User lev glebsky - MathOverflow

Answer by Lev Glebsky for Measures idempotent with respect to addition and multiplication.

2013-02-03T06:22:14Z

I decide to repeat my comments to mark this question as answered. It has the negative answer: If such a measure existed then such a mesure would exist for any homomorphic images of $N$. But ${\mathbb Z}_2$ does not have such a measure.

Measures idempotent with respect to addition and multiplication.

2012-12-05T19:34:08Z

Does there exist a probability finitely additive measure on $\mathbb N$ which is idempotent with respect to addition and multiplication simultaneously?

It is known (due to Hindman) that there is no ultrafilter that is additively and multiplicatively idempotent. But in the closure of the set of additively idempotent ultrafilters there are multiplicatively idempotent ultrafilters.

This question is related to the ones of Justin Moore: http://mathoverflow.net/questions/57903/idempotent-measures-on-the-free-binary-system and http://mathoverflow.net/questions/114996/do-distinct-idempotent-measures-on-finite-binary-systems-have-distinct-supports. There is good background in the first references, but I do repeat some definition...

Consider $l_\infty(\mathbb{N})$ with $\mathbb R$ as a base field. A finitely additive probability measure on $\mathbb{N}$ is $\mu\in l_\infty^*(\mathbb{N})$ which is positive ($\mu(f)\geq 0$ if $f\geq 0$) and $\mu(1)=1$. Denote the set of these measures as $PM(\mathbb{N})$. Let $\star=+$ or $\cdot$. We may define $\star$ on $\mu\in l_\infty^*(\mathbb{N})$ as follows: $$ \nu\star\mu(f)=\nu_x(\mu_y(f(x\star y))), $$ where $\mu_y(f(x\star y))$ means that we apply $\mu$ with respect to $y$ for $x$-shifts of $f$, the result is a function of $x$. It looks that the only algebraic property of $+,\cdots$ conserved by this extension is associativity. The set $PM(\mathbb{N})$ is closed with respect to $\star$. So, in this notation the question is

Does the exist $\mu\in PM(\mathbb{N})$ such that $\mu+\mu=\mu\cdot\mu=\mu$?

The Higman group II

2013-02-01T19:29:38Z

This question is related to the question The Higman group (with a nice answer by M. Sapir). So for background, please, see the above cited question.

The Higman group has an automorphism $h(a_j)=a_{j+1}$ ($j+1$ is mod 4). Does the Higman group have a nontrivial normal subgroup $N$, satisfying $h(N)=N$?

Motivation. It seems to be an open question if the Higman group is hyperlinear. I seem to know how to construct a nontrivial almost representation of it in the sense of hyperlinearity. I don't know if the almost representation is exact. The negative answer on the above question would imply the exactness of my almost representation...

More general groups. Consider $G_{q,r}=\langle a,b,w\;|\;a^q=b^{-1}ab,\;b=w^{-1}aw,\; w^r=1\rangle$. What is known about such a groups? For $q=2,\;r=4$ it is a semidirect product of a cyclic group of order 4 acting on the Higman group by $h$.

Eigenvalues of the products of a fixed unitari matrix with diagonal unitari matrices

2013-01-09T20:46:25Z

How does the spectra of $DU$ change when $D$ runs over all diagonal unitary matrices? Here $U$ is a fixed unitary matrix. Precisely, let spec$(X)$ be a set of eigenvalues of $X$. For a unitary matrix $U$ let $$ S_U=\{ \mbox{spec}(DU)\;|\;D\mbox{ runs over all diagonal unitary matrices}\}. $$ Clearly, $S_U$ depends on $U$. Two examples:

if $U$ is diagonal then spec($DU$) run over all possibilities: $S_U=({\mathbb C}_1)^n$, where ${\mathbb C}_1$ is the set of complex numbers of norm $1$ and $n$ is the size of $U$.
Let $U$ be a cyclic permutation, that is
$$ U=\left(\begin{array}{cccc} 0 & 0 & 0\dots\dots & 1\\ 1 & 0 & 0\dots\dots & 0\\ 0 & 1 & 0\dots\dots & 0\\ \dots & \dots & \dots\dots\dots &\dots \\
\end{array}\right). $$ Then $S_U=\{\{wa,w^2a,w^3a,...,w^na\}\;\;|\;\;a\in{\mathbb C}_1,\;w \mbox{ is primitive root of 1}\}$

Is something known about $S_U$ in general?

What is $S_U$ for

a) random $U$

b)for which $U$ $S_U$ looks like in example 2.

c) some concrete $U$, say $U$ being a finite Fourier transform matrices?

