User oleg eroshkin - MathOverflow

Lattice points in cross-polytopes

2013-06-11T18:58:33Z

Let $E\subset \mathbb{R}^n$ be a cross-polytope: $$E= \left\lbrace x : \frac{|x_1|}{q_1}+\cdots+\frac{|x_n|}{q_n}\leq 1 \right\rbrace, $$ where $q_1,\dots,q_n$ are positive integers. I am interested in estimating the number $L$ of lattice points in $E$.

Conjecture. $$L:=\# \left(E \cap \mathbb{Z}^n\right)>\mathrm{vol}(E)\,.$$

It's obviously true for $n=1$, and very easy to verify (using Pick's formula) for $n=2$. I don't know how to prove it for $n\geq 3$ without some additional constraints.

According to (generalized) Blichfeldt's theorem, there exists $x\in\mathbb{R}^n$ such that $E+x$ has more lattice points than $\mathrm{vol}(E)$. This suggests the following

Conjecture (strong version). For any $x\in\mathbb{R}^n$ $$L\geq \#\left((E+x)\cap\mathbb{Z}^n\right)\,.$$

I don't know how to prove (or disprove) this version even for $n=2$.

Reference request: Samuel's multiplicity and degree

2013-04-27T20:07:18Z

I am looking for references for the following simple facts.

Let $Y\subset \mathbb{P}^n$ be a variety (or pure-dimensional algebraic set). For $P\in Y$ denote by $e_p(Y)$ the (Samuel's) multiplicity of $Y$ at $P$. Then $\deg Y\geq e_p(Y)$.
For simplicity, I will assume that $Y\subset K^n$ is affine variety. Suppose that for any $f\in I(Y)$ derivatives $\partial^a f(P)=0$ for all multi-indices $a=(a_1,\dots,a_n)$ satisfying $a_1/w_1+\cdots+a_n/w_n<1$. Then $e_p(Y)\geq \min w_{i_1} w_{i_2}\dots w_{i_d}$, where $d=\dim Y$.

I assume that the $K$ is algebraically closed field of characteristic 0.

I know how to prove these facts. For example, the first claim follows immediately from the Corollary 12.4 in Fulton's Intersection theory.

Added later: I was surprised to find a nice geometric description of the difference $\deg Y - e_p(Y)$ in the Appendix to Chapter 6 in Mumford's "Algebraic Geometry I: Complex Projective Varieties". I completely forgot about it.

The part 2 is false without some additional conditions on $Y$ (Cohen-Macaulay is sufficient).

Answer by Oleg Eroshkin for Height of algebraic numbers

2011-05-11T18:14:04Z

If you know only heights of $a$ and $b$, you may estimate heights of $a+b$, $a/b$ and $ab$. Assuming that $h$ is an absolute (Weil) height: $$h(ab)\leq h(a)+h(b)$$ $$h(a/b)\leq h(a)+h(b)$$ $$h(a+b)\leq\log 2 +h(a)+h(b)$$ This bounds are sharp. You may find this, for example, in M. Waldschmidt "Diophantine approximation on linear algebraic groups", Chapter 3.

Answer by Oleg Eroshkin for What is the indefinite sum of tan(x)?

2010-10-05T02:38:21Z

There are no "nice" functions with such properties. Every solution is discontinuous at a dense subset of $\mathbb{R}$. Just look on poles of $\tan x$. Let $x=\pi/2+\pi m$, $m\in\mathbb{Z}$. Clearly either $T$ is discontinuous at $x$ or at $x+1$. In the latter case it is also discontinuous at $x+k$ for every positive integer $k$. In the former case, $T$ is discontinuous at $x+k$ for every non-positive integer $k$.

Answer by Oleg Eroshkin for Obstruction for real subvariety to be embedded as complex subvariety

2010-10-04T16:20:38Z

Actually answer is no even when you allow real embedding of $\mathbb{CP}^n$. Consider any null-homotopic embedding of $X$ into $\mathbb{CP}^n$. Such embedding cannot be a pullback of a complex submanifold of $\mathbb{CP}^N$.

