-2}}{{\bf S}_{2n}-1}\right)^c-2 \cos
(\pi c)\right)}{2 (s_{2n-2}-{\bf S}_{2n})}
= -\frac{\pi \csc (\pi c)
\left(\left(\frac{1-s_{2n-2}}{1-s_{2n-2}{\bf S}_{2n-1}}\right)^{-c}+\left(\frac{1-s_{2n-2}}{1-s_{2n-2}{\bf S}_{2n-1}}\right)^c-2 \cos
(\pi c)\right)}{2 (1-s_{2n-2}{\bf S}_{2n-1})}
$$ With these integrals we can perform the integrations over $s_{2n}$ and $s_{2n-1}$ in our generalized integral ${\cal I}_{2n}(\alpha)$, obtaining $$
{\cal I}_{2n}(\alpha)
=\int d^{2n-3}{\bf s} \, {\cal F}_{2n-2}({\bf s})\, \left[\pi^2\csc^2(c\pi)\left({\cal G}(\alpha,s_{2n-2},{\bf S}_{2n-1}) -\frac{ \cos c\pi}{ \sqrt{1-\alpha^2}}\right) \right]
\,
=\int d^{2n-3}{\bf s}{\cal F}_{2n-2}({\bf s})\, \left[\frac{\pi^2}{\alpha^2}\left({\cal G}(\alpha,s_{2n-2},{\bf S}_{2n-1})-1\right) \right]
$$
where to get the second line we just plugged in $\pi c=\sin^{-1}\alpha$. Referring to the definition of ${\cal G}$ in eq.(\ref{gdef}), we can equate powers of $\alpha$ on both sides of the above equation with the result that for every $k\ge 0$,
$$
\int_0^1d^{2n-1}{\bf s}\, {\cal F}_{2n}({\bf s})\, {\cal P}_k(s_{2n},{\bf S}_{2n+1})
= \int_0^1d^{2n-3}{\bf s}\, {\cal F}_{2n-2}({\bf s})\, {\cal P}_{k+1}(s_{2n-2},{\bf S}_{2n-1})
$$
which is a pretty result. The above result allows us to write for the desired $2n$-dimensional integrals as one-dimensional integrals
$$
I_{2n}=\int_0^1d^{2n-1}{\bf s}\, {\cal F}_{2n}({\bf s})\, {\cal P}_0(s_{2n},{\bf S}_{2n+1})
=\int_0^1ds_2 {\cal P}_{n-1}(s_{2},0)\ .
$$
The above results then imply that
$$
\sum_{n=0}^\infty \left(\frac{\alpha}{\pi}\right)^{2n} I_{2n+2} = \int_0^1dx\,\sum_{n=0}^\infty \left(\frac{\alpha}{\pi}\right)^{2n} {\cal P}_n(x,0)
= \int_0^1dx\, {\cal G}(\alpha,x,0)
=\int_0^1dx\frac{1}{2 \sqrt{1-\alpha ^2}}\left[\left(\frac{1-x}{x}\right)^{c}+\left(\frac{1-x}{x }\right)^{-c}\right]
= \frac{\sin^{-1}\alpha}{\alpha\sqrt{1-\alpha^2}}
= \sum_{n=0}^\infty (2\alpha)^{2n} B(n+1,n+1)\ .
$$
Equating powers of $\alpha$ between the first and last expressions answers the posted question. This solution was found in collaboration with E. Mereghetti (we're physicists, so the language might look odd).
|
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"https://mathoverflow.net/questions/129970",
"https://mathoverflow.net",
"https://mathoverflow.net/users/33455/"
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|
130,287 |
What is the intuition behind using compact open topology for eg. in the case of Pontryagin dual ?
|
Given two spaces $X$ and $Y$, how to define the mapping space betweeen them, i.e. what topology should we put on the set of maps between them? If $X$ is compact and $Y$ a metric space, this is quite easy as one can put a metric on $Map(X,Y)$: For $f,g\in Map(X,Y)$ define their distance just to be the maximum of the distances between $f(x)$ and $g(x)$ as $x$ ranging over the points in $X$. If $Y$ is no longer metric, we have to find a replacement what it means for two maps to be close. Say, we have again two maps $f,g\in Map(X,Y)$. Let $K\subset X$ be compact and $U\subset Y$ be open such that $f(K)\subset U$. Assume now that $Y$ is Hausdorff (else, this construction might behave badly anyhow). Then $f(K)\subset Y$ is closed, so you would expect that if you move $f(K)$ a little bit, then it stays inside of $U$. So if $g$ is close to $f$, then $g(K)$ should be still inside of $U$.
Thus, it is sensible to define an open neighborhood of $f$ to be all maps $g$ such that $g(K) \subset U$. And actually, this agrees with the metric definition when $X$ is compact and $Y$ metric! Furthermore, as long as you have "enough" compact subsets in $X$ (i.e. $X$ is compactly generated) and everything is Hausdorff, then this topology has all the pleasant properties one likes.
|
{
"source": [
"https://mathoverflow.net/questions/130287",
"https://mathoverflow.net",
"https://mathoverflow.net/users/30999/"
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|
130,319 |
show the formula always gives an integer $$\frac{(2m)!(2n)!}{m!n!(m+n)!}$$ I don't remember where I read this problem, but it said this can be proved using a simple counting argument (like observing that $\frac{(3m)!}{m!m!m!}$ is just the number of ways of permuting m identical things of type 1, m of type-2 and m of type-3).
|
I found this paper I. M. Gessel, G. Xin, A Combinatorial Interpretation of the
Numbers $6(2n!)/n!(n+2)!$, Journal of Integer Sequences 8 (2005) Article 05.2.3 whose abstract says: It is well known that the numbers $\frac{(2m)!(2n)!}{m!n!(m+n)!}$ are integers, but in
general there is no known combinatorial interpretation for them. When $m = 0$ these
numbers are the middle binomial coefficients ${2n \choose n}$, and when $m = 1$ they are twice the Catalan numbers. In this paper, we give combinatorial interpretations for these
numbers when $m = 2$ or $3$. According to the authors, the first appearance of the problem of whether it's always an integer is in this note by Catalan: E. Catalan, Question 1135, Nouvelles Annales de Mathematiques 13 (1874), 207. Barry's comments to this post below point out an incorrect solution to the problem in a more general form was given by Bourguet here: M. L. Barbier, Solutions des questions proposées dans les Nouvelles annales, Nouvelles Annales de Mathematiques 14 (1875) 66-92, and the error was pointed out (and corrected?) by Catalan here: E. Catalan, Correspondance, Nouvelles Annales de Mathematiques 14 (1875) 178-180. Also in a comment to this post, Karan gave a link to a paper that proves the numbers in question are integers by a recurrence relation: D. Callan, A Combinatorial Interpretation for a
Super-Catalan Recurrence, Journal of Integer Sequences, 8 (2005) Article 05.1.8. Edit: The numbers in question are apparently called super Catalan numbers, and I just found this paper whose abstract says, "we show that the super Catalan numbers are special values of the Krawtchouk polynomials by deriving an expression for the super Catalan numbers in terms of a signed set." E. Georgiadis, A. Munemasa, H. Tanaka, A Note on Super Catalan Numbers, Interdisciplinary Information Sciences , 18 (2012) 23-24.
|
{
"source": [
"https://mathoverflow.net/questions/130319",
"https://mathoverflow.net",
"https://mathoverflow.net/users/33902/"
]
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|
130,355 |
Every elliptic curve $E/\mathbf Q$ is modular, in the sense that there exists a nonconstant morphism $X_0(N) \to E$ for some $N$. It is tempting to extend this definition in a naïve way to an arbitrary projective curve over $\mathbf Q$; if $Y$ is such a curve, we might say that $Y$ is modular if there exists a nonconstant morphism $X_0(N) \to Y$ for some $N$. A necessary condition for $Y$ to be modular is that it should have at least one rational point, since $X_0(N)$ always has a rational point. For an elliptic curve, this condition is satisfied by definition. There are certainly a great deal of curves which are modular in this sense. But is there an example of a curve (with a rational point) which is not modular?
|
One expects that the majority of algebraic curves over number fields having genus $> 1$ should not be modular in this sense. For instance, take a sufficiently general genus 2 curve $C$ over $\mathbf{Q}$ . Then its $\ell$ -adic $H^1$ (which is just the $\ell$ -adic Tate module of its Jacobian) will be a 4-dimensional Galois representation whose image lands inside $\mathrm{GSp}_4(\mathbf{Z}_\ell)$ . If $C$ is sufficiently generic, then the image of this Galois representation should be the whole of $\mathrm{GSp}_4(\mathbf{Z}_\ell)$ for all but finitely many $\ell$ ; in particular, it will be absolutely irreducible. (I don't know if this is known, but certainly one expects it to be the case.) On the other hand, if $C$ admits a non-constant map from $X_0(N)$ , then the its $H^1$ would have to be a quotient of the $H^1$ of $X_0(N)$ , and this can be calculated in terms of modular forms; in particular all its absolutely irreducible subquotients have dimension 2. So most genus 2 curves $C$ will not be modular in your sense, and if you get one that is, you should regard it as a rather unlikely coincidence. (A more high-powered interpretation of this is that $H^1(C)$ should be the Galois representation attached to a degree 2 Siegel modular form. In some very special cases this Siegel modular form will be endoscopic, i.e. describable in terms of lifts from elliptic modular forms, but most Siegel mod forms will not be endoscopic and thus will not have anything to do with $X_0(N)$ for any $N$ .) If you're willing to relax your definition of "modular", though, you can get many more possibilities. There's a very striking result of Belyi stating that any algebraic curve defined over a number field can be obtained as the quotient of the upper half-plane by some subgroup of $PSL(2, \mathbf{Z})$ , although the corresponding group will usually not be a congruence subgroup.
|
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"https://mathoverflow.net/questions/130355",
"https://mathoverflow.net",
"https://mathoverflow.net/users/6779/"
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|
130,623 |
By a drawing of the Fano plane I mean a system of seven simple curves and
seven points in the real plane such that every point lies on exactly three curves, and every curve contains
exactly three points; there is a unique curve through every pair of points, and every two
curves intersect in exactly one point; the curves do not intersect except in the seven points under
consideration. The familiar picture (source) does not count as a drawing, since the last requirement is not satisfied:
there are two "illegal" intersections. In fact, this is easy to fix: (source) However, this drawing is degenerate in the sense that two of the curves
just "touch" each other, without crossing, at some point. And here,
eventually, my question goes: Is every drawing of the Fano plane degenerate? (Although I can give a topological definition of degeneracy, it is a little technical and, may be, not the smartest possible one, so I prefer to suppress it here.)
|
Does this one work? (source)
|
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"source": [
"https://mathoverflow.net/questions/130623",
"https://mathoverflow.net",
"https://mathoverflow.net/users/9924/"
]
}
|
130,883 |
Perhaps the "proofs" of ABC conjecture or newly released weak version of twin prime conjecture or alike readily come to your mind. These are not the proofs I am looking for. Indeed my question was inspired by some other posts seeking for a hint to understand a certain more or less well-establised proof, or some answers to those posts. I am interested in proofs at undergraduate levels. Since the question as asked in the title would be too personal, I suggest a longer and hopefully more positive version: Is there any proof that you feel you didn't understand fully until years later? The reason that I am interested in this question is that we are currently working on an assessment framework for assessing students' understanding of proof. Reading some previous posts on MO, It occurred to me that perhaps we are too naive in our approach just seeking for understanding logical structure, the key point and so on. It would be very informative if you kindly include in your answer the follow-up of the proof you mention.
|
I hate to sound dumb, but I still don't really understand the Pythagorean theorem as well as I like. I've seen lots of proofs but they all feel too clever for me. To me, understanding a theorem or its proof means being able to see why it's true without having to work out the details of the proof.
|
{
"source": [
"https://mathoverflow.net/questions/130883",
"https://mathoverflow.net",
"https://mathoverflow.net/users/29316/"
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|
130,988 |
Given a finite simple group $G$, we can consider the quasisimple extensions $\tilde G$ of $G$, that is to say central extensions which remain perfect. Some basic group cohomology (based on the standard trick of averaging a cocycle to try to make it into a coboundary) shows that up to isomorphism, there are only finitely many such quasisimple extensions, and they are all quotients of a maximal quasisimple extension, which is known as the universal cover of $G$, and is an extension of $G$ by a finite abelian group known as the Schur multiplier $H^2(G,{\bf C}^\times)$ of $G$ (or maybe it would be slightly more accurate to say that it is the Pontryagian dual of the Schur multiplier, although up to isomorphism the two groups coincide). On going through the list of finite simple groups it is striking to me how small the Schur multipliers are for all of them; with the exception of the projective special linear groups $A_{n-1}(q)=PSL_n({\bf F}_q)$ and the projective special unitary groups ${}^2 A_{n-1}(q^2) = PSU_n({\bf F}_q)$, all other finite simple groups have Schur multiplier of order no larger than 12, and even the projective special linear and special unitary groups of rank $n-1$ do not have Schur multiplier of size larger than $n$ (other than a finite number of small exceptional cases, but even there the largest Schur multiplier size is 48). In particular, in all cases the Schur multiplier is much smaller than the order of the group itself (indeed it is always of order $O(\sqrt{\frac{\log|G|}{\log\log|G|}})$). For comparison, the standard proof of the finiteness of the Schur multiplier (based on showing that every $C^\times$-valued cocycle on $G$ is cohomologous to $|G|^{th}$ roots of unity) only gives the terrible upper bound of $|G|^{|G|}$ for the order of the multiplier. In the case of finite simple groups of Lie type, one can think of the Schur multiplier as analogous to the notion of a fundamental group of a simple Lie group, which is similarly small (being the quotient of the weight lattice by the root lattice, it is no larger than $4$ in all cases except for the projective special linear group $PSL_n$, where it is of order $n$ at most). But this doesn't explain why the Schur multipliers for the alternating and sporadic groups are also so small. Intuitively, this is asserting that it is very difficult to make a non-trivial central extension of a finite simple group. Is there any known explanation (either heuristic, rigorous, or semi-rigorous) that helps explain why Schur multipliers of finite simple groups are small? For instance, are there results limiting the size of various group cohomology objects that would support (or at least be very consistent with) the smallness of Schur multipliers? Ideally I would like an explanation that does not presuppose the classification of finite simple groups.
|
The Schur multiplier $H^2(G;{\mathbb C}^\times) \cong H^3(G;{\mathbb Z})$ of a finite group is a product of its $p$-primary parts $$H^3(G;{\mathbb Z}) = \oplus_{ p | |G|} H^3(G;{\mathbb Z}_{(p)})$$ as is seen using the transfer. The $p$-primary part $H^3(G;{\mathbb Z}_{(p)})$ depends only of the $p$-local structure in $G$ i.e., the Sylow $p$-subgroup $S$ and information about how the subgroups of $S$ become conjugate or "fused" in $G$. (This data is also called the $p$-fusion system of $G$.) More precisely, the Cartan-Eilenberg stable elements formula says that $$H^3(G;{\mathbb Z}_{(p)}) = \{ x \in H^3(S;{\mathbb Z}_{(p)})^{N_G(S)/C_G(S)} |res^S_V(x) \in H^3(V;{\mathbb Z}_{(p)})^{N_G(V)/C_G(V)}, V < S\}$$ One in fact only needs to check restriction to certain V above. E.g., if S is abelian the formula can be simplified to $H^3(G;{\mathbb Z}_{(p)}) = H^3(S;{\mathbb Z}_{(p)})^{N_G(S)/C_G(S)}$ by an old theorem of Swan. (The superscript means taking invariants.) See e.g. section 10 of my paper linked HERE for some references. Note that the fact that one only need primes p where G has non-cyclic Sylow $p$-subgroup follows from this formula, since $H^3(C_n;{\mathbb Z}_{(p)}) = 0$. However, as Geoff Robinson remarks, the group $H^3(S;{\mathbb Z}_{(p)})$ can itself get fairly large as the $p$-rank of $S$ grows. However, $p$-fusion tends to save the day. The heuristics is: Simple groups have, by virtue of simplicity, complicated $p$-fusion, which by the above formula tends to make $H^3(G;{\mathbb Z}_{(p)})$ small. i.e., it becomes harder and harder to become invariant (or "stable") in the stable elements formula the more $p$-fusion there is. E.g., consider $M_{22} < M_{23}$ of index 23: $M_{22}$ has Schur multiplier of order 12 (one of the large ones!). However, the additional 2- and 3-fusion in $M_{23}$ makes its Schur multiplier trivial. Likewise $A_6$ has Schur multiplier of order 6, as Geoff alluded to, but the extra 3-fusion in $S_6$ cuts it down to order 2. OK, as Geoff and others remarked, it is probably going to be hard to get sharp estimates without the classification of finite simple groups. But $p$-fusion may give an idea why its not so crazy to expect that they are "fairly small" compared to what one would expect from just looking at $|G|$...
|
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"source": [
"https://mathoverflow.net/questions/130988",
"https://mathoverflow.net",
"https://mathoverflow.net/users/766/"
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|
131,185 |
Yitang Zhang recently published a new attack on the Twin Primes Conjecture . Quoting Andre Granville : “The big experts in the field had
already tried to make this approach
work,” Granville said. “He’s not a
known expert, but he succeeded where
all the experts had failed.” What new approach did Yitang Zhang try & what did the experts miss in the first place?
|
My understanding of this, which is essentially cobbled together from the various news accounts, is as follows: Let $\pi(x;q,a)$ denote the number of primes less than $x$ congruent to $a\bmod q$ , and $\pi(x)$ the number of primes less than $x$ . $\phi(n)$ is the number of positive integers less than or equal to n that are relatively prime to n, and is usually referred to as Euler's totient function . Denote by $EH(\theta)$ the assertion that $\forall A > 0, \exists C > 0, \forall x > 2$ , the following inequality holds: $$\sum_{1\leq q \leq x^{\theta}} \max_{\gcd(a,q)=1} \left| \pi(x;q,a) - \frac{\pi(x)}{\phi(q)} \right| \ll \frac{x}{\log^A(x)} (*)$$ for all large $x$ . The Bombieri–Vinogradov theorem asserts that EH( $\theta$ ) holds for $\theta <1/2$ , and the Elliot–Halberstam conjecture asserts that EH( $\theta$ ) holds for all $\theta<1$ . In the mid 2000's Goldston, Pintz and Yildirim proved that if the Elliott–Halberstam conjecture holds for any level of distribution $\theta>1/2$ then one has infinitely many bounded prime gaps (where the size of the gap is a function of $\theta$ , for $\theta>.971$ they get a gap of size 16). Since Bombieri–Vinogradov tells us that $EH(\theta)$ holds for $\theta < 1/2$ , in some sense the Goldston–Pintz–Yildirim arguments just barely miss giving bounded gaps. On the other hand, in the 1980s Fouvry and Iwaniec were able to push the level of distribution in the Bombieri–Vinogradov theorem above $1/2$ at the expense of (1) removing the absolute values, (2) removing the maximum over residue classes and, (3) weighting the summand by a `well-factorable' function. This was subsequently improved in a series of papers by Bombieri, Friedlander and Iwaniec . For the Goldston–Pintz–Yildirim argument the first two restrictions didn't pose a significant hurdle, however the inclusion of the well-factorable weight appeared to prevent one from using it with the Goldston–Pintz–Yildirim machinery. This is where my knowledge becomes extremely spotty, but my understanding is that Zhang replaces the Goldston–Pintz–Yildirim sieve with a slightly less efficient one. While less efficient, this modification allows him some additional flexibility. He then is able to divide up the quantity that is estimated using (*) into several parts. Some of these are handled using well-established techniques, however in the most interesting range of parameters he is able to reduce the problem to something that involves a well-factorable function (or something similar) which he ultimately handles with arguments similar to those of Bombieri, Friedlander and Iwaniec. Zhang's argument reportedly gives a gap size that is around 70 million. My suspicion is that this gap will quickly decrease. In their theorem, Bombieri, Friedlander and Iwaniec give a level of distribution around $4/7$ , where (according to these notes ) Zhang seems to be working with a level of distribution of the form $1/2 + \delta$ for $\delta$ on the order of $1/1000$ , so there is likely room for optimization. As a reference point if one had the level of distribution $55/100$ ( $< 4/7$ ) in the unmodified Elliot–Halberstram conjecture (without the well-factorable weight), the Goldston, Pintz and Yildirim gives infinitely many gaps of size less than 2956. The parity problem, however, is widely known to prevent approaches of this form (at least on their own) from getting a gap smaller than length 6. Of course, there are more technical obstructions as well and even on the full Elliott–Halberstam conjecture one can only get a gap of 16 at present.
|
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"source": [
"https://mathoverflow.net/questions/131185",
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|
131,221 |
The recent sensational news on bounded gaps between primes made me wonder: what is the status of Yitang Zhang's earlier arXiv preprint On the Landau-Siegel zeros conjecture ? If this result is correct, then (in my opinion) it is even bigger news for analytic number theory. Has anyone checked this paper carefully? Update (November 2022). Yitang Zhang posted a new arXiv preprint Discrete mean estimates and the Landau-Siegel zero .
|
This is not an answer regarding the paper, but I think should be helpful. During a recent interview (in Chinese), he commented: 问：前几天我去北京遇到葛立明，他说当时你在做个大问题，快做出来了。所以找你去新罕布什尔大学。 答： 那是关于Siegel零点的工作 ， 我有一篇网络文章 ， 是不完整的。目前我还不敢说我完全做成 ， 但是的确有很大进展 。孪生质数这个问题我做了三、四年。但希望大家不要误会，这个问题我是想了三、四年，但不是说我所有时间都在做它。一直到去年9月，我因为肯定可以做出来了，才暂时放下别的东西。 The highlighted part can be translated as: ...that is my work on Siegel zeros. I have a paper online, which is incomplete. I cannot say I have finished the work by now, but I did made remarkable progress... So the paper is unfinished, and we can wait until later when it is officially published. Edit: I guess it is public. But in case OP or other people do not know, he has visited IAS and gave some lectures in public on this topic. The videos are available at here: https://www.ias.edu/video/jointiasnts/2013/0926-YitangZhang and a summary can be found at here: http://www.math.ias.edu/node/5320 ( Wayback Machine - the below is copied from here with the MathJax reinstated) Let $\chi$ be a primitive real character. We first establish a relationship between the existence of the Landau-Siegel zero of $L(s,\chi)$ and the distribution of zeros of the Dirichlet $L$ -function $L(s,\psi)$ , with $\psi$ belonging to a set $\Psi$ of primitive characters, in a region $\Omega$ . It is shown that if the Landau-Siegel zero exists (equivalently, $L(1,\chi)$ is small), then, for most $\psi \in \Psi$ , not only all the zeros of $L(s,\psi)$ in $\Omega$ are simple and lie on the critical line, but also the gaps between consecutive zeros are close to integral multiples of the half of the average gap. In comparison with certain conjectures on the vertical distribution of zeros of $\zeta(s)$ , it is reasonable to believe that the gap assertion would fail to hold. In order to derive a contradiction from the gap assertion, we attempt to reduce the problem to evaluating a certain discrete mean; the idea is motivated by the work of Conrey, Ghosh and Gonek on the simple zeros of $\zeta(s)$ . We shall describe the coefficient of the main term and provide some numerical evidences. In some special cases, the problem is further reduced to calculating small positive eigenvalues of linear integral equations with Hermitian kernels.
|
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"https://mathoverflow.net/questions/131221",
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|
131,364 |
This is a somewhat imprecise question, as I am not sure how exactly how to formalise how to do mathematics "without" a certain key tool, but hopefully the intent of the question will still be clear. Let $G$ be a finite group. Traditionally, a character $\chi: G \to {\bf C}$ on $G$ is defined as being the trace of a finite-dimensional unitary representation $\rho: G \to U(V)$ of $G$, and then representation-theoretic tools (including Schur's lemma) can then be used to derive the basic results of character theory, including the following assertions: The irreducible characters form an orthonormal basis of the space $L^2(G)^G$ of class functions, and all other characters are natural number combinations of the irreducible characters. The space of characters form a semiring with identity; in particular, for three irreducible characters $\chi_1,\chi_2,\chi_3$, the structure constants $\langle \chi_1 \chi_2, \chi_3 \rangle$ are natural numbers. For any character $\chi$, $\chi(1)$ is a positive integer. For any character $\chi$, $\chi(g^{-1})=\overline{\chi(g)}$ for all $g$. For any irreducible character $\chi$, the convolution operation $f \mapsto f * \chi(1) \chi$ is a minimal idempotent in $L^2(G)$, and one has the Fourier inversion formula $f = \sum_\chi f * \chi(1) \chi$ for all class functions $f$. Furthermore, the image $I_\chi$ of the convolution operation $f \mapsto f * \chi(1) \chi$ in $L^2(G)$ (or ${\bf C} G$) is an irreducible $G \times G$ representation. If $\chi$ is an irreducible character of $G$, and $\eta$ is an irreducible character of a subgroup $H$ of $G$, then the structure constant $\langle \chi, \operatorname{Ind}^G_H \eta \rangle_{L^2(G)^G} = \langle \operatorname{Res}^H_G \chi, \eta \rangle_{L^2(H)^H}$ is a natural number. Note that representation theory does not make an explicit appearance in the above character theory facts (induction and restriction of characters can be done at a purely character-theoretic level without explicit reference to representations), other than as one of the conclusions to Fact 5. Note also that one can define characters directly, without mention of representations. The space $L^2(G)^G$ of class functions is a finite-dimensional commutative algebra under the convolution operation, and one can then locate the minimal idempotents $f$ of this algebra; $f(1)$ will then be positive, so one can define an "irreducible" character $\chi$ associated to this idempotent by the formula $f = \chi(1) \chi$ with $\chi(1) := f(1)^{1/2}$ being the positive square root of $f(1)$. One can then define an arbitrary character to be a natural number combination of the irreducible characters. This definition of character automatically gives Facts 1, 4, and 5 above (the $G \times G$-irreducibility of $I_\chi$ can be obtained by computing that the dimension of $\operatorname{Hom}^{G \times G}(I_\chi,I_\chi)$ is one), and if one is allowed to use Schur's lemma (or the Artin-Wedderburn theorem), one can also show that this definition is equivalent to the usual representation-theoretic definition of a character which then gives the remaining Facts 2, 3, and 6. My (imprecise) question is whether one can still recover Facts 2, 3, and 6 from this non-representation-theoretic definition of a character if one is "not allowed" to use Schur's lemma or the Artin-Wedderburn theorem. Now, I do not know how to rigorously formalise the concept of not being allowed to use a particular mathematical result; my first attempt was to phrase the problem with the complex numbers replaced by the field of definition of the characters $\chi$ (or the cyclotomic field of order $|G|$), so that the underlying field is not algebraically closed and so Schur's lemma or Artin-Wedderburn do not directly apply. But this is hardly any constraint at all since one can immediately pass to the algebraic closure of these fields in order to bring Schur or Artin-Wedderburn back into play. Another option is to take a constructivist (or maybe reverse mathematics) point of view, and only permit one's mathematical reasoning to work with representations as long as they can be constructed from characters using explicit, "functorial", representation-theoretic constructions (e.g. tensor sum, tensor product, orthogonal complement, isotypic component, Schur functors, induction, or restriction) from concrete representations (e.g. trivial representation, regular representation, or quasiregular representation), but prohibit any argument that requires one to make "arbitrary" or "non-functorial" choices on representations (and in particular, to split an isotypic representation or module into irreducibles). However, I do not know how to formalise this sort of mathematical reasoning (perhaps one has to introduce a suitable topos?). It has also been suggested to me (by Allen Knutson) that perhaps the correct setting for this framework is that of quantum groups over roots of unity rather than classical groups, but I am not familiar enough with quantum groups to formalise this suggestion. Leaving aside the question of how to properly formalise the question, one can make some intriguing partial progress towards Facts 2, 3, 6 without invoking Schur or Artin-Wedderburn. The isotypic representation $I_\chi$ has character $\chi(1) \chi$ by construction, so in particular this shows that $\chi(1)^2$ is a positive integer which is in some sense the "square" of Fact 3. By considering the $\chi_3\otimes \overline{\chi_3}$-isotypic component of $I_{\chi_1} \otimes I_{\chi_2}$, viewed as $G \times G$ representations, one can similarly show that the square $|\langle \chi_1 \chi_2, \chi_3 \rangle|^2$ of the structure constant $\langle \chi_1 \chi_2, \chi_3 \rangle$ is a natural number, and by viewing these representations instead as $G$-representations one also gets that $\chi_1(1)\chi_2(1) \chi_3(1) \langle \chi_1 \chi_2, \chi_3 \rangle$ is a natural number. These two facts don't quite establish Fact 2, but they do at least show that Fact 3 implies Fact 2. Finally, for Fact 6, the Frobenius reciprocity $\langle \chi, \operatorname{Ind}^G_H \eta \rangle_{L^2(G)^G} = \langle \operatorname{Res}^H_G \chi, \eta \rangle_{L^2(H)^H}$ is an easy algebraic identity if one defines induction and restriction in character-theoretic terms, and the same sort of arguments as before show that $|\langle \chi, \operatorname{Ind}^G_H \eta \rangle_{L^2(G)^G}|^2$ and $\chi(1) \eta(1) \langle \chi, \operatorname{Ind}^G_H \eta \rangle_{L^2(G)^G}$ are natural numbers, so again Fact 3 will imply Fact 6. (Conversely, it is not difficult to deduce Fact 3 from either Fact 2 or Fact 6.) So it all seems to boil down to Fact 3, or equivalently that the dimension of any minimal ideal of $L^2(G)$ (or ${\mathbf C} G$, if you prefer) is a perfect square. This is immediate from the Artin-Wedderburn theorem, and also follows easily from Schur's lemma (applied to an irreducible representation of this ideal) but I have been unable to demonstrate this fact without such a representation-theoretic (or module-theoretic) tool. Characters do give a partial substitute for Schur's lemma, namely that the dimension of the space $\operatorname{Hom}^G(V,W)$ of $G$-morphisms between two representations $V,W$ is equal to $\langle \chi_V, \chi_W \rangle_{L^2(G)^G}$, where $\chi_V, \chi_W$ are the characters associated to $V, W$. This gives Schur's lemma when the characters $\chi_V, \chi_W$ are irreducible in the sense defined above (defined through minimal idempotents and square roots rather than through irreducible representations). At the level of $G \times G$-representations, the isotypic components $I_\chi$ are irreducible (in both the character-theoretic and representation-theoretic senses) and so one has a satisfactory theory at this level (which is what is giving the "square" of Facts 2, 3, and 6) but at the level of $G$-representations there are these annoying multiplicities of $\chi(1)$ which do not seem to be removable without the ability to reduce to subrepresentations for which Schur's lemma applies. Somehow the enemy is a sort of phantom scenario in which a group-like object $G$ has an irreducible (but somehow not directly observable) "ghost representation" of an irrational dimension such as (say) $\sqrt{24}$, creating an isotypic component $I_\chi$ whose dimension ($24$, in this case) is not a perfect square. This is clearly an absurd situation, but one which appears consistent with the weakened version of representation theory discussed above, in which Schur's lemma, the Artin-Wedderburn theorem, or "non-constructive" representations are not available. But I do not know if this is a genuine limitation to this sort of theory (e.g. can one cook up a quantum group with such irrational representations?), or whether I am simply missing some clever argument. (My motivation for this question, by the way, is to explore substitutes for character-theoretic or representation-theoretic methods in finite group theory, for instance to find alternate proofs of Frobenius's theorem on Frobenius groups, which currently relies crucially on Fact 6.)
|
I spent a long time writing an answer to this question, but MO did not believe I was a human being ( I did mis-spell one of the test words, but everyone deserves a second chance, I think ), so it seems to have disappeared. I am not sure I have the energy to do it again right now, but here (in precis, though not precise) are three points I thought worth making/suggesting: The group determinant has been mentioned: in a 1991 Proc AMS paper, Formanek and Sibley proved that the group determinant determines the group. Perhaps you could use the analogue of the group determinant to tease out the properties that an algebraic structure $G$, whose "formal character theory" satisfies the properties you can get without Schur and Wedderburn, would have. Such a "group determinant" would not a priori have the property that the multiplicity of an irreducible factor equals its degree. It is possible to get quite far into the structure of the group algebra just using the symmetric algebra structure of the complex group algebra of $G$ induced by the linear form $t$ with $t(\sum_{g \in G} a_{g}g ) = a_{1}.$ Since $t$ vanishes on nilpotent elements, it follows that no non-zero right ideal of $\mathbb{C}G$ consists of nilpotent elements and that for each minimal (two-sided) ideal $A$ of $\mathbb{C}G$, $Z(A)$ is $1$-dimensional.