Polynomial bijection from ZxZ to Z?

2012-12-28T06:39:37Z

It is known that the polynomial $f(n,m)=\frac{1}{2}(n+m)(n+m+1)+m$ defines bijection $\mathbb{N}\times\mathbb{N}\to\mathbb{N}$ (Put pairs of $\mathbb{N}$ into the semi-infinite matrix and count them by diagonals). Does there exist a polynomial bijection $\mathbb{Z}\times\mathbb{Z}\to\mathbb{Z}$? The question is related to the open question about polynomial bijection $\mathbb{Q}\times\mathbb{Q}\to\mathbb{Q}$ here.

Answer by Lev Glebsky for Are almost commuting hermitian matrices close to commuting matrices (in the 2-norm)?

2012-04-14T00:17:16Z

I just found the discussion. In the paper there is the better estimates then the ones of mine and it contains citations on proofs using von Neumann algebras.

(It was a surprise for mi too why my paper is in Algebraic Geometry. Probably it is my error. I have not found an easy way to fix it.)

Answer by Lev Glebsky for Elementary / Interesting proofs of the Nullstellensatz

2012-03-18T19:52:13Z

I have a feeling that some modern proofs hiding from the students the intuition of what is really going on... I am now teaching an undergraduate course "Introduction to algebraic geom. and comm. alg.". I am speaking a lot about Grobner bases and resultant -- they look natural and algorithmic. For the course I am going to use a proof of weak Nullstellensatz using the Groebner bases only.

Let $k$ be a field, $a\in k$. Let ev$_{a}:k[x_1,x_2,\dots,x_n]\to k[x_2,\dots,x_n]$ denote the evaluation homomorphism ev$_{a}:f(x_1,x_2,\dots,x_n)\to f(a,x_2,...,x_n)$. The proof is based on the following lemmas.

Lemma1:

Let $k$ be an algebraically closed field, $I\subseteq k[x_1,\dots,x_n]$ be an ideal, such that $I\cap k[x_1]=\langle p \rangle$ and $p\in k[x]\setminus k$. Then there exists $a\in k$, $p(a)=0$ such that ev$_{a}(I)\neq k[x_2,\dots,x_n]$

Lemma2 (It is valid for any field):

Let $k$ be a field, $I\subseteq k[x_1,\dots,x_n]$ be an ideal, such that $I\cap k[x_1]={0}$. Then there exist non-zero polynomial $q\in k[x_1]$ such that ev$_{a}(I)\neq k[x_2,\dots,x_n]$ for any $a\in k$, $q(a)\neq 0$.

Corollary:

Let $k$ be an infinite field, $I\subseteq k[x_1,\dots,x_n]$ be an ideal, such that $I\cap k[x_1]={0}$. Then ev$_{a}(I)\neq k[x_2,\dots,x_n]$ for some $a\in k$.

The weak Nullstellensatz follows by induction...

Let me show how Groebner bases can be used to prove, say, Lemma 2. Let $I\subset k[x_1,\dots,x_n]$ be an ideal. Let $\langle I\rangle_{k(x_1)}$ denote the ideal generated by $I$ in $k(x_1)[x_2,\dots,x_n]$. Lemma 2 follows from the
statements:

1: $\langle I \rangle _{k(x_1)}=k(x_1)[x_2,\dots,x_n]$ iff $I\cap k[x_1]\neq {0}$.

2: Let $\Gamma={g_1,\dots,g_2}$ be a Groebener basis for $\langle I\rangle_{k(x_1)}$. Suppose, that leading coefficients of $g_i$ are $1$. Let $q\in k[x_1]$ be the common denominator of all $g_i$
(i.e. $qg_i\in k[x_1,...x_n]$ for any $i$). Then ev${a}(\Gamma)$ is a Groebner basis for ev${a}(I)$ for any $a\in k$, $q(a)\neq 0$.

To show (2) it is suffices to note that during reduction of $f\in k[x_1,x_2,\dots,x_n]$ by $\Gamma$ all denominators divide a power of $q$.

Answer by Lev Glebsky for The free group $F_2$ has index 12 in SL(2,$\mathbb{Z}$)

2012-03-13T06:14:47Z

May be the discussion is out of date, but I like to present a simple way to show that the subgroup S generated by $$ \left(\begin{array}{cc} 1 & 2\cr 0 & 1 \end{array}\right) $$ and $$ \left(\begin{array}{cc} 1 & 0\cr 2 & 1 \end{array}\right) $$ has index 12 in $SL(2,Z)$. First of all Mark Sapir noted that by Kargopolov S is a group of matrices $$ \left(\begin{array}{cc} 1+4k_1 & 2n_1\cr 2n_2 & 1+4k_2 \end{array}\right) $$

Lemma. Let $G$ be a group, $H$ be a subgroup of $G$, $N$ be a normal subgroup of $G$, that is subgroup of $H$. Then ind(G:H)=ind(G/N:H/N).

