User poove - MathOverflow

Hopfian property

2011-03-22T17:45:12Z

Let $G$ be a group which is Hopfian and given a short exact sequence $1\to F \to H \to G \to 1$ with $F$ a finite normal subgroup of $H$. Is $H$ Hopfian?

GIT quotient of an algebraic action

2011-12-10T19:14:29Z

Let $G$ be a connected affine algebraic group over an algebraically closed field $K$ which acts algebraically on an affine $K$-variety $V:=\bigcup_{i=1}^{n} V_i$, where $V_i$'s are irreducible $G$-stable pairwise (disjoint) isomorphic as varieties. If one of the restriction $\pi_i:V_i\rightarrow V_i//G$ of the quotient morphism $\pi:V\rightarrow V//G$ is constant, does that imply any $\pi_j:V_j\rightarrow V_j//G, \forall j\ne i,$ is also constant? ($V//G$ is the GIT quotient of $V$).

It may be a trivial question, but not clear to me and I couldn't find out a counter example ( I believe its not true).

Suppose $\Gamma_m$ is a principal congruence subgroup of level m contained in a finite index subgroup $\Gamma$ of $SL(n,\mathbb Z)$. Is $\Gamma_m$ characteristic in $\Gamma$?

2011-09-06T07:00:04Z

We know that principal congruence subgroups are characteristic in $SL(n,\mathbb Z)$. Suppose $\Gamma$ is a finite index subgroup of $SL(n,\mathbb Z)$ and $\Gamma_m$ is a principal congruence subgroup of level m contained in $\Gamma$. Will it be characteristic in $\Gamma$?

Principal congruence subgroups of SL(n,Z)

2011-04-28T18:27:32Z

http://mathoverflow.net/questions/62446/principal-congruence-subgroups-of-sln-mathbbz gives a discription of automorphisms of $SL(n,\mathbb Z)$. Is it true for n even too?. Hau-Reiner's paper gives generators for the group of automorphisms of $SL(n,\mathbb Z)$ which are induced by the automorphisms of $GL(n,\mathbb Z)$.

Principal congruence subgroups of $SL(n, \mathbb{Z})$

2011-04-20T18:33:32Z

I want to know whether the principal congruence subgroups of $SL(n, \mathbb{Z})$ are characteristic? please suggest me a reference.

Comment by Poove

2011-12-11T08:37:10Z

@ Ben Webster: You are right, but here I have fixed a $G$ action on $V$ and so on each $V_i\subset V$, the restricted action of $G$. And $\pi_i:=\pi|V_i$.

Comment by Poove

2011-12-10T19:15:57Z

I had asked this one day back, but deleted for the question was unclear.

Comment by Poove

2011-09-08T21:09:47Z

@Igor Rivin: Any automorphism of $\Gamma$ extends(super rigidity) to an automorphism of $SL(n,\mathbb R)$, for n>2. But it not necessarily restricts as an automorphism to $SL(n,\mathbb Z)$, instead it restricts to the normalizer of $\Gamma$ in $SL(n,\mathbb R)$, Which is again a proper finite index subgroup of $SL(n,\mathbb Z)$. So mathoverflow.net/questions/62446/… doesn't say $\Gamma_m$ is characteristic in $\Gamma$.

Comment by Poove

2011-09-08T20:55:30Z

@Agol: we can simply take $A \in SL(2,\mathbb Q)$. Thanks for the answer.

Comment by Poove

2011-05-05T15:43:55Z

$PGL(n,\mathbb Z)= PSL(n,\mathbb Z)$ for n odd only.

Comment by Poove

2011-05-03T16:22:20Z

@Igor Belegradek: But for n even, the centre of both $ SL(n,\mathbb Z)$ and $GL(n,\mathbb Z)$ is $\{I,-I\}$, thus if $PGL(n,\mathbb Z)= PSL(n,\mathbb Z)$, then by one of the isomorphism theorems it will imply that $GL(n,\mathbb Z)= SL(n,\mathbb Z)$. I guess i am not wrong.

Comment by Poove

2011-05-02T11:34:46Z

@Igor Belegradek:(If I'm not wrong) You have mentioned in your answer above that $PGL(n,\mathbb Z)= PSL(n,\mathbb Z)$ for all $n>2$. Is it true for all such n?.

Comment by Poove

2011-05-02T08:28:55Z

@Ralph: I couldn't see any mistake in your consideration. :).

Comment by Poove

2011-04-29T15:19:56Z

@Ralph: Thanks.

Comment by Poove

2011-04-29T09:31:57Z

See this for answer: mathoverflow.net/questions/63319/…

Comment by Poove

2011-04-27T09:30:44Z

@Ralph: $\Gamma_n$ is characteristic. but what you said about the automorphism group of $SL(n,\mathbb Z)$ is true for n odd, because for n odd, all automorphisms of $SL(n,\mathbb Z)$ are induced by automorphisms of $GL(n,\mathbb Z)$(Hua-Reiner,theorem.3). Is it true if n even too?

Comment by Poove

2011-04-21T05:56:26Z

you are right Ralph

Comment by Poove

2011-04-21T05:55:18Z

Thanks a lot Richard.

Comment by Poove

2011-03-23T10:07:24Z

@ ndkrempel: G has no normal subgroups isomorphic to a free abelian group of finite rank. (dont want to assume G is torsion free.

Comment by Poove

2011-03-22T21:03:28Z

@ ndkrempel: why the math symbols are not readable here? did you mean the group G or H on which i would need stronger assumptions? sorry for disturbing