most recent 30 from http://mathoverflow.net 2013-05-24T05:01:46Z http://mathoverflow.net/feeds/user/4863 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/25220/when-are-algebras-quiver-algebras louis de Thanhoffer de Völcsey 2010-05-19T11:06:55Z 2011-01-19T15:42:44Z

<p>Good Morning from Belgium,</p> <p>I'm no stranger to the mantra that quiver-algebras are an extremely powerful tool (see for example the representation theory of finite dimensional algebras). But what is a bit unclear to me is what is known about what kind of algebras are quiver algebras. The first case I can prove is for graded rings with finite dimensional semisimple $A_0$. But what about the ungraded case (I'm pretty sure that it is also true for finite-dimensional algebras over an algebraically closed field for example). Is there a larger generality possible ?</p> <p>Answers (and references) are as always much appreciated !</p>

http://mathoverflow.net/questions/36536/quiver-mutation louis de Thanhoffer de Völcsey 2010-08-24T09:16:05Z 2011-01-19T10:41:03Z

<p>Hello to all,</p> <p>The phrase "quiver mutation has been invented by Fomin and Zelevinsky and has found numerous applications throughout mathematics and physics" is one that some of us encountered on a near-daily basis. I personally know only of a few applications. I thus propose the following thread: how have you used quiver mutation in your area of expertise ?</p>

http://mathoverflow.net/questions/36084/cohomology-of-cyclic-groups louis de Thanhoffer de Völcsey 2010-08-19T14:57:08Z 2010-08-23T09:22:36Z

<p>Hello to all,</p> <p>While sprucing up my knowledge of group (co)homology,I stumbled onto the following question: The first step you usually take to compute various (co)homologies is to construct the infamous "bar resolution" which resolves $\mathbb{Z}$ by free $\mathbb{Z}[G]$-modules (I'll assume everyone knows which one I mean).</p> <p>Now, in the case of the (co)homology of cyclic groups, one creates a 2-periodic resolution by splicing together certain exact sequences involving the norm element of $\mathbb{Z}[G]$. I was wondering if it was possible to distill this 2-periodic resolution somehow out of the standard bar-resolution above in some natural way ? In the case of $\mathbb{Z}_2$, this is quite trivial, but the higher cases are a mystery to me!</p> <p>Thank you and merry Fields day</p>

http://mathoverflow.net/questions/30419/separated-schemes louis de Thanhoffer de Völcsey 2010-07-03T15:31:41Z 2010-07-03T16:16:15Z

<p>Here's a dangerous question: We all know that a variety is an integral scheme, separated and of finite type over an algebraically closed field. Now, if I remove the separated hypothesis, I get the class of schemes which are made by gluing together (finitely many) classical affine varieties and applying the famous fully faithful functor $Var_k \longrightarrow Schemes_k$. Thus, separatedness must relate somehow to the "way of gluing" together these affine varieties, (which is kind of confirmed when you look at the uber-classical example of the double line). Is there someone here that can explain what way of gluing we ban when restricting to separated schemes ?</p>

http://mathoverflow.net/questions/30067/morphisms-on-quadric louis de Thanhoffer de Völcsey 2010-06-30T16:07:11Z 2010-06-30T21:30:44Z

<p>Hello everyone,</p> <p>A quick question, as I'm not sure I got it right.</p> <p>Let $X=\mathbb{P}^1 \times \mathbb{P}^1$ and let $\mathcal{O}(a,b):=\pi^*_1\mathcal{O}(a)\otimes \pi^*_2\mathcal{O}(b)$. Is there a general expression for $$RHom_X(\mathcal{O}(a,b),\mathcal{O}(x,y))$$</p> <p>Thank you</p> <p>Edit: Whoops, I actually meant derived Hom, but I think I got it now</p>

http://mathoverflow.net/questions/27823/derived-physics louis de Thanhoffer de Völcsey 2010-06-11T15:17:27Z 2010-06-11T16:27:53Z

