All Questions
Tagged with noncommutative-algebra representation-theory
111 questions
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18
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Block decomposition of equivariant maps using Wedderburn-Artin theorem
Let $K$ be a field, $A$ an Artinian simple $K$-algebra with minimal left ideal $M$.
We can view $M$ as a simple $A$-module, so $D^{op}:=\text{End}_A(M)$ is a $K$-division algebra by Schur's lemma. ...
- 2,596
0
votes
0
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39
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Real division algebras with involution
Let $W$ be an irreducible $\mathbb{R}$-representation of finite group $G$. Schur's lemma implies that $\text{End}_{\mathbb{R}G}(W)$ is a division ring. Since its centre is $\mathbb{R}$, $\text{End}_{\...
- 2,596
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0
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31
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Involution and inner product in semisimple group algebras
Let $\mathbb{F}$ be a field with characteristic zero and $G$ be a finite group. The group algebra $\mathbb{F}G$ is semisimple, and so
$$\mathbb{F}G=\mathbb{F}Ge_1\times\cdots \times \mathbb{F}Ge_k,$$
...
- 2,596
2
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1
answer
71
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Finite dimensional Irreps (of algebras) with same traces must be equivalent ('page 136' in Bourbaki)
I look for the reference (or proof) of the following fact which is from appendix (B $27$) of Dixmier's book on $C^*$-algebras.
Claim: Let $A$ be an algebra (not necessarily commutative) over a field $...
- 1,709
1
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0
answers
73
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Conceptual definition of the Auslander-Reiten translate
In homological algebra, we learn to differentiate between
The conceptual definition.
A computation, which is done by choosing efficient resolutions.
The only definition I've seen of the Auslander-...
- 429
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votes
1
answer
45
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If R is semisimple left Artinian, then R is its own quotient ring.
I am trying to do the following exercises in Hungerford.
If $R$ is semisimple left Artinian, then $R$ is its own quotient ring.
By the Wedderburn-Artin theorem we know that $R$ is isomorphic to sum of ...
- 25
1
vote
1
answer
110
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Finite Group's Group Algebra over a Field is a Principal Ideal Ring?
A (left) principal ideal ring (PIR for short) is a ring such that for every left-ideal $I$, there exists a $a \in R$ such that $I=Ra$.
Firstly, similar to the case of PIDs, is that a ring $R$ is a PIR ...
- 403
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1
answer
94
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Linear combination of a character
Hi there I am trying to solve a problem about characters of a finite group $G$. If $\chi$ is a character such that $\langle \chi,\chi \rangle = 2$ and $\chi_1,\dotsc,\chi_n$ the irreducible characters ...
5
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0
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221
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Classification of 4 dimensional real associative unital algebra
I think I have a complete list for all the commutative ones, maybe with possible repeats (I did try my best to make sure none are same up to isomorphism):
$\mathbb{R}^4 \simeq \begin{bmatrix}a&0&...
- 553
1
vote
1
answer
197
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Isotypical decomposition of a semisimple module
I was studying Schur-Weyl duality and the double centralizer theorem and following the proof of the double c.t. by Etingof. Now in the third part he decomposes a module over a semisimple algebra in ...
- 169
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1
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39
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Subalgebra with semisimple centralizer itself semisimple?
Let $V$ be a finite-dimensional $K$-vector space and suppose that $\mathcal{B} \subseteq \operatorname{End}_K(V)$ is a $K$-subalgebra such that the centralizer $$\mathcal{Z}_{\operatorname{End}_K(V)}(\...
- 1,799
4
votes
1
answer
168
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Central-simple algebras, crossed products, and $2$-cocycles
In the following composition is left to right and written exponentially, tensor products over $K$ are written $\otimes$.
The context is as follows: $A$ is a central-simple $K$-algebra, $L/K$ is finite ...
- 1,484
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71
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Proving an algebra is nilpotent
I recently have started studyng about free algebras and I want to know what kind of methods or approaches are there to prove that free algebra $A$ is nilpotent with nilpotency index $n.$
Let me remind ...
