User sebastian burciu - MathOverflowmost recent 30 from http://mathoverflow.net2013-05-22T08:28:56Zhttp://mathoverflow.net/feeds/user/2805http://www.creativecommons.org/licenses/by-nc/2.5/rdfhttp://mathoverflow.net/questions/10487/faithful-characters-of-finite-groupsFaithful characters of finite groups Sebastian Burciu2010-01-02T13:47:52Z2011-04-17T18:33:07Z
<p>Related to a previous <a href="http://mathoverflow.net/questions/10126/reference-for-this-theorem-in-representation-theory" rel="nofollow">question</a> I am asking furthermore a proof
for the following:</p>
<p>Question 1: If $\chi$ is a faithful irreducible character of a
finite group $G$ then the regular character of $G$ is a polynomial
with integer coefficients in $\chi$?</p>
<p>I know this fact is true since there is a generalization of it for
Hopf algebras in Corollary 19 of the paper FSU96-08 from <a href="http://www.math.fsu.edu/~aluffi/eprint.archive.html" rel="nofollow">here</a>.</p>
<p>The proof from that paper is a little complicated using some
(although elementary) results on norms and inner products.</p>
<p>I was wondering if anyone knows a different proof of this.</p>
<p>Using the Stone - Weierstrass method mentioned in the previous
question, I am asking further if the following is true:</p>
<p>Question 2: If $\chi$ is a faithful irreducible character of a
finite group $G$ does any character of $G$ is a complex polynomial
in $\chi$?</p>
http://mathoverflow.net/questions/10126/reference-for-this-theorem-in-representation-theoryReference for this theorem in representation theory?Sebastian Burciu2009-12-30T11:46:23Z2011-04-17T08:41:23Z
<p>Let $G$ be a finite group and $\chi$ be an irreducible character of
$G$ (characteristic zero algebraically closed base field). If $H$ is
the kernel of $\chi$ then the irreducible representations of $G/H$
are exactly all the irreducible constituents of all tensor powers
$\chi^n$.</p>
<ol>
<li><p>Do you know any reference for this theorem?</p></li>
<li><p>Is it also working in positive characteristic?</p></li>
<li><p>Is it also working for some infinite groups? (maybe some special
classes: reductive, Lie type, etc)</p></li>
</ol>
<p>Thank you very much!</p>
http://mathoverflow.net/questions/51582/cocyles-for-abelian-extensionsCocyles for abelian extensionsSebastian Burciu2011-01-09T20:26:03Z2011-01-10T14:22:12Z
<p>Suppose we have an abelian extension of Hopf algebras,</p>
<p>$$k \rightarrow k^G \rightarrow A \rightarrow kF \rightarrow k.$$ </p>
<p>According to the general theory there is a left action of $F$ on $G$ and a $2$-cocycle $\sigma:F\times F \rightarrow k^G$ such that $A=k^G$ #${}_{\sigma} kF$ as algebras.</p>
<p>1)Is it true that $\sigma^n=\mathrm{id}$ for some $n\geq 1$? In other words can one choose $\sigma'$ in the same class with $\sigma$ such that if $\sigma'(f, h)=\sum_{a \in G}\sigma'_a(f,h)p_a$ then $$\sigma\;'_a(f,h)^n=\mathrm{id}$$ for all $a, f, h$. Here $p_a$ is the usual notation for dual basis of the group element basis . </p>
<p>2) A little more general question (that implies the first question) is if the set of all $2$-cocycles of $F$ with values in $k^G$ finite? Equivalently the question is if $H^2(F, k^G)$ is finite with $F$ acting on $k^G$ via the action of $F$ on $G$.</p>
<p>3)A related question for what groups $X$ with an $F$-action the group $H^2(F, X)$ is finite? For example $X=k^*$ gives the usual Schur multiplier. </p>
http://mathoverflow.net/questions/26363/semisimple-hopf-algebras-with-commutative-character-ringSemisimple Hopf algebras with commutative character ringSebastian Burciu2010-05-29T16:07:11Z2010-05-29T20:28:04Z