You might argue that this is using representation theory, since it is not a priori immediately obvious that $t$ vanishes on nilpotent elements until one notes that (up to the multiple $|G|$ ) $t$ is the trace afforded by the regular representation. Frobenius's theorem, and other normal complement theorems of a similar nature suggest
that there might be an analogue (under certain hypotheses) of the transfer homomorphism
but when the target group is not necessarily Abelian. That theorem can be proved in the case that $H$ is solvable by using the usual transfer homomorphism, and what the theorem says in the end (in the general case) is that the identity homomorphism from $H$ to itself extends to a homomorphism from $G$ onto $H.$ If one tries a transfer-type proof, it looks as though it almost would work, except that the order of products matters when the target group is non-Abelian. Nevertheless, the Theorem does in the end say that the homomorphism you would like to define by "transfer" is well-defined after all.
|
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|
131,407 |
Assume for this question that ZF set theory is sound. Now consider the language "PROVELOOP," which consists of all descriptions of Turing machines M, for which there exists a ZF proof that M runs forever on a blank input. It's clear that PROVELOOP is recursively-enumerable, and hence reducible to the halting problem. I can also prove that PROVELOOP is undecidable (details below). But I can't see how to prove that PROVELOOP is Turing-equivalent to the halting problem! (This is contrast to, say, the set of all descriptions of Turing machines that provably halt , which is just the same thing as the set of all descriptions of Turing machines that do halt!) I'm guessing that there's a reduction from HALT that I haven't thought of, though it would be exciting if PROVELOOP were to have intermediate degree like the Friedberg-Muchnik languages. In any case, whatever the answer, I assume it must be known! Hence this question. Proof that PROVELOOP is undecidable. Consider the following problem, which I'll call "Consistent Guessing" (CG). You're given as input a description of a Turing machine M. If M accepts given a blank input, then you need to accept, while if M rejects you need to reject. If M runs forever, then you can either accept or reject, but in either case you must halt. By adapting the undecidability proof for HALT, we can easily show that CG is undecidable also. Namely, suppose P solves CG. Let Q take as input a Turing machine description $\langle M \rangle$, and solve CG for the machine $M(\langle M \rangle)$ by calling P as a subroutine. Then $Q(\langle Q \rangle)$ (i.e., Q run on its own description) must halt, accept if it rejects, and reject if it accepts. Let's now prove that CG is Turing-reducible to PROVELOOP. Given a description of a Turing machine M for which we want to solve CG, simply create a new Turing machine M', which does the same thing as M except that if M accepts, then M' goes into an infinite loop instead. Then if M accepts, then M' loops, and moreover there's a ZF proof that M' loops. On the other hand, if M rejects, then M' also rejects, and there's no ZF proof that M' loops (by the assumption that ZF is sound). If M loops, then there might or might not be a ZF proof that M' loops -- but in any case, by calling PROVELOOP on M', we separate the case that M accepts from the case that M rejects, and therefore solve CG on M. So $CG \le_{T} PROVELOOP$, and PROVELOOP is undecidable as well. One more note. In the comments of this blog post , Andy Drucker supplied a proof that CG is not equivalent to HALT, but rather has Friedberg-Muchnik-like intermediate status. So, the situation is $0 \lt_{T} CG \le_{T} PROVELOOP \le_{T} HALT$ with at least one of the last two inequalities strict. Again, I'm sure this is all implicit in some computability paper from the 1960s or something, but I wouldn't know where to find it.
|
The first thing to notice is that if ZF is consistent, then it is
consistent with ZFC that what you call ProveLoop is actually decidable. The
reason is that if ZF is consistent, then by the incompleteness
theorem, it is consistent with ZFC that $\neg$Con(ZF), in which case everything is provable in
ZF, in which case every program is in ProveLoop. So in the proof that ProveLoop is undecidable, one needs to make
an additional assumption about the reliability of the proofs in ZF to
avoid this issue with the incompleteness theorem. Meanwhile, under such a consistency assumption, ProveLoop is indeed
equivalent to the halting problem. Theorem. Assume Con(ZF). Then ProveLoop is Turing
equivalent to the Halting problem. Proof. Under the Con(ZF) assumption, it follows that whenever ZF proves that a program doesn't halt, then it really doesn't halt, since if it did halt, then this fact would also be provable, contrary to consistency. Clearly ProveLoop is c.e. and hence reducible to the
halting problem, as you pointed out. Conversely, let's reduce the
halting problem to ProveLoop. Given any program $p$, we want to
decide whether $p$ halts on a blank tape, using an oracle for
ProveLoop. Define a computable function $f$, so that $f(q)$ is the program
such that, on trivial input, if $p$ halts on the blank tape, then
$f(q)$ jumps into an immediate infinite loop, and otherwise, while waiting for $p$ to halt, the program $f(q)$ halts just in case it finds a proof that $q$ does not halt.
By the recursion theorem, there is a program $r$ such that $r$ and
$f(r)$ compute the same function, and we can find this $r$
effectively. Furthermore, by using the $r$ from the proof of the recursion theorem, we may also assume that ZF proves that $r$ and $f(r)$ compute the same function. Notice that it can't ever be that $r$ halts on account of finding a proof that $r$ does not halt, by our assumption which ensures the accuracy of proofs of non-halting, and so definitely $r$ does not halt in any case. Meanwhile, if $p$ halts, then $r$ does not halt, but for a trivial reason that will be provable in ZF, namely, the fact that $p$ halted; and otherwise, when $p$ does not halt, then $r$ will run forever, but this fact will not be provable (for if it were provable, then $r$ would halt, contrary to consequence of our assumption that such proofs are reliable). So what we have is exactly a reduction of the halting problem to ProveLoop, as desired. QED
|
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|
131,413 |
I understand that one can give a proof of each of these propositions assuming the truth of the other. But this seems a bit squishy to me, since there is a trivial sense in which any two true theorems are equivalent (to any proof of Theorem A, prepend "Assume Theorem B", and vice versa; the objection "But the proof of Theorem A doesn't really use the assumption that Theorem B holds" seems more psychological than mathematical). One might try to formalize the notion of equivalence by considering the lengths of proofs, saying "There is a derivation of Theorem A from Theorem B that is significantly shorter than any proof of Theorem A from scratch, and vice versa", but this too is squishy, in two distinct ways: the length of a proof depends on the formalization procedure one chooses, and "significantly shorter" is vague. Moreover, it's hard to imagine how one could work with this notion of equivalence, since the totality of all short proofs is going to be hard to get a handle on, for the usual reasons. Can one find some sort of mathematical context (a topos, perhaps?) in which there is a rigorously defined (and not vacuously true) meaning of the equivalence between Sperner and Brouwer? (For a recent article that discusses this equivalence and gives pointers to relevant literature, see "A Borsuk-Ulam Equivalent that Directly Implies Sperner's Lemma" by Nyman and Su in the April 2013 issue of the American Mathematical Monthly, a version of which is available online at http://willamette.edu/~knyman/papers/Fan_Sperner.pdf .) See also the related thread Sperner's lemma and Tucker's lemma .
|
Sperner's lemma is not equivalent to Brouwer's Fixed Point Theorem. All that one can prove directly from Sperner's Lemma is the following weaker statement. Approximate Fixed Point Theorem. Let $K$ be the standard $n$-dimensional simplex and let $f:K \to K$ be a continuous function. For every $\varepsilon \gt 0$ there is an $x \in K$ such that $|f(x) - x| \leq \varepsilon$. A compactness argument is then needed to derive the existence of an actual fixed point for $f$. The analysis is done in §IV.7 of Simpson's Subsystems of Second-Order Arithmetic and the conclusion is that the Brouwer Fixed Point Theorem is equivalent to the Weak König Lemma over the base system RCA 0 . On the other hand, since Sperner's Lemma is a finite combinatorial statement, it is provable in RCA 0 . The Approximate Fixed Point Theorem is also provable in RCA 0 via Sperner's Lemma. The final compactness argument is however beyond the reach of RCA 0 .
|
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|
131,527 |
Suppose I have the symmetric tridiagonal matrix: $ \begin{pmatrix}
a & b_{1} & 0 & ... & 0 \\\
b_{1} & a & b_{2} & & ... \\\
0 & b_{2} & a & ... & 0 \\\
... & & ... & & b_{n-1} \\\
0 & ... & 0 & b_{n-1} & a
\end{pmatrix} $ All of the entries can be taken to be positive real numbers and all of the $a_{i}$ are equal. I know that when the $b_{i}$'s are equal (the matrix is uniform), there are closed-form expressions for the eigenvalues and eigenvectors in terms of cosine and sine functions. Additionally, I know of the recurrence relation: $det(A_{n}) = a\cdot det(A_{n-1}) - b_{n-1}^{2}\cdot det(A_{n-2})$ Additionally, since my matrix is real-symmetric, I know that its eigenvalues are real. Is there anything else I can determine about the eigenvalues? Furthermore, is there a closed-form expression for them?
|
The type of matrix you have written down is called Jacobi matrix and people are still discovering new things about them basically their properties fill entire bookcases at a mathematics library. One of the reasons is the connection to orthogonal polynomials. Basically, if $\{p_n(x)\}_{n\geq 0}$ is a family of orthogonal polynomials, then they obey a recursion relation of the form
$$
b_n p_{n+1}(x) + (a_n- x) p_n(x) + b_{n-1} p_{n-1}(x) = 0.
$$
You should be able to recognize the form of your matrix from this. As far as general properties of the eigenvalues, let me mention two: The eigenvalues are simple. In fact one has $\lambda_j - \lambda_{j-1} \geq e^{-c n}$, where $c$ is some constant that depends on the $b_j$. The eigenvalues of $A$ and $A_{n-1}$ interlace.
|
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|
131,528 |
Recently I posted a conjecture at Math.SE :
$$\int_0^\infty\ln\frac{J_\mu(x)^2+Y_\mu(x)^2}{J_\nu(x)^2+Y_\nu(x)^2}\mathrm dx\stackrel{?}{=}\frac{\pi}{2}(\mu^2-\nu^2),$$
where $J_\mu(x)$ and $Y_\mu(x)$ are the Bessel function of the first and second kind.
It is supported by numerical calculation with hundreds of digits of precision for many different values of $\mu, \nu$. The question is open for several days with +500 bounty on it and is not resolved yet. But my question here is not about if this conjecture true or false. Obviously, several possible closed forms matched my numeric calculations, for example:
$$\left(\frac{\pi}{2}+7^{-7^{7^{7^{7^{7^{\sqrt{5}+\sin \mu\nu}}}}}}\right)(\mu^2-\nu^2).$$
But for some reason that I cannot clearly explain (or even understand) I selected the simpler one, and I am strongly inclined to search for its proof rather than a disproof. I believe most people would feel and behave exactly the same way. (1) Are there any mathematical or philosophical reasons supporting this position? Why when we calculate some sum or integral (which do not contain explicit tiny quantities like $10^{-10^{10^{.^{.^{10}}}}}$) with thousands of digits of precision and it matches some simple closed-form expression, we inclined to believe this is the exact equality rather than an accidental very close value? (2) Are there known cases when such intuition turned out to be wrong? And one more question: (3) Do you believe there can be exact closed forms for some infinite sums or integrals, that cannot be proved in $ZFC$ or any its reasonable extension (like adding some large cardinal axioms) - so to speak, equalities that hold without any reason.
|
Part of what makes this question subtle is that what's intuitive depends on your background knowledge. In particular, the question of what counts as an "explicit tiny quantity" is hard to pin down. For example, Bailey, Borwein, and Borwein pointed out the example
$$
\left(\frac{1}{10^5} \sum_{n=-\infty}^\infty e^{-n^2/10^{10}}\right)^2 \approx \pi,
$$
which holds to over 42 billion digits. If you know something about Poisson summation, it's pretty obvious that this is a fraud: Poisson summation converts this series into an incredibly rapidly converging series with first term $\pi$, after which everything else is tiny. Even if you don't know about Poisson summation, the $10^5$ and $10^{10}$ should raise some suspicions. However, it's a pretty amazing example if you haven't seen this sort of thing before. As for the third question, the truth is more dramatic than that. There are finite identities that are independent of ZFC, or any reasonable axiom system, so you don't even need anything fancy like infinite sums or integrals. This follows from a theorem of Richardson (D. Richardson, Some Undecidable Problems Involving Elementary Functions of a Real Variable , Journal of Symbolic Logic 33 (1968), 514-520, http://www.jstor.org/stable/2271358 ). Specifically, consider the expressions obtainable by addition, subtraction, multiplication, and composition from the initial functions $\log 2$, $\pi$, $e^x$, $\sin x$, and $|x|$. Richardson proved that there is no algorithm to decide whether such an expression defines the zero function. Now, if the function is not identically zero then it can be proved not to be zero (find a point at which it doesn't vanish and compute it numerically with enough accuracy to verify that it is nonzero). If all the identically zero cases could be proved in ZFC too, then that would give a decision procedure: do a brute force search through all possible proofs until you find one that resolves the question. Thus, there exists an expression that actually is identically zero but where there is no proof in ZFC. In fact, if you go through Richardson's proof you can explicitly construct such an expression, although it will be terrifically complicated. (To be precise, you'll be able to prove in ZFC that if ZFC is consistent, then this identity is unprovable, but by Gödel's second incompleteness theorem you won't be able to prove that ZFC is consistent in ZFC.) Being independent of ZFC doesn't mean these identities are true for no reason, and in fact Richardson's proof gives a reason. However, his paper shows that there is no systematic way to test identities. You can indeed end up stuck in a situation where a proposed identity really looks true numerically but you just can't find a proof.
|
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|
131,583 |
My question is: usually, a partial differential equation, for example, those coming from physics, is written in a language of vector calculus in a local coordinate. Is there any way (or any algorithm ) that we can use to rewrite it using language of differential forms, tensor, exterior calculus, Hodge star and other operators which are coordinate independent? An example, the Grad f can be rewritten as a geometric form: (df)#, where # is a sharp operator turning a one-form into a vector. I am currently facing this problem to turn a partial differential equation into its coordinate-independent form, which involves forms, tensors, exterior calculus and other operators. Thank you for anyone who help me about this problem!
|
This is rather standard, though, unfortunately this knowledge is confined to a rather narrow segment of mathematicians working on PDEs. The main reason is that, most of the time, this is not necessary for practical problems. In fact there are multiple ways converting a PDE into invariant form. In mathematical physics, this question comes up when trying to generalized equations from flat space to curved space. The main technical device is to introduce explicit dependence on a metric $g$ and use covariant differentiation, with respect to $g$, to replace ordinary derivatives. The invariant form of the original flat space equations is then obtained by simply replacing $g$ by the Euclidean metric. This method is sometimes colloquially known as the comma goes to semicolon rule , where the punctuation signs represent notation for ordinary and covariant derivatives. More on this can be found in textbooks on GR. For instance, Gravitation by Misner, Thorne & Wheeler would definitely discuss this. I would guess that you might be happy to stop here. But if you are looking for more generality and mathematical sophistication, read on. Another method is to repeatedly introduce auxiliary dependent variables, until the PDE system becomes first order and this can be done in such a way that all dependent variables are actually differential forms and differentiations are only effected through the exterior derivative. Once in this form, the PDE system is manifestly invariant. This approach was pioneered by Cartan and is explained in depth in the well known book Exterior Differential Systems by Bryant et al. Yet another way, which does not involve reduction to a first order system, is to notice that derivatives of arbitrary order are mathematically described in an invariant way by jets . An arbitrary PDE can then be described in terms of jets. Once one unwinds all the relevant definitions, this kind of description is manifestly invariant, but not necessarily always helpful, since all it really does is bundle up the complicated rules for transforming higher derivatives under coordinate changes into the language of jets. Nevertheless, I think it is important to learn about it, as it provides the tools to study PDE systems in a very deep way. An elementary treatment of PDEs in the context of jets can be found in the book Applications of Lie Groups to Differential Equations by Olver. A much more sophisticated treatment can also be found in the aforementioned book by Bryant et al. Another very helpful reference is Natural Operations in Differential Geometry by Kolar, Michor & Slovak.
|
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|
131,657 |
QUESTION They had plenty of time to adopt the theory of categories. They had Eilenberg, then Cartan, then Grothendieck. Did they feel that they have established their approach already, that it's too late to go back and start anew? I have my very-very general answer: World is Chaos , Mathematics is a Jungle , Bourbaki was a nice fluke , but no fluke can last forever, no fluke can overtake Chaos and Jungle. I'd still like to have a much more complete picture. Appendix: CHRONOLOGY 1934: Bourbaki's birth (approximate date); 1942-45: Samuel Eilenberg & Saunders Mac Lane - functor, natural transformation, $K(\pi,n)$; 1946 & 1952: S.Eilenberg & Norman E. Steenrod publish "Axiomatic..." & "Foundations..."; 1956: Henri Cartan & S.Eilenberg publish "Homological Algebra"; 1957: Alexander Grothendieck publishes his "Tohoku paper", abelian category. ( Please, feel free to add the relevant most important dates to the list above ).
|
One thing to keep in mind is that Bourbaki started in the 1930s, so in some sense simply too early to include category theory right from the start on, and foundational matters were rather fixed early on and then basically stayed like this. Since (I think) the aim was/is a coherent presentation (as opposed to merely a collection of several books in similar spirit) to change something like this 'at the root' should be a major issue.
Some 'add on' seems possible but just does not (yet) exist; and it seems the idea to write something like this was (perhaps is?) entertained (see below). To support the above here is a quote from MacLane (taken from the French Wikipedia page on Bourbaki which contains a somewhat longer quote and source): Categorical ideas might well have fitted in with the general program of Nicolas Bourbaki [...]. However, his first volume on the notion of mathematical structure was prepared in 1939 before the advent of categories. It chanced to use instead an elaborate notion of an échelle de structure which has proved too complex to be useful. Apparently as a result, Bourbaki never took to category theory. At one time, in 1954, I was invited to attend one of the private meetings of Bourbaki, perhaps in the expectation that I might advocate such matters. However, my facility in the French language was not sufficient to categorize Bourbaki. There it is also mentioned that (in the context of the influence of the lack of categories on the discussion of homological algebra, only for modules not for abelian categories): On peut lire dans une note de bas de page du livre d'Algèbre Commutative: « Voir la partie de ce Traité consacrée aux catégories, et, plus particulièrement, aux catégories abéliennes (en préparation) », mais les propos de MacLane qui précèdent laissent penser que ce livre « en préparation » ne sera jamais publié. This translates to (my rough translation): One can read in a footnote of the book Commutative Algebra: "See the part of this Treatise dedicated to categories, and, more specificially, to abeliens categories (in preparation)", but the sentiments of Mac Lane expressed above [part of which I reproduced] let one think that this book "in preparation" will never be published. The precise reference for the footnote according to Wikipedia is N. Bourbaki, Algèbre Commutative, chapitres 1 à 4, Springer, 2006, chap. I, p. 55.
|
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|
131,985 |
Is there a way to axiomatize [non-abelian] free groups in first-order logic using the language of groups (which contains the binary operation symbol $\cdot$, and the constant symbol $e$)? Is there one particular axiom, or even a schema, from which we can prove that $G$ is a free group? (Regardless to the cardinality of a generating set.) I should clarify that I'm not interested in augmented languages where we allow additional constant symbols for the generating set (in which case we can just write a schema stating when the various strings are equal).
|
The free groups cannot be axiomized by first order axioms. If the free groups were axiomatizable by first order axioms, then the ultraproduct of free groups would be a free group. However, the group $\mathbb{Z}$ is free, but for every non-principal ultrafilter $\mathcal{U}$ on $\mathbb{N}$, the ultrapower $\mathbb{Z}^{\mathcal{U}}$ is not free since it is an abelian group of cardinality continuum. More generally, any ultrapower $G^{\mathcal{U}}$ by a non $\sigma$-complete ultrafilter $\mathcal{U}$ of any free group $G$ is not free since the ultrapower $G^{\mathcal{U}}$ contains an isomorphic copy of the non-free subgroup $\mathbb{Z}^{\mathcal{U}}$ (recall that the subgroup of a free group is always free).
|
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|
132,191 |
Let $m,n$ be nonnegative integers. The sequence $\{a_{m,n}\}$ satisfies the following three conditions. For any $m$, $a_{m,0}=a_{m,1}=1$ For any $n$, $a_{0,n}=1$ For any $m\ge0, n\ge1$, $a_{m+1,n+1}a_{m,n-1}=a_{m+1,n-1}a_{m,n+1}+a_{m+1,n}a_{m,n}$ Prove that $a_{m,n}$ is an integer for any $m\ge0, n\ge0$. I don't have any good idea. I need your help.
|
This is the knight recurrence (Example 4.1) in Fomin and Zelevinsky's paper on the Laurent phenomenon . They show more generally that if you replace the boundary entries with indeterminates, then all matrix entries are Laurent polynomials.
|
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|
132,244 |
Assume there are two Riemannian metrics on a manifold ( open or closed) with the same set of all geodesics. Are they proportional by a constant? If not in general, what are the affirmative results in this direction?
|
The short answer is no , the geodesics do not determine the metric. For example, in the Cayley-Klein model of hyperbolic geometry the geodesics are straight lines. It is however rather rare for
two Riemannian metrics to have the same geodesics. In two dimensions these metrics were studied by Liouville (see Livre VI of Darboux's Théorie générale des surfaces ). Their geodesic flow is completely integrable and, in fact, admits an additional integral of motion quadratic in the momenta. A basic results is that if locally the geodesics are straight lines (or can be mapped to straight lines), the metric has constant curvature (Beltrami's theorem). Other rigidity results of this kind exist. For more on this topic you can consult the works of Topalov and Matveev.
|
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|
132,268 |
Is there an axiomatic system where the deduction theorem does not hold?
|
Failures of the deduction theorem are one of the more mysterious topics in logic, in my experience. The motto is that axioms are stronger than rules . Here is the simplest nontrivial example that I know. Start with propositional logic with two variables $A$ and $B$. Add the single new rule of inference $A \vdash B$ to the usual Hilbert-style deductive system, with no new axioms. Note that this does not in any way change the collection of formulas that can be derived. (Proof: the first time you use the new rule, you already had to derive $A$ in the original system, but you cannot, because the original system only derives tautologies. So you can never use the new rule.) Thus the new system has the rule $A \vdash B$ but does not derive $A \to B$, and hence the deduction theorem fails. But this new system is not completely trivial. If we add $A$ as a new axiom, then we can derive $B$ in the expanded logic, which we cannot do in ordinary propositional logic. So there is an interplay between the rules of inference and the axioms of a given theory. The deduction theorem for first order logic shows that this interplay is very well behaved in that context: an arbitrary first-order theory $\Delta$ with the usual deductive system has the derived rule $\phi \vdash \psi$ if and only if it has the derived rule $\vdash \phi \to \psi$. In retrospect, there is no reason to expect this to hold for arbitrary sets of deduction rules, because new axioms may give additional strength to the existing rules. As François G. Dorais has mentioned in the comments, more complicated examples are known in proof theory. They are similar to the above example in that they weaken an axiom by replacing it with a rule. The general idea is that an extensionality axiom of the form $x = y \to f(x) = f(y)$ might be replaced with a rule $x = y \vdash f(x) = f(y)$. This suggests immediately how the deduction theorem can fail: if $x$ and $y$ are terms that are not provably equal, but are equal in some interpretation, then the extensionality axiom might fail in that interpretation even if the rule of inference is satisfied in some sense. But this is just a heuristic sketch of the argument. For a short, rigorous explanation, see " A note on Spector’s quantifier-free rule of extensionality " by Ulrich Kohlenbach, Archive for Mathematical Logic 40:2 (2001), pp 89-92.
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|
132,297 |
Consider a country with $n$ families, each of which continues having children until they have a boy and then stop. In the end, there are $G$ girls and $B=n$ boys. Douglas Zare's highly upvoted answer to this question computes the expected fraction of girls in a population formed of complete families and explains why we shouldn't expect it to equal $1/2$. My current question concerns a different statistic, namely the probability that there are more boys than girls (after all families have finished reproducing). This probability turns out to be exactly $1/2$, and I'm looking for an intuitive explanation of why. Indeed, for fixed $n$, it's not hard to see that $$Prob(G=k)=\binom{n+k-1}{k}\cdot {1\over 2^{n+k}}$$
(The binomial coefficient is the number of ways to assign $k$ indistinguishable girls to $n$ distinguishable families.) Therefore
$$Prob(G < B)=\sum_{k=0}^{n-1}\binom{n+k-1}{k}\cdot {1\over 2^{n+k}}$$
It's not hard to check that this sum is exactly equal to $1/2$ (and therefore, in particular, independent of $n$). That is, regardless of the number of families, we always have the surprisingly (to me) simple formula $$Prob(G < B )=1/2$$ (Note that this implies $Prob(G>B)$ is strictly less than $1/2$ --- because there's always some probability weight on the event $G=B$ --- though an application of Stirling's formula shows that $Prob(G>B)$ converges to $1/2$ as $n$ gets large.) My question is: Is there some simple intuitive reason I should have expected this result?
|
Symmetry. Put them all together and tell them to multiply forever. The question then becomes whether the $n$-th boy was born at the $2n$-th birth or later, or not, i.e., who is the majority among the first $2n-1$ births: boys or girls.
|
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|
132,317 |
Given a total function $f \colon A \to B$, by grouping the "points" on $A$ which have the same image on $B$, namely $a,b \in A$ located in the same equivalence class $G_{i}$ iff $f(a) = f(b)$. By that way, I get a partition $G = \{ G_{1},G_{2},\dotsc \}$ on $A$ derived from $f$, and concurrently I can represent $f = g \circ h$ where $h \colon A \to G$ and $g \colon G \to B$. This representation is unique if $h$ is monic, namely if $f = g' \circ h'$ where $g' \colon A \to G$ and monic $h' \colon G \to B$, then $g' = g$ and $h' = h$. I doubt that this simple situation has a more general interpretation in the category theory but I don't know what it is. Can anyone please give me some suggestion ?
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Symmetry. Put them all together and tell them to multiply forever. The question then becomes whether the $n$-th boy was born at the $2n$-th birth or later, or not, i.e., who is the majority among the first $2n-1$ births: boys or girls.
|
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|
132,332 |
Suppose that $F:\mathscr{C}^{op} \to Set$ is a presheaf. For an object $C$, denote by $Aut(C)$ the group of isomorphisms $f:C \to C.$ There is a canonical action $Aut(C)$ of on $F(C).$ Is there a special name for those $F$ for which all of these actions are transitive? Has this situation been studied? Does this imply any nice categorical consequences? If $F$ is also a sheaf for a Grothendiek topology on $\mathscr{C},$ does $F$ satisfy any nice conditions in the associated topos of sheaves?
|
Symmetry. Put them all together and tell them to multiply forever. The question then becomes whether the $n$-th boy was born at the $2n$-th birth or later, or not, i.e., who is the majority among the first $2n-1$ births: boys or girls.
|
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|
132,339 |
This should be such an elementary problem in algebraic topology that I'm almost too embarrassed to ask, but here goes. Let $f: X\to Z$ be a surjective fibration, and let $g: Y\to Z$ be any map. Assume all spaces are path-connected, and base points $x,y,z$ chosen so that $f(x)=g(y)=z$. Form the pullback in the topological category,
$$
\begin{array}{ccc}
E & \to & X \newline
\downarrow & & \downarrow \newline
Y & \to & Z.
\end{array}
$$ Note that $E$ need not be path-connected. Is it possible to express $\pi_1(E,e)$ for a given choice of base point $e\in E$, in terms of $f_\sharp: \pi_1(X,x)\to \pi_1(Z,z)$ and $g_\sharp: \pi_1(Y,y)\to \pi_1(Z,z)$?
|
There is a ``Mayer--Vietoris" sequence
$$\cdots \to \pi_2(Z, z) \to \pi_1(E, e) \to \pi_1(X, x) \times \pi_1(Y, y) \to \pi_1(Z,z) \to \pi_0(E) \to \cdots$$
that can be developed by fitting together the 4 long exact sequences of homotopy groups obtained from the 4 maps in your diagram. With $X \simeq Y \simeq *$ and $Z=S^2$, $E \simeq \Omega S^2$ has fundamental group $\mathbb{Z}$, whereas $X$, $Y$ and $Z$ are simply-connected, so you can't express $\pi_1(E)$ solely in terms of the fundamental groups of the other spaces.
|
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|
132,648 |
This has been bothering me for a while, and I can't seem to find any definitive answer. The following conjecture is well known in additive combinatorics: Conjecture: If $A\subset \mathbb{N}$ and $$\sum_{a\in A} \frac{1}{a}$$ diverges, then $A$ contains arbitrarily long arithmetic progressions. This is often called either "The Erdős-Turán conjecture" or "Erdős's conjecture on arithmetic progressions." I want to make sure I quote things correctly in my writing, and my main question is, who should this conjecture be attributed to? I am asking because I find a lot of conflicting pieces of information regarding how it should be referred to. For example, the Wikipedia page and the two Wolfram pages, page A and page B , are in disagreement. Wikipedia seems to be adamant that calling it the Erdős-Turán conjecture is incorrect, however I am skeptical, as it does not seem to be so clear. Also, both of the choices are used in many papers in the literature. I went back and read the original 1936 paper of Erdős and Turán, " On some sequences of integers ." ( Math Sci Net MR1574918 ). In that paper, they conjecture that $r(N)=o(N)$ , where $r(N)$ is a maximal subset of $\{1,\dots,N\}$ without 3 term arithmetic progressions. They also mention the (incorrect) conjectures of Szekeres on $k$ term progressions, but other than that, they don't seem to mention any conjectures regarding longer progressions at all. As far as I can tell, the above conjecture first appears in the 1972 paper of Erdős, " Résultats et problèmes en théorie des nombres ." ( Math Sci Net MR0396376 ) This suggests that the conjecture should be named after Erdős, however, I believe that there must some reason why this conjecture has often been attributed to Turán as well. I can't imagine that it would be done for no reason, and it is entirely possible that the reason why credit is given to both Erdős and Turán was not written down, or at least, is not easy to find. I am hoping someone can shed some light on this, and provide a clearer answer for what to call the above conjecture. Thanks for your help,
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One of the best places to track these things down is The mathematical coloring book , by Alexander Soifer, Springer 2009. Chapter 35 is on "Monochromatic arithmetic progressions", and section 35.4, "Paul Erdős’s Favorite Conjecture", is on the problem you ask about. As far as I can tell, the question is sometimes called the Erdős-Turán conjecture for two reasons: First, it extends their older conjecture (now Szemerédi's theorem). Second, it was first popularized in connection to Turán, see below. Soifer identifies a problem paper in the premier issue of Combinatorica as the first place in print were the conjecture appears with the attached value of $\\\$3000$ : On the combinatorial problems which I would most like to see solved , Combinatorica, 1 (1981), 25–42, submitted Sep. 15, 1979. Soifer adds that apparently Erdős first offered this amount in a 1976 talk, "To the memory of my lifelong friend and collaborator Paul Turán", in a conference the University of Manitoba, see Problems in number theory and Combinatorics , in Proceedings of the Sixth Manitoba Conference on Numerical Mathematics (Univ. Manitoba, Winnipeg, Man., 1976 ), Congress. Numer. XVIII, 35–58, Utilitas Math., Winnipeg, Man., 1977. (Erdős calls it "A very attractive conjecture of mine.") In Problems and results on combinatorial number theory, II , J. Indian Math. Soc. (N.S.), 40(1–4) (1976), 285–298, he stated the conjecture with an attached prize of $\\\$2500$ . In page 288 of this paper, we read (mildly edited for typos): $2.$ An old conjecture in number theory states that for every $k$ there are $k$ primes in arithmetic progression. This conjecture would immediately follow if we could prove that for every $k$ and $n\gt n_1(k)$ , $r_k (n) \lt \pi (n)$ . Here, as usual, $r_k(n)$ is the size of the largest subset of $\{1,2,\dots,n\}$ without $k$ -term progressions. This method of proof seems very attractive, it tries to prove a difficult property of the primes by just using the facts that the primes are numerous. In some cases I have been successful with this simple minded approach. In this connection I recently formulated the following conjecture for the proof or disproof of which I offer $2500$ dollars: Let $a_l \lt a_2 \lt\dots$ be an infinite sequence of integers satisfying $\sum\frac1{a_i}=\infty$ .
