38

There are two things you need to understand. The first is how to prove sphere packing bounds via harmonic analysis ("linear programming bounds"). My lecture notes from PCMI 2014 give an exposition of this theory, which covers the period up to, but not including, Viazovska's paper on eight dimensions. This reduces the problem to finding radial functions in ...

25

Briefly, this works very nicely when $X$ is locally compact, but not otherwise.
Then the function space carries the compact-open topology.
John Isbell gave a survey of the story and literature in his paper
General Function Spaces, Products and Continuous Lattices,
in Math Proc Cam Phil Soc 100 (1986) 193--205.
It is an ongoing matter in theoretical ...

24

Yes, this is true. There's some fancy number theory that one can apply (the Hasse-Minkowski invariant and embedding of quadratic forms), but one can see this directly without number-theoretic machinery.
First, notice that one can choose a basis for the lattice which is orthogonal. Just start with any basis and apply Gram-Schmidt orthogonalization. The new ...

22

All free Jónsson-Tarski algebras on a finite nonempty set of generators are isomorphic. Thus free objects may not know their rank.
Curiously the automorphism group of this free algebra is the famous Thompson simple group $V$.

22

There are many ways to interpret this question depending on how you "randomly" choose your vectors. Here's one example. Take the set of vectors $v\in\{-1,1\}^N$, where the coordinates of $v$ are chosen independently to be $+1$ half the time and $-1$ half the time. Then an easy calculation shows that the expected value of
$$\frac{|v_1\cdot v_2|^2}{|v_1|^2|...

20

Dear Ronnie,
The example you refer to (April 22) uses the idea of a placement of a body B in an environment E, and notes that a path in the space of placements corresponds uniquely to a placement of B in the space of paths, because both correspond to a lower-order map from IxB to E itself. These correspondences are invertible, as well as smooth, recursive, ...

17

I would quibble with your definition of "arithmetic". There are discrete subgroups of $SL(2, \mathbf{R})$ coming from quaternion algebras, which are not commensurable with $SL(2, \mathbf{Z})$, but are commensurable with $G(\mathbf{Z})$ for some twisted form $G$ of $SL(2)$ which becomes isomorphic to the usual form over $\mathbf{R}$. These are arithmetic ...

16

Gamma(2) is a free group on two generators, so it surjects onto (Z/pZ)^2. For all odd p, the kernel is a non-congruence subgroup, corresponding to a non-congruence cover of the modular curve X(2), and this cover is none other than the Fermat curve x^p + y^p + z^p = 0. So in some sense the "Frey trick" used in the proof of Fermat can be thought of as a ...

16

In my opinion, the most elegant proof is by Björner, Edelman and Ziegler (PDF file). They prove the following generalization: Let $\mathcal{H}$ be a finite set of hyperplanes in $\mathbb{R}^n$. Let $\mathcal{D}$ be the set of connected components of $\mathbb{R}^n \setminus \bigcup_{H \in \mathcal{H}} H$. Choose one region in $\mathcal{D}$; call it $D_0$....

answered Feb 28 '14 at 16:35

David E Speyer

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16

A stronger result is due to R. Wille. See for instance page 3 of http://www.math.uh.edu/~hjm/1973_Lattice/p00512-p00518.pdf.

answered Feb 21 '15 at 20:18

Richard Stanley

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16

Yes, such a configuration exists, even with all inner products positive
(as the title of the question requires, even though the text allows
either sign).
We shall use the standard normalization of the Leech lattice $\Lambda$
that makes it unimodular, so the $196560$ minimal vectors $(v,v) = 4$,
not $32$. Thus we claim that we can choose $24$ vectors $v_i$ ...

answered Mar 17 '17 at 6:51

Noam D. Elkies

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15

It is standard that $K(n)\wedge K(m)=0$ for $n\neq m$. One way to think about this is as follows: if $E$ and $F$ are complex oriented ring spectra then the corresponding formal group laws become isomorphic over $\pi_*(E\wedge F)$, but it is easy to see that formal group laws of different heights can only become isomorphic over the zero ring.
On the other ...

answered Feb 23 '12 at 21:17

Neil Strickland

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15

Yes, such $A$ exists, and the radius can be much smaller than $10$; indeed radius $2^{1/2}$ suffices regardless of the area of $E$.
Write the ellipse $A(E)$ as $ax^2+bxy+cy^2 \leq 1$. By reduction theory of binary quadratic forms, we can choose the ${\rm SL}_2({\bf Z})$ transformation $A$ to obtain coefficients satisfying $|b| \leq a \leq c$. But $a \geq ...

answered Feb 9 '12 at 5:30

Noam D. Elkies

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15

I see that I'm rather late to the party. Here's an answer to the following question that you asked in the comments above:
"[I]s it conceivable that it is a weak AC principle that every set has a compact Hausdorff topology?"
In fact, there is no need for any choice principle at all, if by finite, we mean in bijection with a natural number (or something ...

15

Here is the link to Rota's article in the Notices which makes it clear that he, at least, thought lattice theory is central.
According to the wikipedia article on von Neumann, Garrett Birkhoff writes: "John von Neumann's brilliant mind blazed over lattice theory like a meteor".[25] Von Neumann worked on lattice theory between 1937 and 1939. Von Neumann ...

answered Dec 5 '13 at 1:11

Benjamin Steinberg

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15

McKenzie proved that it is undecidable whether a finite universal algebra has a finite basis of identities.