<p>Suppose that $A$ is a semisimple Hopf algebra with a commutative character ring. Does it follow that $A$ is quasitriangular, i.e $\mathrm{Rep}(A)$ is a braided tensor category?</p>
<p>I think I 've seen this statement in a paper without a proof long time ago. It might be obvious although I don't see how to construct a braiding just knowing non-functorial commutativity of the tensor products.</p>
http://mathoverflow.net/questions/25383/de-equivariantization-by-repgDe-equivariantization by Rep(G)Sebastian Burciu2010-05-20T16:04:43Z2010-05-20T18:59:47Z
<p>I'm trying to understand Proposition 2.9 of this <a href="http://arxiv.org/abs/0809.3031v2" rel="nofollow">paper</a> on weakly group theoretical fusion categories. </p>
<p>First of all I have a problem with understanding the settings for de-equivariantization process. It is written on page 5 of the paper that one needs $\mathcal{E}=\mathrm{Rep}(G)\subset Z(\mathcal{C})$ such that $\mathcal{E}$ embeds into $\mathcal{C}$ via the forgetful functor $Z(\mathcal{C})\rightarrow \mathcal{C}$.</p>
<p><strong>Question 1.</strong>
I would like to know what "embeds" means exactly. </p>
<p>Rephrasing when a functor is an embedding functor is? Is it enough to be injective on objects and to send simple objects into simple objects. Obviously this is necessary.<br>
<strong>Question 2.</strong> In a concrete example where $\mathcal{C}=\mathrm{Rep}(A)$ for a Hopf algebra $A$ what would be the conditions to be imposed for the corresponding composition to be an embedding. </p>
<p>Is the corresponding category $\mathcal{C}_G$ the category of representations of a Hopf algebra, i.e does it posses a fiber functor? </p>
<p><strong>Question 3.</strong> The third question I have is regarding the de-equivariantization $\mathcal{E}'_G $ from the second part of the proof proposition. </p>
<p>I understood the construction of $\mathcal{E}$ and clearly $\mathcal{E}\subset \mathcal{E}'$ since $\mathcal{E}$ is symmetric. The question I have is why after composing to the restriction functor $Z(\mathcal{E}') \rightarrow \mathcal{E}'$ one still has an inclusion. </p>
http://mathoverflow.net/questions/24122/subgroups-of-direct-product-of-groupsSubgroups of direct product of groups Sebastian Burciu2010-05-10T16:31:45Z2010-05-10T17:13:49Z
<p>I am interested in the following question on products of finite groups. Let $\Gamma$ be a subgroup of $U_1\times U_2$ such that the compositions with the canonical projections $\Gamma \subset U_1\times U_2 \rightarrow U_1$ and $\Gamma \subset U_1\times U_2 \rightarrow U_2$ are both surjective.</p>
<p>Does it follow that there is a group $G$ such that $\Gamma$ is isomorphic to the fiber product $U_1 \times_G U_2$? This means that there are surjections $\pi_1:U_1\rightarrow G$ and $\pi_2:U_2\rightarrow G$ such that $\Gamma$ is the set of pairs $(u_1,u_2)$ with $\pi_1(u_1)=\pi_2(u_2)$.</p>
<p><a href="http://en.wikipedia.org/wiki/Goursat_lemma" rel="nofollow">Goursat's</a> Lemma mentioned <a href="http://mathoverflow.net/questions/23692/what-are-the-normal-subgroups-of-a-direct-product" rel="nofollow">in this question</a> proves the statement in the case $\Gamma$ is a normal subgroup of $U_1\times U_2$.</p>
<p>If the statement is not true without the normality assumption, then what would be a general characterization of these subgroups $\Gamma$?</p>
http://mathoverflow.net/questions/17631/group-adjoint-and-hopf-algebra-adjoint-maps/17735#17735Answer by Sebastian Burciu for Group-Adjoint and Hopf-Algebra-Adjoint MapsSebastian Burciu2010-03-10T16:36:29Z2010-03-10T16:36:29Z<p>The coadjoint action should be defined in such way that the quantum Fourier transform becomes an isomorphism between the adjoint and co-adjoint actions.</p>