- 19
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57
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Is it known the $S_n$-module structures of free anti-commutative algebra
I recently started studiyng representation theory, especially I am interested in $S_n$-module structures of free algebras over some variety. I know that the $S_n$-module structures are known for some ...
- 19
2
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2
answers
153
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Division algebras are frobenius algebras
I am following the book Frobenius Algebras I by Andrzej Skowronski and Kunio Yamagata to learn about Frobenius algebras. The goal of Chapter IV, section 5 is to show that finite dimensional semisimple ...
- 652
2
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1
answer
133
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Are Weyl algebra $A_1$ and it's opposite algebra isomorphic?
Let $A$ be a noncommutative ring and $ab=c$ in $A$. $A'$ is it's opposite ring if $ba=c$ in $A'$. If $A$ is a Weyl algebra $A_1$, are $A$ and $A'$ isomorphic?
I have an idea, but I think it's wrong. ...
- 21
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0
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73
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A generalization of the Clifford algebra
A minimal example of the Clifford algebra is the $\mathbb{C}$-algebra (unital, associative) generated by $x,y$ quotient over the relations
\begin{eqnarray}
x^2&=&1,\tag{1}\\
y^2&=&1,\...
- 876
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votes
0
answers
46
views
Subalgebras which are semisimple
Let $k$ be a field.
Let $A$ be a $k$-algebra and $B \subset A$ be a subalgebra of $A$ which is a semisimple algebra.
Note that we do not assume that $A$ is commutative.
We also assume that $A$ is a ...
- 435
5
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240
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Jacobson density theorem, and its relations to Artin-Wedderburn, and double centralizer theorems
$\newcommand{\End}{\operatorname{End}}\newcommand{\Hom}{\operatorname{Hom}}$
On pg. 647 of Lang's Algebra, Lang proves the Jacobson density theorem by doing some stuff with $\End_{\End_R(V^n)}(V^n)$ ...
- 9,195
1
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0
answers
82
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I've proven Krull-Schmidt for arbitrary decomposition into indecomposables. What's wrong in my proof?
My question is probably stupid and I'm likely committing a very trivial mistake. It's well known that the uniqueness of decomposition of modules into indecomposable submodules ${}_A M = \bigoplus_{I \...
- 5,423
2
votes
1
answer
60
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Sum on product of two charcteres, which runs on symmetric generating set
Let $G$ be a finite (not necessarily abelian) group and let $S$ be a symmetric generating set of $G$, i.e. if $s\in S$ then $s^{-1} \in S$.
Let $\chi$ be an irreducible character of $G$. I have ...
- 31
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0
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108
views
Global dimension of a quiver with relations that do not overlap
I am reading Green, Hille & Schroll's paper Algebras and Varieties and find myself a bit confused with Proposition 5.1, which says:
Consider a field $K$, a quiver $Q$ and a tip-reduced nonempty ...
- 121
1
vote
0
answers
19
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$T(z) e^{-\partial_z} $ for Yangian is a Manin matrix
Let $T(u)$ be the generating matrix of the Yangian $Y(\mathfrak{gl}_n)$ of $\mathfrak{gl}_n$. So we have the identity $[T_{ij}(u),T_{kl}(v)]=\frac{1}{u-v}(T_{kj}(u)T_{il}(v)-T_{kj}(v)T_{il}(u))$.
We ...
- 155
1
vote
1
answer
125
views
Casimir element of a semisimple Lie algebra is "additive"
Definitions
Let $\mathfrak{g}_1$ and $\mathfrak{g}_2$ be semisimple Lie algebras, and suppose that $(V,\rho)$ is a representation of $\mathfrak{g} = \mathfrak{g}_1 \oplus \mathfrak{g}_2$. Given a ...
- 3,931
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1
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200
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Finitistic dimension conjeture for $A^{op} $ implies the strong Nakayama conjecture for A
I have some trouble with some detail in the proof of the following theorem. Assume that the Finitistic dimension conjecture is true for $ A^{op} $ that is $ sup\{ proj.dim(M) \vert M \in mod(A^{op}) ~...
- 83
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32
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Fitting's Lemma version for pseudocompact modules or linearly compact modules
Let $R$ be a pseudocompact ring or a linearly compact ring. Is there a version of Fitting's Lemma for pseudocompoact or linearly compact $R$-modules?
