User natalie - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T13:03:26Z http://mathoverflow.net/feeds/user/23104 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/105881/brauer-homomorphism-and-simple-modules Brauer homomorphism and simple modules Natalie 2012-08-29T20:40:19Z 2012-09-03T16:06:11Z <p>Hey there, </p> <p>several weeks ago, there was a discussion on the Brauer hom (see <a href="http://mathoverflow.net/questions/99106/is-the-brauer-correspondence-injective/99117#99117" rel="nofollow">http://mathoverflow.net/questions/99106/is-the-brauer-correspondence-injective/99117#99117</a>). I like to investigate this hom when being applied to simple modules: Let $k$ be an alg. closed field of positive characteristic and let $G$ be a finite group. Assume that $P$ is a $p$-subgroup of $G$. What can be said about the image of the Brauer hom on a simple $kG$-module $S$ with vertex $P$, i.e. what can be said about <code>$Br_P(S):=\frac{S^P}{\sum_{Q&lt;P}tr^P_Q(S^Q)}$</code>, where $S^Q$ denotes the $Q$-invariants of $S$ and $tr^P_Q:S^Q\rightarrow S^P, x\mapsto \sum_{g\in P/Q}gx$ denotes the trace map. Does the fact that $S$ is simple add any properties to the image as for instance indecomposable modules with trivial source do?</p> http://mathoverflow.net/questions/100165/number-of-double-cosets/100181#100181 Answer by Natalie for Number of double cosets Natalie 2012-06-20T20:57:35Z 2012-06-20T20:57:35Z <p>Take $K=H$ and consider the diagonal action of $G$ on $\Omega\times\Omega$, where $\Omega$ is the set of the right cosets $H\backslash G$. Let the number of orbits of this action be $d$. Then $d$ is the number of double cosets $H\backslash G/H$:</p> <p>If we denote the orbits by $\Omega_1,\Omega_2,\ldots, \Omega_d$, let $g_1,\ldots,g_d$ be representatives of the respective orbits. Then the map sending $(Hg,Hg^{\prime})$ to $Hg^{\prime}g^{-1}H$ is easily shown to be a bijection between the set of orbits and double cosets.</p> <p>More details can be found under the topic of Schur bases.</p> http://mathoverflow.net/questions/99453/searching-for-text-for-studying-representation-theory/99495#99495 Answer by Natalie for searching for text for studying representation theory Natalie 2012-06-13T20:06:52Z 2012-06-13T20:06:52Z <p>I'd also recommend</p> <p>Erdmann, Karin; Wildon, Mark J. Introduction to Lie algebras.</p> http://mathoverflow.net/questions/99106/is-the-brauer-correspondence-injective/99146#99146 Answer by Natalie for Is the Brauer correspondence injective ? Natalie 2012-06-08T20:34:46Z 2012-06-08T20:34:46Z <p>I carefully disagree with the above answers.</p> <p>Take two $kG$-modules with trivial source (i.e. p-permutation) modules $M_1,M_2$ which are not isomorphic and whose vertices are $Q_i$ such that $Q_i$ is a proper subgroup of $P$ for $i=1,2$. Then $Br_P(M_i)=0$ for $i=1,2$ (see (1.3) in Broue's Paper you mentioned or Cor. (27.7) in Thevenaz' book "$G$-algebras and modular representation theory"). So the answer to your question is no. You might also look at M. Cabanes Paper "Brauer Morphisms between Modular Hecke Algebras", Section A.</p> <p>But I think the above arguments hold if you consider modules with the same vertex (up to conjugation) such that their images under $Br$ are isomorphic (and maybe not zero?).</p> http://mathoverflow.net/questions/95063/modules-with-trivial-source-and-projective-modules Modules with trivial source and projective modules Natalie 2012-04-24T19:50:47Z 2012-06-08T09:19:56Z <p>This is settled in $p$-modular representation theory of finite groups and finite dim. algebras. So we let $k$ be an algebraically closed field of characteristic $p>0$ and $G$ a finite group. A module with trivial source (TS module for short) is an indecomposable direct summand of the induced module $k_Q^G$ where $Q$ is any $p$-subgroup of $G$. Note that $k_Q^G$ is the tensor product $k_Q\otimes kG$ .</p> <p>Projective indecomposable $kG$-modules (PIMs for short) can be shown to have trivial source (namely as direct summands of $k_1^G$). My question is if they just ''happen'' to have trivial source as well, or if PIMs and other modules with trivial source have even more structure in common. </p> <p>To put the question from another point of view: In other finite dimensional algebras over $k$, can we find analogues of modules with trivial source in the sense that they are connected to projective modules in a similar (to me unknown) way as they are in group rings? </p> http://mathoverflow.net/questions/62218/german-mathematical-terms-like-nullstellensatz/98985#98985 Answer by Natalie for German mathematical terms like "Nullstellensatz" Natalie 2012-06-06T20:27:07Z 2012-06-06T20:27:07Z <p>Anzahl-theorems is one I have recently read in Wan's book on classical groups.