bio | website | math.tifr.res.in/~venky |
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location | India | |

age | ||

visits | member for | 2 years, 5 months |

seen | 2 hours ago | |

stats | profile views | 2,354 |

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 |
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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 |
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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 |
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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 |
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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 |
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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 |
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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 |
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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 |
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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 |
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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 |
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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 |
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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 |