Then for every $k$ there are $k$ $a$ 's in an arithmetic progression. (The truth of this conjecture would imply that for every $k$ there are $k$ primes in arithmetic progression). One can put this problem in a finite form as follows: Put $$g_k (n) = \max\sum_{a_i>n}\frac1{a_i},\quad G (k) = \lim_{n=\infty} g_k (n),$$ where the maximum is extended over all sequences $\{a_i\}$ not exceeding $n$ which do not contain an arithmetic progression of $k$ terms. The $2500$ dollar conjecture is equivalent to $G(k) \lt\infty$ for every $k$ (it is not quite trivial to show that $G (k) = \infty$ implies the falseness of the conjecture). By the way, the prize currently attached to the conjecture is actually $\\\$5000$ . Soifer indicates that the update occurred in the paper Some of my favorite problems and results , in The Mathematics of Paul Erdős, vol. I , Graham, R. L., and Nesetril, J., (eds), 47–67, Springer, Berlin, 1997. (The paper appeared posthumously, and was written in 1996.) I offer $5000 for a proof (or disproof) of this [problem]. Neither Szemerédi nor Furstenberg’s methods are able to settle this but perhaps the next century will see its resolution. Soifer then tries to identify when the conjecture was first stated. I searched for evidence in the ocean of his writings, and found three indicators. First, in a paper submitted on September 7, 1982 to Mathematical Chronicle (now called New Zealand
Journal of Mathematics ) that appeared the following year [E83.03], Paul writes: "This I conjectured more than forty years ago." In the same year, 1982, Paul spoke at the Conference on Topics in Analytic Number Theory in Austin, Texas. I read in the proceedings (published in 1985 [E85.34],
p. 60): "I conjectured more than 40 years ago that if $a_1 \lt a_2\lt \dots$ is a sequence of integers for which $\sum_{i=1}^\infty \frac1{a_i}=\infty$ then the $a$ ’s contain arbitrarily long arithmetic progressions." Thus, both of these publications indicate that the conjecture was posed before 1942. On the other hand, in the 1986 Jinan, China, Conference proceedings (published in 1989 [E89.35]), Paul writes (p. 142): "About 30 years ago I conjectured that if $\sum_{n=1}^\infty 1/{a_n}=\infty$ , then the $a$ 's contain arbitrarily long arithmetic progressions." This would date the birth of the conjecture to about 1956. The cited papers are: [E83.03]: Combinatorial problems in geometry , Math. Chronicle 12 (1983), 35–54. [E85.34]: Some solved and unsolved problems of mine in number theory , in Topics in
Analytic Number Theory (Austin, Tex., 1982) , Univ. Texas Press, Austin, TX, 1985, 59–75. [E89.35]: Some problems and results on combinatorial number theory , in Graph theory
and its applications: East and West (Jinan, 1986) , Ann. New York Acad. Sci. 576 , 132–145, New York Acad. Sci., New York, 1989. For the finitude of $G_k$ , see for example ArxiV:math/0703467 . By the way, the conjecture in the 1972 paper you identify seems to be restricted to $3$ -term progressions. Even this version remains open (and carries an attached prize of $\\\$250$ .)
|
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|
132,675 |
I am interested in the following: Let $G$ be a finite group of order $n$. Is there an explicit function $f$ such that
$|s(G)| \leq f(n)$ for all $G$ and for all natural numbers $n$, where $s(G)$ denotes
the set of subgroups of $G$?
|
A non-identity subgroup of a group of order $n$ can be generated by $\log_{2}(n)$ or fewer elements. There is quite a lot of duplication, but if you count the number of subsets of $G$ of cardinality at most $\log_{2}(n),$ you will have an upper bound (which could be made more precise with more care) and the case of elementary Abelian $2$-groups show that such a bound is of the right general shape if it is to cover all groups of all orders. To be more precise, this gives an upper bound of approximately $\log_{2}(n)n^{\log_{2}(n)}$ for the total number of subgroups of a group of order $n,$ which is generically rather generous. On the other hand, the number of subgroups of an elementary Abelian group of order $n= 2^{r}$ is close to $\sum_{i=0}^{r}2^{i(r-i)},$ so generally larger that $n^{log_{2}(n)/4}.$
|
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|
132,795 |
I have downloaded from this link a quite poor quality scan of the letter dating May 1966 that Grothendieck sent to Tate mentioning his ideas about generalizing Monsky-Washnitzer cohomology. I am trying to put it in LaTeX, at least for my own reference. So, first of all, do you know if it ever appeared in print somewhere or if the typing has already been done by someone? And, secondly, does anyone has a better-quality file? The big problem with this is that many top and bottom lines are missing, sometimes making it impossible to reconstruct the original sentence.
|
In the light of past events (" Les Archives Grothendieck "), we now have an "ameliorated" version of the letter: Cote n° 6 . Lettre Tate (mai 66) (Cristaux) : lettre (1966), tapuscrit, notes manuscrites . 1966 - [à partir de 1970]. [scan] , and project of transcription [pdf] See [here] for more.
|
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|
132,838 |
In general, it seems not known which finite abelian groups are class groups of quadratic number fields.
For imaginary quadratic number fileds, I read that $(\mathbb{Z}/3\mathbb{Z})^3$ is the smallest
abelian group which does not occur. It does occur as the class group of $\mathbb{Q}(\sqrt{188184253})$.
What other groups are known not to be the class groups of imaginary quadratic number fields ?
How does one prove this ? What is the situation for class groups of real quadratic number fields ?
|
The only proof that I know that $(\mathbb{Z}/3\mathbb{Z})^3$ does not appear as the class group of a quadratic imaginary number field is by brute force search. Roughly, the idea is that since class numbers of quadratic fields increase with the discriminant, there are only finitely many such fields of any given class number. We can then enumerate all class groups of a given size, and check to see which class groups appear. Most recently, Watkins's computation of all quadratic imaginary number fields with class number up to 100 provides in particular a complete list of all class groups of order 27, and so by simple enumeration one can observer that $(\mathbb{Z}/3\mathbb{Z})^3$ does not appear. I should also mention in this regard the tremendous amount of data available in a recent article of Jacobson, Ramachandran, and Williams. [ Edit : As to another of your questions, the same technique shows 6 other class groups of order less than 100 do not appear as quadratic imaginary number fields, of orders 27,32,54,64,64,81,81 respectively, a fact I have stolen from https://math.stackexchange.com/a/11025/20762 (and now updated based on ABC's answer below).] Lacking any particularly compelling arithmetic reasons which proscribe certain finite abelian groups from appearing as class groups of quadratic imaginary number fields, it's worth thinking about some heuristics. An argument of Soundararajan gives that, on average, the number of quadratic imaginary number fields of class number $p^n$ is roughly $p^n$ (or better, asymptotically between $p^n/n$ and $np^n$). On the other hand, the number of finite abelian groups of order $p^n$ is governed by the partition theory of $n$, and hence a standard asymptotic suggests there are $\frac{1}{4\sqrt{3}n}e^{\pi\sqrt{2n/3}}$ such groups. So roughly, there are increasingly more number fields of order $p^n$ than there are possible class groups of that size, so one might perhaps expect that (again, without any obvious arithemtical obstruction) that there are only finitely many examples like $(\mathbb{Z}/3\mathbb{Z})^3$ that do not appear, at least among $p$-groups (and especially for large $p$). For real quadratic fields, there is no such bound pushing class numbers higher, and I can't see any reason not to expect every finite abelian group to appear as a class group of such a field. Of course, this result is nowhere near known.
|
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|
132,879 |
This question is motivated by the physical description of magnetic monopoles. I will give the motivation, but you can also jump to the last section. Let us recall Maxwell’s equations: Given a semi-riemannian 4-manifold and a 3-form $j$. We describe the field-strength differential form $F$ as a solution of the equations $\mathrm{d}F=0$ $\mathrm{d}\star F=j$ (where $\star$ denotes the Hodge star). If the second de-Rham-cohomology vanishes (for example in Minkowski space), $F$ is exact and we can write it as $F=\mathrm{d}A$, where $A$ denotes a 1-form. Now let us consider monopoles: We use two 3-forms $j_m$ (magnetic current) and $j_e$ (electric current) and consider the equations $\mathrm{d}F=j_m$ $\mathrm{d}\star F=j_e$. Essentially, it is described in this paper , but the author Frédéric Moulin (a physicist) uses coordinates. Now he assumes that (in Minkowski space) $F$ can be decomposed using two potentials — into an exact (in the image of the derivative) and a coexact (in the image of the coderivative) form: $F=\mathrm{d}A-\star\mathrm{d}C$. Is there a mathematical justification for this assumption (maybe it is just very pragmatic)? The actual question Given a 2-form $F$ on 4-dimensional Minkowski space (more generally: semi-riemannian manifolds)—are there any known conditions such that $F$ decomposes into an exact and a coexact form: $F=\mathrm{d}A+\star\mathrm{d}C$)? For compact riemannian manifolds there is the well-known Hodge decomposition: There is always a decomposition into an exact, a coexact and a harmonic form. In the non-compact case you might be able to get rid of the harmonic form by only considering “rapidly decaying” forms ( Wikipedia suggests that , but I do not have a good reference, in euclidean space there is the Helmholtz decomposition, and non-trivial (smooth) harmonic 1-forms do not vanish at infinity). That is why I also ask: Are there “rapidly decaying” harmonic 2-forms in Minkowski space? Any references where I could see what is known about harmonic forms and Hodge theory in the semi-riemannian case are also welcome.
|
There are no "rapid decaying harmonic 2-forms" in Minkowski space. Consider the expression
$$ 0 = \mathrm{d} \star \mathrm{d} F = \partial^i \partial_{[i}F_{jk]} = \frac13 (\Box F_{jk} + \partial_j \partial^i F_{ki} + \partial_k \partial^i F_{ij})$$
The second and third terms in the parentheses vanish by $\mathrm{d}\star F = 0$, so we see that over Minkowski space, Maxwell's equation implies that the components of the Faraday tensor must solve a linear wave equation. (In the case you have source, you also see that the equations are equivalent to $\Box F_{jk} = J_{jk}$ where $J_{jk}$ is built out of the magnetic and electric currents in the appropriate way.) Now, the linear decay estimates for the wave equation (that $|F| \sim \frac{1}{1+|t|}$) is sharp. In fact, by the finite speed of propagation + strong Huygens principle, if $F$ decays too rapidly in time, conservation of energy immediately implies that $F$ must vanish identically! (If you don't want to go through the wave equation, you can also argue through the Maxwell equation by considering the energy integrals.) Conservation of energy also implies that over Minkowski space $\mathbb{R}^{1,3}$ there cannot exist a non-vanishing "harmonic" form with finite space-time $L^2$ norm. So harmonic forms cannot exist, even in a class that is not necessary "rapidly" decaying. Now, under the influence of external sources $j_e$ and $j_m$, you can pose the ansatz (by linearity of the equations) $F = G + H$, where
$$ \begin{align}
\mathrm{d} G & = j_e & \mathrm{d} H & = 0 \newline
\mathrm{d} \star G & = 0 & \mathrm{d} \star H &= j_m
\end{align} $$
Then the cohomology result you mentioned implies that $H = \mathrm{d}A$ and $\star G = \mathrm{d}C$ for $A,C$.
The difference of the true solution with your ansatz is $\tilde{F} - F$ a harmonic two form, which can be completely absorbed into the electric potential $A$ if you want... So the claimed decomposition is available whenever it is possible to solve the equations (possibly nonuniquely) for $G$ and $H$. For compatible source terms (you need $\mathrm{d} j_e =0 = \mathrm{d} j_m$ as a necessary condition) which are $L^1$ in space-time, the existence of a solution follows from the well-posedness of the corresponding Cauchy problem, for example. Much of "Hodge theory" is false for "compact semi-Riemannian manifolds" in general. For example, there is no good connection between the cohomology and the number of harmonic forms. (Simplest example: on any compact Riemannian manifold the only harmonic functions are constants. But take, for example, the torus $\mathbb{T}^2 = \mathbb{S}^1\times\mathbb{S}^1$ with the Lorentzian metric $\mathrm{d}t^2 - \mathrm{d}x^2$. Then for any $k$ the function $\sin (kx) \sin(kt)$ is "harmonic", and they are linearly independent over $L^2$. ) Number 4 above also implies that the Hodge decomposition is not true for general compact semi-Riemannian manifolds. Consider the exact same $\mathbb{T}^2$ as above. Let $\alpha$ be the one form $\sin(t) \cos(x) \mathrm{d}t$. We have that
$$ \mathrm{d}\alpha = \sin(t) \sin(x) \mathrm{d}t\wedge \mathrm{d}x \qquad \mathrm{d}\star \alpha = \cos(t) \cos(x) \mathrm{d}t \wedge \mathrm{d}x$$
To allow a Hodge decomposition requires that there exists $A$ and $C$, scalars, such that $$\mathrm{d}\star\mathrm{d}C = \mathrm{d}\alpha \implies \Box C = \pm \sin(t)\sin(x) $$
and
$$ \mathrm{d}\star \mathrm{d}A = \mathrm{d}\star\alpha \implies \Box A = \pm \cos(t) \cos(x) $$
where $\Box$ is the wave operator $\partial_t^2 - \partial_x^2$. Now taking Fourier transform (since we would desire $C$ and $A$ be in $L^2$ at least), we see that we have a problem. If $f\in L^2(\mathbb{T}^2)$ we have
$$ \widehat{\Box f} = -(\tau^2 - \xi^2)\widehat{f} = 0 \qquad\text{ whenever } |\tau| = |\xi| $$
But the Fourier support of $\sin(t)\sin(x)$ and $\cos(t)\cos(x)$ are precisely when $\tau = \xi = 1$. So there in fact does not exist a Hodge-like decomposition for the $C^\infty$ form $\alpha$. The above just goes to say that You are not going to find anything in the literature concerning Hodge decomposition for general compact semi-Riemannian manifolds, as such as theory does not exist You are also unlikely to find literature that studies the Hodge decomposition in the non-compact case in general: on non-compact manifolds causality violations (closed time-like curves, for example) can, in principle, also cause problems with solving the wave equation (both homogeneous and inhomogeneous), which will prevent an analogue for Hodge decomposition to hold. Your best bet, in so far as two-forms are concerned, are precisely the mathematical-physics literature concerning Maxwell equations over globally hyperbolic Lorentzian manifolds. For these manifolds, at least in principle, some version of the Hodge decomposition can be had by solving the appropriate initial value problems (something like $\Box C = j_e, \Box A = j_m, \mathrm{d}F_0 = 0 = \mathrm{d}\star F_0$ where $F_0$ is the harmonic part, with initial data prescribed so that the data for $C$ and $A$ vanishes on the initial surface and $F_0$ is equal to $F$ on the initial surface).
|
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|
133,005 |
Hello everybody. I am a Ph.D student in North America looking for advice about my prospective research area. My supervisor works in a research area, let's say area A, so as soon as I was accepted as his student I started to learn the background in the area A. At some point I started to study by my own the connection of area A with a nearby area, let's say area B. I found that area B match much better my interest, I can use my former background in area B and definitely I have more mathematical intuition in that area. I have spent the last six months studying really hard several advances books, recent papers and lectures notes to get the necessary advanced background to start doing research in area B. Some days ago I had a meeting with my supervisor, I was really enthusiastic explaining the connection of area B with what he does in his research (area A), however my supervisor told me that it is not a good idea to continue working in that direction because it is difficult to be accepted in the circle of people who work in area B (several professors at top U.S. universities and their former/current students), that there is a lot competition between them to publish results and because neither my supervisor nor me belong to that circle it would be difficult to publish or even find a specific problem for my research. I did not understand what my supervisor means when he said " it is difficult to be accepted in the circle of people who work in area B", does that mean that if I submit a paper it would not be published even if it is correct, well-written and meets the standard of quality and originality of the journal? or if I try to submit a talk to a conference in that area my talk would be always rejected? I find difficult to believe that there are areas of mathematics that are closed to people who does not belong to a certain circle of leading researchers and their students. Despite this, I consider that is possible to perform my research in area B, I could contact by email people working in that area, several of them posed open questions and further directions of research in recent papers, moreover at the end of this year I plan to attend an important conference specialized in area B, a lot of junior and senior researchers of "the circle" are attending and I am interested and familiarized with the research of several of them, so the connections I could make there would be very useful. To sum up my questions are: i) Is is true that there are areas of mathematics that are closed to people who does not belong to a certain group of professors and students at certain universities? ii) If I make the right connections with people working in area B, do you think is feasible to perform my Ph.D research in area B with my supervisor, whose research is just nearby area B, and the advice of a specialist in area B mainly by email? Some additional information: I am not very far from those universities where the people of "the circle" are, so I could travel from time to time to those universities to meet those people or give talks at the seminars. Thanks for your answers.
|
Perhaps your advisor had meant merely that the group who work in area B are very strong and have a rich knowledge, and it is difficult for outsiders to enter into or compete with that group because they won't have risen to the high expertise to which that group had brought itself? You seem to present the issue as one of political intrigue and exclusion, but it may not be like this at all. There are surely many mathematical groups, who by working intensely on a focused topic bring themselves to a high level of expertise on that topic. If this is the situation, then it would seem by your other remarks that you can make contacts with that group and begin to study with them and thereby involve yourself in their expertise. That said, I also believe that it is wise to listen to one's advisor's suggestions about topics of investigation. It may be that your advisor simply feels that he will not be able to help you as much in area B, simply because he doesn't himself have the knowledge necessary to guide you in that area. Thus, your plan to work in area B is essentially amounting to not working with your current advisor, and instead having only an email advisor, who may not ultimately give you the attention that you will want and need later on, and that may not be the best situation. But if there is someone in that group who can server as your mentor, then it may work out. Regarding questions (i), I think this kind of concern is likely misplaced. In my experience, any mathematician with talent will eventually be recognized for it, regardless of whatever connections they may or may not have.
|
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|
133,028 |
The inequality is
\begin{equation*}
\sum_{k=1}^{2d}\left(1-\frac{1}{2d+2-k}\right)\frac{d^k}{k!}>e^d\left(1-\frac{1}{d}\right)
\end{equation*}
for all integer $d\geq 1$. I use computer to verify it for $d\leq 50$, and find it is true, but I can't prove it. Thanks for your answer.
|
[ Edited mostly to fix a typo noted by David Speyer ] The following analysis simplifies and completes the
"routine but somewhat unpleasant" task of
recovering the actual inequality from the asymptotic analysis. The idea is that once we've obtained the asymptotic expansion
$$
\sum_{k=1}^{2d} \left( 1 - \frac1{2d+2-k} \right) \frac{d^k}{k!}
\sim e^d \left( 1 - \frac1{d} + \frac1{d^2} - \frac2{d^3} \cdots \right)
$$
by expanding $1 - 1/(2d+2-k)$ in a power series about $k=d$,
we should be able to replace $1 - 1/(2d+2-k)$ by something smaller
that can be summed exactly and is close enough that the result
is within a small enough multiple of $e^d$ to maintain the
desired inequality. Because it takes about $2m$ terms of the power series in $k$
to get within $O(1/d^m)$, I had to match the power series to within $O(k-d)^6$.
Let
$$
A_6(k) =
\frac{d+1}{d+2}
- \frac{k-d}{(d+2)^2}
- \frac{(k-d)^2}{(d+2)^3}
- \frac{(k-d)^3}{(d+2)^4}
- \frac{(k-d)^4}{(d+2)^5}
- \frac{(k-d)^5}{(d+2)^6}
- \frac{(k-d)^6}{2(d+2)^6},
$$
so the final term has denominator $2(d+2)^6$ instead of $(d+2)^7$.
Then
$$
1 - \frac1{2d+2-k}
= A_6(k) + \frac{(k-d)^6(2d-k)}{2(d+2)^6(2d+2-k)} \geq A_6(k)
$$
for all $k \leq 2d$. Hence
$$
\sum_{k=1}^{2d} \left( 1 - \frac1{2d+2-k} \right) \frac{d^k}{k!} >
\sum_{k=1}^{2d} A_6(k) \frac{d^k}{k!}.
$$
On the other hand, since $A_6(k)$ is a polynomial in $k$,
the power series $\sum_{k=0}^\infty A_6(k) d^k/k!$ is elementary
(see my earlier answer for the explanation; David Speyer implicitly
used this too in the calculation "with the aid of Mathematica ").
I find
$$
\sum_{k=0}^\infty A_6(k) \frac{d^k}{k!}
= \frac{2d^6 + 22d^5 + 98d^4 + 102d^3 + 229d^2 + 193d + 64}{2(d+2)^6} e^d
$$ $$
= \left(
1 - \frac1d + \frac{2d^5 + 5d^4 + 69d^3 + 289d^2 + 320d + 128}{2d(d+2)^6}
\right) \cdot e^d > \left(1 - \frac1d\right) e^d.
$$
We're not quite finished, because we need a lower bound on
$\sum_{k=1}^{2d} A_6(k) d^k/k!$, not $\sum_{k=0}^\infty$.
However, once $d$ is at all large the terms with $k=0$ and $k>2d$
are negligible compared with our lower bound
$$
\sum_{k=0}^\infty A_6(k) d^k/k! - \left(1 - \frac1d\right) e^d
\geq \frac{2d^5 + 5d^4 + 69d^3 + 289d^2 + 320d + 128}{2d(d+2)^6} e^d >
\frac{d^4}{(d+2)^6} e^d.
$$
Indeed the $k=0$ term is less than $1$, and for $k>2d$ we have
$A_6(k) < A_6(2d) = 1/2$ while $d^k/k!$ is exponentially smaller than $e^d$:
$$
\sum_{k=2d+1}^\infty \frac{d^k}{k!} <
2^{-2d} \sum_{k=2d+1}^\infty \frac{(2d)^k}{k!} <
2^{-2d} \sum_{k=0}^\infty \frac{(2d)^k}{k!} = (e/2)^{2d}.
$$
So we're done once
$$
1 + \frac12 \left(\frac{e}{2}\right)^{2d} < \frac{d^4}{(d+2)^6} e^d,
$$
which happens once $d \geq 14$. Since the desired inequality
has already been verified numerically up to $d=50$, we're done. QED
|
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|
133,262 |
Update: The answer to the title question is no, as pointed out by Tapio and Willie. I would be more interested in lower bounds. Monsky's famous theorem with amazingly tricky proof says that if we dissect a square into an odd number of triangles, they cannot have the same area. However, this proof does not give any quantitative bounds, while from the compactness it follows that there should be some bound, e.g. on the size of the largest triangle that is well separated from the reciprocal of the number of triangles. I don't think the original proof gives any bounds, and I am not aware of any (substantially) different proofs. So if a unitsquare is partitioned into n (odd) triangles, then what is the smallest $f$ such that the largest will have area at least $f(n)$? Monsky says that $f(n)>1/n$, but how much bigger? If we partition to $n-1$ (even) equal area triangles and then cut one of them, then this gives $f(n)\le 1/(n-1)$, which is not sharp as shown by Tapio and Willie. But can anyone give a better lower bound than $1/n$?
|
As suggested in a comment above, I had asked a version of this question years ago. It makes sense to look at upper and lower bounds for the quantity $f(n)-\frac1n$. It is easily seen that
$$ f(n) - \frac1n \le \frac1{n^2-n} $$
and indeed has shown that
$$ f(n) - \frac1n = O\big(\frac1{n^3}\big), $$
see Bernd Schulze's paper " On the area discrepancy of triangulations of squares and trapezoids ", The Electronic Journal of Combinatorics 18(1) (2011), #P137, 16 pp. Computational experiments from an unpublished Diploma thesis (Katja Mansow, Ungerade Dreieckszerlegungen, TU Berlin 2003, 49 pages, in German) suggest that the true order of this difference is, however, singly-exponentially small. From gap-theorems in semi-algebraic geometry there is a doubly-exponential lower bound (details to be worked out).
|
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|
133,342 |
There's a close relationship between curvature and the holonomy group; the holonomy theorem of Ambrose and Singer, for example. It seems to me that there should be an analogous result for torsion. I believe that torsion measures the extent to which certain geodesic parallelograms don't close. Is there a theorem that makes this more precise?
|
Here is another way to think of the relation between torsion and parallel transport, one that some may find more congenial than many of the other interpretations that have been proposed: Start with a manifold $M$ , and a connection $\nabla$ on $TM$ . Consider the bundle $\hat TM=\mathbb{R}\oplus TM$ (where, by $\mathbb{R}$ , I really mean the trivial bundle $M\times\mathbb{R}$ ). Define a connection $\hat\nabla$ on $\hat TM$ by the rule $$
\hat\nabla_X\ \begin{pmatrix}a \\\\ Y\end{pmatrix}
= \begin{pmatrix}da(X) \\\\ aX + \nabla_XY\end{pmatrix}
$$ for any function $a$ on $M$ and any vector fields $X$ and $Y$ on $M$ . I like to think of $\hat TM$ as a sort of 'extended' tangent bundle to $M$ . Of course, it canonically contains $TM= 0\oplus TM\subset \mathbb{R}\oplus TM$ as a subbundle, and one sees that, under this identification of $\begin{pmatrix}0\\ Y\end{pmatrix}$ with $Y$ , one has $\hat\nabla_X Y = \nabla_XY$ . Also, by its definition, $\hat\nabla$ -parallel translation in $\hat TM$ along a curve in $M$ will keep the $\mathbb{R}$ -component of a section constant, so that the 'affine hyperplane subbundles' $a=const$ in $\hat TM$ are preserved under $\hat\nabla$ -parallel translation. Now, the curvature endomorphism of $\hat\nabla$ , namely $$
R^{\hat\nabla}(X,Y) = \hat\nabla_X\hat\nabla_Y-\hat\nabla_Y\hat\nabla_X - \hat\nabla_{[X,Y]},
$$ is just $$
R^{\hat\nabla}(X,Y)\begin{pmatrix}a \\\\ Z\end{pmatrix}
= \begin{pmatrix}0&0\\\\T^{\nabla}(X,Y) & R^{\nabla}(X,Y)\end{pmatrix}
\begin{pmatrix}a \\\\ Z\end{pmatrix}.
$$ Thus, the condition that $T^\nabla(X,Y)=0$ is the condition that $\hat\nabla$ -parallel translation around closed loops preserve the splitting $\hat TM = \mathbb{R}\oplus TM$ , at least to lowest (i.e., second) order. In particular, if one thinks of the hyperplane subbundle $a=1$ as a sort of 'shadow copy' of the actual tangent plane, the torsion measures how its 'zero', i.e., $\begin{pmatrix}1\\0\end{pmatrix}$ , moves in that 'shadow tangent plane' when one uses $\hat\nabla$ to parallel translate around in a loop. In particular, note that the curvature of $\hat\nabla$ incorporates both the torsion and curvature of $\nabla$ , and the vanishing of the torsion of $\nabla$ says precisely that $\hat\nabla$ -parallel translation preserves, to lowest order, the origins in the 'shadow copies' of the tangent plane. NB: It is a notational accident that, in this interpretation, $T(X,Y)$ represents 'infinitesimal translation' while $R(X,Y)$ represents 'infinitesimal rotation'. Remark 1: The addition of a line bundle to the tangent bundle may seem somewhat artificial. However, there is a way to present this construction more naturally in a dual formulation: First, note that the vector bundle $J^1(M,\mathbb{R})$ of $1$ -jets of smooth functions on $M$ has a natural identification with the vector bundle $\mathbb{R}\times T^*\!M$ by identifying the element $j^1_xf$ with $(f(x), \mathrm{d}f_x)\in \mathbb{R}\times T^*\!M$ .
Second, note that a connection $\nabla$ on $TM$ naturally extends to a connection on $T^*\!M$ , uniquely defined by the conditions that $X\bigl(\eta(Y)\bigr) = (\nabla_X\eta)(Y) + \eta(\nabla_XY)$ for all $1$ -forms $\eta$ and vector fields $X,Y$ on $M$ . Using this identification of $J^1(M,\mathbb{R})$ with $\mathbb{R}\times T^*\!M$ , we can define a connection $\hat\nabla$ on $J^1(M,\mathbb{R})$ by $$
\hat\nabla_X(a,\alpha) = \bigl(Xa + \alpha(X),\ \nabla_X\alpha\bigr)
$$ for all functions $a$ , $1$ -forms $\alpha$ , and vector fields $X$ on $M$ .
Then one easily computes that $$
R^{\hat\nabla}(X,Y)(a,\alpha) = \bigl(\alpha\bigl(T^\nabla(X,Y)\bigr),\ R^{\nabla}(X,Y)(\alpha)\bigr),
$$ which is the dual of the above formula on the 'extended tangent bundle'. Remark 2: Finally, going back to the OP's original request for how to think about torsion in terms of failure of 'quadrilaterals' to close, I checked on the coordinate expressions and looked at the Taylor series. This is, of course, very classical; it's in Schouten's work, but it might be worth making explicit in the following way: Let $\nabla$ be a connection on $TM$ , fix a point $p$ and $p$ -centered coordinates $x=(x^i)$ . Define the connection coefficients $\Gamma^i_{jk}$ by the usual rule $$
\nabla_{\partial_i}\partial_j = \Gamma^k_{ij}\ \partial_k\ ,
$$ where $\partial_i$ are the dual vector fields, i.e., $dx^i(\partial_j) = \delta^i_j$ . Choose $v,w\in T_pM$ be tangent vectors and write $v=v^i\ \partial_i(p)$ and $w=w^i\ \partial_i(p)$ . Let $a(t)=\exp_p(tv)$ be the $\nabla$ -geodesic starting at $p$ with initial velocity $v$ , let $b(t)\in T_{a(t)}M$ be the $\nabla$ -parallel translate along $a$ of $w$ , and set $c(s,t) = \exp_{a(t)}(sb(t))$ . Then the functions $c^i= x^i(c(s,t))$ have Taylor expansions in $s$ and $t$ of the form $$
c^i(s,t) = tv^i+sw^i - \tfrac12\Gamma^i_{jk}(0)\bigl(t^2v^jv^k+2st\ v^jw^k+s^2w^jw^k\bigr) + R^i_3(s,t).
$$ Thus, if switching $(t,v)$ and $(s,w)$ in this construction is to yield the same result up to second order in $s$ and $t$ for all $v$ and $w$ , one must have $\Gamma^i_{jk}(0) = \Gamma^i_{kj}(0)$ . In particular, if all the 'attempted parallelograms' close to second order at all points, the torsion of $\nabla$ must vanish.
|
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|
133,597 |
The power set of every infinite set is uncountable. An infinite set (as an element of the power set) cannot be defined by writing the infinite sequence of its elements but only by a finite formula. By lexical ordering of finite formulas we see that the set of finite formulas is countable. So it is impossible to define all elements of the uncountable power set. The power set axiom seems doubtful. Therefore my question.
|
This is a real question. You're not the only one who find the power set axiom dubious. At the time of the great foundational controversies, Russell and Weyl both expressed a similar view. It is now known, from work of Weyl, Wang, Feferman, the reverse mathematics school, and others, that the vast bulk of mainstream mathematics can be developed without power sets. You can effectively treat objects like $\mathcal{P}(\mathbb{N})$ and $\mathbb{R}$ as proper classes. You might be interested in looking at my paper on "Mathematical conceptualism" at arxiv:math/0509246. Here's a direct link: http://arxiv.org/pdf/math/0509246.pdf
|
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|
134,057 |
For probably twenty years, category theorists have known of some objects in the Platonic universe called "(weak) $\infty$-categories", in which there are $k$-morphisms for all $k\in \mathbb N$, with some weak forms of composition and associativity and .... It was recognized a bit later that it is worth recording two numbers for each category: an $(m,n)$-category for $m,n \in \mathbb N\cup\lbrace \infty\rbrace$ with $n\leq m$ has morphisms up to dimension $m$ (and above that only equalities), but everything above dimension $n$ is invertible (in some weak sense). Thus an $(\infty,0)$-category is a "higher groupoid", which Grothendieck's homotopy hypothesis says should be the same as a homotopy type. (Many people take the homotopy hypothesis to be a definition; others prefer to take it as a check, so that a definition is good if the homotopy hypothesis is a theorem.) Just to confuse the language, an $(\infty,1)$-category is what many people now call by the name "$\infty$-category". In any case, for quite a while these objects were expected to exist but definitions were not available (or perhaps too available, but without enough known to decide which was the correct one). More than a decade ago, Leinster provided A survey of definitions of n-category (Theory and Applications of Categories, Vol. 10, 2002, No. 1, pp 1-70.). Much more recently, Bergner and Rezk have begun a Comparison of models for (∞,n)-categories (to appear in Geometry and Topology). These works tend to focus on $(\infty,n)$-categories with $n<\infty$ (although some of Leinster's definitions do include the $n=\infty$ case). My impression, almost certainly very biased by the people that I happen to hang out with, is that by now it is known that "$(\infty,0)$-category" might as well mean "Kan simplicial set" and "$(\infty,1)$-category" might as well mean "simplicial set in which all inner horns are fillable". By "might as well mean", I mean that there are many reasonable definitions, but all are known to be Quillen-equivalent. For $n< \infty$, "$(\infty,n)$-category" can mean "iterated complete Segal space", for example. It is not clear how to take $n=\infty$ in this approach. A very tempting definition of "$(\infty,n)$-category", which I believe is discussed in Lurie's (to my knowledge not yet fully rigorous) paper On the Classication of Topological Field Theories , is that an $(\infty,n)$-category should be an $(\infty,1)$-category enriched in $(\infty,n-1)$-categories. This is based on the observation that the coherence axioms of being an enriched category only require invertible morphisms above dimension $1$. Thus as soon as you've settled on a notion of "$(\infty,1)$-category" (this seems to have been settled) and a notion of "enriched" (settled?), you get a notion of "$(\infty,n)$-category". This approach is developed in Bergner–Rezk, for example. In any case, it is less clear to me how to use these inductive definitions to get to $(\infty,\infty)$. Let $\mathrm{Cat}_n$ denote the $(\infty,n+1)$-category of $(\infty,n)$-categories. Then there are clearly inclusion functors $\mathrm{Cat}_n \to \mathrm{Cat}_{n+1}$, but the colimit of this sequence does not deserve to be called $\mathrm{Cat}_\infty$, since any particular object in it has morphisms only up to some dimension. I have seen it proposed that one should instead take the (inverse) limit of the "truncation" functors $\mathrm{Cat}_{n+1} \to \mathrm{Cat}_n$; I am unsure exactly how to define these, and even less sure why this deserves the name $\mathrm{Cat}_\infty$. Thus my question: Is there by now an accepted / consensus notion of $(\infty,\infty)$-category? For instance, if I have decided on my favorite meaning of $(\infty,1)$-category, is there some agreed-upon procedure that produces a notion of $(\infty,\infty)$-category? Or are there known comparison results that can push all the way to $n=\infty$? A closely related MathOverflow question was asked a little less than a year ago (or via Wayback Machine here ), but received no answers.