<p>Hello to all,</p> <p>This question will probably be closed down as being off-topic faster than one can say "string theory", but here it goes: I've noticed that the problems I'm working on -the structure of derived categories of some interesting surfaces- have a very important physical meaning. Unfortunately, I have in no way any idea as to why. Is there anyone out there who could give a mathematical explanation -or a link to a paper- as to why physicists would be interested in such highly abstract gizmos like derived categories, mutation, orbifolding, tilting...(the list goes on and on)</p>

http://mathoverflow.net/questions/21852/tensor-product-of-sheaves-and-modules louis de Thanhoffer de Völcsey 2010-04-19T16:24:52Z 2010-04-19T16:46:14Z

<p>Hello to all,</p> <p>I have been looking quite recently at the following theorem: Let $X$ be a projective variety and $T$ a tilting object for $X$. If $A:=End(T)$ is the associated endomorphism algebra, then the functor $RHom(T -): D^b(X) \rightarrow D^b(A)$ is in fact an equivalence. Now, this is proven (as in the claasical Bondal paper) by showing that the functor is fully faithful and essentially surjective. But in I have noticed another version where one defines a functor $-\otimes^L_A T: D^b(A) \longrightarrow D^b(X)$. My question is could someone maybe give a definition of this functor (I of course know what all types of tensor-products are, but I'm not really sure how to "tensor" a module over a noncommutative ring $A$ together with a sheaf to obtain another sheaf.</p>

http://mathoverflow.net/questions/19193/divisors-on-projufd louis de Thanhoffer de Völcsey 2010-03-24T14:22:24Z 2010-04-19T16:34:45Z

<p>Hello to all,</p> <p>I have been perusing Harthorne for some time, and I noticed something: it is well known that the class group on $\mathbb{P}^n_k$ is $\mathbb{Z}$. But as I look at Harthorne's proof it seems to mee that it works in much greater generality. Namely if I consider any projective scheme $X=Proj(A)$, where $A$ is a graded $UFD$ in such a way that there exists an irreducible element of degree $1$, then the exact same reasoning shows that the class group of $X$ is also $\mathbb{Z}$, and generated by the prime divisor $(a)$. Is this true ?</p>

http://mathoverflow.net/questions/27823/derived-physics louis de Thanhoffer de Völcsey 2010-06-11T15:37:32Z 2010-06-11T15:37:32Z

community wiki done

http://mathoverflow.net/questions/25220/when-are-algebras-quiver-algebras/25234#25234 louis de Thanhoffer de Völcsey 2010-05-19T16:53:15Z 2010-05-19T16:53:15Z

The reason for this restriction $(I \subset k[Q]^2)$ being that the quiver is uniquely determined in this case.

http://mathoverflow.net/questions/25220/when-are-algebras-quiver-algebras/25234#25234 louis de Thanhoffer de Völcsey 2010-05-19T15:56:21Z 2010-05-19T15:56:21Z

Professor, I guess my terminology wasn't very clear (I thought there was a standard way to talk about these things, but as MO proves, there isn't) What a quiver algebra is to me is a quotient of a path algebra by an ideal generated by relations $I \subset k[Q]^2$. This restriction might be trivial, but I don't think we can just simply apply the free-algebra trick. PS. thanks for the related material on path-algebras, I wasn't aware that the situation was so intricate and subtle !

http://mathoverflow.net/questions/25220/when-are-algebras-quiver-algebras/25222#25222 louis de Thanhoffer de Völcsey 2010-05-19T12:52:32Z 2010-05-19T12:52:32Z

By the ungraded case, I meant a quotient-algebra of a path algebra by relations which need not be homogeneous,a construction in which you lose the grading

http://mathoverflow.net/questions/21852/tensor-product-of-sheaves-and-modules/21856#21856 louis de Thanhoffer de Völcsey 2010-04-19T16:51:50Z 2010-04-19T16:51:50Z

great, thanks, that was the only paper I couldn't get my hands on, I'll see if I can find it

http://mathoverflow.net/questions/19193/divisors-on-projufd/19283#19283 louis de Thanhoffer de Völcsey 2010-03-26T10:03:40Z 2010-03-26T10:03:40Z

Thank you for that answer, that's a relief. Although it still seems kind of weird to me that Hartshorne would impose such an unnecessary restriction like considering only polynomial rings over fields