15

You can write a similar expression to the usual formula for the remainder term in the circle problem by using the Poisson summation formula. The shift of the center of the circle simply means that one gets a variant of the usual formula for the remainder term modified by suitable exponential terms.
Indeed Huxley has considered a more general problem and ...

14

Another extended comment. The best bound on the number of subgroups is $|G|^{(1/4+o(1))\log_2 |G|}$ proved in http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.100.913&rep=rep1&type=pdf by Borovik, Pyber and Shalev.
For the number of maximal solvable subgroups they get $|G|^c$ for some constant. They don't estimate c but conjecture it is 1. ...

14

@Edward, here is a short proof which I wrote awhile ago for my notes on group theory.
Lemma. If a 2nd countable locally compact group $G$ contains a lattice $\Gamma$ then $G$ is unimodular.
Proof. For arbitrary $g \in G$ consider the push-forward $\nu=R_g(\mu)$ of the (left) Haar measure $\mu$ on $G$; here $R_g$ is the right multiplication by $g$:
$$
\nu(...

14

For the "simpler question": yes, in the decade 1930-1940 the (not many) pioneers of lattice theory had big hopes; one can read their hopes in the Bulletin AMS of 1938 for the first symposium in lattice theory, and the introduction to the first edition to Birkhoff's lattice theory book. And then compare these hopes with the subsequent admissions that such ...

14

In a somewhat different direction from Alireza: the conjecture is true for a large family of groups, including all abelian groups and many supersolvable groups.
Let me start with the abelian case. Pick an element $g$ of highest possible prime-power order in an abelian group $G$. Then $\langle g \rangle$ has a complement: that is, there is a subgroup $K$ ...

14

No, not in general: for instance, the real interval $([0,1],{\le})$, or any non-atomic complete Boolean algebra, do not have such an embedding. This follows from the following characterization:
Proposition: Let $H$ be a complete Heyting algebra. The following are equivalent:
$H$ has a complete embedding into $\mathrm{Up}(X)$ for some poset $X$.
$...

answered Mar 8 '18 at 16:48

Emil Jeřábek 3.0

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14

No, there is no result in this form because in dimension 3 or higher it is allowed to have some non-first minima relatively small even when the first minimum is very small.
For example, for any $N>0 $ consider the lattice $$ \Lambda_N = \frac 1 N \mathbb Z \times \mathbb Z \times \mathbb Z^{n-3}\times N\mathbb Z$$
with unit determinant. Then the set $$ ...

13

As it was noticed by Abhinav Kumar, you only can hope for equality up to $\mathrm{GL}_n(\mathbb Z)$ transformations.
The following picture shows two $\mathrm{GL}_n(\mathbb Z)$-distinct plane figures which have the same number of integer points after any rescaling.
It still might be true that such bodies are $\mathrm{GL}_n(\mathbb Z)$-equidecomposable.
P.S....

13

The natural generalization for your inequality is the setting of distributive lattices. The inequality is then known as the Fortuin–Kasteleyn–Ginibre (FKG) inequality, and has a long history. See for example Graham's article "Applications of the FKG inequality and its relatives". A generalization of the FKG inequality is the Ahlswede-Daykin inequality. For ...

answered Dec 8 '13 at 0:59

Gjergji Zaimi

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13

A deep theorem of Oates and Powell shows that any finite group has a finite basis for its identities. One might think that the same is true for semigroups. But Perkins showed the 6-element semigroup consisting of the $2\times 2$ identity matrix, the zero matrix and the four matrix units $E_{ij}$ is not finitely based.
Mark Sapir, in a tour-de-force work ...

13

Let's show that $SL(n,Z)$ ($n\ge 3$) contains no free groups (and surface groups) of finite index. The same argument shows that it contains no finite index hyperbolic subgroups; in other words, each $SL(n,Z)$ (for $n\ge 3$) has nonlinear Dehn function. Consider the subgroup $N$ of strictly upper triangular matrices in $SL(3,Z)$ (clearly, $N$ is also a ...

13

The results from a large number of repeated trials can be viewed as a random point in a high-dimensional space, so we can use our intuition about probability in the long run as a substitute for our inability to visualize geometry in high dimensions. The "trend towards the mean" after many repeated trials of the same experiment explains why you should expect ...

13

Your question is related to the study of (generalized) numerical polynomials: If $R$ is an integral domain and $K$ the field of fractions of $R$, then the set ${\rm Int}(R) := \{f \in K[x]: f(R) \subseteq R\}$ is a subdomain of $K[x]$, whose elements are called the numerical polynomials over $R$ (in one variable $x$).
The domain ${\rm Int}(R)$ has been the ...

12

Most of this is classical, starting with the memoir by Alexandrov and Urysohn, in which they introduced their notion of the compact Hausdorff space (as bicompact), and also of the absolutely closed space (closed in any Hausdorff superspace) including an extensive discussion of them. This and the minimal Hausdorff spaces, and the related stuff, is very nicely ...

answered Jun 27 '13 at 8:24

Włodzimierz Holsztyński

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