most recent 30 from http://mathoverflow.net 2013-05-22T12:31:11Z http://mathoverflow.net/feeds/user/19229 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/115310/matrices-over-finite-prime-fields/115316#115316 labirintas 2012-12-03T17:03:53Z 2012-12-03T17:03:53Z

<p>There are $\displaystyle \sum_{i=0}^n \frac{N(n)}{N(i)N(n-i)}$ solutions to your problem, where $N(r)$ is the number of elements in $GL_r(\mathbb{F}_p)$.</p> <p>Any such $X$ is invertible, and $X\in GL_n(\mathbb{F}_p)$ is a solution if and only if it is conjugate in $GL_n(\mathbb{F}_p)$ to a diagonal matrix with the first $i$ entries equal to $1$ and last $n-i$ entries equal to $-1$. This explains the number of summands. Now $g\in GL_n(\mathbb{F}_p)$ commutes with such diagonal matrix if and only if it preserves both eigenspaces. Hence, there are $\frac{N(n)}{N(i)N(n-i)}$ elements in the conjugacy class. </p> <p>The formula can be made even more explicit by substituting a formula for $N(r)$.</p>

http://mathoverflow.net/questions/110998/linear-algebra-over-principal-rings/111003#111003 labirintas 2012-10-29T14:36:39Z 2012-10-29T14:36:39Z

<p>I think that I have a counterexample. Let $R=\mathbb Z$, $S=\mathbb Q$ and $M= 1/2 \mathbb{Z}$. The induced map $R/2 R \rightarrow M/2M$ is zero. If $R$ is a direct summand of some $N$ the $R/2R\rightarrow N/2N$ is non-zero. Hence, $R\subset M\subset N$ implies that $R$ cannot be a direct summand of $N$, since $R/2R \rightarrow N/2N$ factors through $R/2R \rightarrow M/2M$.</p>

http://mathoverflow.net/questions/102593/group-index-formula-in-lang-algebraic-number-theory/102611#102611 labirintas 2012-07-19T02:21:24Z 2012-07-19T02:21:24Z

<p>If there is a real embedding then $n=2$, because all the solutions to the equation $x^n-1$ in $\mathbb R$ are $\pm 1$ and the field contains the $n$-th roots of unity. Hence, there are 2 possibilities $r_1=0$ and $n$ arbitrary and the formula is ok, or $r_1>0$, $n=2$ and the formula is also ok.</p>

http://mathoverflow.net/questions/82381/on-the-comparison-of-linear-topologies-on-a-local-ring/82469#82469 labirintas 2011-12-02T14:50:30Z 2011-12-02T15:20:22Z

<p>If $R$ is not noetherian then it is not true, e.g $R=k[[X_1,...]]$ modulo all the monomials of degree $2$, so that $m^2=0$, $a_i= (X_i, X_{i+1}, ....)$. Then none of the $a_i$ is contained in $m^2$.</p>

http://mathoverflow.net/questions/80821/a-small-question-on-commutative-algebra/80836#80836 labirintas 2011-11-13T18:00:05Z 2011-11-14T17:14:59Z

<p>I think it is right. Write down the Koszul complex for the regular sequence, which generates $I$. THe regularity of the sequence, is equivalent to vanishing of all the homology groups of the complex, except for $H_0$, which is isomorphic to $B/I$. Tensoring with $C$ over $A$, you obtain the Kozsul complex for the same sequence but in $C\otimes_A B$. Since $B$ is $A$-flat, then again by looking at the homology, you deduce that the sequence remains regular in $C\otimes_A B$. The flatness of $B/I$ over $A$ implies that the ideal generated by the sequence in $C\otimes_A B$ is equal to $C\otimes_A I$. </p> <p>Edit: unknown is right, i have been spending too much time with local rings. However, the argument together with matsumura Thm 16.8 on p.131 shows that the length of a maximal regular sequence in $C\otimes_A I$ is equal to the length of a regular sequence in $I$. </p> <p>Edit2: After a while, I had actually decided to look at what Fulton writes. In his appendix A.5 he talks not of regular sequences, but of regular sections, which are defined in terms of vanishing of Kozsul complex, and basically my proof is his Lemma A.5.3.</p>

http://mathoverflow.net/questions/120096/explicit-casselman-theory-reference-needed labirintas 2013-01-28T17:13:33Z 2013-01-28T17:13:33Z