Proof. Indeed, let $x\not\in H$. Then $xN\cap HN\subseteq XH\cap H=\emptyset$. So the natural homomorphism $G\to G/N$ sends different classes to different classes.

Now consider the natural homomorphism $\phi:SL(2,Z)\to SL(2,Z_4)$. $G=SL(2,Z)$, $H=S$ and Ker$(\phi)=N$ satisfy the Lemma. So, ind$(SL(2,Z):S)$=ind$(SL(2,Z_4):\phi(S))=(2*4^2+2*2*2+4*2):4=12$

Comment by Lev Glebsky

2013-02-03T05:52:21Z

@Ashot. This answers my first question! Thank you.

Comment by Lev Glebsky

2013-02-02T15:15:37Z

@Yves and @Derek. Oops, you are right. I will make the corresponding corrections. Thank you.

Comment by Lev Glebsky

2013-01-24T18:56:06Z

@ Michael I have not notice your comment before. I just consider it as a set... About b). Let $n$ be "very large" Then in example 2) al matrices $DU$ have "very small" spectral gaps. So, the question: for which $U$ all $DU$ have small maximal spectral gap? As $\\{DU\\}$ may be considered as a point in the flag manifold, one could try to relate this gap with a Reimann distance on the flag manifold....

Comment by Lev Glebsky

2013-01-09T00:29:13Z

Yes, a free group seems to have a large subgroup. It may be constructed inductively adding $g$ to $<g_1,g_2,...,g_k>$ for each $g$ with $g^F\cap <g_1,...,g_k>=\emptyset$. (We need to start with a good initial $<g_1,g_2>$ to avoid getting all $F$.)

Comment by Lev Glebsky

2012-12-28T18:32:13Z

@Dicman and @Boumol: Thank you for interesting references. Interesting, AMM6028 asks for polynomials with integer coefficients. In fact, the bijection $\mathbb{N}\times\mathbb{N}\to\mathbb{N}$ I know has rational coefficients. Does there exist polynomial $\mathbb{N}\times\mathbb{N}\to\mathbb{N}$ bijection with integer coefficients?

Comment by Lev Glebsky

2012-12-08T16:53:55Z

If one could convert it to a proof, when it probably worked for $\mathbb{Z}$. So, is there a polynomial bijection $\mathbb{Z}\times\mathbb{Z}\to\mathbb{Z}$? (The diagonal counting of pairs of $\mathbb{N}$ using infinite matrix gives a polynomial bijection $\mathbb{N}\times\mathbb{N}\to\mathbb{N}$.)

Comment by Lev Glebsky

2012-12-06T04:28:54Z

May be I was to quick to post the question. It has the negative answer: $\mathbb{Z}_2$ has no such measure: it has unique idempotent for $\cdot$: the measure of $1$ is $1$. This is not an idempotent for $+$. The homomorphism $\mathbb{N}\to\mathbb{Z}_2$ solves the problem. So, why the approach of Hindman was so complicated? Probably he thought about ultrafilters only....

Comment by Lev Glebsky

2012-12-01T00:42:40Z

I am wrong as it was explained to mi by Justin.

Comment by Lev Glebsky

2012-11-26T14:43:07Z

Thank you for the answer. Yes I have to think more. Sorry. But what is true that SAGE did not find this root. That is strange but another story. Normally I believe SAGE in such a simple situation...

Comment by Lev Glebsky

2012-11-23T23:53:17Z

It looks like $\mu^\mu=\mu$ is impossible for the free 1-generated magma $\mathbb T$. Indeed, suppose that such a measure $\mu$ exists. Then for any homomorphic image of $\mathbb T$ should exists an idempotent measure. Using SAGE I was found that magma $\{0,1,2,3\}$ with multiplication matrix $$ \left(\begin{array}{rrrr} 1 & 3 & 2 & 2 \\ 1 & 2 & 1 & 3 \\ 0 & 2 & 3 & 3 \\ 0 & 2 & 1 & 2 \end{array}\right) $$ has no idempotent measure. It is also generated by $0$: $0^0=1$, $1^0=3$, $3^0=2$. So, $\mathbb T$ does not have an idempotent measure. Am I right?