most recent 30 from http://mathoverflow.net 2013-05-24T22:10:39Z http://mathoverflow.net/feeds/user/21254 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/93572/applications-of-descent Grant Rotskoff 2012-04-09T14:23:01Z 2012-08-08T04:07:05Z

<p>The technique of faithfully flat descent, and, in the case of vector spaces, Galois descent has been used quite a bit in Algebraic Geometry. However, the question of whether, say, a given $k$-vector space $V$ arises from some $L$-vector space $W$ seems like it could be asked in a wide variety of settings. I'm wondering, in particular, if anyone has seen descent in modular representation theory.</p>

http://mathoverflow.net/questions/98513/is-there-any-way-to-generalize-the-laplacian-to-finite-groups Grant Rotskoff 2012-05-31T19:02:28Z 2012-06-01T02:12:11Z

<p>The group theoretic interpretation of harmonic analysis was born out of the observation that the discrete Fourier transform on a signal of length $n$ was precisely the Fourier transform of the finite group $\mathbb{Z}/n\mathbb{Z}.$ I was thinking about the extent to which this analogy holds, but I was having trouble with the Laplacian, which, of course, occupies a central role in harmonic analysis. In particular, there doesn't seem to be a generalization of the operator to finite groups (though certainly there is for compact connected Lie groups). </p> <p>Does anyone know of such an object?</p>

http://mathoverflow.net/questions/95106/which-groups-have-strictly-rational-representations Grant Rotskoff 2012-04-25T02:06:28Z 2012-04-25T11:04:42Z

<p>It can be shown, via the construction of the representations of the symmetric group, that every representation of $S_n$ is equivalent to a representation with values in $\mathbb{Q}.$ Presumably, this is a fairly rare phenomenon: it clearly doesn't hold for cyclic groups ($\mathbb{Z}/p\mathbb{Z}$ has one-dimensional representations given by the $p$th roots of unity, hence $p-1$ of its representations lie outside of $\mathbb{Q}$).</p> <p>Moreover, there is a formula which constructs the rational characters of a group (due to Artin: see Curtis and Reiner section 15), but it doesn't seem to give an answer to the following question: </p> <p>Are there other "classes" of groups such that every irreducible representation is realizable over $\mathbb{Q}$?</p> <p>Take "classes" to mean whatever you think appropriate (so long as it doesn't mean the collection of all groups with only rational irreps).</p>

http://mathoverflow.net/questions/93069/when-are-conformal-maps-holomorphic Grant Rotskoff 2012-04-04T04:05:48Z 2012-04-04T04:21:11Z

<p>It is a standard fact from elementary complex analysis that a holomorphic function $f:\mathbb{C}\to \mathbb{C}$ is a conformal mapping. Now, suppose I have a map $f':\mathbb{R}^2\to \mathbb{R}^2$ which is a conformal mapping of the plane onto itself. Write $$f'(x,y) = (f_1(x,y),f_2(x,y)).$$ Is $f_1 + if_2$ holomorphic?</p>

http://mathoverflow.net/questions/90661/the-rabinowitz-trick Grant Rotskoff 2012-03-09T05:15:11Z 2012-03-09T07:15:00Z

<p>The recent question about problems which are solved by generalizations got me thinking about the Rabinowitz trick, which is used to prove a statement of Hilbert's Nullstellensatz, specifically, the inclusion of the ideal generated by an affine variety $V(J)$ over an algebraically closed field into the radical of $J.$</p> <p>Let $0\neq f\in J,$ as above. In the course of the proof, one extends the given polynomial ring by a single indeterminate and writes its elements as, $$\sum_{i=1}^l h_ig_i + h(X_n\cdot f - 1),$$ where $h_i,h\in k[X_1,\dots,X_{n+1}]$ and $g_i\in k[X_1,\dots,X_n].$ One then applies the weak Nullstellensatz, to see that, indeed, every element of $k[X_1,\dots,X_{n+1}]$ can be written in the above form. Then, mapping back to the smaller polynomial ring, via $X_{n+1} \mapsto \frac{1}{f}$ yields the result, by simply clearing denominators. </p> <p>My question is this: While the trick uses some exceedingly clever algebra, does it have some sort of deeper geometric meaning? Why does it make sense to try this in the first place?</p>

http://mathoverflow.net/questions/90532/any-software-to-compute-multivariable-resultant/90536#90536 Grant Rotskoff 2012-03-08T03:49:14Z 2012-03-08T03:49:14Z

<p>Maple will do it.</p> <p><a href="http://www.maplesoft.com/support/help/AddOns/view.aspx?path=Algebraic/Resultant" rel="nofollow">http://www.maplesoft.com/support/help/AddOns/view.aspx?path=Algebraic/Resultant</a></p> <p>You can also do it in C, or Matlab with MARS:</p> <p><a href="http://gamma.cs.unc.edu/MARS/" rel="nofollow">http://gamma.cs.unc.edu/MARS/</a></p>

http://mathoverflow.net/questions/48771/proofs-that-require-fundamentally-new-ways-of-thinking/90134#90134 Grant Rotskoff 2012-03-03T18:55:56Z 2012-03-03T18:55:56Z