Also, I am certainly unaware of many papers in the literature, and likely I have misrepresented even those papers I mentioned above. My apologies to all parties for doing so. It is precisely because of these gaps in my knowledge of the literature that I pose this question.
|
One thing that might interest you is my result with Clark Barwick which gives an axiomatiation + uniqueness result for the homotopy theory of higher categories: arXiv:1112.0040 (i.e. $(\infty,n)$-categories).
This axiomatization includes several variants of $(\infty,n)$-category for finite n, such as what you mention in your question. In particular, when also taken in context with the comparison results of Bergner-Rezk (which you mentioned) and Lurie arXiv:0905.0462 , I would say that there is a clear consensus for what the homotopy theory of $(\infty,n)$-categories should be, and that it is fairly rigid (few automorphisms). It is at least fair to say that such a consensus is forming. Once you pin down the theories of $(\infty,n)$-categories, I would say that there are two distinct reasonable choices for what the (homotopy) theory of $(\infty,\infty)$-categories should be. So in that sense the answer to your question is no, there is not a single theory of $(\infty,\infty)$-categories; there are exactly two. But this "no" is a far cry from saying that there is a vast uncharted landscape of possibilities. What are these theories? As Charles Rezk mentions, the inclusion of $(\infty,n)$-categories into $(\infty, n+1)$-categories has both a left and right adjoint and this gives rise to two towers of homotopy theories of higher categories. The limits of these towers give the two potential models of $(\infty,\infty)$-categories. They are not equivalent. Let's call the limit using right adjoints $Cat_{(\infty,\infty)}$ and the limit using left adjoints $LCat_{(\infty,\infty)} $. One consequence of the unicity result is that these towers are essentially uniquely defined and are essentially model independent (provided the models satisfy our axioms). So in that sense there are these two canonical (established?) choices for the homotopy theory of $(\infty,\infty)$-categories. I know Clark and I have discussed this idea with many people, but the idea is certainly not new. I think one of the important parts of this story is Eugenia Cheng's theorem from her paper "An omega-category with all duals is an omega groupoid". Her result applies in the tower using the left adjoints, $LCat_{(\infty,\infty)} $, the "coinductive" version. There an $(\infty, \infty)$-category can be tought of as a sequences of $(\infty, n)$-categories, where each truncates to the previous theory. Cheng's result implies that in such a higher category if you have all duals you are an $\infty$-groupoid. I prefer the other limit, the limit of right adjoints. There an $(\infty,\infty)$-category is a sequence of $(\infty,n)$-categories where the previous is the maximal $(\infty,n-1)$-category. For some reason this seems more natural to me, though I know of others who disagree. This version includes one of my favorite examples: the infinite cobordism category. There are cobordisms and cobordisms between cobordisms and cobordisms between these and so on forever. This is an object in the tower of right adjoints which has duals for all objects, but which is not an $\infty$-groupoid. The limit of right adjoints fails Cheng's theorem. It also has a sort of inductive notion of equivalence, rather than coinductive. Your idea of taking the union of the theories of $(\infty,n)$-categories is actually not as far off as you might think. One problem with the naive union is that it is not cocomplete. There are diagrams which increase category number and have no colimits, or at least not the colimit which should exist. One way out of this is to note that you probably want the homotopy theory of $(\infty, \infty)$-categories to be presentable. So instead of taking the colimit naively you should take the colimit in presentable $(\infty,1)$-categories. Now since the inclusion functors preserve both limits and colimits, you can take this colimit in either $Pr^L$ or $Pr^R$. In either case you can use the equivalence $Pr^R \simeq (Pr^L)^{op}$ and Higher Topos Theory 5.5.7.6 or 5.5.3.13 to compute these colimits. In short you compute the colimit by taking adjoints and computing the limit naively. Thus you find: Taken in $Pr^L$:
$$ \mathrm{colim} \; Cat_{(\infty,n)} = Cat_{(\infty,\infty)} $$ taken in $Pr^R$:
$$ \mathrm{colim} \; Cat_{(\infty,n)} = LCat_{(\infty,\infty)} $$ Which is the "right" notion of $(\infty,\infty)$-category probably depends on taste and what you want to do with the notion. Both are useful. The above descriptions gives you a variety of universal properties for these two theories. I would also love to know if there are any theories in between these two. I also strongly believe that $LCat_{(\infty,\infty)} $ is a localization of $Cat_{(\infty,\infty)} $, but I haven't written down a proof. Perhaps it is easy to see this from the above description? You can be more explicit about these models too. Dominic Verity has a model of higher categories based on "weak complicial sets". There are $(\infty,n)$-versions of this and also $(\infty, \infty)$-versions.
One of the conjectures that Clark and I made at the end our paper is that some variant of Dominic's $(\infty,n)$-theory satisfies our axioms (Dominic, Emily Riehl, and I have a partial sketch of this, so hopefully the truth of this conjecture will be known... soon?). If that is true, then it is straighforward to show that the Dominic's $(\infty,\infty)$-version of weak complicial sets is a model of the tower of right adjoints, the one which includes the infinite bordism category. So there are also concrete models of these theories. There are also others whose work will yield explicit models of this. Rune Haugseng's work was already mentioned. Jeremy Hahn's (upcoming?) work will provide nice model of both limits. I am sure there are many ways to model these two theories. So to summarize: There are now a variety of uniqueness and comparison results which pin down the theory of $(\infty,n)$-categories as well as it is pinned down for $(\infty,1)$-categories. It actually is clear how to send $n \to \infty$; there are two ways to do it, giving two such infinite theories. We can describe and study these theories explicitly and they have interesting properties and possess interesting examples. Does this constitute a "consensus"? Of course not, but I hope it goes a little towards answering the real question.
|
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134,581 |
The Symmetric groups $S_n$ has interesting property that all complex irreducible characters are rational (i.e. $\chi(g)\in \mathbb{Q}$ for all $\mathbb{C}$-irreducible characters $\chi$,$\forall g\in S_n$). Question: What are other families of (finite) groups where all complex irreducible characters are rational? Are such (finite) groups characterised?
|
Here's one characterization that I learned from Serre
(see Definition 7.1.1 in his Topics
in Galois Theory (p.65)): an element $g$ of a finite group $G$
satisfies $\chi(g) \in {\bf Q}$ for all characters $\chi$ iff $g$ is conjugate in $G$ to $g^m$ for all $m$ relatively prime to
the exponent $e(g)$. [If $m$ is not coprime to $e(g)$ then
$e(g^m) \lt e(g)$ so $g^m$ cannot possibly be conjugate to $g$.]
It is enough to check this for all $m$ relatively prime to $\left| G \right|$. In particular, all character values are rational iff every group element
is conjugate to its $m$-th power for all $m$ coprime to $\left| G \right|$.
|
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|
135,012 |
How to prove (or to disprove) that all the roots of the polynomial of degree $n$ $$\sum_{k=0}^{k=n}(2k+1)x^k$$ belong to the disk $\{z:|z|<1\}?$ Numerical calculations confirm that, but I don't see any approach to a proof of so simply formulated statement. It would be useful in connection with an irreducibility problem.
|
Let $f$ denote your polynomial. Roots of the polynomial
$$g(x)=\sum_{k=0}^nx^{2k+1}=x\frac{x^{2(n+1)}-1}{x^2-1}$$
are $0$ and roots of unity. By the Gauss–Lucas theorem , the roots of $g'(x)=f(x^2)$ lie in their convex hull, and a fortiori in the disk $\{z:|z|\le1\}$. In order to get a strict inequality, it suffices to show that $g$ is square-free.
|
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|
135,189 |
When dealing with moduli spaces of, say connections or metrics, I am using the notions of Frechet spaces/manifolds/groups. I have become familiar with Banach manifolds (I think), but Frechet manifolds less so. In my studies, we want to apply the Inverse and Implicit Function Theorems , which work on Banach spaces, but apparently fail for Frechet spaces (and as a result we pass to Sobolev completions). Why does it fail (intuitively), or why should I expect it to fail? What is the hinge that allows it to work for Banach but not Frechet? Do I have to do a lot more to get an analogous result in the land of Frechet? This should also help me see the differences between a Banach and Frechet manifold, because right now it seems subtle.
|
The usual proof of the inverse function theorem in the setting of Banach spaces uses the Banach fixed point theorem. We cannot make sense of the Banach fixed point theorem in a Fréchet space, since a Fréchet space is merely metrizable: there is no preferred choice of metric. I do not know exactly how crucial the Banach fixed point theorem actually is for proving the inverse function theorem in Banach spaces. Hopefully someone can come along and explain whether this is really the "hinge" that makes the inverse function theorem work. Since you say IFT "apparently" fails for Fréchet spaces, I'm guessing you haven't seen a counterexample. Let $C^\infty[-1,1]$ denote the Fréchet space of smooth real-valued functions on $[-1,1]$. Consider the smooth map
$$P: C^\infty[-1,1] \longrightarrow C^\infty[-1,1],$$
$$f \mapsto f - xff'.$$
Now
$$DP(f)g = g - xgf' - xfg',$$
so $DP(0) = I$. Since $P(0) = 0$, if the inverse function theorem holds then $P$ should be invertible in some neighborhood of $f = 0$, with $P^{-1}(0) = 0$. To see that this is not true, consider the family of functions
$$g_n(x) = \frac{1}{n} + \frac{x^n}{n!}.$$
Now $g_n \longrightarrow 0$ in $C^\infty[-1,1]$, but we claim that $g_n$ is not in the image of $P$ for any $n$. To see this, write the power series expansion
$$f(x) = a_0 + a_1x + a_2x^2 + a_3 x^3 + \cdots.$$
Then we have
$$Pf(x) = a_0 + (1 - a_0)a_1 x + (a_2 - a_1^2 - 2a_0a_2)x^2 + (a_3 - 3a_1a_2 - 3a_0a_3)x^3 + \cdots.$$
Suppose $Pf(x) = g_n(x)$. Then $a_0 = \tfrac{1}{n}$, and an inductive argument shows that $a_1 = a_2 = \cdots = a_{n-1} = 0$, giving
$$Pf(x) = \frac{1}{n} + (1 - na_0)a_nx^n + \cdots.$$
The order $n$ term is then zero, contradicting the fact that $Pf(x) = g_n(x)$. Hence $g_n$ does not lie in the image of $P$ for any $n$. Since $P(0) = 0$, $DP(0) = I$ is invertible, and $g_n \longrightarrow 0$, this provides a counterexample to the inverse function theorem. The generalization of the inverse function theorem to Fréchet spaces is the Nash-Moser theorem. You need some much better conditions on the map $f$ you want to invert in this case: First, $f$ must be invertible in a neighborhood , not just at a single point. Second, the Fréchet spaces involved and $f$ must satisfy the technical condition of "tameness." Finally, the inverses of $f$ in the neighborhood of interest must also be tame. There are counterexamples to show that each of these extra conditions is necessary. For an overview of the Nash-Moser theorem, see Hamilton's The inverse function theorem of Nash and Moser .
|
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|
135,272 |
If $G$ is a discrete or topological group, $G$ is a closed subgroup of $EG$, and normal iff $G$ is abelian, according to Segal, Cohomology of topological groups , Symposia Mathematica IV (1970) (a reference I found from two answers by Chris Schommer-Pries: Classifying Space of a Group Extension and Good functorial model for BG ). Hence for $G$ not abelian, $G$ is not normal in $EG$, hence $BG=EG/G$ is not a group. On the other hand, we also learn from Segal that the reason $EG$ is a group is that the universal bundle functor $E$ is monoidal with respect to the Cartesian monoidal structure. Any monoidal functor takes group objects in one category to group objects in another simply by functoriality. But the classifying space functor $B$ is also monoidal (see Good functorial model for BG or Peter May's answer at Group structure on Eilenberg-MacLane spaces ). Hence $BG$ should be a group?
|
The classifying space functor may be a monoidal functor out of $\text{Grp}$, but nonabelian groups aren't group objects in $\text{Grp}$. (The group objects in $\text{Grp}$ are precisely the abelian groups. This is also a corollary of the Eckmann-Hilton argument.)
|
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|
135,669 |
Recall that two $k$-algebras $A, B$ are Morita equivalent iff their categories of left modules are equivalent. However, this relation turns out to be rather fine and one introduces a coarser equivalence relation of derived Morita equivalence by using (bounded) derived categories of modules, along with their triangulated structure. (Note that I am no expert in this matters and I was mostly exposed to this viewpoint through algebraic geometry, where one instead works with bounded derived categories of coherent sheaves on a variety.) However, as I understand it, the derived category of an algebra arises as a homotopy category of the stable model category of chain complexes. The latter may be seen as presenting a homotopy theory (ie. an $(\infty, 1)$-category), for example through the process of simplicial localization (and probably also some more direct, dg-theoretic methods?). One could then say that two algebras are "higher derived Morita equivalent" if their $(\infty, 1)$-categories of (bounded?) complexes are equivalent as higher categories. The question is as follows: What can we say about this new equivalence relation? How far is this relation from derived Morita equivalence? How far is it from ordinary Morita equivalence? I have no intuition about this and I can imagine answers that completely equate "higher derived Morita equivalence" with either of these two, although it would be probably most interesting if it was somewhere between them. Note that one can imagine that somehow the derived category of an algebra remembers all the "higher homotopy", as it happens to be the case for some other homotopy categories of stable model categories. For example, in "The stable homotopy category is rigid" by S. Schwede it is proven that any stable model category $\mathcal{C}$ that satifies $ho(\mathcal{C}) \simeq \mathcal{SHC}$ (as triangulated categories, where the latter is the stable homotopy category) is in fact Quillen equivalent to model category of spectra, so they present the same homotopy theory. I ask the question since I am currently studying higher categories and this led me to wonder what is their possible strength as invariants of other mathematical objects.
|
Derived Morita equivalence is the same as higher derived Morita equivalence. Clearly, $\Leftarrow$ is obvious, and $\Rightarrow$ follows from Theorem 2.6 in Dugger-Shipley's 'K-theory and derived equivalences' Duke Math. J. 124 (2004), no.3, 587--617. This is surprising at a first glance, it follows from the fact that algebras are concentrated in degree $0$. The analogous result for DG-algebras is false, compare 'A curious example of triangulated-equivalent model categories which are not Quillen equivalent' Algebraic and Geomtric Topology 9 (2009), no. 1, 135-166, by the same authors. There're still some interesting open questions connected to this, though.
|
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|
135,820 |
I want to know whether there exist a non-square number $n$ which is the quadratic residue of every prime.
I know it is very elementary, and I think those kind of number are not exist, but I don't know how to prove.
|
This is actually an elementary consequence of quadratic reciprocity,
generalizing the familiar proof à la Euclid that that are infinitely many
primes of the form $4k+3$
(i.e. primes of which $-1$ is not a quadratic residue).
We want to show that there exists $p$ such that $(n/p) = -1$.
[As stated Paul's question asks only for $(n/p) \neq +1$, but this is
trivial (if $n = -1$, take $p=3$; else let $p$ be a factor of $n$),
so we'll exclude the finitely many prime factors $p$ of $n$.]
By QR there exists a nontrivial homomorphism
$\chi: ({\bf Z} / 4n{\bf Z})^* \rightarrow \lbrace 1, -1 \rbrace$
such that $(n/p) = \chi(p)$ for all primes $p \nmid 2n$.
Let $a$ be any positive integer coprime to $4n$ such that $\chi(a) = -1$.
Then we have a prime factorization $a = \prod_j p_j$,
and $\prod_j \chi(p_j) = \chi(a) = -1$.
Therefore $\chi(p_j) = -1$ for some $j$, QED . As in Euclid we can iterate this argument to construct
infinitely many distinct $p$ for which $(n/p) = -1$.
|
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|
135,911 |
Liouville's theorem gives such a proof for antiderivatives of functions like $e^x/x$ or $e^{x^2}$ , and differential Galois theory extends that to Bessel functions, say. But what tools exist for implicit functions like Lambert's W?
|
It seems that the non-linear differential equation satisfied by the Lambert $W$ function is simple enough for this question to have already been answered. This paper proves that the Lambert $W$ is non-elementary by appealing to a result of Rosenlicht (1969): Bronstein, M., Corless, R. M., Davenport, J. H. and Jeffrey, D. J.
(2008) Algebraic properties of the Lambert W Function from a
result of Rosenlicht and of Liouville. Integral Transforms and
Special Functions , 19 (10). pp. 709-712. https://dx.doi.org/10.1080/10652460802332342 The argument runs roughly as follows. First note that $W(x)$ satisfies a two-variable algebraic-differential system, where the differential part has a special form: \begin{equation*}
W e^W = x \implies W'\!/W + W' = 1/x \implies \{Y = x/W,\; Y'\!/Y = W'\} .
\end{equation*} Now, consider the differential field $\mathbb{C}(x)(Y,W)$ where $x'=1$ , $Y'/Y = W'$ and $f(x,Y,W)=0$ , with $f$ polynomial in $Y$ and $W$ over $\mathbb{C}(x)$ . According to a more general result of Rosenlicht, the field $\mathbb{C}(x)(Y,W)$ is Liouvillian only if $Y$ and $W$ are algebraic over $\mathbb{C}(x)$ . So, if $W(x)$ were elementary (elementary is a special case of Liouvillian) then it would be algebraic. It remains to check that $W(x)$ is not algebraic. The above article shows that as well. I'll just leave it at saying that it follows from the fact that it satisfies a transcendental equation.
|
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|
135,949 |
Let $p\geq 3$ be a prime number, and let $u:\mathbb{Z}/p\mathbb{Z}\to \mathbb{Z}/p\mathbb{Z}$ be a map such that, for all $l\in \mathbb{Z}/p\mathbb{Z}$,$l\neq 0$, the map $k\mapsto u(k+l)-u(k)$ is a permutation. Is $u$ a polynomial of degree $2$? Note that the property clearly holds when $u$ is a polynomial of degree $2$. Explicit computations seem to show that the converse holds -- that is, the answer is positive -- for $p$ at most $13$. This is (in a non-obvious way) a special case of this other question , but presumably the statement here is much easier. The question came up quite naturally when thinking about some aspects of complex Hadamard matrices.
|
Yes, $u$ must be a polynomial of degree $2$.
I had to draw on a few unexpected ingredients to prove this;
perhaps there's a simpler proof.
[ EDIT Or maybe not: Peter Mueller's answer reports that
this was "an open problem on planar functions for many years",
and gives links to three independent papers c.1990 that independently
solved it. Two of them give the same argument that I found 23 years later,
and the third, by Hiramine, either avoids or re-proves Segre's theorem
but is even more complicated. ] Let $\kappa$ be the finite field ${\bf Z}/p{\bf Z}$
(usually this would be called $k$, but that letter's already taken).
Fix a nontrivial $p$-th root of unity $\rho \in {\bf C}$,
say $\rho = e^{2\pi i/p}$; for any $n \in \kappa$
we shall naturally use $\rho^n$ to mean $\rho^{\tilde n}$
for any lift $\tilde n$ of $n$ to ${\bf Z}$.
Let $K$ be the $p$-th cyclotomic field ${\bf Q}[\rho]$,
and $A = {\bf Z}[\rho]$ its ring of algebraic integers,
which contains the Gauss sum
$\gamma := \sum_{n \in \kappa} \rho^{n^2} \in A$,
with $\gamma^2 = \pm p$ according as $p \equiv \pm 1 \bmod 4$.
For any $p$-th roots of unity $\omega,\zeta \in A$ with $\zeta \neq 1$,
define
$$
G(\omega,\zeta) = \sum_{k \in \kappa} \omega^k \zeta^{u(k)}.
$$
I claim that $G(\omega,\zeta)$ is $\pm\gamma$ times some $p$-th root of unity
(as it must be if $u$ is quadratic).
We prove this by mimicking the usual proof of $\left|\gamma\right|^2 = p$: write
$$
\left|G(\omega,\zeta)\right|^2
= \mathop{\sum\sum}_{k,k' \in \kappa}
\omega^{k'-k} \zeta^{u(k')-u(k)}
= \sum_{l \in \kappa} \left[
\omega^l \sum_{k \in \kappa} \zeta^{u(k+l)-u(k)}
\right]
$$
where $l=k'-k$; now for $l=0$ the inner sum is $\sum_k 1 = p$,
and for $l\neq 0$ the inner sum vanishes by the hypothesis on $u$
(it is a permutation of $\sum_{n\in\kappa} \zeta^n = 0$), so
$\left|G(\omega,\zeta)\right|^2 = p$. This holds for every
Galois conjugate of $G(\omega,\zeta)$, so the algebraic norm of
$G(\omega,\zeta) \in K$ is $p^{(p-1)/2}$.
Because there's a unique prime of $K$ above $p$, it follows that
$\gamma^{-1} G(\omega,\zeta)$ is an algebraic integer
all of whose Galois conjugates have absolute value $1$.
By a theorem of Kronecker this integer must be a root of unity.
This proves the claim that $G(\omega,\zeta)$ is of the form
$\pm\zeta^s\gamma$, because the only roots of unity in $A$
are powers of $\rho$ and their negatives.
[ EDIT Gluck's paper (Discrete Math. 80 (1990) 97$-$100)
cites Theorem 1 of "Cavior, S.: Exponential sums related to polynomials
over GF(p), Proc. A.M.S. 15 (1964) 175$-$178" for the result that
$\pm\zeta^s\gamma$ are the only elements of absolute value $\sqrt p$ in $A$. ] Now for any $c \in \kappa$ we have
$G(\rho^c,\rho) = \sum_{k \in \kappa} \rho^{u(k)+ck}$,
which is a representation of some $\pm\rho^a\gamma$ as
a sum of $p$ powers of $\rho$.
This representation is unique because the cyclotomic polynomial
$\sum_{n=0}^{p-1} X^n$
is irreducible and does not vanish at $X=1$.
We already know one such representation,
$\pm\rho^s\gamma = \sum_{n\in\kappa} \rho^{an^2+s}$,
where $a$ is a quadratic residue or nonresidue of $p$
according to the choice of plus or minus sign.
Therefore $u(k)+ck$ must take the same values and multiplicities as
$an^2+s$ when $k$ varies over $\kappa$.
In particular each $b \in {\bf Z}/p{\bf Z}$ occurs at most twice as $u(k)+ck$.
(Could this conclusion have been reached without the foray into
algebraic number theory?) This strongly suggests that $u$ must be quadratic,
but the implication is still not obvious.
To reach that conclusion we use a theorem
of Segre on ovals in algebraic projective planes of odd order.
Recall that an oval in a projective plane $\Pi$ of order $q$
is a $(q+1)$-element set of points of $\Pi$ that meets each line
in at most $2$ points. For example, a conic in an algebraic
projective plane is an oval. Theorem (Segre 1955). If $F$ is a finite field of odd order
then every oval in ${\bf P}^2(F)$ is a conic. Now we have just proved that the subset
$\lbrace (x,y) = (k,u(k)) : k \in \kappa \rbrace$
of the affine plane $\kappa^2$ meets every line $cx+y=b$ in at most two points;
it also meets every line $x=x_0$ in exactly one point. Thus we can construct
an oval ${\cal O}$ in ${\bf P}^2(\kappa)$ consisting of these points
$(k:u(k):1)$ together with the point at infinity $(0:1:0)$.
By Segre's theorem $\cal O$ is a conic.
Since it meets the line at infinity at just the one point $(0:1:0)$,
this conic ${\cal O}$ consists of that point together with the graph of
a quadratic polynomial, QED .
|
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|
136,233 |
The study of algebraic geometry usually begins with the choice of a base field $k$. In practice, this is usually one of the prime fields $\mathbb{Q}$ or $\mathbb{F}_p$, or topological completions and algebraic extensions of these. One might call such fields $0$-dimensional . Then one could say that a field $K$ is $d$-dimensional if it has transcendence degree $d$ over a $0$-dimensional field. But is there a way to make this less ad hoc? Is there a reason I've never seen any such definition of $0$-dimensional fields ? Am I missing something? To what extent has the classification of abstract fields been considered?
|
I think you are lumping too many disparate kinds of fields together under the heading "zero-dimensional". As Jason says in his answer, there are some precise definitions of dimensions of fields (e.g. cohomological dimension but also other definitions of a field-arithmetic nature). Another important comment is that in modern algebraic / arithmetic geometry there is no restriction on the kind of field that can be taken as a "ground field", i.e., over which to define algebraic varieties. Although you have indeed listed some common ground fields, some people think about other special cases as well as the general case. (For instance I have recently been thinking about elliptic curves defined over transfinitely iterated function fields.) The basic classification of fields, as I understand/view it, is as follows: Step 1: Every field has a characteristic, either $0$ or a prime number. There are no homomorphisms between fields of different characteristics, so fields of a given characteristic are somehow different worlds (one can think of the characteristic as a connected component in the category of fields, essentially). Step 2: For a fixed characteristic $p \geq 0$ there is a unique minimal field, called the prime subfield, say $k_0$. This is precisely the initial object in the category of fields of characteristic $p$: it's $\mathbb{Q}$ in characteristic $0$ and $\mathbb{F}_p$ in characteristic $p$. This means that the absolute theory of fields is reduced to the theory of field extensions , and one can define "absolute invariants" on a field $K$ by giving invariants of $K/k_0$. In particular: Step 3: The absolute transcendence degree of a field $K$ is the cardinality of a transcendence basis for $K/k_0$. This means that there is a subextension $k_0 \subset F \subset K$ such that $F/k_0$ is purely transcendental : a rational function field, perhaps in infinitely many variables, and $K/F$ is algebraic. Step 4: The algebraic extension $K/F$ is in many ways the most interesting part. For instance, as Qiaochu Yuan mentions, algebraic geometry in dimension $n$ is the study of finite degree field extensions of $\mathbb{C}(t_1,\ldots,t_n)$. When $n = 1$ we get precisely the compact Riemann surfaces, or (replacing $\mathbb{C}$ with any field $k$ and looking at finitely generated field extensions of transcendence degree $1$) of complete, regular, integral algebraic curves over $k$. So a classification here is the (rich) classical story of the genus, the moduli spaces $\mathcal{M}_g$, and so forth. When $k$ is not algebraically closed, Galois cohomology enters the picture. When $n \geq 2$ we are studying birational algebraic geometry only, but that is still very rich. When $n = 2$ there is a "classification of algebraic surfaces" which ends up tossing most of them into a very large box called "general type". To the best of my knowledge there is no explicit description of the connected components of the (infinite type) moduli space of all complex algebraic surfaces like there is in dimension one. In fact, I believe that probably starting even in dimension $2$ it is in some sense hopeless to try to give an algorithmic classification, although I cannot recall having seen a precise impossibility theorem along these lines analogous e.g. to the algorithmic impossibility of classifying compact $4$-manifolds because this problem -- via the fundamental group -- contains the word-problem for finitely presented groups which Novikov proved is algorithmically unsolvable. Bjorn Poonen discussed similar (open) problems in the last of a series of three lectures on undecidability that he gave at UGA a few years ago; notes are available on his webpage. Step 5: A vaguely dual role to purely transcendental extensions of the prime subfield is played by the algebraically closed fields . Here there is a precise classification: the characteristic and the absolute transcendence degree determine an algebraically closed field up to isomorphism. If the field is uncountable, then the absolute transcendence degree is just its cardinality, so that classifies the field: e.g. the only algebraically closed field of characteristic $0$ and continuum cardinality is $\mathbb{C}$. This has been used for some sneaky purposes: e.g. the algebraic closure of $\mathbb{Q}_p$ must then also be isomorphic to $\mathbb{C}$. (However for countable fields cardinality is not enough: e.g. $\overline{\mathbb{Q}}$ is not isomorphic to $\overline{\mathbb{Q}(t)}$.) This uncountable categoricity means that the first order theory of algebraically closed fields of given characteristic $p \geq 0$ is complete , which has various pleasant consequences. Algebraically closed fields of infinite transcendence degree have some further nice properties which makes them suitable for use as "ground fields" in algebraic geometry, as was exploited by Weil in his pre-(scheme-theoretic) foundations. From a model-theoretic perspective, these large algebraically closed fields enjoy good saturation properties. Fields which are "close" to being algebraically closed tend to be better understood than fields which are farther away from being algebraically closed. One can measure this (at least in characteristic $0$) by the size / complexity of the absolute Galois group $\operatorname{Aut}(\overline{K}/K)$. Thus for instance the absolute Galois group of $\mathbb{Q}_p$ is known completely -- i.e., it is a topologically finitely generated compact totally disconnected Hausdorff group, and explicit generators and relations are known, which in some sense means we understand every algebraic extension of $\mathbb{Q}_p$. This needs to be taken with a grain of salt: a certain filtration on the absolute Galois group provides much coarser information -- e.g. you can break the group up into three pieces, two of which are commutative and one of which is pro-$p$, so the group is pro-solvable. That's useful: the pro-$p$-part is the hard part and knowing generators and relations for it does not in practice seem to take away the mystery. For instance, the Local Langlands Correspondence is a deep theorem about representations of the absolute Galois group of a $p$-adic field, and it was certainly not proved by looking at its explicit structure as a topologically finitely presented group! Let me note finally that every compact totally disconnected topological group occurs up to isomorphism as the automorphism group of an algebraic Galois extension of fields (the Leptin-Waterhouse Theorem; I discovered this independently as a graduate student, and at the time I knew neither that others had published the result before nor even that such a result would be publishable)...so algebraic field extensions can be awfully complicated.
|
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|
136,236 |
Reverse mathematics, as I mean here, is the study of which theorems/axioms can be used to prove other theorems/axioms over a weak base theory. Examples include Subsystems of Second Order Arithmetic (SOSOA). Theorems expressible in second order arithmetic are compared over $\mathsf{RCA}_0$ (which is roughly the theory of computable sets of natural numbers). Constructive Reverse Mathematics. Bridges and others use Bishop-style constructive mathematics as a base theory to compare various nonn-constructive principles. Here are my questions. Would it be possible to do reverse mathematics over Homotopy Type Theory (HoTT)? Would HoTT make for a good base theory (roughly corresponding to computable objects in mathematics)? My understanding is that HoTT is a constructive theory, this makes me believe the answer is yes. Would the univalence axiom need to be removed from the base theory? My understanding is that the univalence axiom is incompatible with the law of excluded middle. This would be problematic. Would the results be similar to those in Constructive Reverse Mathematics? (Would they be exactly the same?) Would computability theoretic ideas be of use as they are in SOSOA reverse mathematics? Computability theorists are drawn to SOSOA because of its computability theoretic nature. Would the computer implementations of HoTT, in say Coq, be helpful in proving reverse-mathematics-like results. It would be nice to be able to do reverse mathematics with the "safety net" of a proof checker. Has reverse mathematics in HoTT been done already in any sense (formally or informally)? Note:
There may be some relationship with this question: Forcing in homotopy type theory
|
Before I attempt to address your specific questions, let me give a thought provoking non-answer: Reverse mathematics is impossible (and irrelevant) in HoTT! This is because HoTT fully supports proof-relevant mathematics , so when you refer to a theorem you necessarily refer to a proof of that theorem. The question whether the hypotheses are necessary doesn't make sense in this context since the hypotheses are part of the theorem itself! Now that your thoughts have been provoked, let me amend the above statement: HoTT is not the end of reverse mathematics! Actually, it just makes the reverse (and constructive) mathematics questions even more obvious. The most fundamental reason why reverse mathematics exists is the incredible power of the existential quantifier. In HoTT, that is immediately obvious. A plain existential statement is interpreted as a dependent sum type in HoTT: "$\sum_{x:A} P(x)$ is inhabited" is the right way to say that "there is an $x$ of type $A$ such that $P(x)$." By definition, every inhabitant $x:A$ of this type is equipped with a justification of $P(x)$. To get the usual existential quantifier, one must truncate $\sum_{x:A} P(x)$ to a proposition. This raises the question: can this truncation be reversed? What are the necessary hypotheses to reverse this truncation? This is what reverse mathematics questions get turned into when translated into the language of HoTT. Note how "mathematical" the reverse mathematics question has become! This is no surprise to practicing reverse mathematicians but it is very interesting how reverse mathematics becomes less mythical in this context. I will now add what I know about each of your questions. Since I'm still learning about HoTT, these answers are far from complete or definitive. I hope that experts will chime in at some point. The above sort of addresses this question. An additional difficulty is that it seems that there is still some work to be done in understanding models of HoTT. The soundness and completeness theorems obtained by Awodey & Warren and Gambino & Garner are almost there but Awodey points out some subtle issues. I don't know if these issues are problematic enough to make it difficult to establish non-provability results for HoTT. HoTT is perhaps too strong for use as a base system for classical reverse mathematics. The reason is basically the same as why ZF is often too strong for that purpose. Note, however that even ZF is not completely useless. For example, as witnessed by a great deal of literature, ZF is a perfectly fine base theory for the analysis of choice principles. HoTT is more promising as a base system for constructive reverse mathematics but the constructivity of the univalence axiom is currently an open problem. I don't think the univalence axiom is that problematic. The law of excluded middle always clashes with the propositions-as-types interpretation. The correct way to formulate the law of excluded middle in HoTT is to restrict it to propositions — types with at most one element. This version of the law of excluded middle does not clash with univalence and captures all of the normal uses. I don't think anyone has addressed the question whether HoTT is a conservative extension of BISH (say) or how far it is from being a conservative extension. It's difficult to make a comparison. The base system RCA 0 was explicitly designed to capture basic computability theory. HoTT wasn't designed that way but other aspects of computability were important design components. I don't see much gain in using proof assistants for reverse mathematics but I might be nearsighted. Yes, some additional axioms such as propositional resizing, the law of excluded middle, and the axiom of choice have been analyzed to some extent. HoTT is so young that very little of this has been done yet.
|
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|
136,314 |
As part of a more complex algorithm, I need a fast method to find random points of the n-sphere, $S^n$, starting with a RNG (random number generator). A simple way to do this (in low dimensions at least) is to select a random point of the (n+1)-ball and normalize it. And to get a random point of the (n+1)-ball select a random point of the (n+1)-cube $[-1,1]^{n+1}$ (by selecting (n+1) points of $[0,1)$ with the RNG and scaling using $x \mapsto 2x-1$) and then use "rejection", i.e., just ignore a point if it is not in the (n+1)-ball. This works fine if n is reasonably small, however for large n the volume of the ball is such a tiny fraction of the volume of the cube that rejection is enormously inefficient. So what is a good alternative approach.
|
The usual approach is to generate $n+1$ i.i.d. mean zero Gaussian random variables $X_1, \dotsc, X_{n+1}$ to get a random point $X$ in $(n+1)$-space with rotationally invariant distribution and normalize. Incidentally, if you ever actually need to generate a random point in an $n$-ball, the best way is probably to generate a random point in the $(n+1)$-sphere as above and drop the last two coordinates.
|
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"source": [
"https://mathoverflow.net/questions/136314",
"https://mathoverflow.net",
"https://mathoverflow.net/users/7311/"
]
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|
137,114 |
First, I will explain my situation.