Answer by Oleg Eroshkin for Equidistant points in negatively curved metric spaces

2010-09-26T20:10:23Z

Hello Dave,

Three disks of equal radius in Euclidean plane with centers on a circle of sufficiently large radius seems to be an easy counter-example.

Answer by Oleg Eroshkin for Diameter of m-fold cover

2009-12-09T15:12:33Z

I can show that $diam(\widetilde{M})\leq (m+1)diam(M)$. It follows from the fact that the fundamental group of $M$ is generated by "short" loops of length at most $2diam(M)$ (this is proved in Gromov's book "Metric structures ..."). Lets show that if $p$ and $q$ in $\widetilde{M}$ have the same projection $x$ in $M$, then $dist(p,q)\leq [m/2]\cdot diam(M)$. Consider the following graph. Vertices are $m$ preimages of $x$, edges correspond to short loops in $M$. This graph is 2-connected, therefore the diameter of the graph is at most $[m/2]$.

To improve the estimate it is sufficient to show, that for every two points $x,y\in M$ short loops with based point $x$ through $y$ generates the fundamental group.

Answer by Oleg Eroshkin for When is A : C(X) --> C(Y) a composition operator?

2009-12-04T20:44:51Z

For hemicompact k-space $X$ the space of continuous homomorphisms of algebra $C(X)$ to ℂ is $X$ (up to the obvious isomorphism). The proof can be found, for example, in H. Goldmann "Uniform Frechet Algebras". Then the same construction as for compact spaces give you the map $T$.

Answer by Oleg Eroshkin for Minimal surface in a ball

2009-11-15T03:11:52Z

This result (and several similar) proved in a nice paper Alexander, H.; Osserman, R. "Area bounds for various classes of surfaces." Amer. J. Math. 97 (1975), no. 3, 753--769.

Comment by Oleg Eroshkin

2013-06-11T20:09:45Z

Great! That is quite surprising. I assumed, that is better to look for co-prime $q_i$ for counterexamples.

Comment by Oleg Eroshkin

2011-05-12T00:48:05Z

@Felipe Good point. Than even this conjecture (wide open) is too weak for vanvu.

Comment by Oleg Eroshkin

2011-05-12T00:28:38Z

Roth's theorem gives bound for $|q\alpha-p|$. This is the linear polynomial. Vanvu need such bounds for polynomials of degree $d$. Conjecturally for algebraic number $\alpha$ and $\epsilon>0$ there are only finitely many polynomials $P$ of degree $d$ with integer coefficients of absolute values at most $N$ with $0<|P(\alpha)|<N^{-d-1-\epsilon}$.

Comment by Oleg Eroshkin

2011-05-12T00:09:06Z

This is Liouville's inequality. For $d=1$, Roth's theorem gives required bounds. The generalization of Roth theorem for polynomials of higher degree is open. I doubt that the special case of $\alpha$ is significantly easier.

Comment by Oleg Eroshkin

2010-10-02T15:29:07Z

Colin, do you mean to embed $\mathbb{CP}^n$ as a real submanifold?

Comment by Oleg Eroshkin

2010-06-09T15:29:19Z

Greg, clearly this condition is the same as normality. Just add different prefixes to the string.

Comment by Oleg Eroshkin

2009-12-05T01:29:50Z

I don't remember if there are examples of hemicompact spaces (without k-space propery) such that $M(C(X))\neq X$ (I don't have Goldmann's book at home). Here, $M(C(X))$ is the space of all continuous homomorphisms of the algebra $C(X)$ to ℂ. It seems, that $C(M(C(X)))=C(X)$ (is it?). Then try $Y=M(C(X))$ and identity maps between $C(X)$ and $C(Y)$.