</p> http://mathoverflow.net/questions/96705/computer-package-for-representation-theory-of-the-symmetric-group/96713#96713 Answer by Natalie for Computer package for representation theory of the symmetric group Natalie 2012-05-11T21:20:32Z 2012-05-11T21:20:32Z <p>You could try GAP. This is far from being elegant, just if you are very desperate ;-)</p> <p>gap> g := SymmetricGroup(7);;</p> <p>gap> regmod := RegularModule(g,GF(19));;</p> <p>//This not elegant: I chose GF(11) to make the group ring semisimple (as if working over a field of characteristic 0. This is necessary here, as GAP uses the MeatAxe which needs finite fields). Moreover it would be better to get the simple constituents from another source than the regular module as the regular module soon gets too large in dimension. Maybe one could consider an action of $g$ on cosets of a subgroup where your simples of interest occur as composition factors. Ask for more details if you are interested. For $g=S_7$ this is too large to handle. So I take the natural PermutationModule, which has two constituents, a module of dimension 1 and a module of dimension 6.//</p> <p>gap> permod := PermutationGModule(g,GF(11));; </p> <p>gap> comps := MTX.CompositionFactors(permod);</p> <p>[ rec( field := GF(11), isMTXModule := true, dimension := 1, generators := [ [ [ Z(11)^0 ] ], [ [ Z(11)^0 ] ] ], smashMeataxe := rec( algebraElement := [ [ [ 1, 2 ], [ 3, 1 ] ], [ Z(11)^0, Z(11)^0, Z(11)^9, Z(11) ] ], algebraElementMatrix := [ [ Z(11)^5 ] ], characteristicPolynomial := x_1+Z(11)^0, charpolFactors := x_1+Z(11)^0, nullspaceVector := [ Z(11)^0 ], ndimFlag := 1 ), IsIrreducible := true ), rec( field := GF(11), isMTXModule := true, dimension := 6, generators := [ &lt; immutable compressed matrix 6x6 over GF(11) >, &lt; immutable compressed matrix 6x6 over GF(11) > ], smashMeataxe := rec( algebraElement := [ [ [ 2, 1 ], [ 2, 3 ] ], [ Z(11)^6, Z(11)^9, Z(11)^8, Z(11) ] ], algebraElementMatrix := &lt; immutable compressed matrix 6x6 over GF( 11) >, characteristicPolynomial := x_1^6+Z(11)^6*x_1^5+Z(11)^7*x_1^3+Z(11)^3*x_1^2+Z(11)^4*x_1+Z(11)^8, charpolFactors := x_1-Z(11)^0, nullspaceVector := [ Z(11), Z(11)^8, Z(11)^5, Z(11)^3, Z(11)^7, Z(11)^0 ], ndimFlag := 1 ), IsIrreducible := true ) ]</p> <p>gap> c1 := comps[1];;</p> <p>gap> c2 := comps[2];;</p> <p>gap> TensorProductGModule(c1,c2);</p> <p>rec( field := GF(19), isMTXModule := true, dimension := 6, generators := [ &lt; immutable compressed matrix 6x6 over GF(19) >, &lt; immutable compressed matrix 6x6 over GF(19) > ] ) gap> </p> <p>From this point on you can do whatever MTX in GAP allows you to do with modules. See also <a href="http://www.gap-system.org/Manuals/doc/htm/ref/CHAP067.htm#SSEC002.2" rel="nofollow">http://www.gap-system.org/Manuals/doc/htm/ref/CHAP067.htm#SSEC002.2</a> for more information on GAP.</p> http://mathoverflow.net/questions/96652/cosets-and-conjugacy-classes/96706#96706 Answer by Natalie for Cosets and conjugacy classes Natalie 2012-05-11T20:36:29Z 2012-05-11T20:36:29Z <p>This is not a solution for your questions but a remark which might help you: The number of elements in one conjugacy class, which lie in a coset is constant over all cosets which lie in a fixed double coset. By this I mean the following:</p> <p>Let $Hx_1$ and $Hx_2$ be cosets which both lie in the same double coset $HgH$, let $C$ be a conjugacy class and fix $g_0\in C\cap Hx_1$. We like to show that $|Hx_1\cap C|=|Hx_2\cap C|$.</p> <p>By assumption there are elements $h_l,h_l^{\prime}$ for $l = 1,2$ such that $x_l = h^{\prime}_lgh_l$. Thus, $g_0∈C∩Hx_1 =C∩Hgh_1$, so that $(h_1^{-1}h_2)^{−1}g_0(h_1^{-1}h_2)\in C∩Hgh_2 =C∩Hx_2$.</p> <p>So at least for cosets which lie in the same double coset you get an answer for the first question.</p> <p>If you define for each representative $g$ of the double cosets of $H$ in $G$ a valency to be the number $k_g=|H|^{-1}|D_g|$, then by the above, we have that </p> <p>$|C\cap D_g|/k_g$ is a natural number for all conjugacy classes $C$. Maybe this helps.