In my University most of the careers are doing videos to explain what we do and try to attract more people to our careers. I am in a really bad position, because the people who are in charge of the video want me to explain what a pure mathematician does and how it helps society. They want practical examples, and maybe naming some companies that work with pure mathematicians, and what they do in those companies. All this in only 5 or 10 minutes, so I think that the best I can do is give an example. Another reason that I am in a bad position: In my University we have the career "Mathematical engineering" and they do mostly applications and some research in numerical analysis and optimization. (*) I know that pure mathematics is increasing its importance in society every year.
Many people in my country think that mathematics has stagnated over time and now only engineers develop science.
I think that the most practical thing I can do is give some examples of what we are doing with mathematics today (since 2000). If some of you can help me, I need the following: A subject in mathematics that does not appear in (*). Preferably dynamical systems, logic, algebraic geometry, functional analysis, p-adic analysis or partial differential equations. A research topic in that subject. Practical applications of that research and the institution that made the application. Extra 1. If you know an institution (not a University) that contracts with pure mathematicians and you know what they do there, please tell me also. Extra 2. If you have a very good short phrase explaining "what a mathematician does" or "how mathematics helps society" I will appreciate it too. Thanks in advance.
|
Here are two recent speakers from our departments lecture series on applied mathematics (the Wing Lectures at U. Rochester). We've had a lot of great lecturers, but these two stick out to me as having an impact on society. Adrien Treuille -- he works in computer graphics, and his research purely in this areas includes algorithms realistic modeling of crowds, and real time fluid mechanics (e.g. with basically no delay, they can add digital trails of flames behind a race cars on TV). Also, he can construct initial data to make (digital) smoke form specified shapes at specified times. But, what's even cooler, is that he's collaborated with biologists on an interactive game (called FoldIt ) for finding the optimal conformation of proteins. Players get points for moving the protein into better conformations. Humans are way better at this than computers, and they've actually published papers based on conformations discovered by players; in one case, the answer they found had eluded scientists working in the field for at least 10 years! They call this "crowd source science". They've also created another game for engineering shapes with RNA, and the players of that game have discovered things as well. Gunnar Carlsson -- he is an algebraic topologist who originally worked in K-theory, but shifted to using algebraic topology to understand data. Specifically he was one of (the?) pioneers of persistent homology, which is a way of using topology to understand data. The really great thing about it is that it discovers structure for you--instead of fitting the data to a model, persistent homology discovers the model for you. For instance, they used PH to find the most commonly occurring 9-bit patterns in black and white images, which in theory would allow better image compression that JPEG (but in practical, JPEG is highly optimized, so it would take a lot of work to benefit from this discovery). In another example, they analyzed genomic information from cancer patients and linked it up with survival information; PH discovered the threshold for the expression of a certain gene at which survival drops significantly. Besides discovering structure in data, PH is also able to incorporate data collected at different times and from different sources (perhaps a reflection that the method is "metric free" --I'm not sure about that). A good reference to check is their paper in Nature . Also, Tony DeRose from Pixar gave a really great talk on the methods they've developed in computer graphics. I don't remember the details so well (maybe because I was distracted by the entertaining presentation :) so here's a link to a good article .
|
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"https://mathoverflow.net/questions/137114",
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|
137,249 |
I seek a metric $d(\cdot,\cdot)$ between pairs of (infinite) lines in $\mathbb{R}^3$.
Let $s$ be the minimum distance between a pair of lines $L_1$ and $L_2$.
Ideally, I would like these properties: If $L_1$ and $L_2$ are parallel, then $d(L_1,L_2) = s$. $d(L_1,L_2)$ increases with the degree of skewness between the lines, i.e., the angle $\theta$ between their projections onto a plane orthogonal to a segment that realize $s$, where $\theta=0$ for parallel lines and $\theta=\pi/2$ for orthogonal lines. A natural definition (suggested in an earlier MSE question ) is $d(L_1,L_2) = s + |\theta|$.
But this is not a metric.
For example, a sufficiently large value of $a$ below ensures the triangle inequality will be violated: I would be interested to learn of metrics defined on lines in space, and whether or not any such metric satisfies the properties above.
Perhaps the properties cannot be achieved by any metric? Update .
Here is my current understanding of the rich variety of the erudite answers provided.
Apologies in advance if my summary is inaccurate. First, there is no such metric, interpreting my second condition as (naturally) demanding invariance under Euclidean motions, as convincingly demonstrated by Robert Bryant, Vidit Nanda, and Pierre Simon.
Second, a looser interpretation requires only that if we fix $L_1$, then $d(L_1,L_2)$ is monotonic with respect to $\theta$ as $L_2$ is spun "about their intersection point in the plane that contains them [Yoav Kallus]." Then, Will Sawin's metric satisfies this condition. Here is an example of why this metric fails the more stringent condition—it depends on the relationship between the lines and the origin: The right lines could be further apart than the left lines (depending on $a$ and $b$).
|
There is no hope of making any such metric continuous with respect to $\theta$ at $s=0$ without violating the triangle inequality even in the plane. Consider parallel lines $L_1$ and $L_2$ which are some huge distance $s$ apart and a third line $L_3$ which intersects both at a very tiny angle $\theta$. Here's a picture: In order to make $d(L_1,L_2) = s$ huge, you would have to make $d(L_1,L_3) + d(L_2,L_3)$ also huge, even for arbitrarily small $\theta$ values. Thus there is no way, when two lines intersect, to force their distance to shrink to $0$ continuously as the angle between them reduces to $0$.
|
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"https://mathoverflow.net/questions/137249",
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|
137,270 |
Take a differential operator with elliptic symbol, consider just the principal part of the operator. Can one invert this principal part with some parametrix type construction (at least construct a left inverse)? Background: the concern is on linear PDE systems $$\mathcal L(x, \frac{\partial}{\partial x})u=\mathcal S(x), \qquad \mathcal B(x,\frac{\partial}{\partial x})u|_{\partial\Omega}=\varphi(x)$$ whose symbol coincides with their principal symbol,
$$ \mathcal L_0(x,i\xi)=\mathcal L(x,i\xi),\qquad \mathcal B_0(x,i\xi)=\mathcal B(x,i\xi),$$
so there are no terms of lower order. The operator is considered on a bounded domain $\Omega\in\mathbb{R}^n$. There can be more equations in the systems than unknown functions $u=(u_1,\ldots, u_J)$. The symbol is assumed to be elliptic/full rank in $\Omega$ for $|\xi|\neq 0$. The concern is not about existence, but only about regularity estimates given existence of the solution. To this purpose, some general theory asserts the existence of a left regularizer to $\mathcal A$, where $\mathcal Au=(\mathcal L\times\mathcal B)u=\left(\begin{array}\mathcal S\\\varphi\end{array}\right)$,
that is $$ \mathcal R\mathcal A=\mathcal{Id}-\mathcal T,$$
where $\mathcal T$ is compact. The question is whether one can achieve that $\mathcal T=0$ because there are no lower order terms, as $\mathcal L_0(x,i\xi)=\mathcal L(x,i\xi)$. Is invertibility of the operator easy to analyze when the elliptic symbol has no lower order terms? The aim is to get a regularity estimate for the solution $u$ just with $\mathcal S$ and $\varphi$ on the right hand side, without Lebesgue norm of $u$. I recall that in the theory of $\Psi DO$ in $\mathbb{R}^n$, I saw a construction of a parametrix for an elliptic principal symbol which acted as inverse but could not find the source again. A subquestion: is there literature on (redundant) PDE systems apart from that Russian article from the 70ies ( link ), which has its own peculariar notation and proofs which are at sometimes difficult to follow, relying on Russian articles not available in English. If the task is promising, I want to study the mentioned problem but lack other literature which construct regularizers with more perspicuous proofs. Any comments are appreciated. Thanks.
|
There is no hope of making any such metric continuous with respect to $\theta$ at $s=0$ without violating the triangle inequality even in the plane. Consider parallel lines $L_1$ and $L_2$ which are some huge distance $s$ apart and a third line $L_3$ which intersects both at a very tiny angle $\theta$. Here's a picture: In order to make $d(L_1,L_2) = s$ huge, you would have to make $d(L_1,L_3) + d(L_2,L_3)$ also huge, even for arbitrarily small $\theta$ values. Thus there is no way, when two lines intersect, to force their distance to shrink to $0$ continuously as the angle between them reduces to $0$.
|
{
"source": [
"https://mathoverflow.net/questions/137270",
"https://mathoverflow.net",
"https://mathoverflow.net/users/37439/"
]
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|
138,081 |
Consider the real algebraic group $SO(p,q)$, this is the automorphism group of the vector space $\mathbb{R}^n$ of dimension $n=p+q$ over $\mathbb{R}$, endowed with the diagonal quadratic form with $p$ pluses and $q$ minuses on the diagonal and with a nonzero skew symmetric $n$-form. Now consider the corresponding spinor group $G=Spin(p,q)$. Can one say that it is the group of automorphisms of a some object over $\mathbb{R}$, which is not very difficult to describe?
|
To expand on my comment to the question, we have the following algebraic construction (I think originally due to Robert Brown, 'A characterization of spin representations'): Let $V$ be a quadratic space over a field $k$ of characteristic $\neq 2$. Attached to this is the Clifford algebra $C=C(V)$: it is equipped with a $\mathbb{Z}/2\mathbb{Z}$-grading $C=C^+\oplus C^-$ and an embedding $V\hookrightarrow C^-$. The general Spin group $GSpin(V)$ consists of units in $C^+$ that preserve $V$ under conjugation. The group $Spin(V)$ is the sub-group of elements that have trivial spinor norm. So, to describe $Spin(V)$ as an automorphism group, it suffices to do so for $GSpin(V)$. Let $H$ be the graded vector space $C$ viewed as a representation of $GSpin(V)$ via left multiplication: it is also a right $C$-module via right multiplication. Then $GSpin(V)$ clearly lies within the group $U(H)$ of $C$-equivariant, grading preserving automorphisms of $H$. Set $E=End(H)$: this is a representation of $GSpin(V)$ via conjugation. Define a bilinear form $\{,\}:E\times E\to k$ by $$\{f,g\}=\frac{1}{2^{dim(V)}}trace(fg).$$ Now choose a basis $\{v_i\}$ for $V$, and let $A=(v_i\cdot v_j)_{i,j}$ be the inner product matrix attached to this basis. Set $(b_{i,j})=B=A^{-1}$. Define an endomorphism $\pi:E\to E$ by the formula: $$\pi(f)(h)=\sum_{i,j}b_{i,j}\{v_i,f\}v_jh.$$ Clearly, the image of $\pi$ is $V\subset E$, where $V$ acts on $H$ via left multiplication. Let $G'\subset U(H)$ be the stabilizer of the endomorphism $\pi$. Then $G'$ preserves $V$ via conjugation, and is therefore contained in $GSpin(V)$. On the other hand, it is not hard to see that $GSpin(V)$ stabilizes $\pi$. So we see that $GSpin(V)$ can be described as the group of $C$-equivariant, grading preserving automorphisms of $H$ that also stabilize $\pi$.
|
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"https://mathoverflow.net/questions/138081",
"https://mathoverflow.net",
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|
138,106 |
Is there an algorithm which, given a polynomial $f \in \mathbb{Q}[x_1, \dots, x_n]$,
decides whether the mapping $f: \mathbb{Q}^n \rightarrow \mathbb{Q}$ is surjective,
respectively, injective? --
And what is the answer if $\mathbb{Q}$ is replaced by $\mathbb{Z}$? The motivation for this question is Jonas Meyer's comment on the question Polynomial bijection from $\mathbb{Q} \times \mathbb{Q}$ to $\mathbb{Q}$ which says that the explicit determination of an injective polynomial mapping
$f: \mathbb{Q}^2 \rightarrow \mathbb{Q}$ is already difficult, and that
checking whether the polynomial $x^7+3y^7$ is an example is also. Added on Aug 8, 2013: SJR's nice answer still leaves the following 3 problems open: Is there at all an injective polynomial mapping from $\mathbb{Q}^2$ to $\mathbb{Q}$? Would a positive answer to Hilbert's Tenth Problem over $\mathbb{Q}$ imply that
surjectivity of polynomial functions $f: \mathbb{Q}^n \rightarrow \mathbb{Q}$ is
algorithmically decidable? Hilbert's Tenth Problem over $\mathbb{Q}$.
|
We treat all four problems in turn. In all that follows $n>1$. Surjectivity over $\mathbb{Q}$: If there is an algorithm to test whether an arbitrary polynomial with rational coefficients is surjective as a map from $\mathbb{Q}^n$ into $\mathbb{Q}$ then Hilbert's Tenth Problem for $\mathbb{Q}$ is effectively decidable. Proof: Let $g(x_1,\ldots,x_n)$ be any nonconstant polynomial with rational coefficients. We want to construct a polynomial $H$ that is surjective if and only if $g$ has a rational zero. First we define an auxillary polynomial $h$ as follows;
$$h(y_1,\ldots,y_6):=y_1^2+(1-y_1y_2)^2+y_3^2+y_4^2+y_5^2+y_6^2.$$
The point of the definition is that $h(\mathbb{Q}^6)$ is precisely the set of positive rationals. This follows from Lagrange's four-square theorem and from the fact that $y_1^2+(1-y_1y_2)^2$ is never 0 but takes on arbitrarily small positive values at rational arguments. Next, let $a$ be any positive rational such that $a$ is not the square of a rational, and such that for some tuple $b\in \mathbb{Q}^n$, it holds that $g(\bar{b})^2<a$. Define the polynomial $H$ as follows:
$$H(x_1,\ldots,x_n,\bar{y}):=g(\bar{x})^2(g(\bar{x})^2-a)h(\bar{y}).$$
Of the three factors that make up $H$, the only one that can vanish is $g(\bar{x})^2$. Therefore if $H$ is surjective then $g$ has a rational zero. Conversely if $g$ has a rational zero then $H$ is surjective: Obviously $H$ takes on the value 0. To obtain any rational $r\ne 0$ as a value of $H$, choose $\bar{b}\in \mathbb{Q}^n$ such $g(\bar{b})^2-a$ has the same sign as $r$ and such that $g(\bar{b})\ne 0$, and then choose values for the tuple $\bar{y}$ so that $h(\bar{y})$ is whatever positive rational it needs to be. Surjectivity over $\mathbb{Z}$: There is no algorithm to test surjectivity of a polynomial map $f:\mathbb{Z}^n\to \mathbb{Z}$. The proof is by reduction to Hilbert's Tenth Problem.
Let $g(x_1,\ldots,x_n)$ be a polynomial with integer coefficients. Then $g$ has an integral zero if and only if $h:=x_{n+1}(1+2g(x_1,\ldots,x_n)^2)$ is surjective. For if $g$ has an integral zero $\bar{a}$, then $h(x_1,a_1\ldots,a_n)=x_1$: therefore $h$ is surjective. Conversely, if $h$ is surjective then choose $\bar{a}\in \mathbb{Z}^n$ such that $h(\bar{a})=2.$ Then $a_{n+1}(1+2g(a_,\ldots,a_n)^2)=2$, which is possible only if $g(\bar{a})=0$. Injectivity over $\mathbb{Z}$: There is no algorithm to test injectivity (also by reduction to HTP). We shall make use of the non-obvious fact that there are polynomials $\pi_n$ mapping $\mathbb{Z}^n$ into $\mathbb{Z}$ injectively. Such maps are constructed in a paper by Zachary Abel here. Let $g(x_1,\ldots,x_n)$ be a polynomial with integer coefficients. Let $h$ be the polynomial $gg_1$, where $g_1$ is obtained by substituting $x_1+1$ for $x_1$ in $g$. The point of this definition is that $g$ has an integral zero if and only if $h$ has at least two different integral zeros. Define the polynomial $H(x_1\ldots,x_n)$ as follows: $$H(\bar{x}):=\pi_{n+1}(x_1h(\bar{x}),\ldots,x_nh(\bar{x}),h(\bar{x})).$$ We claim that $g$ has an integral zero if and only if the polynomial $H(\bar{x})$ does not define an injective map from $\mathbb{Z}^n$ into $\mathbb{Z}$. This gives the reduction of the injectivity problem to Hilbert's Tenth Problem. To prove the claim, suppose, for the left-to-right implication, that $g$ has an integral zero $\bar{a}$. Then $h(\bar{a})=0$, and $h$ has a different integral zero, call it $\bar{b}$. But then
$$H(\bar{a})=H(\bar{b})=\pi_{n+1}(\bar{0}),$$
so $H$ is not injective. For the right-to-left implication, suppose that $H$ is not injective, and fix two different tuples $\bar{a},\bar{b}\in \mathbb{Z}^n$ such that $H(\bar{a})=H(\bar{b})$. Since $\pi_{n+1}$ is injective, the following equations hold:
\begin{align*}
a_1h(\bar{a})&=b_1h(\bar{b})\\
&\,\vdots\\
a_nh(\bar{a})&=b_nh(\bar{b})\\
h(\bar{a})&=h(\bar{b})
\end{align*}
If $h(\bar{a})$ was not 0, then by dividing each of the first $n$ equations by $h(\bar{a})$, it would follow that the tuples $\bar{a}$ and $\bar{b}$ were identical, a contradiction. So $h(\bar{a})=0$, hence $g$ has an integral zero. Injectivity over $\mathbb{Q}$: The same technique that we used over $\mathbb{Z}$ works perfectly well, assuming that we have polynomials $\pi_n$ mapping $\mathbb{Q}^n$ into $\mathbb{Q}$ injectively. The existence of such polynomials is, it seems, an open question. But if there are no such polynomials then the decision problem for injectivity disappears!
|
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|
138,218 |
I want to know the historic reasons behind singling out Cohen-Macaulay rings as interesting algebraic objects. I'm reviewing my previous lecture notes about Cohen-Macaulay rings because now I'm studying about Stanley-Reisner rings and I think I need to have a better general understanding about why I need to study CM rings.
|
I think there are many reasons. Here are a few. Practical reasons Cohen-Macaulay rings are just plain easier to work with. Computations in local cohomology For example, any number of computations in local cohomology modules become much easier in the Cohen-Macaulay case (see for example Bruns and Herzog's book on the topic). Explicitly, it's much easier to determine if a class in $z \in H^{\dim R}_{\mathfrak{m}}(R)$ is zero or not in the case that $R$ is Cohen-Macaulay. Duality Both Grothendieck-local and Grothendieck-Serre duality work much better in Cohen-Macaulay rings. The dualizing complex (assuming it exists) is a complex whose first non-zero cohomology is the canonical module and which is equal to this (shifted) canonical module if and only if the ring is Cohen-Macaulay. Without this hypothesis one frequently needs to work in the derived category and do numerous computations with spectral sequences. It is convenient to not have to. Vanishing and exactness If $R$ is Cohen-Macaulay and $I$ is a height-one ideal (and suppose the rings are quotients of Gorenstein/regular rings so they have dualizing complexes). Then we have a surjection of canonical modules $\text{Hom}_R(I, \omega_R) = \omega_R(I^{-1}) \to \omega_{R/I}$. This is surjective because the next term is zero when $R$ is Cohen-Macaulay. This sort of vanishing applies to more general situations and is really useful (there is a local dual version involving local cohomology). There are lots of other vanishing results that you can deduce from this kind too. Ubiquity of Cohen-Macaulay rings Ok, if Cohen-Macaulay rings weren't so common, the above nice properties would be less interesting. But Cohen-Macaulay rings are really common. Here are some examples. Summands of regular (or Cohen-Macaulay) rings If $R \subseteq S$ is a extension of rings and $R \to S$ splits as a map of $R$-modules, then if $S$ is Cohen-Macaulay, so is $R$ (the point is $H^i_m(R) \to H^i_{mS}(S)$ injects and the latter term is zero, at least after a little localization on $S$ if necessary). Lots of rings coming from representation theory for instance are summands of regular rings. Complete intersections Complete intersection rings are Cohen-Macaulay. Rational/log terminal/F-regular singularities A lot of classes of singularities which are most useful today are Cohen-Macaulay. One of their most useful properties is their vanishing properties (see above). Pithy quotes "Life is really worth living in a Noetherian ring $R$ when all the local rings
have the property that every s.o.p. is an $R$-sequence. Such a ring is called
Cohen-Macaulay (C-M for short)." [Page 887 of Hochster, Some applications of the Frobenius in characteristic 0 ]
|
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|
138,310 |
Lusztig and James provided conjectures for dimensions of simple modules (or decomposition numbers) for algebraic groups and symmetric groups in characteristic $p$. These conjectures have been considered almost certainly true and have guided a lot of the research in these areas for a long time. A new preprint by Geordie Williamson, (with a classily understated title: "Schubert calculus and torsion") shows that the Lusztig conjecture is false (and therefore that James' conjecture is also false). Amongst other things, this certainly implies that the existing cases in which the conjectures have been proved become more interesting (very large $p$ for Lusztig's, RoCK blocks and certain defects for James'). But my question is: what next? Do we just accept that we'll only understand algebraic groups in very large characteristic? Do we direct focus on non-abelian defect blocks for symmetric groups? Or... there's a lovely conjecture due to Doty which (though less explicit than Lusztig's conjecture) could, in principle, give a character formula for the characters of simple modules for $GL_n$ and hence symmetric groups. It states that: The modular Kostka numbers are defined as follows: $K′= [Tr^λ(E) : L(μ)]$, for
$Tr^\lambda(E)$ the truncated symmetric power and $L(\mu)$ the irreducible polynomial $GL_n$-module
of highest weight $\mu$. Then Doty’s Conjecture states that the modular Kostka matrix $K′ = (K′_{\mu,\lambda} )$, with rows and columns indexed by the set of all partitions $\lambda$ of length $\leq n$, and bounded by $n(p−1)$ (fixed in some order), is non-singular for all $n$ and all primes $p$. Could this conjecture be the new "big problem" for algebraic and symmetric groups in positive characteristic? Are there any other conjectures out there of a similar flavour? References G. Williamson, Schubert Calculus and Torsion. http://people.mpim-bonn.mpg.de/geordie/Torsion.pdf Doty, S., Walker, G., Modular Symmetric Functions and Irreducible Modular Representations of General Linear Groups, Journal of Pure and Applied Mathematics, 82, (1992), 1-26. See also page 105 of S. Martin "Schur algebras and Representation theory".
|
The questions raised here will probably need some substantial research papers to answer, inventing new approaches and methods. In any case, the question of what to do about "small" primes has been around for decades without any clear program emerging. Three logical outcomes are possible: these range from least satisfactory (no general method, just a lot of ad hoc case-by-case and prime-by-prime results) to highly satisfactory (a single all-encompassing theoretical framework, though requiring recursive calculations as at present). The third possibility lies of course somewhere in between. Meanwhile maybe I can fill in a little more of the background, community-wiki style. The general problem is to understand the $p$-modular representation theory (mainly finite dimensional) of a (simple or perhaps reductive) algebraic group $G$, in conjunction with the study of its Lie algebra $\mathfrak{g}$ and various finite subgroups of $G$ (Chevalley or twisted groups). Historically, the search for simple modules has been a natural starting point, but there are many other questions involving indecomposables, projectives, blocks, extensions, cohomology, etc. In the narrower case at hand, the focus is on the interaction of modular representations of general (or special) linear
groups and symmetric groups. But one might prefer to work in the full generality of simple algebraic groups in spite of the lesser symmetry in some root systems. (Here Jantzen's 2003 edition provides most of the needed foundations.) At the risk of oversimplification and with apologies to those whose work is slighted, I'll sketch briefly how the ideas have evolved: (1) In the early period (late 1950s into 1960s), Chevalley parametrized the simple modules $L(\lambda)$ for $G$ as in the classical theory by dominant integral weights $\lambda$ (realizing them in effect as submodules of the global
sections of suitable line bundles on a flag variety). Further study by Curtis and then Steinberg related these modules to those for $\mathfrak{g}$ when the weights are "$p$-restricted" and with those for related finite groups of Lie type. Here Steinberg's twisted tensor products account for arbitrary weights, while there is more emphasis on realizing the $L(\lambda)$ as quotients of modules obtained from characteristic 0 by reduction mod $p$ (later called Weyl modules ). The two realizations are essentially dual, a consequence of Kempf's Kodaire-type vanishing theorem for dominant line bundles proved in the 1970s. (2) In the middle period (1970s), more details and examples were filled in, along with some general theory which
often imitated the infinite dimensional representation theory of Lie algebras rather than Cartan-Weyl theory. At first some people had expected closed formulas like Weyl's for characters or weight multiplicities. Instead a version of Harish-Chandra's action of the Weyl group on weights (shifted by $\rho$) got combined with reduction mod $p$. In my emphasis on $\mathfrak{g}$ I had to omit primes dividing the index of connection, but soon Jantzen developed much better versions for $G$ and Andersen removed conditions on $p$ for the group notion of "linkage". My approach yielded an early form of BGG reciprocity for $\mathfrak{g}$ but overlooked the appearance of an affine Weyl group $W_p$ relative to $p$ (for the Langlands dual root system). That was developed by Verma around the time of the 1971 Budapest summer school on Lie groups. He also conjectured that most of the theory should be independent of the prime $p$. Improved results and examples by Jantzen and Andersen in the 1970s were complemented by cohomology results of Cline-Parshall-Scott and others. The subject became broader and more active, but still lacked a good conjecture on the character of $L(\lambda)$. (Jantzen did however realize that a solution in characteristic $p$ would contain a solution of the hard open problem for Verma modules in characteristic 0.) (3) Initiating the modern period, the landmark 1979 paper by Kazhdan-Lusztig on Hecke algebras and Coxeter groups, with its explicit conjecture on the composition factor multiplicities of Verma modules, led Lusztig to a parallel conjecture for Weyl modules in characteristic $p$. Here he recognized the need to avoid "small" $p$, which has continued to be a source of concern in the question asked here. But the main breakthrough was the realization that Kazhdan-Lusztig polynomials for an affine Weyl group (which is a Coxeter group) should supply key multiplicities. It took many years for a partial proof to be developed by Andersen-Jantzen-Soergel, which used an indirect comparison with quantum groups at a root of unity and left bounds on $p$ uncertain. (Soergel's former students Fiebig and Williamson have gone further.) Playing off general linear and symmetric groups is especially tricky for small primes, since even the optimistic lower bound on $p$ given by the Coxeter number gets arbitrarily large here. This difficulty has been appreciated by those working in the modular theory of symmetric groups. In the algebraic group situation, the built-in problem with reliance on the single affine Weyl group $W_p$ has been the involvement of higher powers of $p$ when weights are relatively small. More systematic study of small primes is needed here, to create a database. But even Jantzen's early examples are instructive. For instance, in type $B_2$ when $p=2$, one fundamental weight reflects to the lower 0 weight across a $p$-hyperplane which is also a $p^2$-hyperplane. Relative to $p$, the weight is singular, but not relative to $p^2$. This suggests the interaction of a hierarchy of affine Weyl groups for increasing powers of $p$. How complicated will this be to formulate? And can it be enough to predict all character formulas? (Maybe not.)
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|
138,321 |
Let $p(n)$ denote the number of partitions of a positive integer $n$. It seems to me that we have for all $n>25$
$$
p(n)^2>p(n-1)p(n+1).
$$
In other words, the sequence $(p(n))_{n\in \mathbb{N}}$ is log-concave, or satisfies $PF_2$, with
$$
\det \begin{pmatrix} p(n) & p(n+1) \cr p(n-1) & p(n) \end{pmatrix}>0
$$
for $n>25$. Is this true ? I could not find a reference in the literature so far. On the other hand, the partition function is really studied a lot.