http://mathoverflow.net/questions/10819/module-categories-over-repgModule categories over $Rep(G)$.Sebastian Burciu2010-01-05T16:50:21Z2010-01-05T18:15:05Z
<p>Related to this <a href="http://mathoverflow.net/questions/3309/are-there-two-groups-which-are-categorically-morita-equivalent-but-only-one-of-wh" rel="nofollow">question</a> I also had some troubles to understand the classification of module categories over $Rep(G)$. Specifically, on page 12 of Ostrik's <a href="http://arxiv.org/PS%5Fcache/math/pdf/0111/0111139v1.pdf" rel="nofollow">paper</a> what is the category $\mathrm{Rep}^1(\tilde{H})$? $k^*$ acting as "identity character on V" means $a.v=av$ for all $a \in k^*$ and $v \in V$? Then what is the structure of module category over $Rep(G)$? Tensor product should be after restricting representations of $G$ to $H$ and then inducing back to $\tilde{H}$?</p>
<p>Concretely, I was thinking about the following example. Let $H$ be a subgroup of $G$. Then $Rep(H)$ is a module category over $Rep(G)$ via tensor product as $H$-modules. What is the decomposition of $Rep(H)$ in indecomposable module categories and what are the corresponding subgroups $H$ and cocyles $\omega \in H^2(H,\;k^*)$ for each indecomposable subcategory?</p>
http://mathoverflow.net/questions/9749/characterising-extendable-automorphismsCharacterising extendable automorphismsSebastian Burciu2009-12-25T18:01:36Z2009-12-26T05:26:04Z
<p>Suppose H is a subgroup of a finite group G. Can the group of all automorphisms of H that extend to G can be characterized somehow? What condition on the group extension would guarantee that any automorphism of H can be extended to G? </p>
http://mathoverflow.net/questions/51582/cocyles-for-abelian-extensions/51653#51653Comment by Sebastian BurciuSebastian Burciu2011-01-10T16:03:48Z2011-01-10T16:03:48ZCesar, thank you very much! I knew that should be true but couldn't find a proof written somewhere.http://mathoverflow.net/questions/51582/cocyles-for-abelian-extensionsComment by Sebastian BurciuSebastian Burciu2011-01-10T11:41:53Z2011-01-10T11:41:53ZThanks for taking care of the Tex!http://mathoverflow.net/questions/51582/cocyles-for-abelian-extensionsComment by Sebastian BurciuSebastian Burciu2011-01-09T21:36:05Z2011-01-09T21:36:05ZNo they are all Hopf algebras. $kF$ is a group algebra and $k^G$ is the dual of a group algebra. It should be thought as an analogue to an extension of groups. http://mathoverflow.net/questions/26363/semisimple-hopf-algebras-with-commutative-character-ring/26381#26381Comment by Sebastian BurciuSebastian Burciu2010-05-30T06:56:20Z2010-05-30T06:56:20ZThank you for answer! I also thought that might not be true in general.http://mathoverflow.net/questions/25383/de-equivariantization-by-repg/25406#25406Comment by Sebastian BurciuSebastian Burciu2010-05-20T19:25:00Z2010-05-20T19:25:00ZThank you very much! The confusion probably came from the fact I am not very acquainted with the de-equivariantization process. http://mathoverflow.net/questions/24122/subgroups-of-direct-product-of-groups/24124#24124Comment by Sebastian BurciuSebastian Burciu2010-05-10T17:23:37Z2010-05-10T17:23:37ZThank you very much!http://mathoverflow.net/questions/24122/subgroups-of-direct-product-of-groupsComment by Sebastian BurciuSebastian Burciu2010-05-10T17:02:40Z2010-05-10T17:02:40Z@ Jack: Could you please make your comment comment as an answer? Otherwise I should perhaps answer it by myself.
