3,829 reputation
11023
bio website math.tifr.res.in/~venky
location India
age
visits member for 2 years, 5 months
seen 2 hours ago

interested in representations of groups, algebraic groups, arithmetic groups, rigidity, and related automorphic forms. Recently interested in braid groups and configuration spaces.


Sep
27
reviewed Approve suggested edit on Question about total derivative matrix
Sep
27
comment Subgroups of $SL_3(\mathbb{Z})$ that are finitely generated, Zariski-dense, infinite index, and torsion-free
@Rivin: ha! ha! more like a rat!
Sep
27
comment Subgroups of $SL_3(\mathbb{Z})$ that are finitely generated, Zariski-dense, infinite index, and torsion-free
@Rivin: thanks!! How did you find out it was I? (I had used a "false name").
Sep
26
comment Can any reductive $k$-group be written as a semidirect product of $k$-linear groups?
As to the original question, take $G$ to be a finite group (e.g the group of upper triangular unipotent matrices matrices with entries in $Z/3Z$. Then the commutator is the centre and the extension above does not split. If it did, the group would be abelian.
Sep
26
comment Can any reductive $k$-group be written as a semidirect product of $k$-linear groups?
@anon: what you say is not right: take the torus inside the diagonals in $GL_n$ all of whose entries except the first one are $1$. This maps iso to $GL_n/SL_n=G_m$.
Sep
25
comment Irreducible action of an algebraic group
Take the standard representation $V$ of $G=N(T)$ the normalizer of $T$, the group of diagonals in $GL(V)$. Then $V$ is irreducible for $G$; as a rep of $G^0=T$, $V$ decomposes as a direct sum of one dimensional distinct characters.
Sep
25
comment Irreducible action of an algebraic group
Itis easy to see that if $V$ is any (say finite dimensional ) module over a group $H$ which is a sum $\sum W_i$ of irreducible $H$ modules $W_i$, then $V$ is a direct sum of irreducibles. To see this, let $W$ be an $H$ subspace of $V$ which is of the largest dimension, which is a direct sum of irreducibles; then $W\neq 0$ since $W$ contains one of the $W_i$. If some $W_j$ does not lie in $W$, then $W+W_j$ is the direct sum of $W$ and $W_j$, since $W_j$ is irreducible; this contradicts maximality of $W$ and hence $W=V$.
Sep
25
comment Irreducible action of an algebraic group
of course; as a rep of the trivial group, $V$ is a direct sum of one dimensional trivial representations of the trivial group
Sep
25
comment Irreducible action of an algebraic group
It is true: take any irreducible $G^0$ submodule $W$ of the irreducible $G$-module $V$; then $V$ is the sum of $g(W)$ for $g\in G$. Hence we have a surjection from the direct sum of $g(W)$ onto $V$. This means that $V$ is a direct sum of irreps of the same dimension. Of course, isotypical is false.
Sep
24
reviewed Approve suggested edit on Why is there a connection between enumerative geometry and nonlinear waves?
Sep
24
awarded  Autobiographer
Sep
23
comment on the center of a Lie group
Does this mean what the OP asks for is true? That the list there is a complete set of representatives for the centre?
Sep
23
comment is $x_{n}\ll \overline{x}_{n}^{2}$?
I suppose $x_n$ is not strictly increasing
Sep
23
revised query about quasi-simple algebraic groups over local fields
added 290 characters in body
Sep
23
revised how many Q-forms of SL_n(R) are there for a given Q-rank
added 2 characters in body
Sep
23
revised how many Q-forms of SL_n(R) are there for a given Q-rank
addes some words on $\mathbb Q$-rank
Sep
23
awarded  Necromancer
Sep
23
comment Subgroups of $SL_3(\mathbb{Z})$ that are finitely generated, Zariski-dense, infinite index, and torsion-free
I am sorry; I saw this (very interesting post) just now. What Yves Cornulier says is correct; if a finite index subgroup of the Heisenberg group embeds in a Zariski dense subgroup $\Gamma$ of $SL_3({\mathbb Z}$, then $\Gamma$ does have finite index in $SL_3({\mathbb Z})$. I proved this ( a long time ago) but it is an easy consequence of a result of Tits, on unipotent generators of arithmetic groups.
Sep
23
comment Numerical integration of legendre polynomials
I am quite unfamiliar with approximations and the like. But, let me add my "one rupee worth": the integral may be converted to one of the form $\int _0 ^1 f(x) \frac{d^n}{dx^n}(\frac{x^n(1-x)^n}{n!})$. Up to $\pm 1$ this is (by integration by parts), the same as $\int _0 ^1 f^{(n)}(x)\frac{x^n(1-x)^n}{n!}$, which may be more tractable.
Sep
23
awarded  Revival