So it seems likely that this is known. Similarly, property $PF_3$, with the corresponding $3\times 3$ determinant, seems to hold for all $n>221$, too, and also
$PF_4$ for all $n>657$. The question is also motivated from the study of Betti numbers for nilpotent Lie algebras, in particular filiform nilpotent Lie algebras.
|
The first two terms of the Hardy-Ramanujan formula give
$$p(n) = \frac{1}{4 \sqrt{3} n} \exp(\pi \sqrt{2n/3}) + O \left(\exp(\pi \sqrt{n/6} ) \right)$$
so
$$\log p(n) = \pi \sqrt{2/3} \sqrt{n} - \log n - \log (4 \sqrt{3}) + O(\exp(-\pi \sqrt{n/6} ) ).$$
So
$$\log p(n+2) - 2 \log p(n+1) + \log p(n) = $$
$$ \pi \sqrt{2/3} \left( \sqrt{n+2} - 2\sqrt{n+1} + \sqrt{n} \right) - \left( \log(n+2) - 2 \log(n+1) + \log n \right) + O(\exp(-\pi \sqrt{n/6} ) )$$
$$= \left[ \left( \frac{- \pi \sqrt{2/3}}{4} \right) n^{-3/2} + O(n^{-5/2}) \right] + O(n^{-2}) + O(\exp(-\pi \sqrt{n/6} ) ).$$
So this quantity is negative for $n$ sufficiently large. The larger determinants seem harder; there is probably a smarter way to do this. With the help of Mathematica, I set $q(n) = a \exp(c \sqrt{n})/n$ and computed that
$$\det \begin{pmatrix} q(n) & q(n+1) & q(n+2) \\ q(n-1) & q(n) & q(n+1) \\ q(n-2) & q(n-1) & q(n) \end{pmatrix} = q(n)^3 \left( \frac{c^3}{32 n^{9/2}} + O(n^{-10/2}) \right).$$
The error in approximating $p(n)$ by $q(n)$ (for $a = 1/(4 \sqrt{3})$ and $c = \pi \sqrt{2/3}$) will be exponentially smaller than $n^{-9/2}$, so the $3 \times 3$ determinant is positive for $n$ large. The $4 \times 4$ determinant vanishes to order at least $n^{-12/2}$, and I gave up waiting for the computation to finish when I asked for more terms.
|
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|
138,348 |
I am interested to know other examples vacuously true statements that are non-trivial. My starting example is Turan's result in regards to the Riemann hypothesis, which states Suppose that for each $N \in \mathbb{N}_{>0}$ the function $\displaystyle \sum_{n=1}^N n^{-s}$ has no zeroes for $\mathfrak{R}(s) > 1$. Then the function
$$\displaystyle T(x) = \sum_{1\leq n \leq x} \frac{\lambda(n)}{n}$$
is non-negative for $x \geq 0$. In particular, this would imply the Riemann hypothesis. Here $\lambda(n) = \lambda(p_1^{a_1} \cdots p_r^{a_r}) = (-1)^r$ is the Liouville function. The interesting thing about this statement is that both the hypothesis and the consequence can be proven false independently. In particular, Montgomery showed in 1983 that for all sufficiently large $N$ the above sums have zeroes with real parts larger than one, and Haselgrove showed in 1958 that $T(x)$ is negative for infinitely many values of $x$. Peter Borwein et al. found the smallest such $x$ in 2008. I find this result fascinating because it relates to a well-known conjecture, and both the hypothesis and consequence were plausible. Are there any other mathematical facts of this nature, perhaps in other areas?
|
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|
138,454 |
Let $(X,d)$ be a compact connected metric space with the property that for any distinct points $a,b$, $X\backslash \lbrace a,b\rbrace$ is disconnected. Clearly the unit circle has this property. Is there any other example (up to isomorphism) ??
|
This property indeed characterizes the circle, but this is not obvious. This was shown by R. L. Moore, according to Sam Nadler's Continuum Theory p. 156. Added: the precise reference is [522] in this historical survey of continuum theory.
|
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|
138,478 |
Let $R$ be a commutative ring with identity and let $f \in R[x]$. There are well known characterizations for $f$ to be a nilpotent element of $R[x]$ or to have a multiplicative inverse in $R[x]$. Is there any characterization for idempotent elements in $R[x]$ ?
|
Let $f = a_0 + a_1x + ... + a_nx^n$ be idempotent. Then $a_0^2 = a_0$. Also
$a_0a_1 + a_1a_0 = a_1$. Multiply by $a_0$ to get $a_0a_1 = 0$ which
means that $a_1 = 0$ and by induction it is easy to show that $a_2 = ... = a_n = 0$
Therefore $f$ is idempotent iff its constant term is idempotent and other coefficients are
zero. Note that this is not true if we drop the commutativity condition. For example
consider the polynomial $f(x) = \begin{pmatrix}
1 & 0 \\
0 & 0 \\
\end{pmatrix} + \begin{pmatrix}
0 & 1 \\
0 & 0 \\
\end{pmatrix}x $ in $M_2(\Bbb{R})[x]$ which is clearly an idempotent polynomial.
|
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|
139,105 |
Can a (finite) collection of disjoint circle arcs in $\mathbb{R}^3$ be interlocked in the sense in that they cannot be separated, i.e. each moved arbitrarily far from one another while remaining disjoint (or at least never crossing) throughout?
(Imagine the arcs are made of rigid steel; but infinitely thin.)
The arcs may have different radii; each spans strictly less than $2 \pi$ in angle, so each has a positive "gap" through which arcs may pass: Of course, if one could prove that in any such collection, one arc can be removed to infinity, the result would follow by induction.
But an impediment to that approach is that sometimes there is no arc than can be removed while all the others remain fixed. Another approach would be to reduce the piercing number of the configuration:
the number of intersections of an arc with the disks on whose boundary the arcs lie. If the piercing number could always be reduced in any configuration, then it would "only" remain to prove that if there are no disk-arc piercings at all, the configuration can be separated. Intuitively it seems that no such collection can interlock, but I am not seeing a proof.
I'd appreciate any proof ideas—or interlocked configurations!
|
I believe there is no such locked configuration. The proof is by induction, as you suggest. Pick any arc and imagine moving it to infinity. Of course, to do this, it will have to pass through some other arcs, and thus this is not a valid motion. We can, however, by picking our motion "generically", ensure that there are just finitely many times when our arc passes through another arc, and that at each of these times, it passes through exactly one other arc at exactly one point. But now if we rotate (in the plane of the circle) the arc during the motion, we can ensure that it's "gap" is moved to each of the points where it used to pass through another arc. Thus we have turned our invalid motion into a valid one.
|
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|
139,140 |
The original post is below. Question 1 was solved in the negative by David Speyer, and the title has now been changed to reflect Question 2, which turned out to be the more difficult one. A bounty of 100 is offered for a complete solution. Original post. It follows from the prime number theorem and the periodicity properties $f(n+p) \equiv f(n) \mod{p}$ that for each $A < e$ there are only finitely many integer polynomials $f \in \mathbb{Z}[x]$ such that $|f(n)| < A^n$ for all $n \in \mathbb{N}$. On the other hand, for each $k \in \mathbb{N}$ the binomial coefficient $\binom{n}{k}$ is an integer- valued polynomial in $n$ bounded by $2^n$. Question 1. Are there infinitely many integer polynomials with $|f(n)| < e^n$ for all $n \in \mathbb{N}$? Question 2. Given $A < 2$, are there only finitely many integer-valued polynomials $f \in \mathbb{Q}[x]$ with $|f(n)| < A^n$ for all $n \in \mathbb{N}$?
|
Question 2:
The constant $A$ can be brought down to $\sqrt 3$, and probably
a bit below that but not all the way down to $1+\epsilon$. Instead of the polynomial $f(n) = {n \choose m}$, use a
finite difference of such polynomials,
$$
f(n) = \sum_{i=0}^m (-1)^i {m \choose i} {n \choose m+i}.
$$
This is the $x^n$ coefficient of
$$
\frac1{1-x} \left( \frac{x(1-2x)}{(1-x)^2} \right)^m
$$
and can be estimated by contour integration on $|x| = 3^{-1/2}$;
the maximum of $|f(n)|^{1/n}$ occurs near $n=3m$, for which
the critical points are at $x = (3 \pm \sqrt{-3}) / 6$. Note that for $f(n) = {n \choose m}$ the generating function was
$\frac1{1-x} (x/(1-x))^m$, and the maximum occurred near $n=2m$,
for which the critical point was at $x = 1/2$.
The factor $1-2x$ kills that maximum, and it seems that
using $x/(1-x)$ and $(1-2x)/(1-x)$ to the same power is optimal.
To reduce $A$ further, try to include also some power of $(3x^2-3x+1)/(1-x)^2$
to kill the new critical point.
|
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|
139,267 |
Since Boole it is known that probability theory is closely related to logic. According to the axioms of Kolmogorov, probability theory is formulated with a (normalized)
probability measure $\mbox{Pr}\colon \Sigma \to [0,1]$ on a Boolean
$\sigma$-algebra $\Sigma$ (of events). Realizing these data by a set $X$ (sample space of elementary events) and a corresponding $\sigma$-algebra $\Sigma(X)\subseteq P(X)$ of subsets of $X$, one obtains a probability space $(X,\Sigma(X),\mbox{Pr})$. The $\sigma$-homomorphisms $f \colon {\cal B}({\mathbb R})\to \Sigma$ (real $\Sigma$-valued measures) are defined on the Borel $\sigma$-algebra
${\cal B}({\mathbb R})$ of the real Borel sets. They can be realized by real-valued measurable functions $F\colon X\to {\mathbb R}$ (random variables). I wonder how this theory extends from the classical to the intutionistic logic i.e. from the Boolean to the Heyting ($\sigma$-)algebras and what the major differences between the two theories are. Where can I find precise descriptions of the following topics: Definition and properties of probability measures on a Heyting algebra ${\cal H}$. Definition and properties of real ${\cal H}$-valued measures $f \colon {\cal B}({\mathbb R})\to {\cal H}$. (Already the discrete case would be of interest.) (BTW: Boole 1815–1864; Heyting 1898–1980; Kolmogorov 1903–1987)
|
The title of the question is a bit of a misnomer, or at least has the potential to cause confision. "Intutionistic probability theory" means to me "theory of probability developed in intuitionistic logic". But you seem to be asking whether we can replace $\sigma$-algerbas (which are Boolean algebras) with Hetying algebras. Regarding your first question, the relevant notion is that of a valuation . This is like a real-valued measure, but it is defined only on the open sets instead of Borel sets. The open sets form a complete Heyting algebra, so this should provide you with a starting point. Regarding your second question about Heyting-algebra-valued measures on Borel sets, I would just like to observe that a homomorphism $f : \mathcal{B}(\mathbb{R}) \to \mathcal{H}$ preserves complements (because it preserves $0$, $1$, $\land$ and $\lor$), which means that it factors through the regularization of $\mathcal{H}$ (the Boolean algebra consisting of those elements of $\mathcal{H}$ that are closed under double complement). So you might as well replace $\mathcal{H}$ with its Boolean algebra of regular elements, or else consider instead $\mathcal{H}$-valued valuations $\mathcal{O}(\mathbb{R}) \to \mathcal{H}$.
|
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|
139,388 |
Can anyone give an example of an unnatural isomorphism? Or, maybe, somebody can explain why unnatural isomorphisms do not exist. Consider two functors $F,G: {\mathcal C} \rightarrow {\mathcal D}$. We say that they are unnaturally isomorphic if $F(x)\cong G(x)$ for every object $x$ of ${\mathcal C}$ but there exists no natural isomorphism between $F$ and $G$. Any examples? Just to clarify the air, $V$ and $V^\ast$ for finite dimensional vector spaces ain't no gud: one functor is covariant, another contravariant, so they are not even functors between the same categories. A functor should mean a covariant functor here.
|
For a simpler, but arguably more artificial, example than Mark's, take $\mathcal{C}$ to be the category with one object and two morphisms. Then the identity functor $\mathcal{C}\to\mathcal{C}$ is "unnaturally isomorphic" to the functor that sends both morphisms to the identity map.
|
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|
139,575 |
I use Magma to calculate the L-value, yields E:=EllipticCurve([1, -1, 1, -1, 0]);
E;
Evaluate(LSeries(E),1),RealPeriod(E),Evaluate(LSeries(E),1)/RealPeriod(E); Elliptic Curve defined by y^2 + x*y + y = x^3 - x^2 - x over Rational Field
0.386769938387780043302394751243 3.09415950710224034641915800995
0.125000000000000000000000000000 $#torsionsubgroup = 4, c_{17}(E)=1.$ But the strong BSD predicts that $L(E,1)/\Omega_{\infty}$ = $(#Sha(E)/#tor(E)^2)*c_{17}(E)$ We will get $L(E,1)/\Omega_{\infty}=1/16$ , not $1/8$ .
Why does that happen? Thanks a lot.
|
To expand my comment, there are at least 3 subtle ways to get BSD wrong: 1) The BSD period over ${\mathbb Q}$ is the real period $\Omega_\infty$ when $E({\mathbb R})$ is connected ($\Delta(E)<0$) and $2\Omega_\infty$ when it has two connected components ($\Delta(E)>0$). The same thing happens over number fields, at every real place. So in your example you have to divide the $L$-value by $2\Omega_\infty$ to get $1/16$. Another way (alternative) of phrasing this: is that the Tamagawa number at an Archimedean place is $2$ if real and split ($\Delta>0$), and $1$ otherwise. It is less easier to forget to include it then, and you can always use $\Omega_\infty$. Magma does not have Tamagawa numbers at infinite places directly. The other two concern BSD over number fields: 2) The height pairing in BSD depends on the ground field. For instance, $E=37A1$ has $E({\mathbb Q})={\mathbb Z}\cdot P$ and $E({\mathbb Q(i)})={\mathbb Z}\cdot P$, where $P$ is the same point $(0,0)$. The regulator is $\approx 0.051$ over ${\mathbb Q}$ but $\approx 2\cdot 0.051=0.102$ over ${\mathbb Q(i)}$; the factor $2$ is $[{\mathbb Q(i)}:{\mathbb Q}]$. 3) Over ${\mathbb Q}$ there is this luxury of having a global everywhere minimal model, so there is a canonical differential to integrate. Over number fields you cannot do this, so one usually takes any invariant differential and introduces a correction term that measures its failure to be minimal at all primes. The point is that if you start with a curve over ${\mathbb Q}$ with additive reduction at $p$ and go up to a number field $K$ where $p$ ramifies, e.g. $50A3$ over $K={\mathbb Q}(\sqrt 5)$, the minimal model might stop being minimal, and this correction factor comes in for BSD over $K$. The functions ConjecturalRegulator and ConjecturalSha in Magma take care of these normalizations - it's actually quite nice to experiment with them. Hope this helps. P.S. You would not believe how many times each of these mistakes was made!
|
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|
139,607 |
All mathematicians are used to thinking that certain theorems are deep, and we would probably all point to examples such as Dirichlet's theorem on primes in arithmetic progressions, the prime number theorem, and the Poincaré conjecture. I am planning to give a talk on the history of "depth" in mathematics, and for
that reason I would like to have a longer list of examples and, if possible, some thoughts about what makes them deep. Most examples, I expect, will be from after 1800, but I am also interested in examples before that date. When it comes to the meaning of "depth," I am interested in both specific and general explanations. In specific cases, one might point to the introduction of unexpected methods, such as analysis in Dirichlet's theorem, or differential geometry in the Poincaré conjecture, which are not implicit in the statement of the theorem.
In most cases, it is probably not provable that these methods are necessary (e.g. there are "elementary" proofs of Dirichlet's theorem), but in some cases it is provable, by general theorems of logic. Both types of explanation are welcome. Update. I am a little surprised that nobody mentioned reverse
mathematics, which seems to offer a precise sense in which certain
theorems are "equally deep." For example, on pages 36--37 of
Simpson's Subsystems of Second Order Arithmetic there is a list
of 14 theorems, including the Brouwer fixed point theorem and
Riemann integrability of continuous functions, which are equally
deep in a precise sense. Admittedly, these are not the deepest
theorems around, but they're not shallow either. Later in the book
one finds other results of equal, but greater, depth. How do MO members
view such results?
|
There are many possible meanings of word "deep" one can detect in the common speech. I'll list three good and three bad but I do not pretend the list is anywhere near complete. 1) Very difficult (Fermat-Wiles, Carleson, Szemeredi, etc.) These theorems usually stand as testing tools for our methods and we can measure the development of the field by how easily they can be derived from the "general theory". Their "depth" in this sense deteriorates with time albeit slowly. 2) Ubiquitous (Dirichlet principle, maximum principles of all kinds). They may be easy to prove but form the very basis of all our mathematical thinking. This depth can only grow with time. 3) Influential (Transcedence of $e$, Furstenberg's multiple recurrence.) This meaning relates not as much to the statement as to the proof. Some new connection is discerned, some new technical tool becomes available, etc. 1') With an ugly proof (4 color, Kepler's conjecture). They usually reflect our poor understanding of the matter 2') Standard black boxes used without understanding ("By a deep theorem of..." something trivial and requiring no such heavy tool follows.) They are used to produce junk papers on a conveyor belt and create high citation records. 3') Hot (I'll abstain from giving an example here to avoid pointless discussions). They reflect the current fashions and self-promotion.
|
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139,668 |
Let $X$ be a non-compact metric space (though if the answer to the question is positive, then it probably also holds for more general spaces like, e.g., paracompact Hausdorff) and $E \to X$ a vector bundle over it. Suppose that over every compact subset $K \subset X$ the restricted bundle $E|_K$ is trivial. Can we conclude that $E$ is globally trivial?
|
As Igor Belegradek showed in the comments, one could find an example by finding a CW-complex $X$ and a map $X \to BO(n)$ which is not nullhomotopic, but where the restriction to every finite subcomplex is nullhomotopic. Such a map is called a phantom map. The question "is this map nullhomotopic?" has the same answer whether or not we are asking our maps to preserve the basepoint, and so I will take some steps that are casual about basepoints. For our example, we're going to take $n = 3$ and $X = \Sigma \mathbb{CP}^\infty$, the suspension of $\mathbb{CP}^\infty$. This is a CW-complex whose finite subcomplexes are $\Sigma \mathbb{CP}^n$. These spaces are simply connected, so $[\Sigma \mathbb{CP}^n, BO(3)] = [\Sigma \mathbb{CP}^n, BSO(3)]$ for all $n \leq \infty$. Then $[\Sigma \mathbb{CP}^n,BSO(3)] = [\mathbb{CP}^n, SO(3)]$ for all $n \leq \infty$ by the loop-suspension adjunction. ("A vector bundle on a suspension is determined by a clutching function.") We can also identify $SO(3)$ with $\mathbb{RP}^3$, which has $S^3$ as a double cover. Again because $\mathbb{CP}^n$ is simply connected, $[\mathbb{CP}^n,SO(3)] = [\mathbb{CP}^n, S^3]$ for all $n \leq \infty$. One of the famous examples of phantom maps is a map constructed by Brayton Gray: a map $\mathbb{CP}^\infty \to S^3$ which is not nullhomotopic, but where the restriction to $\mathbb{CP}^n$ is nullhomotopic for any $n$. (I believe that this is in his paper "Spaces of the same $n$-type, for all $n$", and that a proof can be given using Milnor's $\lim^1$ sequence.) Pushing this back, we get a vector bundle on $\Sigma \mathbb{CP}^\infty$ whose restriction to any finite subcomplex is trivial.
|
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140,327 |
Arnold, in his paper
The underestimated Poincaré, in Russian Math. Surveys 61 (2006), no. 1, 1–18
wrote the following: ``...Puiseux series, the theory which Newton, hundreds of years before Puiseaux,
considered as his main contribution to mathematics (and which he encoded as a second,
longer anagram, describing a method of asymptotic study and solution of all equations,
algebraic, functional, differential, integral etc.)...'' Arnold says this is several other places as well. As I understand, the "first anagram" is this 6accdae13eff7i3l9n4o4qrr4s8t12ux You can type this on Google to find out what this means. Or look in Arnold's other popular books and papers. Question: what is the "second anagram" Arnold refers to? P.S. This was my own translation from Arnold's original. The original is available free
on the Internet, but the translation is not accessible to me at this moment. I hope my translation is adequate. P.P.S. I know the work of Newton where he described Puiseux series, probably it was unpublished. But there is no anagram there.
|
Newton's anagram on his method to solve differential equations is contained in his letter to Leibniz dated October 24, 1676, as described here At the end of his letter Newton alludes to the solution of the
"inverse problem of tangents," a subject on which Leibniz had asked
for information. He gives formulae for reversing any series, but says
that besides these formulae he has two methods for solving such
questions, which for the present he will not describe except by an
anagram which, being read, is as follows, "Una methodus consistit in
extractione fluentis quantitatis ex aequatione simul involvente
fluxionem ejus: altera tantum in assumptione seriei pro quantitate
qualibet incognita ex qua caetera commode derivari possunt, et in
collatione terminorum homologorum aequationis resultantis, as eruendos
terminos assumptae seriei." You can read the anagram in the published letter: 5accdæ10effh11i4l3m9n6oqqr8s11t9y3x:
11ab3cdd10eæg10ill4m7n6o3p3q6r5s11t8vx,
3acæ4egh5i4l4m5n8oq4r3s6t4v,
aaddæcecceiijmmnnooprrrsssssttuu. as well as the translation of the latin text: "One method consists in extracting a fluent quantity from an equation at the same time involving its fluxion; but another by assuming a series for any unknown quantity whatever, from which the rest could conveniently be derived, and in collecting homologous terms of the resulting equation in order to elicit the terms of the assumed series". A curiosity: the anagram is flawed, there are two i 's too few and one s too many. Errare humanum est.
|
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140,459 |
There is a long tradition of mathematicians remarking that FLT in itself is a rather isolated claim, attractive only because of its simplicity. And people often note a great thing about current proofs of FLT is their use of the modularity thesis which is just the opposite: arcane, and richly connected to a lot of results. But have there been uses of FLT itself? Beyond implying simple variants of itself, are there any more serious uses yet? I notice the discussion in Fermat's Last Theorem and Computability Theory concludes that one purported use is not seriously using FLT.
|
Corollary 3.17 in this paper of Stefan Keil uses FLT for exponent 7 to show that if $E/\mathbb{Q}$ is an elliptic curve with a rational 7-torsion point $P$, and $E\rightarrow E'$ is the 7-isogeny with kernel $\langle P\rangle$, then $E'(\mathbb{Q})[7]=0$. There are of course lots of ways of proving this, but the paper does it by writing down a parametrisation of all elliptic curves over $\mathbb{Q}$ with 7-torsion and of their rational 7-isogenies, and then playing with parameters to get a contradiction to FLT.
|
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140,655 |
what are the examples of elliptic curves defined over $\mathbb{Q}$ with supersingular reduction at a prime $p$ and having a $p$-isogeny over $\mathbb{Q}$ ?
|
In fact, this cannot happen: an elliptic curve over $\mathbb{Q}_p$ is supersingular if and only if its associated mod $p$ Galois representation is irreducible, but if it is irreducible as a representation of $\mathbb{F}_p[G_{\mathbb{Q}_p}]$ then it is certainly irreducible as a representation of $\mathbb{F}_p[G_{\mathbb{Q}}]$ and thus it has no $p$-isogenies. This argument doesn't work over larger fields: if $K$ is an extension of $\mathbb{Q}_p$ it's no longer true that $E/K$ is supersingular if and only if $E[p]$ is irreducible as a $\mathbb{F}_p[G_{K}]$-module, cf. this MSE question . In particular, one can have an elliptic curve $E / \mathbb{Q}$ with a $p$-isogeny and bad reduction at $p$, and a number field $F / \mathbb{Q}$ such that $E$ has good supersingular reduction at all primes of $F$ above $p$; my previous suggestion to look for points on $X_0(p)$ gives lots of examples of this.
|
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|
140,768 |
I just ran across this delightful paper by an amazing triumvirate: Paul Erdős, Frank Harary, and William Tutte. "On the dimension of a graph." Mathematika 12.118-122 (1965): 20.
( Cambridge link )
( PDF download link ) They prove that $K_n$ has dimension $n-1$, $K_{n,n}$ has dimension $\le 4$,
and the dimension of the $n$-cube $Q_n$ is $2$. And the Petersen graph has dimension $2$: Surely there must have been advances on characterizing graphs according to this concept in the last ~half-century(!).
Can anyone provide some updates the status on this notion? Update . Here is a modern, metrically accurate drawing of the Petersen graph: (Image from Wikipedia: Unit Distance Graph )
|
There are several well-known and hard problems in discrete geometry that concern the dimension of a graph. For example, the unit distance problem asks for the maximum number of edges of a graph on $n$ vertices of dimension 2. Erdős conjectured in the 1940s that the answer is $n^{1+o(1)}$, but the best known bound is only $O(n^{4/3})$. The chromatic number of the plane is another famous problem about graphs of dimension $2$. This question asks for the maximum chromatic number of any graph with dimension $2$. The answer is only known to be between $4$ and $7$, and it has been stuck that way for more than five decades. Shelah and Soifer speculate in a series of papers that the answer might depend on the axioms for set theory. This is indeed the case for some other distance graphs they construct. A related result from Paul O'Donnell's thesis is that there are graphs of dimension $2$, chromatic number $4$, and arbitrarily large girth. It is unlikely that graphs of dimension $2$ will be characterized in the near future.
|
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|
140,954 |
Not so long ago I took a class called "Discrete analysis". I remember that I couldn't find any "novice" level material on Mobius functions in combinatorics. So then I went to the roots and read Rota's original paper "On the foundations of combinatorial theory I" and it really impressed me. So I wonder is there other mathematical subjects that it would be better for novice to get started with by reading rather original papers than actual books? ADDED: Thanks for your answers. That's really interesting!
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Very recently I and Misha Sodin had a strong incentive to learn the Ito-Nisio lemma (which, roughly speaking, says that weak convergence in probability of a series of symmetric independent random variables with values in a separable Banach space implies almost sure norm convergence to the same limit). The textbooks we could find fell into 2 categories: those that didn't present the proof at all and those presenting it on page 2xx as a combination of theorems 3.x.x, 4.x.x, 5.x.x, etc. The original paper is less than 10 pages long, essentially self-contained, and very easy to read and understand. The moral is the same as Boris put forth: the books are there to optimize the time you need to spend to learn the whole theory. However, for every particular implication A->B the approach they usually take is something like E->F->G, G->F, (F and Q)->B; since A->E, then A->G; once we know G, we have F, so it suffices to prove that A->Q to show that A->B; we show that Q,R,S,T,U are equivalent, with the trivial implication S->Q left to the reader as an exercise; finally, we prove that A->S. So if all you need is A->B, you may be much better off reading the paper whose only purpose is to prove exactly that.
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141,100 |
Let $p_1\ p_2\ \ldots$ be the sequence of all natural prime numbers. There is a slight (just slight) but clear tendency for imitating the number of primes in an interval $(p_k;\ p_n)$ by the number of primes in the double interval $(p_k\!+p_{k+1};\ p_{n-1}\!+p_n)$; possibly by $(2\cdot p_k; 2\cdot p_n)$ too. Let me ask two open questions along this line. The first one will be most likely hopeless while the second one may lead to a discussion and at least to numerical computations. P1. Does there exist a natural number $d$ such that for every natural number $n$ the real interval $$ (2\cdot p_n;\ 2\cdot p_{n+d})$$ contains at least one prime? P2. (when P1 fails): Given a natural number $d$, let $w(d)$ be the least natural number such that the interval of P1 (see above) does not contain any prime number. What is the growth of the sequence $$w(1)\ \ w(2)\ \ w(3)\ \ldots$$ The above notions got shifted from my original definition by a half of a prime. The question Q1 below is still equivalent to question P1 above: Q1. Does there exist a natural number $d>1$ such that for every natural number $n$ the real interval $$ (p_n\!+p_{n+1};\ p_{n+d-1}\!+p_{n+d})$$ contains at least one prime? Q2. (when Q1 fails): Given a natural number $d>1$, let $v(d)$ be the least natural number such that the interval of Q1 (see above) does not contain any prime number. What is the growth of the sequence $$v(1)\ \ v(2)\ \ v(3)\ \ldots$$ EXAMPLE Consider the consecutive primes $$p_{360} = 1901 \qquad p_{361}=1907 \qquad p_{362}=1913$$ Then the real interval $$(p_{360}\!+p_{361};\ p_{361}\!+p_{362})\ \ =\ \ (3802; 3820)$$ contains no primes, i.e. $v(2)\le 360$. In general, I'd be interested in similar relative properties of primes, where primes are studied in relations to other primes, and the relation is not trivial, meaning not reduced to general properties between integers.
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The answer to P1 is negative, thanks to the recent work of Maynard on bounded gaps between primes. What Maynard shows is that given any $d$, there exists a k-tuple $h_1,\dots,h_k$ such that for infinitely many $n$, at least $d+1$ of $n+h_1,\dots,n+h_k$ are prime. In fact, the argument shows that for sufficiently large $x$, the number of $n \in [x,2x]$ such that at least $m$ of $n+h_1,\dots,n+h_k$ are prime, and the rest are almost prime in the sense that they have no prime factor less than $x^\varepsilon$ for some small fixed $\varepsilon>0$, is $\gg \frac{x}{\log^k x}$. (Such strengthenings of Zhang/Maynard type theorems are discussed in this paper of Pintz , and also in this Polymath8b preprint .) A standard upper bound sieve (e.g. Selberg sieve or beta sieve) then shows that after removing about $O( x/\log^{k+1} x)$ of these $n$, one can also ensure that none of the numbers between $2(n+h_1)$ and $2(n+h_k)$ are prime. If we let $p_i$ be the first prime greater than or equal to $n+h_1$, then we have $2(n+h_1) \leq 2p_i \leq 2p_{i+d} \leq 2(n+h_k)$, and so we obtain a counterexample to P1 for any $d$. Unfortunately we don't get an effective rate on P2 this way due to the reliance on the Bombieri-Vinogradov theorem in the work of Maynard etc. However this can likely be removed by following the ideas mentioned near the end of this paper of Pintz , though I did not attempt this. [EDIT: it looks likely that the quantitative version of Maynard's results in this recent paper of Banks, Freiberg, and Maynard will do the trick, after some small modification.]
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141,157 |
After having read Gunnar Carlsson's Topology and Data I feel enthusiastic to use some topological data analysis (TDA) methods in my current research, mostly in social sciences. We often handle huge databases and I think it can be an interesting exercise to do TDA. I wonder two things about the current state and purpose of TDA (I do understand TDA's main advantages are not for the social sciences though): Are there TDA methods than can be used to establish a relationship among a variable of interest and a set of (possible) explanatory variables? That is, to do some sort of statistical inference? Can we make predictions based on TDA tools? Any bibliographical reference or explanation about why this is not possible is appreciated.
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Let me answer the broad question first: depending on what you actually want to do, the barcode-type invariants extracted by topological data analysis could be quite useful in your work. And it doesn't take too much prior knowledge to use the TDA tools. For instance, if all you want to do is show that two datasets are qualitatively different, you can just compute their barcodes (I've written software to do this, as have others ) and calculate the difference between them. It's easy, fast and free so why not try something orthogonal and complementary to your existing techniques? The usual pipeline for TDA is as follows: starting with your data, you impose the structure of a filtered cell complex, compute the persistent homology, and output the barcode. The reason you may find it difficult to get precise answers to your questions is quite simple: everything depends on how the filtration is concocted! It is a bit of an art form to know exactly what to compute the persistent homology of, given the features that you actually care about. Here's a typical TDA approach to your first question: let the dependent variable be $x$ and the independent variables $y_1,\ldots,y_n$. Assume, by thresholding and binning if necessary, that each $y_j$ attains only finitely many states. Construct the flag complex on the $n$-partite graph whose vertex-bins correspond to values attained by the $y_j$s. Each simplex is weighted by the minimum $x$-value corresponding to the $y$-values fixed by its vertices. These weights give you a filtered simplicial complex, and generators of the $0$-dimensional persistence intervals of the super-levelset filtration tell you which configuration of $y$s correspond to which $x$ values. Regarding your second question, a lot depends on what you mean by "predictions". For two examples of using persistent homology for predictions, consider Liz Munch's PhD thesis available here . It is possible to predict -- up to an extent -- coverage failure in sensor networks modeled as devices with a given failure probability. Perhaps a modification of this model could make predictions in situations that are of interest to you.
|
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|
141,173 |
I am an undergraduate student. I am not sure if it's OK to ask this question here. I want to learn Hodge theory. But I do not know how to start it, and how much mathematics I should need before I read Deligne's paper. Is there an elementary book or note on Hodge theory for undergraduate students? Is it worthy to read Hodge's book The theory and applications of harmonic integrals ?
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I would recommend Voisin's "Hodge Theory and Complex Algebraic Geometry" as an introduction to Hodge theory--Volume I should suffice for your purposes. The book also does a bit of the Hodge theory of non-compact varieties, in Section 8.4. The relevant sections in Griffiths-Harris aren't bad either. Deligne's papers also require some familiarity with spectral sequences and homological algebra, but this can be obtained in tons of places. That said, the actual proofs of the Hodge decomposition, etc. are not so relevant for understanding Deligne's work--in my opinion, it would be a much better use of your time to understand the statements and compute some examples if this is really your goal. For example, I found Hodge theory for smooth projective curves (which works over an arbitrary base!) very illuminating--see e.g. Theorem 2 in these notes (which have relatively heavy homological algebra preliminaries). Once you have a good understanding of the statements of Hodge theory for compact Kahler manifolds, you can try using the formal functoriality properties of mixed Hodge structures to compute some examples for some open varieties (by finding nice compactifications) and singular varieties (by finding nice resolutions). From your question, you seem to view Deligne's papers as the apotheosis of Hodge theory--I would argue rather that you should focus your efforts on understanding the case of compact Kahler manifolds, and wait until you have a good feel for them before moving on to the general situations with which Deligne deals. If you are more algebraically than analytically inclined, the Deligne-Illusie paper establishing the degeneration of the Hodge-to-de Rham spectral sequence is quite beautiful and very readable--two more readable accounts are these notes of Piotr Achinger, and this account by Illusie. You may also find the comparison theorem between analytic and algebraic de Rham cohomology interesting.
|
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|
141,501 |
Definition. A topological space $X$ has the Fixed Point Property (FPP) if every continuous self-map $X\to X$ has a fixed point. Question. If $X$ and $Y$ are homotopy-equivalent compact metrizable spaces and $X$ has the FPP, does it follow that $Y$ also has FPP? Another way to put it: Can one force a fixed point for a self-map of a compact by a "non-homological" argument? I do not know an answer to this even for finite simplicial complexes, but my primary interest is in locally connected finite-dimensional compacts.