http://mathoverflow.net/questions/24122/subgroups-of-direct-product-of-groupsComment by Sebastian BurciuSebastian Burciu2010-05-10T16:51:30Z2010-05-10T16:51:30ZOh, thanks a lot! So the statement is always true. Basically $G\cong U_1/N_2 \cong U_2/N_1$ where $N_1$ and $N_2$ are the kernels of the projections $\pi_1$ and $\pi_2$. I thought that normality of $\Gamma$ is used to prove that that $N_1$ and $N_2$ are normal but from the proof from wikepedia it is clear that is not the case. http://mathoverflow.net/questions/18008/is-there-a-relative-version-of-tannakian-reconstructionComment by Sebastian BurciuSebastian Burciu2010-03-13T14:32:16Z2010-03-13T14:32:16ZA little more general question than the present one would be under what conditions on F one gets $\mathbb{C}G$ as a normal Hopf subalgebra of H? Then of course under these conditions it follows that H is a "byproduct" between $\mathbb{C}G$ and the corresponding quotient Hopf algebra.
@YBL In the affine group scheme settings is it known when $G$ is a normal subsecheme of $G'$?http://mathoverflow.net/questions/18008/is-there-a-relative-version-of-tannakian-reconstructionComment by Sebastian BurciuSebastian Burciu2010-03-13T14:31:30Z2010-03-13T14:31:30ZIn the case of Hopf algebras, probably the same condition mentioned by YBL is necessary and sufficient to get an inclusion, although I didn't check that.
http://mathoverflow.net/questions/18008/is-there-a-relative-version-of-tannakian-reconstructionComment by Sebastian BurciuSebastian Burciu2010-03-13T11:20:49Z2010-03-13T11:20:49ZDoes it follow that $\mathcal{C}G$ is a Hopf subalgebra of H? I guess one should impose additional properties on the tensor functor $\mathcal{C} \to Rep(G)$ to get an inclusion of Hopf algebras.http://mathoverflow.net/questions/10487/faithful-characters-of-finite-groups/13723#13723Comment by Sebastian BurciuSebastian Burciu2010-02-02T14:30:20Z2010-02-02T14:30:20ZThank you for the answer. Corollary 19 from the paper FSU96-08, mentioned above, states indeed that it should be a polynomial in \chi with rational coefficients, not integers. Sorry about that!http://mathoverflow.net/questions/12213/ftr-quantization-for-any-subalgebra-of-glnComment by Sebastian BurciuSebastian Burciu2010-01-18T20:05:32Z2010-01-18T20:05:32ZIn Kassel's book, VIII.7, FRT construction is illustrated for GLq(2) and SLq( 2) . I don't know a reference for SLq(n). One has to start with a given solution for QYBE and, as far as I know, there are very few that can be written down explicitly. Do you have a reference for SUq(n)?
http://mathoverflow.net/questions/12213/ftr-quantization-for-any-subalgebra-of-glnComment by Sebastian BurciuSebastian Burciu2010-01-18T19:04:53Z2010-01-18T19:04:53ZAre you referring to the- FRT construction?http://mathoverflow.net/questions/11298/what-are-some-realizability-problems-besides-the-inverse-galois-problemComment by Sebastian BurciuSebastian Burciu2010-01-10T12:53:14Z2010-01-10T12:53:14ZIn algebraic topology there are various questions of realization for groups. Most of them don't have exact answers although partial results are obtained. The main examples are about realization as fundamental groups of all kinds of varieties (projective, hyperplane arrangements etc.)
A topologist might give more details about this if you are interested in them.