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 Yearling
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11h
comment Quiver algebra as a wreath product?
It seems to me the first important issue I can clarify for you is that the wreath product of an algebra $A$ is given as follows. Take the $n$-fold tensor product $A^{\otimes n}$ of $A$ over the underlying field (or some nice ring), then the symmetric group of rank $n$ acts on $A^{\otimes n}$ by permuting the coordinates. So now you can define the semi-direct product of $A^{\otimes n}$ with $k\mathfrak{S}_n$, which is what we call the wreath product $A\wr\mathfrak{S}_n$. For more details, you can refer to J.Chuang-K.M.Tan's Reprsentations of wreath product algebras
11h
comment Quiver algebra as a wreath product?
I am having trouble understanding your terminologies and notation (as well as the question). First, it seems to me you are asking about quiver Hecke algebras, not quiver algebra; they are two completely different thing. A quiver algebra is the algebra with basis given by compositions of arrows in a quiver. On the other hand, a quiver Hecke algebra, is an algebra defined by Hecke-like relation determined the structure of the quiver. Second, what is $k[x]^{\Omega}$? Third, what do you mean by $k[x]$ indexed by $\Omega^n$ or $\Omega$?
Jul
14
comment Question concerning a self-injective algebra and a faithful module
Faithfulness of $M$ means we have ses $0\to A\to M^r \to C\to 0$ (where $C$ is the cokernel of the injection). Self-injectivity of $A$ means that this ses splits and in particular $add(M)=add(A)$. The final question follows from the theorem (Double centraliser property of $A$ and $B$)
May
1
comment How much information about $R-\mathrm{Mod}$ can be extracted from $\underline{R-\mathrm{Mod}}$ and $K_0(R)$?
In that case, I think it is quite difficult to just use $K_0(R)$ to reconstruct $R$ - describing $R$ is the same as asking describing morphisms (in the ordinary module cat.) between the indec. projectives. $K_0(R)$ effectively forgets morphisms between indec projectives, how can we use information from $\underline{\mathrm{mod}}-R$ to get back the morphisms between them? The problem when we try to reconstruct $R$ (or any stably equiv. ring) from $\underline{\mathrm{mod}}-R$ is that many maps got killed in the stable category, and you don't know which one got killed.
Apr
30
comment How much information about $R-\mathrm{Mod}$ can be extracted from $\underline{R-\mathrm{Mod}}$ and $K_0(R)$?
Do you have any condition on your ring? Stable module categories are rather crazy things usually. For example, no method of constructing stable equivalences is known, apart from some special cases; in particular, reconstructing the ring behind such equivalence. When we stablise, many structures of the category get destroyed - most notably exactness. Anyway, I want to know what $K_0(R)$ means here, is it the group spanned by simple $R$-module or is it the one spanned by indec. projective $R$-module? And what is the relation?
Apr
11
awarded  Yearling
Apr
4
comment References request: are there some references about simple modules of group algebras?
Which field are you working on? For characteristic zero, use ATLAS of finite simple group (+basic character theory) extensively. Otherwise, it's an extremely difficult question in general. People tackle it case by case, as far as I know.
Mar
24
awarded  Nice Answer
Mar
4
answered global dimension of bounded path algebra
Feb
23
comment Is tilting theory extended also to arbitrary derived categories?
Yes, instead of defining what tilting objects are, one should define what tilting subcategories are. Also, it turns out that tilting is a little bit strong condition, one should use "silting" instead. See, for example, arXiv: 1009.3370
Jan
24
comment Merge operation in homological algebra?
nevertheless, if you want to play around with the combinatorics in the example above. Syzygy $\Omega(M)$ of $M$ is: $Q=(1\to 2)$, $\Omega(1)=2$, and $2,\substack{1\\2}$ are projective, so no syzygy. For $Q=1\to 3\leftarrow 2$, $\Omega(1)=\Omega(2)=\Omega(\substack{1 2 \\ 3})=3$, the rest are projectives, so no syzygy.
Jan
24
comment Merge operation in homological algebra?
as I said, I don't understand your idea, just as Tobias commented - we don't see why what you want to do has anything to do with homological algebra. To do homological algebra, one needs to start with some abelian category, which means one want to start with some geometric space or a ring, then consider sheaves or modules (respectively), then one can start doing homological algebra. in particular, you can only talk about syzygy of one particular sheave/module. I don't see why you are obsessed with a translation in hom. alg. (and in particular) syzygy, so I can't really answer your question.
Jan
22
answered Merge operation in homological algebra?
Dec
31
awarded  Tumbleweed
Dec
27
asked Is there any study on noncrossing partitions for cyclic polytopes?
Dec
24
asked Is the notion of strongly graded algebra a Morita invariant?
Dec
22
comment Understanding the Definition of the Tensor Product of Chain Complexes
for explanation of Koszul sign rule, read this: math.umn.edu/~tlawson/papers/signs.pdf
Dec
22
comment Extension of nonisomorphic simple objects
If you have a cycle, then the path algebra becomes infinite dimensional, and representation theory on that algebra may not work with usual quiver representation machinery. For example, $Q$ with one vertex one arrow gives path algebra $\Bbbk[x]$ (the polynomial ring).
Dec
21
comment Extension of nonisomorphic simple objects
finite oriented tree relaced by finite oriented acyclic graph is OK too (if you don't like the word (finite acyclic) quiver). Also, this example is actually the algebra of upper triangular matrix over $\Bbbk$ if one doesn't like quiver representation. The two simple modules correspond to having only entry in $(1,1)$ and $(2,2)$ (with obvious action), then the (unique) non-split extension is associated to the algebra itself.
Dec
9
comment Is there a non-projective submodule of a free module?
I like to think in finite dimensional algebras, and the example I would give is take $R$ the dual number $k[x]/(x^2)$. Obviously $R$ itself is free and projective, and the 1-dim $k$-space $k.x$ spanned by $x$ is a $R$-submodule of $R$, and $k.x\cong R/k.x$ as $R$-module. So we have non-split exact sequence $0\to k \to R\to k\to 0$ which says that $k$ is non-projective submodule of free module $R$.