- 21
3
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63
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$\mathbb{Q}$-representation and Galois theory
Let $G=\mathbb{Z}/n\mathbb{Z}$, we know that all irreducible $\mathbb{Q}$-representation of $G$ is all the subfields of the cyclotomic extension $\mathbb{Q} [\zeta_n]/\mathbb{Q}$, but we know that $\...
- 900
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1
answer
107
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Prove $\mathrm{Hom}_{R}(M_{1}, M_{2})=K$ and $\mathrm{Hom}_{R}(M_{2},M_{1})= 0$ for $M_{1}$ and $M_{2}$ modules over a matrix ring and $K$ a field.
Filling a gap of an example I got the following problem:
Let $K$ be a field. If we consider the ring $$R=\begin{pmatrix} K & 0\\ K^{(\mathbb{N})} & K \end{pmatrix}$$ and the idempotents $\...
- 1,975
1
vote
1
answer
157
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Computing the the lattice of left ideals of the lower triangular matrix ring and proving it is left hereditary.
Let $K$ be a field and $$R=\begin{pmatrix} K & 0\\ K & K \end{pmatrix}$$ the ring of lower matrix with coefficients in $K$. I want to find the left ideals of $R$ and also prove that $R$ is an ...
- 1,975
2
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1
answer
348
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Induced module; what is the $K[G]$-action?
Given a finite group $G$, $H$ a subgroup, a field $K$, and a $K[H]$-module $M$, define
$$\operatorname{ind}_H^G M := \lbrace f : G \to M : f(gh) = h^{-1}f(g)\ \text{for all}\ g \in G, h \in H\rbrace....
- 4,688
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0
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39
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Dimension of the dual of a simple module over a simple $\mathbb{C}$-algebra.
I am given that A is a simple finite-dimensional associative unital algebra over the $\mathbb{C}$, and $M$ is a simple $A$-module. Furthermore, $V_M = \text{Hom}_A(M, A)$ is a right A-module with the ...
- 3,931
3
votes
2
answers
777
views
Uniqueness of Artin-Wedderburn decomposition
I am studying Artin-Wedderburn structural decomposition theorem for semisimple rings. I understand that it says that any semisimple ring, $R$ is isomorphic (as rings) to $M_{n_1}(D_1) \times M_{n_2}(...
- 517
2
votes
1
answer
546
views
Tensor product of matrix algebras over field
I want to prove that $Mat_{n_1}(k) \otimes_k Mat_{n_2}(k) \cong Mat_{n_1n_2}(k) $ (as $k$-algebras) where $k$ is a field by checking the universal property.
Namely, I need to inclusion $Mat_{n_{1,...
- 2,678
0
votes
1
answer
110
views
Describing all the irreducible and f.g. completely reducible modules over $\mathbb{C}[x,y]$
I couldn't even find all the irreducible modules over $\mathbb{C}[x,y]$ yet.
I know that modules over $\mathbb{C}[x,y]$ is the same as $\mathbb{C}$-vector spaces $V$ equipped with two commuting ...
- 2,678
1
vote
1
answer
337
views
Categorical duals for finitely-generated projective modules
For a not necessarily commutative ring $R$, its category of bimodules $_BMod_B$ has a monoidal structure given by $\otimes_R$. Consider now an object $M$ in the subcategory whose objects are finitely-...
- 269
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vote
1
answer
91
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Example of a finite group for which $\mathbb Q$ is not injective.
I had to prove that the cohomology with rational coefficients of any finite group is trivial. I tried to prove it by proving that $\mathbb Q$ is an injective $G$-module, however it doesn't seem to me ...
- 1,210
2
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1
answer
114
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Is a (non-commutative) quasi-Frobenius regular ring semisimple?
Let $R$ be a non-commutative ring which is quasi-Frobenius and regular. Is $R$ a semisimple ring?
Recall:
quasi-Frobenius means a right $R$-module is projective if and only if it's injective (among ...