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Lovely question! Sadly, the answer is "no" in the sense that the fixed point property is not homotopy-invariant even in the category of finite polyhedra. In fact, it is also not invariant under the operations of taking products or suspensions. See the three page paper of W Lopez called " An example in the fixed point theory of polyhedra " for the construction of an explicit counterexample to your desired property as well as the two properties listed above. Basically, Lopez's construction involves two finite polyhedra $X$ and $Y$ whose wedge product has the fixed point property but whose union along an edge does not (!!). The Corollary to Theorem 3 on the second page of the linked pdf is of interest. Update (4th Oct 2015): I have also been looking for positive results lately, and one good source is Robert Brown's Handbook of Topological Fixed Point Theory (the Google book is here . Theorem 8.11 in the book is this cool result of Jerrard: Suppose $X$ and $Y$ are compact polyhedra so that the Lefschetz numbers of every self-map $X \to X$ and $Y \to Y$ are nonzero, and so that every composite $H_n(X) \to H_n(Y) \to H_n(X)$ is trivial for $n > 0$. Then $X \times Y$ has the fixed point property. So if you can decompose your space as a nice enough product of fixed point spaces, then there is some hope depending on their homology...
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|
142,205 |
I was reading about the monster group, and how hard it was to do calculations in it, and I wondered: Is there a known presentation of the monster group? I know that it is a hurwitz group, but other than that I don't know. If we have two generators a and b such that $a^2=b^3=(ab)^7=1$, what are the possible orders of $[a,b]$? I believe that the conjugacy classes are known, so if you are given the conjugacy classes of two elements x and y, can we determine the possible conjugacy classes of $xy$?
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There's a 12-generator 80-relator presentation for the Monster group. Specifically, we have 78 relators for the Coxeter group Y443: $12$ relators of the form $x^2 = 1$, one for each node in the Coxeter-Dynkin diagram; $11$ relators of the form $(xy)^3 = 1$, one for each pair of adjacent nodes; $55$ relators of the form $(xy)^2 = 1$ (commutators), one for each pair of non-adjacent nodes; together with a single 'spider' relator, $(a b_1 c_1 a b_2 c_2 a b_3 c_3)^{10} = 1$, which results in the group $M \times C_2$. We can get rid of the $C_2$ by quotienting out by an eightieth relation, $x = 1$, where $x$ is the unique non-identity element in the centre of the group. See http://www.maths.qmul.ac.uk/~jnb/web/Pres/Mnst.html for the explicit Coxeter-Dynkin diagram.
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|
142,220 |
Fermat proved that $x^3-y^2=2$ has only one solution $(x,y)=(3,5)$ . After some search, I only found proofs using factorization over the ring $Z[\sqrt{-2}]$ . My question is: Is this Fermat's original proof? If not, where can I find it? Thank you for viewing. Note: I am not expecting to find Fermat's handwritings because they may not exist.
I was hoping to find a proof that would look more ''Fermatian''.
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Fermat never gave a proof, only announced he had one (sounds familiar?). Euler did give a proof, which was flawed, see Franz Lemmermeyer's lecture notes, or see page 4 of David Cox's introduction. For a discussion why a proof along the lines set out by Fermat is unlikely to work, see this MO posting . ---- trivia ---- As a curiosity, I looked up Fermat's original text (reproduced below from his collected works), written in the margin of the Arithmetica of Diophantus: Can one find in whole numbers a square different from 25 that, when
increased by 2, becomes a cube? This would seem at first to be
difficult to discuss; and yet, I can proof by a rigorous demonstration
that 25 is the only integer square that is less than a cube by
two units. For rationals, the method of Bachet would provide an infinity
of such squares, but the theory of integer numbers, which is very
beautiful and subtle, was not known previously, neither by Bachet,
nor by any author whose work I have read.
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|
142,243 |
Terms like "impractical" and "unfeasible" are used to say the Robertson, Sanders, Seymour, and Thomas proof of the four color theorem needs computer assistance. Obviously no precise measure is possible, for many reasons. But is there an informed rough estimate what a graph theorist would need to verify the 633 reducible configurations in that proof? $10^4$ hours? $10^8$ years? I am not asking if other proofs are known. I want to know if graph theorists have an idea what scale of practicality we are talking about when we say the Robertson, Sanders, Seymour, and Thomas proof is impractical without machine assistance.
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To answer the question it is important to disentangle the proof as follows. Theorem 1. Every minimum counterexample to the 4CT is an internally 6-connected triangulation. Theorem 2. If $T$ is a minimum counterexample to the 4CT, then no good configuration appears in $T$. Theorem 3. For every internally 6-connected triangulation $T$, some good configuration appears in $T$. See the actual paper for the definitions of these terms. Theorem 1 does not require computer assistance, while Theorem 2 and Theorem 3 both do require computer assistance. According to this version of the paper , Theorem 3 can in principle be checked by hand. Indeed it is explicitly mentioned that It can be checked by hand in a few months , or a few minutes by computer (this was about 15 years ago though). I quote more on Theorem 3: For each of these five cases we have a proof. Unfortunately, they are very long (altogether about 13000 lines, and a large proportion of the lines take some thought to verify), and so cannot be included in a journal article. Theorem 2, on the other hand really requires a computer. From the same paper, The proof of Theorem 2 takes about 3 hours on a Sun Sparc 20 workstation and the proof of Theorem 3 takes about 20 minutes. Thus, given that it took a computer 9 times longer to verify Theorem 2 than Theorem 3, and Theorem 3 apparently can be verified by hand in a few months (let us define few=3), then under some very dubious assumptions we have the ballpark answer of Ballpark Answer. 30 months.
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142,395 |
By the field of constructible numbers I mean the union of all finite towers of real quadratic extensions beginning with $\mathbb{Q}$. By decidable I mean the set of first order truths in this field, in the language of 0,1, + and $\times$, is recursive. Is this field either known to be decidable, or known not to be? As of 1963 Tarski's question of whether this field is decidable was open -- so i doubt any simple adaptation of his result on real closed fields can settle this question. He conjectured that the only decidable fields were finite, real closed, or algebraically closed. See Julia Robinson, The decision problem for fields, Theory of Models (Proc. 1963 Internat. Sympos. Berkeley), North-Holland, Amsterdam, 1965, pp. 299–311. especially pages 302 and 305. Much has gone on since 1963, and Tarski's general conjecture is well refuted, but I do not find a solution to this problem.
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According to the following paper: Carlos R. Videla, On the constructible numbers, Proceedings of the American Mathematical Society Vol. 127, No. 3 (Mar., 1999), pp. 851-860. The problem has remained open at least until 1999. I think the problem is still open. In the above paper the author proves that the ring of constructible algebraic integers is first-order definable in the field of constructible numbers. The author hopes that $\mathbb{Z}$ should be definable in the ring of constructible algebraic integers and therefore his result would be a partial result towards resolving the problem negatively.
|
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142,808 |
Let us consider the following operation on positive integers: $$n=\prod_{i=1}^{k}p_i^{\alpha_i} \qquad f(n):= \prod_{i=1}^{k}\alpha_ip_i^{\alpha_i-1}$$ (Is it true that if we apply this operation to any integer multilpe times, it will eventually get into a finite cycle?) Is there a constant $K$ such that any integer will fall into a cycle after $K$ steps? Edit4: We managed to settle affirmatively the question of Mark Sapir, whether a cycle of arbitrary length exists: http://www.math.bme.hu/~kovacsi/Pub/arithmetic_derivation_v04.pdf Edit3: I proposed two questions (in retrospect, it was a minor mistake), one of them was answered. To appreciate this, i accept Mark Sapir's answer, and alter the original text by putting the unanswered stuff into parentheses. Making the answered one the main question. Edit2: István Kovács pointed out that there is a nice formula for $f(n)$ using the 'number of divisors' function: $$ f(n):= d \left( \frac{n}{ \prod_{i=1}^{n}p_i } \right) \frac{n}{ \prod_{i=1}^{n}p_i } $$ from which it fillows that for any $\varepsilon >0 , \quad f(n)=o(n^{1+\varepsilon})$. I think that the answer to the first question is yes, but to the second no. We tested the first $10000$ integers and every integer fell into a cycle after at most $6$ steps. Edit: @MarkSapir proved that the answer to the second question is no. His proof raises the (third) question: How long can such a cycle be?
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I will show that the answer to the second question is "no". Note that if the answer to the first question is "no", we are done. Hence assume that the answer is "yes" and for every number $n$, the chain eventually turns into a cycle. Take any number $n$ and consider a sequence of numbers $n, f(n)(n-1), f(f(n)(n-1))(n-2),...$, that is $a_1=n, a_{m+1}=f(a_m)(n-m)$, for every $m=1,...,n-1$. Let $A=a_n$. Let $p$ be a prime that is bigger than any number that occurs in the chain for $A$. Consider $p^n$. Then the chain for $p^n$ looks like $p^n\to a_1p^{n-1}\to ...\to a_n\to...$ and the chain for $A=a_n$ follows. That chain will never hit the first $n$ of the numbers in the chain for $p^n$, hence the chain for $p^n$ does not go into a cycle before the step number $n$. Since $n$ was arbitrary, we are done. This answers the second question. The first question seems more difficult. Edit. As @DanielSoltész pointed out, I answered a harder question about bounding the length of a pre-cycle in the chain $m\to f(m)\to\ldots$. If we want to show that there is no bound for the number of different elements in a chain, then assuming $p>n^{2^n-1}$ is enough. This leads to another reasonable question about bounding the lengths of cycles $m\to...\to m$. That question is open.
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142,841 |
Is it possible to appreciate the geometric/polytopal properties of the amplituhedron without delving into the physics that gave rise to it?
All the descriptions I've so far encountered assume familiarity with quantum field theory,
but perhaps there exist more purely geometric explications...?
If so, I would appreciate a reference—Thanks! (Image from Quanta Magazine ) (Added) . Here is a snapshot from p.64 of the paper which jc cited,
" Scattering Amplitudes and the Positive Grassmannian ."
(I am guessing that what they called the "positroid" in this paper
is either closely related or perhaps identical to what
was later named the "amplitudedron"?)
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Nima Arkani-Hamed had a series of talks at JHU roughly 6 months ago which I attended related to this topic. He discussed it at Stony Brook a little bit over a week ago (pointed out by Emilio Pisanty in the comments above in which he used the term "amplituhedron", but my understanding of this comes mostly from his earlier talks. Update: I managed to track Nima down today and get him to explain the details. Surprisingly almost everything was correct, but the loop level description has some changes. I've also made some aesthetic changes, attempting to follow the notation in Trnka's slides as closely as possible while being readable by mathematicians. Also note that Nima claimed that their paper on this would appear "very soon". Update 2: Their paper has appeared on the arXiv: http://arxiv.org/abs/1312.2007 . At first glance everything seems consistent with what I've written below. I think it should be readable to mathematicians simply by skipping the few sections which require knowledge of physics. They do find some combinatorial results therein which may be of interest, but as this answer is already rather large I'll just direct those interested to the paper directly. Everything here is over $\mathbb R$. We know that the quotient of the subset $\tilde{M}(k,n) \subset M(k,n)$ of matrices with full rank by the $GL(k)$ left action gives the Grassmannian $G(k,n)$, in which each matrix is mapped to its row space. Define $M_+ (k,n)$ to be the subset of $M(k,n)$ in which all $k \times k$ minors are positive (ordering the rows in the process). The image of $M_+(k,n)$ under this quotient is the positive Grassmannian $G_+(k,n)$. Fix a matrix $Z \in M_+(k+m,n)$ as input data, which comes from physics, but won't really affect the combinatorial structure* at all. Then there is a map $Y_{n,k,m}: G_+(k,n) \rightarrow G(k,k+m)$ given by $([C],Z) \mapsto [CZ^T]$, where brackets denote equivalence classes under the $GL(k)$ action (the requirement that $CZ^T$ has full rank is automatically satisfied if both are in their respective positive pieces). Its image $\mathcal P_{n,k,m}$ is called the tree-level amplituhedron. The combinatorial structure does depend on $n,k,$ and $m$. This case is apparently fairly well understood thanks to their work with Alexander Postnikov (according to Nima). (One of the interesting aspects of this positive-real story is that while the map $Y_{n,k,m}$ is only a rational map of algebraic varieties, its base locus doesn't intersect $G_+(k,n)$.) *I've seen it claimed that $Z$ can be taken to live in $G_+(k+m,n)$. I must admit this doesn't make sense to me because the map described above is not invariant under the right action by $GL(k+m)$. If I'm missing something obvious though than feel free to correct me. At the very least, I'm pretty sure that the above description works, but it might be a bit more redundant than necessary. As I alluded to above, $\mathcal P_{n,k,m}$ is not the full amplituhedron. Rather, it's just the tree-level case. The full amplituhedron has another nonnegative integer parameter $l$ which determines the loop order. This gives additional coordinates to points in the amplituhedron. The subregion of $M(k+2l, k+m)$ of interest which the $C$ vary through satisfies somewhat more stringent positivity constraints than those of the ordinary positive Grassmannian that we had above in the case $l=0$. I will call this $l$-positivity, though this is my own terminology. $C' = \left( \begin{matrix} C \\ C^{(1)} \\ \vdots \\ C^{(l)} \end{matrix} \right)$ with $C$ as $k \times n$ and each $C^{(i)}$ as $2 \times n$ is $l$-positive iff for any $I = \{i_1 , \ldots, i_r\} \subseteq \{1, \ldots, l\}$ (with $i_1 < i_2 < \cdots < i_r$), the submatrix $\left( \begin{matrix} C \\ C^{(i_1)}\\ \vdots \\ C^{(i_r)} \end{matrix} \right) \in M_+(k+2r,n)$ (including the case $I=\phi$), and each $C^{(i)}$ is only well-defined up to addition of elements of $C$. For convenience of notation let $\mathcal A_{n,m} = Y_{n,2,m}$. The $l$-loop amplituhedron is then the image of $\{[C'] | C'\text{ is }l\text{-positive}\} \times \{Z\}$ in $G(k,k+m) \times (G(2,k+m))^l$ by just applying $Y_{n,k,m}$ to $([C],Z)$ and $\mathcal A_{n,m}$ to each $([C^{(i)}],Z)$ (all with the same $Z$). This is called $\mathcal P_{n,k,l,m}$ (Trnka drops the $m$, presumably since $m=4$ for physics). The space that this is embedded in has no significance either combinatorially or physically. We could take it to be in the $l$-fold product of the $G(2,m)$ bundle over $G(k,k+m)$ such that the fiber at each point is those $2$-planes orthogonal to that $k$-plane. It is important to realize that the amplituhedron itself is not so much the object of interest: rather, that is a meromorphic volume form defined on it (and on the Grassmannian, or bundle over Grassmannian, in which the amplituhedron is Zariski-dense). The principal job of the amplituhedron is to help nail down this form: the form is required to be well-defined on the interior of the amplituhedron. It seems very hard to make this statement mean anything without talking about positive real parts of varieties. The parameters are important for physics, so I'll list them, but of course if you're not interested in the physics you can set them to be whatever you like. $k$ is the order in perturbation theory. $l$, the loop order. $m=4$ is the case for physics, but in principle $m$ can be any even positive integer ($m=2$ makes the loop part trivial, so $m=4$ is in some sense the first interesting case). $n$ is the number of momenta in the scattering process. $Z$ is a positive matrix that represents all of the momenta, but at least for the purpose of combinatorics the structure doesn't really depend on the choice of $Z$. There are of course cases in which the construction does not make sense; these are irrelevant for physics (e.g. $n-k < m$ is unphysical). Also, while I'm talking about physics, to get physical predictions out of the amplituhedron, there is a particular volume form which is simply integrated over the region. This volume gives the amplitude for the process.
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143,031 |
In the book "Cohomology of Groups" of Kenneth S. Brown, it is told in the introduction that Teichmüller arrived to $H^3$ in an algebraic context, i.e. that Teichmüller worked with an algebraic interpretation of $H^3$ in group cohomology. I have tried to find the original article Über die sogenannte nichtcommutative Galoische Theorie und die Relation $\xi_{\lambda,\mu,\nu}\xi_{\lambda,\mu\nu,\pi}\xi^\lambda_{\mu,\nu,\pi}=\xi_{\lambda,\mu,\nu\pi}\xi_{\lambda\mu,\nu,\pi}$ of Teichmüller, but I haven't found it. So here is my question, which is this algebraic interpretation of $H^3$ in group cohomology? Or, in another way, which was the algebraic context in which Teichmüller was working to arriving to $H^3$ ?
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Teichmüller's article appeared in 1940 in the wartime journal $\mathfrak{Deutsche}$ $\mathfrak{Mathematik}$ (of which I've seen copies in the Tata Institute library in Bombay). He is trying to extend the Galois correspondence (between subextensions of $E\mid F$ on the one hand, and subgroups of the Galois group $\mathrm{Gal}(E\mid F)$ on the other, where $E\mid F$ is a Galois extension of fields) to a non-commutative context, where the extension $E$ is replaced by an associative (but not necessarily commutative) $F$-algebra $A$. So given such an $F$-algebra $A$, he is looking for a correspondence between sub-$F$-algebras of $A$ and subgroups of the group $\mathrm{Aut}_F(A)$. An account of Teichmüller's paper can be found in the doctoral thesis ( La naissance de la cohomologie des groupes ) of Nicolas Basbois available at the archives ouvertes . Here is a quotation from this thesis: Mac Lane connaissait ce travail de Teichmüller où apparaissait un système
de facteurs d'un type inconnu jusqu'alors. Dès les premières publications
d'Eilenberg et Mac Lane sur l'homologie et la cohomologie des groupes, le
travail de Teichmüller est cité, et il devient vite manifeste qu'un des buts de
leur collaboration sera l'interprétation des résultats de Teichmüller à l'aide
de la cohomologie. Ils la mettent au point dans un papier ([73]) soumis aux
Transactions of the American Mathematical Society en mai 1947. Ce papier
est à l'origine d'une branche de la cohomologie des groupes, la cohomologie
galoisienne, qui utilise les méthodes cohomologiques pour étudier l'action
d'un groupe de Galois sur certains groupes. The Eilenberg-MacLane paper [73] can be found in the Transactions, 64 (1948) 1, 1--20 . Teichmüller's contribution is explained in the first paragraph of the introduction as follows:
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143,144 |
I hope you don't shoot me for this question.
I try to understand among other things the Ricci flow. However I have no idea of the intuition behind the definition. So my questions is: What is the intuition behind the Ricci flow? I would appreciate any illustrative example.
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See J. Hyam Rubinstein and Robert Sinclair. "Visualizing Ricci Flow of Manifolds of Revolution", Experimental Mathematics v. 14 n. 3, pp. 257–384. ( Journal link ) (Image from that paper, via Wikipedia)
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143,263 |
For any $m\in\mathbb N$, let $S(m)$ be the digit sum of $m$ in the decimal system. For example, $S(1234)=1+2+3+4=10, S(2^5)=S(32)=5$. Question 1 :Is the following true?
$$\lim_{n\to\infty}S(3^n)=\infty.$$ Question 2 :How about $S(m^n)$ for $m\ge 4$ except some trivial cases? Remark : This question has been asked previously on math.SE without receiving any complete answers, where spin proved that $\lim_{n\to\infty} \sup S(m^n )=\infty$ when $m$ is not a power of $10$. https://math.stackexchange.com/questions/501019/letting-sm-be-the-digit-sum-of-m-then-lim-n-to-inftys3n-infty Motivation : I've got the following :
$$\lim_{n\to\infty}S(2^n)=\infty.$$ Proof : The point of this proof is that there exists a non-zero number between the ${m+1}^{th}$ digit and ${4m}^{th}$ digit. If
$$2^n=A\cdot{10}^{4m}+B, B\lt {10}^m, 0\lt A,$$
then $2^n\ge {10}^{4m}\gt 2^{4m}$ leads $n\gt 4m$.
Hence, the left side can be divided by $2^{4m}$. Also, $B$ must be divided by $2^{4m}$ because ${10}^{4m}=2^{4m}\cdot 5^{4m}$. However, since
$$B\lt {10}^m\lt {16}^m=2^{4m},$$
$B$ can not be divided by $2^{4m}$ if $B\not=0$. If $B=0$, then the right side can be divided by $5$ but the left side cannot be divided by $5$. Hence, we now know that there is a non-zero number between the ${m+1}^{th}$ digit and ${4m}^{th}$ digit. Since $2^n$ cannot be divided by $5$, the first digit is not $0$. There exists non-zero number between the second digit and the fourth digit. Again, there exists non-zero number between $5^{th}$ digit and ${16}^{th}$ digit. By the same argument as above, if $2^n$ has more than $4^k$ digits, then $S(n)\ge {k+1}$. Hence,
$$n\log {2}\ge 4^k-1\ \ \Rightarrow \ \ S(n)\ge k+1.$$
Now we know that
$$\lim_{n\to\infty}S(2^n)=\infty$$
as desired. Now the proof is completed. However, I've been facing difficulty for the $m=3$ case. I've got $\lim\sup S(3^n)=\infty$. Proof : Suppose that $3^n$ has $m$ digits. Letting $l=\varphi({10}^m)+n$, then
$$3^l-3^n=3^n(3^{\varphi({10}^m)}-1).$$
Since this can be divided by ${10}^m$, we know that the last $m$ digits of $3^l$ are equal to those of $3^n$. Hence, we get $\lim\sup S(3^n)=\infty$. However, I can't get $\lim_{n\to\infty}\inf S(3^n)$. Can anyone help?
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This follows from W. M. Schmidt's Subspace theorem, which is a deep theorem in diophantine approximations generalizing Roth's to several variables. A full account of this theorem and its proof, as well as some of its striking applications, can be found in chapter 7 of Heights in Diophantine Geometry by Bombieri and Gubler. The following result, the finiteness of the number of non-degenerate solutions to the so-called "$S$-unit equation," is a straightforward application of Schmidt's theorem. (See Theorem 7.4.2 in [HIDG]): Let $S$ be a finite set of prime numbers, and fix $n \in \mathbb{N}$. Consider $\mathcal{X}$ the set of solutions to $x_1 + \cdots + x_n = 1$ in rational numbers $x_i$ of the form $\pm \prod_{p \in S} p^{a_p}$, $a_p \in \mathbb{Z}$, such that no proper subsum of $x_1+\cdots+x_n$ vanishes. Then $\mathcal{X}$ is a finite set. This implies your question immediately upon considering $S := \{2,3,5,7\}$. However, the proof of the Subspace theorem is not effective, and this only shows $S(3^n) \to +\infty$ without any lower estimate on the rate of growth. An effective lower bound on $S(3^n)$ (going to infinity with $n$) is available through Baker's method; it is due to Stewart, and the google search led me to the old MathOverflow post linked to in my comment below. I will just copy the references which Gerry Myerson supplied there: C. L. Stewart, On the representation of an integer in two different bases , J Reine Angew Math 319 (1980) 63-72, MR 81j:10012; H G Senge, E G Straus, PV-numbers and sets of multiplicity , Proceedings of the Washington State University Conference on Number Theory (1971) 55-67, MR 47 #8452.
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|
143,309 |
I am trying to not forget my old math. I finished my PhD in real algebraic geometry a few years ago and then switched to the industry for financial reasons. Now I get the feeling that I want to do a postdoc and I face the dilemma that I actually have forgotten a lot of my algebraic geometry classics. Sometimes they are minor things and if I browse a book I recall everything, and sometimes they are major (I don't really think I have a mental lapse though!). So I see myself reading a lot of books in order to recall some of my old math. It gets frustrating that I have to repeat reading 80% of the article that I once used to read and understand. Maybe the new math that I have been feeding myself should be blamed too (I tried learning more differential geometry and fractal theory after doing algebraic geometry and hardly looked back at algebraic geometry after that). I have never tried avoiding to forget old math, especially parts that I do not use in a daily basis (esp. now that I work in the industry). But this can and will be fatal if I do apply for a postdoc. So now I want to read again, yes, but I don't want to forget again. Is there a magic recipe for this? Usually I do find it helpful to always connect even the most abstract of mathematics with something that is tangible as an example, either in real life or in easier math (e.g. connect invertible sheaves and Picard group to line bundles, vector bundles to tangent bundles and tangent spaces .. etc.). This usually helps me not to forget things, but some of the math that I used to learn is too abstract to make such a connection, or maybe I just didn't learn correctly to apply such a connection. So my approach now, when I start reading something new or old, is to find a practical example ASAP, or ask myself why the originator of the theory first thought of developing this in the first place, before even getting any deeper into the subject. I must be honest though, sometimes this is very difficult to do (esp. if you read references for which such connection is not made).
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You do forget things you are not working on. Nothing can be done about it. I could read German easily by the end of 8th grade and now I can hardly spell "Entshuldigen Sie mir bitte". There are several math. papers I read as a student of which I remember next to nothing. The most frustrating and shameful thing is that I don't remember the details of my own papers written 20 years ago with a few exceptions. After age 40 I also started to lose the ability I always took for granted: to get to the board at any time and start lecturing on some subject in my field with full proofs without any preparation. Now I have to sit for half an hour and to prepare my lectures now and then (thanks God this concerns only advanced graduate courses yet). And I work as a professional mathematician in academia full time! The only way to cope with this loss of memory I know is to do some reading on systematic basis. Of course, if you read one paper in algebraic geometry (or whatever else) a month (or even two months), you may not remember the exact content of all of them by the end of the year but, since all mathematicians in one field use pretty much the same tricks and draw from pretty much the same general knowledge, you'll keep the core things in your memory no matter what you read (provided it is not patented junk, of course) and this is about as much as you can hope for. Relating abstract things to "real life stuff" (and vice versa) is automatic when you work as a mathematician. For me, the proof of the Chacon-Ornstein ergodic theorem is just a sandpile moving over a pit with the sand falling down after every shift. I often tell my students that every individual term in the sequence doesn't matter at all for the limit but somehow together they determine it like no individual human is of any real importance while together they keep this civilization running, etc. No special effort is needed here and, moreover, if the analogy is not natural but contrived, it'll not be helpful or memorable. The standard mnemonic techniques are pretty useless in math. IMHO (the famous "foil" rule for the multiplication of sums of two terms is inferior to the natural "pair each term in the first sum with each term in the second sum" and to the picture of a rectangle tiled with smaller rectangles, though, of course, the foil rule sounds way more sexy). Since it is a "general" question, I suggest making it community wiki (and mark my answer as such).
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|
143,334 |
I found the following formula in a book without any proof: $$\sum_{k=0}^{2m}(-1)^k{\binom{2m}{k}}^3=(-1)^m\binom{2m}{m}\binom{3m}{m}.$$ This does not seem to follow immediately from the basic binomial identities. I would like to know how to prove this, and any relevant references. Remark : This question has been asked previously on math.SE without receiving any complete answers.
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Here is a short proof of the more general identity $$ \sum_{k=0}^{2m} (-1)^k \binom{2m}{k} \binom{x}{k}\binom{x}{2m-k} = (-1)^m \binom{2m}{m} \binom{x+m}{2m}. $$ Considered as polynomials in $x$, both sides have degree $2m$. If $x = m$ then $\binom{x}{k}\binom{x}{2m-k}$ is non-zero only when $k=m$, and so both sides equal $(-1)^m \binom{2m}{m}$. If $x \in \{0,1,\ldots,m-1\}$ then both sides are zero. If $x = -r$ where $r \in \{1,\ldots, m\}$ then, using that $\binom{-r}{k} = \binom{r+k-1}{r-1}$,
the left-hand side becomes $$ \sum_{k=0}^{2m} (-1)^k \binom{2m}{k} f(k) $$ where $f(y) = \binom{r+y-1}{r-1} \binom{r+2m-y-1}{r-1}$. Since $f$ is a polynomial of degree $2(r-1) < 2m$ in $y$, its $2m$-th iterated difference is zero. So both sides are again zero. This shows that the two sides agree at $2m+1$ values of $x$, and so they must be equal as polynomials in $x$. This identity is a specialization of (2) in the Gessel-Stanton paper linked to above.
It is (6.56) in Volume 4 of Gould's tables. To get Dixon's identity, take $x=2m$ and use that $\binom{2m}{2m-k} = \binom{2m}{k}$.
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143,569 |
In his blog , Jeff Shallit asks, what was the first occurrence of the exact phrase, "by the usual compactness arguments," in the mathematical literature? He reports that the earliest appearance he has found was in a paper from 1953: it's on page 400 of John W. Green, Pacific Journal of Mathematics 3 (2) 393-402. I found another example from that same year, on page 918 of F. A. Valentine, Minimal sets of visibility, Proc Amer Math Soc 4, 917-921. So, is this year the 60th anniversary of the first appearance of that phrase? If there was an earlier occurrence of the equivalent phrase in a language other than English, that, too, would be of interest.
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There is a paper, written in German, called Ueber Räume mit verschwindender erster Brouwerscher Zahl. by Urysohn and Alexandroff from 1928. (Notice that Urysohn drowned while swimming with Alexandroff in 1924) The following quote is from page 810 (emphasis added): Der Beweis dieser Tatsache ist wörtlich derselbe wie im Falle, wo $R$ der $R^n$ ist: es genügt zu zeigen, dass die erwähnte Trennungseigenschaft im Brouwerschen Sinne induktiv ist ( was aus der Kompaktheit von $R$ in der üblichen Weise folgt ), und dann den Phragmén-Brouwerschen Satz anzuwenden. And here is my translation of that quote: The proof of this fact is literally the same as in the case, where $R$ is $R^n$: it is sufficient to show that the mentioned separation property is inductive in the sense of Brouwer ( which follows from the compactness of $R$ in the usual fashion ), and then apply the Phragmén-Brouwer Theorem . I suspect, that as soon as there was a good notion of compactness, people used it in a routine way as an standard argument.
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143,739 |
Let $n \in \mathbb{N}$, then the order of the Galois Group
of $x^n-2$ coincide with $n \phi(n)$ for $n\in \{ 1 , \dots , 36 \}$
except for $n=\{ 8, 16, 24, 32 \}$ where this order is $\frac{ n \phi (n)}{2}$
and is easy to prove that for $p$ prime we have that the order
of the Galois Group of $x^p-2$ is $p(p-1)$. What is the order of the Galois Group of $x^n-2 : n\in \mathbb{N}$ is general? Thanks in advance.
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The splitting field of $x^n-2$ over $\mathbb{Q}$ is $K.L$ where
$K=\mathbb{Q}(\zeta_n)$ and $L=\mathbb{Q}(\sqrt[n]{2})$, so the order of the Galois group
is
$$
[K.L:\mathbb{Q}] = \frac{[K:\mathbb{Q}]\cdot[L:\mathbb{Q}]}{[K\cap L:\mathbb{Q}]}
= \frac{n \phi(n)}{[K\cap L:\mathbb{Q}]}.
$$
It remains to compute $m:=[K\cap L:\mathbb{Q}]$. First show that $K\cap L = \mathbb{Q}(\sqrt[m]{2})$. For this, note that the norm
$N_{L/(K\cap L)}(\sqrt[n]{2})$ is in $K\cap L$. This norm is the product
of the $n/m$ conjugates of $\sqrt[n]{2}$ over $K$, so it is the product of $n/m$ of the conjugates of $\sqrt[n]{2}$ over $\mathbb{Q}$, and each of these conjugates has the form
$\zeta_n^i \sqrt[n]{2}$. Hence the norm has the form $\zeta_n^i \sqrt[n]{2}^{n/m}
= \zeta_n^i \sqrt[m]{2}$.
Since this is in $K\cap L$, and $\zeta_n^i\in K$,
it follows that $\sqrt[m]{2}\in K$, so $\sqrt[m]{2}\in K\cap L$.