</p> http://mathoverflow.net/questions/96022/totally-singular-subspaces-in-orthogonal-vector-spaces Totally singular subspaces in orthogonal vector spaces Natalie 2012-05-04T20:35:41Z 2012-05-06T21:04:07Z <p>This is for all that are interested in classical groups and their representations. We are investigating the following situation:</p> <p>Let $V$ be $d$-dimensional $k$-vector space (where $k$ is a finite field) equipped with a non-degenerate quadratic form $Q$, so that $V$ is an orthogonal space. Let $GO$ be the group of similarities (which are a bit more general than isometries; we use the language of Kleidman's and Liebeck's book 'The Subgroup Structure of the Finite Classical Groups') and fix $g_1,g_2\in GO$. Suppose we have the following setting:</p> <p>There are non-degenerate subspaces $U_i$ which are irreducible as $\langle g_i\rangle$-modules and which have large dimension $e_i>d/2$ for $i=1,2$. (we call $g_i$ 'fat' in this case, a generalization of ppd-elements as described in A. Niemeyer and C. Praeger 'A recognition algorithm for classical groups over finite fields'; our project actually deals with finding the proportions of such elements within the classical groups)</p> <p>What we are interested in here is a third subspace:</p> <p>Let $W$ be an irreducible $\langle g_1,g_2\rangle$-module which intersects trivially with $U_i$ and assume that $W$ is totally singular i.e. $Q_W=0$. Is it possible that $W$ - as a subspace of $U_i^{\perp}$ - is maximal totally singular in $U_1^{\perp}$ or $U_2^{\perp}$, which is means that the dimension of $W$ equals the Witt index of $U_i^{\perp}$? </p> <p>We can control the case where $W$ is non-degenerate or generated by a non-singular element $v$ (i.e. $Q(v)\neq 0$). The bad guys are such $W$ as above, as the group $\Omega\le GO$ as described in (Kleidman and Liebeck) does not act transitvely on maximal totally singular subspaces in general. The best result would be if such subspaces cannot exist in the above setting.</p> <p>P.S. For simplicity we could assume that the characteristic of $k$ is odd.</p> http://mathoverflow.net/questions/94907/applications-of-group-theory-to-math-biology-pharmacology/94969#94969 Answer by Natalie for Applications of group theory to math. biology (pharmacology) ? Natalie 2012-04-23T18:46:24Z 2012-04-23T18:46:24Z <p>Maybe it is too far fetched. But I heard of the so called Conley index theory which deals with the question of existence/non-existence of equilibria in dynamical systems. It involves homology groups of the occurring manifolds. So I think of dynamical systems which one might find in some situation in biology/pharmacology etc. and apply the Conley index theory to it. Check also <a href="http://wwwb.math.rwth-aachen.de/~barakat/MTNS2010/Conley.pdf" rel="nofollow">http://wwwb.math.rwth-aachen.de/~barakat/MTNS2010/Conley.pdf</a></p> http://mathoverflow.net/questions/102233/orthogonal-groups-over-finite-fields/102234#102234 Comment by Natalie Natalie 2012-07-14T20:46:37Z 2012-07-14T20:46:37Z you should also have a look at Kleidman and Liebeck's book The Subgroup Structure of the Finite Classical Groups&quot;. $\Omega$ is the group you are looking at and which is explained there in detail. http://mathoverflow.net/questions/95063/modules-with-trivial-source-and-projective-modules/99105#99105 Comment by Natalie Natalie 2012-06-09T12:30:26Z 2012-06-09T12:30:26Z I like the last paragraph of your answer in particular. Thanks. http://mathoverflow.net/questions/99106/is-the-brauer-correspondence-injective Comment by Natalie Natalie 2012-06-09T12:27:20Z 2012-06-09T12:27:20Z I'd also love to know in what context you are studying this paper. http://mathoverflow.net/questions/99106/is-the-brauer-correspondence-injective/99117#99117 Comment by Natalie Natalie 2012-06-08T20:40:14Z 2012-06-08T20:40:14Z Is it true that &quot;projective indecomposable $kN_G(P)/P$-module are precisely the indecomposable $kN_G(P)$ modules with vertex $P$&quot; as you say? What about the indecomposable $kN_G(P)$-modules, which do not have $P$ in their kernel? Even though $P$ is normal in $N_G(P)$, we just know that irreducible $kN_G(P)$-modules have $P$ in their kernel (Feit p102) but not indec. modules in general. http://mathoverflow.net/questions/96022/totally-singular-subspaces-in-orthogonal-vector-spaces Comment by Natalie Natalie 2012-05-06T21:17:17Z 2012-05-06T21:17:17Z @Derek, you are absolutely right, $W$ is a subspace of $U_i^{\perp} - sorry for that. http://mathoverflow.net/questions/95952/a-question-on-zsigmondys-lemma Comment by Natalie Natalie 2012-05-04T17:47:05Z 2012-05-04T17:47:05Z Maybe I don't understand the question, but if you take$p=5$and$q=19$, then 19 divides$5^9-1$, but$p$is not a Mersenne prime. http://mathoverflow.net/questions/95972/endomorphismrings-of-maximal-submodules Comment by Natalie Natalie 2012-05-04T17:32:13Z 2012-05-04T17:32:13Z Which assumptions of the$k\$-algebra do you make or allow?