A similar question has been asked: <a href="http://mathoverflow.net/questions/115933/what-is-the-support-of-the-whittaker-function-of-a-new-vector-on-gl2" rel="nofollow" title="what is the support of the whittaker function of a new vector on gl2">mathoverflow.net/questions/115933/&hellip;</a>

http://mathoverflow.net/questions/116624/serre-functor-of-an-injective-module labirintas 2012-12-17T18:09:02Z 2012-12-17T18:09:02Z

He wants to derive the functor of global sections in the category of sheaves of abelian groups, and not in the category of quasi-coherent sheaves. His argument shows that the two notions coincide. Let me give you an example, where they don't: Let $G$ be a $p$-group, let $R$ be the category of all $\mathbb{F}_p$-representations of $G$ and let $V$ be the full subcategory of $R$, consisting of vector spaces with trivial $G$ -action. If you derive the functor $r\mapsto r^G$ (the invariants), then the derived functors in $V$ are zero, and non-zero in $R$.

http://mathoverflow.net/questions/115988/does-the-linear-representations-of-an-finite-group-on-an-k-vector-space-forms-a-r labirintas 2012-12-10T15:52:12Z 2012-12-10T15:52:12Z

What is the additive inverse of the trivial representation?

http://mathoverflow.net/questions/9154/how-to-think-about-parabolic-induction/56598#56598 labirintas 2012-12-10T09:58:51Z 2012-12-10T09:58:51Z

I guess the argument for p-adic groups is also an induction-restriction argument, which OP doesn't like.

http://mathoverflow.net/questions/115933/what-is-the-support-of-the-whittaker-function-of-a-new-vector-on-gl2 labirintas 2012-12-10T09:49:31Z 2012-12-10T09:49:31Z

Why is W non-zero on the centre? All you know that W is non-zero on some coset $U g K_1(c)Z$. Paskunas-Stevens compute some Whittaker functions for supercuspidals, see <a href="http://arxiv.org/abs/math/0603051" rel="nofollow">arxiv.org/abs/math/0603051</a>, but the fucntions there will not be new vectors in general, i think.

http://mathoverflow.net/questions/9154/how-to-think-about-parabolic-induction labirintas 2012-12-10T09:37:29Z 2012-12-10T09:37:29Z

In the previous comment I consider representations of $GL_2(F_p)$ on $F_p$-vector spaces.

http://mathoverflow.net/questions/9154/how-to-think-about-parabolic-induction labirintas 2012-12-10T09:36:08Z 2012-12-10T09:36:08Z

Let me just note that this fails if you consider modular representations. So if you induce a regular character $\chi$ from the upper triangular matrices in $GL_2(F_p)$ and from the lower triangular matrices, you get non-isomorphic representations, which have the same semi simplification. Regular means $\chi^s\neq \chi$, and $s$ is the matrix (0 &amp; 1// 1 &amp; 0).

http://mathoverflow.net/questions/115311/injective-and-projective-module-on-a-integral-domain-is-zero-module labirintas 2012-12-03T18:41:29Z 2012-12-03T18:41:29Z

Not if it has cows on it.

http://mathoverflow.net/questions/115320/prime-divisor-of-finite-group labirintas 2012-12-03T18:31:33Z 2012-12-03T18:31:33Z

What do you sum over in the last sum? Over all the conjugacy classes?

http://mathoverflow.net/questions/110680/constituents-of-induced-representation/110692#110692 labirintas 2012-10-25T18:03:14Z 2012-10-25T18:03:14Z

Do you assume H normal in G in your answer? Since otherwise the typical direct summand of the restriction to H will be $Ind_{H \cap H^g}^H M^g$ and this need not be semi-simple, and the restriction of $M^g$ to $H \cap H^g$ need not be irreducible.