<p>I am always impressed by proofs that reach outside the obvious tool-kit. For example, the proof that the dimensions of the irreducible representations of a finite group divide the order of the group relies on the fact that the character values are algebraic integers. In particular, given a finite group $|G|$ and an irreducible character $\chi$ of dimension $n,$ $$\frac{1}{n} \sum_{s \in G} \chi(s^{-1})\chi(s) = \frac{|G|}{n}.$$ However, since $\frac{|G|}{n}$ is an algebraic integer (it is the image of an algebra homomorphism) lying in $\mathbb{Q},$ it in fact lies in $\mathbb{Z}.$</p>

http://mathoverflow.net/questions/102297/how-to-get-this-inequality Grant Rotskoff 2012-07-15T17:36:28Z 2012-07-15T17:36:28Z

You should try asking this on math.stackexchange.com.

http://mathoverflow.net/questions/102175/how-do-i-log-out-of-mathoverflow Grant Rotskoff 2012-07-13T20:24:03Z 2012-07-13T20:24:03Z

Ask this on meta.

http://mathoverflow.net/questions/98513/is-there-any-way-to-generalize-the-laplacian-to-finite-groups/98523#98523 Grant Rotskoff 2012-06-01T02:16:34Z 2012-06-01T02:16:34Z

Thank you for the recommendations. Terras' book looks promising.

http://mathoverflow.net/questions/98513/is-there-any-way-to-generalize-the-laplacian-to-finite-groups/98525#98525 Grant Rotskoff 2012-06-01T02:15:51Z 2012-06-01T02:15:51Z

Thanks, very helpful perspective.

http://mathoverflow.net/questions/98513/is-there-any-way-to-generalize-the-laplacian-to-finite-groups Grant Rotskoff 2012-06-01T02:12:36Z 2012-06-01T02:12:36Z

@Qiaochu Indeed, thanks.

http://mathoverflow.net/questions/96705/computer-package-for-representation-theory-of-the-symmetric-group Grant Rotskoff 2012-05-12T06:12:19Z 2012-05-12T06:12:19Z

I'm writing modules in Haskell to do this sort of thing. See www.github.com/rotskoff. It won't be of use for this sort of question yet, but if other people have computations like examples (1) and (2) that they'd like to see implemented please let me know!

http://mathoverflow.net/questions/95106/which-groups-have-strictly-rational-representations/95149#95149 Grant Rotskoff 2012-04-25T14:20:27Z 2012-04-25T14:20:27Z

Artin's Theorem actually provides an explicit formula for the values of rational characters of a general group (see: Curtis and Reiner sect. 15) but I'm more interested in groups of the type F. Ladish mentions, i.e., those which are realizable over $\mathbb{Q}$

http://mathoverflow.net/questions/93931/partitions-limit-shape-and-ldp Grant Rotskoff 2012-04-13T15:19:46Z 2012-04-13T15:19:46Z

I edited your question to reflect your correction. You should delete your answer when you get a chance. Also, very interesting question.

http://mathoverflow.net/questions/93576/relation-between-the-representation-of-a-group-and-the-representation-of-one-subg Grant Rotskoff 2012-04-09T16:03:16Z 2012-04-09T16:03:16Z

The general idea that you're looking for is the notion of an induced representation. Suppose you have an $H$-representation $W,$ then $\mathbb{C}G \otimes_{\mathbb{C}H} W$ gives you the induced representation of $W$ in $G.$ By the universal property of the tensor product, this induced representation exists and is unique. That said, I'm not sure the question is appropriate for math overflow. I suggest you take a look at Fulton &amp; Harris (or any other introductory rep theory text) as induction is fairly basic material.

http://mathoverflow.net/questions/90655/what-are-the-relation-between-repg-and-reps-n Grant Rotskoff 2012-04-05T17:40:15Z 2012-04-05T17:40:15Z

I agree that this question is too general. If $G\leq S_n$ then there's an adjoint pair of functors between $Rep(G)$ and $Rep(S_n)$ given by $Res$ and $Ind.$ It is not the case that every representation of $G$ arises as the restriction of a representation of $S_n$ for some $n.$ Take, for example, either of the non-trivial representations of $\mathbb{Z}/3\mathbb{Z}.$ It has character values which are not rational, thus it cannot arise as the restriction of any $S_n$ rep. However, inducing the regular representation of $G$ gives the regular rep of $S_n,$ hence all of its representations.

http://mathoverflow.net/questions/90700/where-is-number-theory-used-in-the-rest-of-mathematics Grant Rotskoff 2012-03-09T17:37:33Z 2012-03-09T17:37:33Z

Number theory is used in the representation theory of finite groups to address rationality questions. Algebraic integrality seems to come up just about everywhere.