But $[\mathbb{Q}(\sqrt[m]{2}):\mathbb{Q}]=m=[K\cap L:\mathbb{Q}]$,
so indeed $K\cap L=\mathbb{Q}(\sqrt[m]{2})$. Next, since $\sqrt[m]{2}\in K$, and $K/\mathbb{Q}$ is abelian, it follows that $\mathbb{Q}(\sqrt[m]{2})/\mathbb{Q}$ is abelian and hence is Galois. Since $\zeta_m\sqrt[m]{2}$ is a conjugate of $\sqrt[m]{2}$ over $\mathbb{Q}$, it follows that $\zeta_m\in\mathbb{Q}(\sqrt[m]{2})$, so in particular $\zeta_m\in\mathbb{R}$, whence $m\le 2$. The final step is to determine when $\sqrt{2}\in\mathbb{Q}(\zeta_n)$. For this, note
that $\sqrt{2}\in\mathbb{Q}(\zeta_8)$ but $\sqrt{2}\notin\mathbb{Q}(\zeta_4)$, and that
$\mathbb{Q}(\zeta_r)\cap\mathbb{Q}(\zeta_s)=\mathbb{Q}(\zeta_{(r,s)})$. Thus
$\sqrt{2}\in\mathbb{Q}(\zeta_n)$ if and only if $8\mid n$. So the conclusion is that
the splitting field of $x^n-2$ over $\mathbb{Q}$ has degree $n\phi(n)$ if $8\nmid n$,
and has degree $n\phi(n)/2$ if $8\mid n$. Question: is there a way to do this without showing that $K\cap L=\mathbb{Q}(\sqrt[m]{2})$, by using from the start that $K\cap L$ is a subfield of $\mathbb{Q}(\sqrt[n]{2})$ which is Galois over $\mathbb{Q}$, and then somehow going directly to the final step?
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143,827 |
Here's a question out of idle curiosity. Let $G$ be a topological group. Is it possible for both $G$ and (a model of) $BG$ to be finite CW complexes? (Apart from the obvious example of $G$ being [up to homotopy] the trivial group.) A comment is that the fibration sequence $G \to EG \to BG$ shows that $\chi(G)\chi(BG) = 1$ in this case, so $\chi(G) = \pm 1$. This rules out plenty of examples, e.g. Lie groups, finite groups...
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To expand my comment: No it is not possible. First suppose that $G$ is connected and that both $G$ and $BG$ have the homotopy types of finite complexes. If $G$ is not contractible, then let $k>0$ be minimal such that $\pi_k(G)$ is nontrivial. We have $H_k(G)=\pi_k(G)=\pi_{k+1}(BG)=H_{k+1}(BG)$. Choose a prime $p$ such that the latter finitely generated abelian group is nontrivial mod $p$. From now on let homology be with mod $p$ coefficients. Let $d$ be maximal such that $H_d(BG)$ is nontrivial. Let $e$ be maximal such that $H_e(G)$ is nontrivial. The spectral sequence of the fibration has a nontrivial group at $E^2_{d,e}$ that will not go away. Contradiction. So $G$ connected implies $G$ contractible. If $G$ is not connected then $\pi_0G$ is finite. Let $G_0$ be the component of the identity in $G$. Then $BG_0$, the universal cover of $BG$, is also a finite complex, so by the previous paragraph $G_0$ is contractible. That is, $G$ is (up to homotopy) discrete and $BG$ is the classifying space of a finite group. For any cyclic subgroup $C$ of the finite group $G$, $BC$ is a finite cover of $BG$ and therefore finite. The cohomology of $BC$ is such that it cannot be equivalent to a finite complex unless $C$ is trivial. So $G$ is trivial (contractible).
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144,316 |
This question is indeed very important for me. Thus I hope you bear with my subjective explanations for a few minutes. I am an "excellent" lecturer, at least according to course evaluation forms filled by students. More often than not, I use the so-called problem method in the courses I teach, and I advocate a particular philosophy of student-centered teaching. Yet, when I evaluate myself, something bothers me. As a professional mathematics educator, the best I can do is to help my students to learn the concepts and the techniques of the course internally , i.e. bounded to the syllabus of the course. What if I could see beyond the course? What if I was an active mathematician who indeed works with those concepts and techniques, and knows a more advanced and perhaps more general version of those ideas? I was faced with these questions years ago when people started to compare my teaching with the teaching of a mathematician who is indeed an excellent "traditional" lecturer. To my own view, in a sense he could give to his students "more", since he could also see beyond the course . I had forgotten the whole issue until the current term; for the first time I am teaching a course in linear algebra and matrices for mathematics undergraduate students. That excellent colleague of mine is not around now (!), but the question is badly with me: If I could see beyond the course what ideas (concepts, techniques,
theorems, proofs, problems) would I stress more? To keep the question suitable for MO, please do not "argue", and just give one piece of concrete advice to a person who now teach to potentially some of your future colleagues! PS. In this paper ( Moore and Less ; PRIMUS ) you may find the story of the course that the comparison mentioned above started with.
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In my opinion, what you should stress in a course on linear algebra depends more on what the particular students in your class want and/or need, and less on what you can "see beyond the course." However, since you asked this on MathOverflow, you are presumably asking for some insight into how professional mathematicians think about linear algebra, so I will try to address that question. I would say that one the main hallmarks of those who have truly mastered linear algebra is that they can see how linear algebra is applicable in situations where the less well-trained do not . They are able to detect the presence of the "abstract structure" of linear algebra lying under the surface, even when it is not immediately evident from the statement of the problem. Here are some examples. Sound waves can be decomposed into a weighted sum of pure tones. "Weighted sum" signals "linear algebra" to the cognoscenti. It doesn't matter that what you're adding together are functions and not finite sequences of numbers, and it doesn't matter that there are infinitely many possible pure tones. What matters is that you can take weighted sums, and that there is a precise sense in which different pure tones are "orthogonal" to each other. That means that linear algebra is applicable, and the concepts of eigenvalues and eigenvectors (or eigenfunctions) are applicable. The Netflix Prize competition asked for an algorithm to recommend new movies based on your ratings of movies you've already seen. Where's the linear algebra? Start by writing down a large matrix with rows representing people, columns representing movies, and entries representing ratings. Experienced mathematicians know that the biggest singular values of this matrix capture most of the relevant information in it, and provide a good start to constructing the desired algorithm. An old Putnam problem asked whether two matrices $A$ and $B$ with the property that $ABAB=0$ must also satisfy $BABA=0$. The obvious approach is to start playing around with examples, and there's nothing wrong with that. However, a more insightful approach is to build an abstract vector space with the basis $e_\emptyset, e_A, e_{BA}, e_{ABA}, e_{BABA}$ and define the linear transformations $Ae_S := e_{AS}$ and $Be_S := e_{BS}$, where $S$ is any string of $A$'s and $B$'s, $AS$ and $BS$ denote concatenation, and $e_{S} = 0$ if $S$ is not one of the strings $\emptyset$, $A$, $BA$, $ABA$, $BABA$. This is admittedly a very clever proof and even professional mathematicians might not think of it right away, but this example underlines the power of understanding that anything can be used as the basis of a vector space, even strings of symbols . Suppose you have a large system of polynomial equations in $x$, $y$, and $z$, containing equations such as $xyz + 4x^2y - z^3 + 7 = 0$ and $y^2z^2 - xyz + 3 = 0$ and many others. At first glance we might think that linear algebra does not help here because we have variables multiplied together, and multiplication is nonlinear. However, if we have enough equations, and if the same terms appear often enough (in this example, $xyz$ appears in both equations), then we might be able to solve the system by using linear algebra, by treating each term as a separate variable and think of the system as a giant system of linear equations in a much larger number of variables. This might seem like a hopelessly optimistic approach, but in fact it is the basis for a general technique for solving systems of polynomial equations. Again, my point is that with a practiced eye, you can learn to see an entity such as $xyz$ not only as a product of three variables, but as a basis vector in a very large vector space. These examples may not translate directly into useful material for your teaching. However, I do believe that they give a good taste of how mathematicians think about linear algebra. They have internalized what "linear structure" means in the abstract and are able to detect it everywhere, to their advantage. Ideally, one would like to train students to think the same way. Of course, that may be more easily said than done.
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144,392 |
Suppose that I have two matrices $A$ and $B$, and I want them to share a common eigenvector $x$. For simplicity let's just assume that the eigenvalue associated with $x$ is $1$ for both matrices, so $Ax=x$ and $Bx=x$. Is there a simple condition on $A$ and $B$ which is both necessary and sufficient for this to occur? Edit: loup blanc's answer covers the case where the eigenvalues are not known, which is generally much more interesting than the case I was asking about, which is when both eigenvalues are 1. The solution to my case is just that $\ker(A-I) \cap \ker(B-I) \ne 0$. I would still be interested if someone found an even simpler condition which is equivalent to this, though.
|
Let $A,B$ be two $n\times n$ matrices with entries in a field $K$.
Then $A,B$ have a common eigenvector iff
$\cap_{k,l=1}^{n-1}\ker([A^k,B^l])\not=\{0\}$. This result is due to
D. Shemesh. Common eigenvectors of $2$ matrices. Linear algebra and appl., 62, 11-18, 1984.
|
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144,498 |
I know that iterating the following incircle construction approaches an equilateral triangle in the limit: Starting with any triangle $T$, one forms $T'$ by connecting the three points
of tangency of the circle inscribed inside $T$. Does the analogous process for tetrahedra approach a regular tetrahedron?
And the same question may be asked for simplices in $\mathbb{R}^d$.
A reference would be appreciated—Thanks!
|
Here is a partial answer: in $\mathbb R^3$ there are cycles of length $2$. To show this, it is enough to find non-trivial systems of unit vectors $x_i, y_j$ ($i,j=1,\dots,d+1$) such that $(x_i,y_j)=t$ for some $t>0$ and all $i\ne j$. If $d=3$, we can take the normalized copies of $(1,0,4),(1,0,-4),(-1,4,0),(-1,-4,0)$ and $(-2,0,-1),(-2,0,1),(2,-1,0),(2,1,0)$. However, in dimension $d=4$, I do not see how to construct anything like that, so we have a chance for the affirmative answer again. Edit: OK, let's give it a shot. Unfortunately, the answer is negative in every dimension $d>2$. If we believe dimensional considerations, that is easy to predict: the 2 cycle in $\mathbb R^d$ will require constructing $d+1$ unit $x$ vectors and $d+1$ unit $y$ vectors in a non-trivial decent position so that all scalar products $(x_i,y_j)$, $i\ne j$ are equal. $(d+1)$ vectors on the sphere require $2(d^2-1)$ parameters. We have $d(d+1)-1$ equations plus the action of the orthogonal group of dimension $\frac{d(d-1)}2$, so as soon as
$$
2(d^2-1)>d(d+1)-1+\frac{d(d-1)}2\,
$$
which happens for $d>2$, we have a chance. Of course, we need to check that there is no stupid degeneracy anywhere, so the exact analysis is a bit more tiresome. Let's try to perturb the standard configuration in which $x_i=y_i$ are the vertices of a regular simplex. Let $z_i$ and $w_i$ be the infinitesimal perturbations of $x_i$ and $y_i$ correspondingly. To keep the norms, we need $(z_i,x_i)=(x_i,w_i)=0$. Once we have that, the only linear restrictions are $\sum_j (z_i,x_j)=0$ and $sum_i (x_i,w_j)=0$. The perturbations of the scalar products are $(z_i,x_j)+(x_i,w_j)$, i.e., we are allowed sums of two matrices with zero diagonals, the first of which has row sums $0$ and the second one has column sum $0$. For $d\ge 3$, this allows us to create any linear perturbation of the scalar product matrix with zero diagonal and total sum $0$, and, moreover, to do it with the dimension excess $2(d+1)(d-1)-d(d+1)+1$. Note that the addition of a matrix with total sum $0$ is not enough to achieve arbitrary scalar products, but is enough to balance the scalar products if they are slightly disbalanced. Now we just play the usual implicit function game: create linear perturbations of order $t$ so that they don't change the scalar products in the first order but the resulting configuration is $ct$ away from any rotation of the initial configuration (this is pure dimension story). Now, the scalar products differ by about $t^2$. By the implicit function theorem (which should rather be called "sufficient linear range" theorem in this case), we can now use another perturbation of size $t^2$ to balance the scalar products perfectly. This will still leave us away from rotations of the regular simplex but create a cycle of length $2$. The interesting feature here is that the incenter coincides with the circumcenter. I do not have any counterexample to the general conjecture that we always have convergence to this position (which is always a $2$-cycle). Probably, this is how the question should be posed now: Is it true that no matter what the starting simplex was, the distance between the incenter and the circumcenter of the rescaled to size $1$ iterated simplices tends to $0$? It looks like you cannot get away with the length $2$ cycle anymore, and to analyze longer options seems quite a bit less trivial task. Another edit (the penultimate one, I hope :-)) As I said above, if we have a simplex whose circumcenter coincides with its incenter, it generates a cycle of length $2$. Joseph's experiment suggests that we always get attracted to a cycle like this. I think I have an idea how to prove it though one "little detail" is still missing. Here is the sketch. Let us normalize all iterations so that the circumradius is $1$. Then the problem can be restated as follows: Given a system of unit vectors $x_i$ such that $0$ can be written as their convex combination with positive coefficients, define the new system $y_j$ of unit vectors by
$(x_i+u,y_j)=t$ ($i\ne j$) where $u$ is some vector (whose physical meaning is the vector going from the incenter to the circumcenter) and $t>0$ is some number (the radius of the inscribed sphere). The next step will be given by the equations $(z_i,y_j+v)=s$. What I want to show is that $|v|<|u|$ and, moreover, if we have some quantitative bound on the non-degeneracy of $x_i$, we can improve it to $|v|\le q|u|$ with $q<1$. If we can then show that our configurations are uniformly non-degenerate (the part that is still missing), we'll have geometric convergence of the "correction vectors" $u$ to $0$, which is enough to ensure the geometric convergence of the vector systems themselves to a $2$-cycle. One key point is that both $x_i+u$ and $z_i$ are orthogonal to $y_j-y_k$ for $j,k\ne i$, which ensures that they are collinear. With a little more thought, we see that $x_i+u=|x_i+u|z_i$ (i.e., the directions are the same). The other key point is that if we have two systems of vectors $X_i,Y_j$ such that $(X_j,Y_j)=t$ for all $i\ne j$ and $0$ can be written as convex combinations
$$
0=\sum_i a_i X_i=\sum b_j Y_j\,,
$$
then
$$
a_k(X_k,Y_k)+t\sum_{i:i\ne k}a_i=\left(\sum_i a_iX_i, Y_k\right)=0=
\left(X_k,\sum_j b_jY_j\right)=b_k(X_k,Y_k)+t\sum_{j:j\ne k}b_j\,,
$$
which is normally enough to conclude that $a_k=b_k$, the "normal" case including our geometric situation. Now we get $\sum_i a_i(x_i+u)=\sum_i a_i y_i=0$ and $\sum_i b_iz_i=\sum_ib_i(y_i+v)=0$.
Since $x_i+u=|x_i+u|z_i$, we get $b_i=\lambda|x_i+u|a_i$ with some $\lambda>0$ chosen so that $\sum_i a_i=\sum_i b_i=1$. Thus, for every $\mu$, we have
$$
|v|=\left|\sum_i (b_i-\mu a_i)y_i\right|\le \sum_i|b_i-\mu a_i|\,.
$$
Now notice that $1-|u|\le |x_i+u|\le 1+|u|$ and $|u|<1$ (the incenter is inside the circumsphere). Hence the inequality $|v|\le|u|$ will be obtained if we prove the following Lemma: Let $a_j>0$, $\sum a_j=1$. Let $\delta\in(0,1)$. Assume that $|u_j|\le\delta$ and $b_j=\frac{a_j(1+u_j)}{1+\sum_i a_i u_i}$. Then
$$
\min_{\mu}\sum_j|b_j-\mu a_j|\le \delta
$$ Proof Multiplying both sides by $1+\sum_i a_i u_i$ and adjusting $\mu$, we can rewrite the conclusion of the lemma as
$$
\min_{\mu}\sum_j a_j|u_j-\mu|\le \delta\left(1+\sum_i a_i u_i\right)
$$
Now the left hand side is convex in $u_j$ and the right hand side is linear in $u_j$, so it is enough to consider the case when each $u_j$ is either $+\delta$ or $-\delta$.
Let $A_\pm=\sum_{j:u_j=\pm \delta}a_j$. Then the left hand side is $2\delta\min(A_+,A_-)$ while the right hand side is $\delta[1+\delta(A_+-A_-)]$. Canceling $\delta$, we get the inequality $2\min(A_+,A_-)\le 1+\delta(A_+-A_-)=(1-\delta)A_-+(1+\delta)A_+$, which is obvious. We need to be a bit more careful to get the strict inequality, of course (the point is that if the system $x_i$ is non-degenerate, the inequalities $1-|u|\le |x_i+u|\le 1+|u|$ are strict most of the time and when we have a quantitative bound for non-degeneracy, the vectors $x_i$ and $u$ cannot align well). However, for now what we've done is good enough. Last edit (I hope): OK, let's prove the uniform non-degeneracy claim. We will show that the inradius does not decrease at each step. Since the circumradius is fixed, this will finish the story. In the above notation, we need to show that $s\ge t$. Let's again talk about general systems SX_i,Y_jS satisfying $(X_i,Y_j)=t$ for all $i\ne j$. Let $\sum_i a_iX_i=\sum_i a_i Y_i=0$. For each $j$, we have
$0=\sum_i a_i(X_i,Y_j)=a_j(X_j,Y_j)+(1-a_j)t$, whence, summing over $j$,
$$
dt=-\sum_j a_j(X_j,Y_j)\,.
$$
Note that we can add any fixed vector to each $X_j$ or to each $Y_j$ (but not both) in the scalar products on the right hand side without affecting the identity. Now, we get
$$
ds=-\sum_j b_j(z_j,y_j+v)=-\sum b_j(z_j,y_j)=-\lambda\sum_j a_j(x_j+u,y_j)=\lambda dt\,,
$$
so all we need is to show that the normalizing factor $\lambda$ is at least $1$, i.e., that $\sum_j a_j|x_j+u|\le 1$. However,
$|x_j+u|=\sqrt{1+2(x_j,u)+|u|^2}\le 1+(x_j,u)+\frac 12|u|^2\le 1+(x_j+u,u)$. Thus
$$
\sum_j a_j|x_j+u|\le 1+\sum_ja_j(x_j+u,u)=1
$$
as needed. The proof may need some minor combing to put everything together in a neat and completely rigorous way, but all that final polishing is totally routine, so, unless somebody requests it explicitly, I'll leave it to the reader. The upshot is the following Proposition: Normalized in any meaninful way, the sequence of iterations always converges to a $2$-cycle generated by a simplex whose incenter and circumcenter coincide. Moreover, the convergence speed is geometric,
and, if the normalization fixes the circumradius, at each step the distance between the incenter and the circumcenter decreases and the inradius increases. Note that this also encompasses the classical result for $d=2$ because the only triangle on the plane whose circumcenter coincides with the incenter is the equilateral triangle. I guess we have figured out everything worth figuring out here by now unless Joseph gives this problem some new and unexpected twist again :-).
|
{
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"https://mathoverflow.net/questions/144498",
"https://mathoverflow.net",
"https://mathoverflow.net/users/6094/"
]
}
|
144,529 |
This question arises from an issue in my post on Ashutosh's excellent question on Restrictions of the null/meager ideal . Question 1. Does every set of reals contain a measure-zero subset
of the same cardinality? In other words, if $A\subset\mathbb{R}$, is there a measure-zero set $B\subset A$ with $|B|=|A|$? Is this assertion at least consistent? Does it follow from the
continuum hypothesis? Does it follow from some other cardinal characteristic hypothesis?
In the intended application, what is needed is that the assertion is consistent with the additivity number for measure being equal to the continuum. Is this consistent? Can anyone prove the consistency of the failure of the property? Similarly, in the case of category rather than measure: Question 2. Does every set of reals contain a meager subset of the same cardinality? And similarly, is this statement consistent? Does it follow from CH or other cardinal characteristic hypotheses? Is it consistent with the additivity number for the meager ideal being large? Can anyone show the consistency of the failure of the property? The questions arise in my post on Ashutosh's question, where I had proposed as a solution idea the strategy of a back-and-forth construction of length continuum, where the domain and target remain measure-zero during the course
of the construction. But in order for this strategy to succeed, we seem to need to know in the context there that one may extend a given
measure-zero set inside another non-measure-zero set to a larger
measure-zero set with the same cardinality (and the same with meagerness). I had thought at first
that this should be easy, but upon reflection I am less sure about it, and so I ask these questions here.
|
It is consistent that both of the questions have a negative answer. Indeed, this happens if MA holds. A set $E$ of reals is called a Luzin set if $E$ has size continuum and for every meager set $X$ the intersection $E\cap X$ has size less than continuum. A set of reals $E$ is called a Sierpiński set if $E$ has size continuum and for every measure zero set $X$ the intersection $E\cap X$ has size less than continuum. Theorem: MA implies that there are Luzin and Sierpiński sets. Proof: To construct a Luzin set, list all Borel nowhere dense sets in order type continuum: $\langle F_\alpha;\alpha<\mathfrak{c}\rangle$. For each $\alpha$ choose some
$e_\alpha\notin \bigcup_{\beta<\alpha}F_\beta$; this is possible since MA implies that the union of less than continuum meager sets is meager.
$E=\{e_\alpha;\alpha<\mathfrak{c}\}$ has size continuum and its intersection with every closed nowhere dense set has size less than continuum by construction. But since a meager set is contained in a union of countably many closed nowhere dense sets, $E$ must be a Luzin set. To construct a Sierpiński set, replace "Borel nowhere dense" above by "Borel of measure zero" and "meager" by "measure zero". $\square$ In particular, this shows that assuming cardinal characteristics are large is not helpful for this problem. The same avoidance idea seems to also show: Theorem: If $V$ was obtained from $W$ by adding more than $\mathfrak{c}^W$ many Cohen (or random) reals to $W$, then the set of generic reals is Luzin (or Sierpiński) in $V$.
|
{
"source": [
"https://mathoverflow.net/questions/144529",
"https://mathoverflow.net",
"https://mathoverflow.net/users/1946/"
]
}
|
144,689 |
The product topology is the categorical product, and the disjoint union topology is the categorical coproduct. But the arrows in the characteristic diagrams for the subspace and quotient topologies point the same way as in the diagrams for the product and disjoint union topologies, respectively (but there are different conditions on the "constructor" arrow). This leads me to wonder: Are there categorical constructions that generalize the subspace and quotient topologies?
|
Inclusions of subspaces are precisely the regular monomorphisms, and projections of quotients are precisely the regular epimorphisms.
|
{
"source": [
"https://mathoverflow.net/questions/144689",
"https://mathoverflow.net",
"https://mathoverflow.net/users/29961/"
]
}
|
144,773 |
A cumulant is defined via the cumulant generating function $$ g(t)\stackrel{\tiny def}{=} \sum_{n=1}^\infty \kappa_n \frac{t^n}{n},$$ where $$
g(t)\stackrel{\tiny def}{=} \log E(e^{tX}).
$$ Cumulants have some nice properties, including additivity - that for statistically independent variables $X$ and $Y$ we have $$
g_{X+Y}(t)=g_X(t)+g_Y(t)
$$ Additionally, in a multivariate setting, cumulants go to zero when variables are statistically independent, and so generalize correlation somehow. They are related to moments by Moebius inversion . They are a standard feature in undergraduate probability courses because they feature in a simple proof of the Central Limit Theorem (see for example here ). So the cumulants are given by a formula and have a list of good properties. A cumulant is clearly a fundamental concept, but I'm having difficulty figuring out what it is actually measuring, and how it is more than just a computational convenience. Question : What are cumulants actually measuring? What is their conceptual meaning? Are they measuring connectivity or cohesion of something? I apologise that this question is surely completely elementary. I'm in low dimensional topology, and I'm having difficulty wrapping my head around this elementary concept in probability; Google did not help much. I'm vaguely imagining that perhaps they are some kind of measure of "cohesion" of the probability distribution in some sense, but I have no idea how.
|
Cumulants have many other names depending on the context (statistics, quantum field theory, statistical mechanics,...): seminvariants, truncated correlation functions, connected correlation functions, Ursell functions...
I would say that the $n$ -th cumulant $\langle X_1,\ldots,X_n\rangle^{T}$ of random variables $X_1,\ldots,X_n$ measures the interaction of the variables
which is genuinely of $n$ -body type.
By interaction I mean the opposite of independence. Denoting the expectation by $\langle\cdot\rangle$ as in statistical mechanics, independence implies the factorization $$
\langle X_1\cdots X_n\rangle=\langle X_1\rangle\cdots\langle X_n\rangle\ .
$$ If the variables are jointly Gaussian and centered then for instance $$
\langle X_1 X_2 X_3 X_4\rangle=\langle X_1 X_2\rangle\langle X_3 X_4\rangle
+\langle X_1 X_3\rangle\langle X_2 X_4\rangle
+\langle X_1 X_4\rangle\langle X_2 X_3\rangle
$$ so the lack of factorization is due to $2$ -body interactions: namely the absence of factorization
for $\langle X_i X_j\rangle$ .
The $4$ -th cumulant for variables with vanishing moments of odd order would be
the difference $LHS-RHS$ for the previous equation. Thus it would measure
the "interaction" between the four variables which is due to their conspiring all together
instead of being a consequence of conspiring in groups of two at a time.
For higher cumulants, the idea is the same. Cumulant are definitely related to connectedness. For instance for variables whose joint probability density is a multiple of a Gaussian by a factor $\exp(-V)$ where $V$ is quartic, one can at least formally write moments as a sum of Feynman diagrams.
Cumulants are given by similar sums with the additional requirement that these diagrams or graphs must be connected. Some references: Chapters 1 and 4 of the book "Path Integrals in Quantum Mechanics" by Zinn-Justin. My article "Feynman diagrams in algebraic combinatorics" . Although it applies to a somewhat different context, it explains the combinatorics of Feynman diagrams in terms of Joyal's theory of species. The review "Feynman diagrams for pedestrians and mathematicians" by Polyak. However the discussion (e.g. in Section 4.5) of the issue of convergence does not give an accurate idea of this research area.
|
{
"source": [
"https://mathoverflow.net/questions/144773",
"https://mathoverflow.net",
"https://mathoverflow.net/users/2051/"
]
}
|
145,077 |
Why are optimization problems often called programs ? linear programming geometric programming convex programming Integer programming ...
|
It may be that this question had been answered here before, but I couldn't find the answer. Anyway, the answer is given by the person who coined the name itself: George Dantzig wrote in " LINEAR PROGRAMMING ": Here are some stories about how various linear programming terms arose. The military refer to their various plans or proposed schedules of training, logistical supply and deployment of combat units as a program. When I first analyzed the Air Force planning problem and saw that it could be formulated as a system of linear inequalities, I called my paper Programming in a Linear Structure. Note that the term ‘program’ was used for linear programs long before
it was used as the set of instructions used by a computer.
In the early days, these instructions were called codes.
In the summer of 1948, Koopmans and I visited the
Rand Corporation. One day we took a stroll along the
Santa Monica beach. Koopmans said: “Why not shorten
‘Programming in a Linear Structure’ to ‘Linear Programming’?” I replied: “That’s it! From now on that will be its
name.” Later that day I gave a talk at Rand, entitled “Linear
Programming”; years later Tucker shortened it to Linear
Program.
|
{
"source": [
"https://mathoverflow.net/questions/145077",
"https://mathoverflow.net",
"https://mathoverflow.net/users/41438/"
]
}
|
145,097 |
Deligne's Weil I has been published under the title "La conjecture de Weil: I" in 1974, and Weil II in 1980. So did Deligne know in 1974 that there would be a Weil II, and can one explain the period between the two publications?
|
To complete Carlo's answer, I think that one thing that can explain the long
gap (in addition of the amount of difficult material in Weil II) is that Deligne
felt the need to consolidate his result of Weil I before going further. It should be reminded that Weil I was criticized from various directions for relying on results that were not yet formally published. Those results were essentially the theory of Grothendieck of application of Etale Cohomology to L-functions (SGA 5) and the theory of Lefschetz as generalized to schemes by Grothendieck (SGA 7). At the time of the publishing of Weil I, Grothendieck had left the IHES and was not working anymore on the SGA's (nor the EGA's), and his students and colleagues were left with notes of his talks in various
states of redaction. Serre tells, for example, in a letter to Grothendieck (published by the SMF in the volume "correspondence Serre-Grothendieck"),
that Illusie, who was in charge to prepare for publication some crucial parts of SGA 5, told him he was not able to check the commutativity of certain diagrams,
commutativity considered as obvious in the notes. So something that occupied Deligne for quite a while between 1974 and 1980 was this huge work of publishing/completing the work that Grothendieck left abruptly in 1970 (how much this work was just cleaning, proofreading and publishing, and how much it involved original mathematical work is the subject of polemics in Grothendieck's "Récoltes et Semailles"). For instance, Deligne (with others) published SGA 4+1/2 in 1977 with the stated intent to be a partial substitute for the then missing SGA 5. Deligne also worked with Katz and the second part of SGA 7, etc.
|
{
"source": [
"https://mathoverflow.net/questions/145097",
"https://mathoverflow.net",
"https://mathoverflow.net/users/-1/"
]
}
|
145,174 |
I'm trying to write mathematics in English and I'm clearly missing something : linking words. I'm writing "so, we get", "Observe that" too many times and I'm afraid to use some expressions : "it implies" sound weird for me (and I don't know if I'm right here...) for example. So I'm looking for good AND bad linkings words for writing Mathematics in order to diversify my mathematical language. Thank you
|
I can recommend you the booklet Writing Mathematical Papers in English , Jerzy Trzeciak, EMS Publishing, http://www.ems-ph.org/books/book.php?proj_nr=34 . It contains many lists of useful expressions to borrow from.
|
{
"source": [
"https://mathoverflow.net/questions/145174",
"https://mathoverflow.net",
"https://mathoverflow.net/users/37238/"
]
}
|
145,193 |
Question : Is the following true for any $n,N\in\mathbb N$? $$\sum_{k_1+k_2+\cdots+k_N=n,\ k_i\ge0\in\mathbb Z}\frac1{\prod_{j=1}^{N}\{(N-1)k_j+1\}}\le 1$$ Motivation : I've known the $N=3$ case : $$\sum_{k_1+k_2+k_3=n,\ k_i\ge0\in\mathbb Z}\frac1{(2k_1+1)(2k_2+1)(2k_3+1)}\le 1$$ I proved this inequality by estimating the left hand side with integral. After proving this, I reached the above expectation by using computer. The above expectation seems true, but I'm facing difficulty. I would like to know how to prove this (if it's true) and any relevant references. Remark : This question has been asked previously on math.SE without receiving any answers. Update : I'm going to show the proof for $N=3$ case without using integral. This is because it seems that this idea can be generalized (though I'm facing difficulty). For any non-negative integer $n$,
$$\sum_{k_1+k_2+k_3=n,\ k_i\ge0\in\mathbb Z}\frac1{(2k_1+1)(2k_2+1)(2k_3+1)}\le 1$$ Proof : Let $A_n$ be the left hand side, and suppose that $\sum$ represents $\sum_{k_1+k_2+k_3=n,k_i\ge 0\in\mathbb Z}$. Noting that $(2k_1+1)+(2k_2+1)+(2k_3+1)=2n+3$, we get
$$\begin{align}A_n & =\sum\frac{(2k_1+1)+(2k_2+1)+(2k_3+1)}{(2n+3)(2k_1+1)(2k_2+1)(2k_3+1)}\\
& =\frac{1}{2n+3}\sum\left\{\frac{1}{(2k_1+1)(2k_2+1)}+\frac{1}{(2k_2+1)(2k_3+1)}+\frac{1}{(2k_3+1)(2k_1+1)}\right\}\\
& =\frac{3}{2n+3}\sum\frac{1}{(2k_1+1)(2k_2+1)}\\
& =\frac{3}{2n+3}\sum_{j=0}^n\sum_{k_1+k_2=j,k_i\ge 0\in\mathbb Z}\frac{1}{(2k_1+1)(2k_2+1)}\\
& \le \frac{3}{2n+3}\left(1+\frac 23 n\right)=1\end{align}$$
Here, I used
$$B_0=1, B_j\le \frac 23\ (j=1,2,\cdots,n)$$
where
$$B_j=\sum_{k_1+k_2=j,k_i\ge 0\in\mathbb Z}\frac{1}{(2k_1+1)(2k_2+1)}.$$
|
I can recommend you the booklet Writing Mathematical Papers in English , Jerzy Trzeciak, EMS Publishing, http://www.ems-ph.org/books/book.php?proj_nr=34 . It contains many lists of useful expressions to borrow from.
|
{
"source": [
"https://mathoverflow.net/questions/145193",
"https://mathoverflow.net",
"https://mathoverflow.net/users/34490/"
]
}
|
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