- 43
7
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1
answer
200
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Socle of sum is the sum of socles
Let $k$ be a field and $A$ a finite-dimensional unital associative $k$-algebra. For a finite-dimensional left $A$-module $V$, the socle of $V$, written $\operatorname{soc}V$, is defined to be the sum ...
- 1,892
1
vote
1
answer
149
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Module is $\mathbb{F}_2[S_3]$-projective iff it's $\mathbb{F}_2[S_2]$-projective.
Let $S_2\subset S_3$ be the subset of the symmetric group that fixes 3, i.e. $S_2=\{1, (12)\}$. Consider the group algebras $A:= \mathbb{F}_2[S_2]\hookrightarrow \mathbb{F}_2[S_3]=: B$. If $M$ is a (...
- 1,090
7
votes
1
answer
344
views
Jacobson radical of a subring
Let $A$ be any ring and $B$ be its subring.
Is it true that $J(A)\cap B\subseteq J(B)$?
If not, is it true that for a finitely generated algebra $A$ over a local commutative noetherian ring $\...
- 533
1
vote
1
answer
67
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A certain decomposition of a semisimple Hopf algebra
$\newcommand{\Irr}{\mathrm{Irr}}$
Let $H$ be a finite-dimensional semisimple Hopf algebra over an algebraically closed field $k$, and let $\Irr(H)$ be the set of (choices of representatives of) ...
- 2,115
0
votes
1
answer
501
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Artin-wedderburn and simple modules
Given a finite-dimensional semisimple algebra $A$ over algebraically-closed field $k$, Artin-Wedderburn says it is isomorphic to the direct sum of matrix rings $M_{n_i}(k)$. If $\dim(A)=n$ then $\...
- 479
3
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0
answers
926
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Artin-Wedderburn theorem and representation theory
How to apply the Artin-Weddernburn theorem to representation theory? The Artin-Weddeburn theorem says that if $R$ is finite dimensional $k$ algebra( where $k$ is an algeraically closed field), then $R=...
- 53
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votes
0
answers
95
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Computing Ext functor when no projective modules are known.
I have a $p^3$-dimensional (not semisimple) algebra $\mathcal{U}_{q}(sl_2)$ over $\mathbb{C}$ and i know how all its simple modules look like (there are $p$ of them, each $M_i$ has dimension $i$ for $...
- 679
3
votes
1
answer
92
views
Why is $\Omega (M)$ a superfluous submodule?
I'm struggling with a result that seems intuitive and that authors don't even bother to prove, so I think there's something stupid that I'm not seeing.
Let $M$ be a finitely generated module over $...
- 44.4k
4
votes
1
answer
830
views
Representation of an algebra is absolutely irreducible if and only the representation map is surjective
This should be well known but I can't seem to locate a reference:
Let $k$ be a field, $V$ a $n$-dimensional vector space over $k$ with an action of a $k$-algebra $A$. We say that $V$ is an absolutely ...
- 8,369
3
votes
1
answer
171
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Showing a certain endomorphism algebra is basic
Let $A$ be a finite dimensional algebra over $\mathbb{C}$. Let $P_1, \dots P_n$ be the projective indecomposable $A$-modules, pairwise non-isomorphic, and define $P=P_1\oplus\dots\oplus P_n$.
Let $B=\...
- 334
7
votes
0
answers
1k
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Deriving the projection onto the isotypic component
Let $k$ be a field, and $G$ a finite group such that the characteristic of $k$ does not divide $|G|$. Then $kG$ is a semisimple $k$-algebra, and the representation theory of $G$ over $k$ is semisimple....
- 13.5k
0
votes
1
answer
79
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The possible representations of a monoid derived from four "standard" or "regular" presentations and their relation
Given some monoid $M$ we can form the algebra $\mathbb C[M]$ by considering all formal sums and lineary extending the given multiplication in $M$.
Let $f = \sum_{x\in M} \lambda_x x$ and $m\in M$. ...
- 18.4k
1
vote
0
answers
117
views
Classifying irreducible real representations
Let $G$ be a group and say $V$ is an irreducible representation over $\mathbb{R}$. Then $End_G(V) = End_{\mathbb{R}[G]}(V)$ must be a division algebra, since $V$ is a simple $\mathbb{R}[G]$-module. ...
- 1,090