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Questions tagged [super-algebra]

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Irreducible representations of $\mathfrak{sl}(m|n)$

It is well known that the irreducible representations of the Lie algebra $\mathfrak{sl}(n)$ are symmetric powers of a vector space of dimension $n$. This can be viewed for instance using Young ...
User43029's user avatar
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Shouldn't $\mathrm{End}_{C(TM)}(E)$ be defined differently in Heat Kernels and Dirac Operators?

The first four chapters of the book lead up to the proof of theorem 4.1. Its main consequence is that it provides the local index theorem for Dirac operators. The statement of theorem 4.1 involves a ...
Filippo's user avatar
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smooth super scheme which is not smooth

I am following the very nice "Notes on fundamental algebraic supergeometry. Hilbert and Picard superschemes" by Bruzzo, Ruiperez and Polishchuk. I am having some problem in order to give ...
User43029's user avatar
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5 votes
1 answer
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Real forms of the general linear Lie superalgebra

I'm interested in a classification of the real forms of the general linear Lie superalgebra $\mathfrak{gl}_{m|m}(\mathbb{C})$. The real forms of the simple complex Lie superalgebras were classified by ...
Alistair Savage's user avatar
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Existence of a minimal ideal with a specific property

Suppose that $R$ is a super-commutative ring (i.e. it is a unital $\mathbb{Z}_2$-graded ring satisfying $xy=(-1)^{|x|\cdot |y|}yx$ where $|x|$ denotes the grading degree of a homogeneous element $x\in ...
FNH's user avatar
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Even and odd part of the Hochschild and cyclic homology of a super-algebra

Let $A$ be a $\mathbb Z_2$-graded $k$-algebra, where $k$ is a field of characteristic $0$. Then we know that the tensor product of $A$ with itself is also $\mathbb Z_2$-graded by $$(A\otimes_k A)_0:=...
Flavius Aetius's user avatar
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Formulation of matrix representation of morphisms between free super modules

I asked this question in MathStackExchange 9 days ago but get no response (not a vote nor a comment), so I'm copying it here below. The link to the original question is: https://math.stackexchange.com/...
Shana's user avatar
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C^*-algebra theory with all the Koszul signs

I was wondering if someone knows of a reference in which $\mathbb{Z}_2$-graded $C^*$-algebra theory is developed using the sign convention $(ab)^* = (-1)^{|a||b|}b^* a^*$. I would be most enthusiastic ...
Luuk Stehouwer's user avatar
8 votes
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200 views

Cohomology algebra of the maximal nilpotent subalgebra of a semisimple Lie algebra

The answer to this question may be well-known, but I failed to locate it in any obvious source. From the results of Bott (Ann. Math. 66 (1957), 203-248) and Kostant (Ann. Math. 74 (1961), 329-387), it ...
Vladimir Dotsenko's user avatar
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necessary or sufficient condition for super commutation of matrices

We have many results on commutativity of two complex matrices. For example, it two matrices are simultaneously diagonalisable then they commute. I would like to know a similar result for super ...
GA316's user avatar
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Constructions with Superschemes via Kan extensions

Let $\operatorname{CAlg}$ be the category of commutative rings (with unit) and $\operatorname{S-CAlg}$ the category of supercommutative $\mathbb{Z}/2$-graded rings. Then we have an adjoint triple (as ...
Markus Zetto's user avatar
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Does there exist a type of discriminant not only for irreducible polynomials but also for exponential functions, logarithm functions?

I think discriminant is the strongest tool that I've used_ https://math.stackexchange.com/q/4035405/822157, however, does there exist a type of discriminant not only for irreducible polynomials but ...
user avatar
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1 answer
207 views

Cayley-Hamilton over super rings

If $R$ is a commutative ring, then the Cayley-Hamilton theorem states that any endomorphism $\phi: R^{n} \rightarrow R^{n}$ of a rank $n$ free module satisfies its own characteristic polynomial, in ...
Piotr Pstrągowski's user avatar
3 votes
2 answers
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Polynomial identities of supercommutative-gradable algebras

All algebras below are associative, and not assumed unital, and, to fix ideas, over the complex numbers. An algebra $A$ is supercommutative-gradable if it admits a grading $A=A_0\oplus A_1$ in $\...
YCor's user avatar
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12 votes
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Is the category $\operatorname{sVect}$ an "algebraic closure" of $\operatorname{Vect}$?

$\DeclareMathOperator\sVect{sVect}\DeclareMathOperator\Vect{Vect}$The category $\sVect_k$ of (let's say finite-dimensional) super vector spaces can be obtained from the category $\Vect_k$ of (finite-...
Tim Campion's user avatar
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How to formulate supercommutativity in a characteristic free way?

I would never dare posting this here, but the question https://math.stackexchange.com/q/3019853/214353 on math.SE did not receive any feedback (except for 13 views and one upvote) since November 30, ...
მამუკა ჯიბლაძე's user avatar
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1 answer
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Classification of finite-dimensional real super C*-algebras

The title says it all. I feel like one should be able to find this somewhere, but every time I try to google, I just get results for "super Lie algebras". Does anybody know a reference? I am not so ...
Andi Bauer's user avatar
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20 votes
1 answer
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Odd primary dual Steenrod algebra

My question is related to this, this, and this older questions. Let $\mathcal A_*$ be the dual Steenrod algebra. This is a super-commutative Hopf algebra, and so its $Spec$ is an algebraic super-group....
André Henriques's user avatar
1 vote
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symmetric polynomials for Super Hecke Clifford algebra

Fix a natural number $n$. In https://arxiv.org/abs/1107.1039, §3.5, Kang/Kashiwara/Tsuchioka define a (version of a) Hecke Clifford superalgebra. It is the superalgebra with the following generators: ...
Bubaya's user avatar
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2 votes
1 answer
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Noncommutative cohomology of flag varieties

Consider the Grassmannian $Gr(n, N)$ of $n$-dimensional subspaces of $\mathbf C^N$. Its cohomology ring is isomorphic to $\mathbf C[x_1, \ldots, x_n, \bar x_1, \ldots \bar x_{N-n}]/I_{n,N}$, wherethe ...
Bubaya's user avatar
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3 votes
1 answer
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Generators of odd polynomial superalgbras

I am getting myself acquainted to superalgebras. One often comes across odd polynomial rings of the form $$k\langle x_i \rangle_{i\in I} / (x_ix_j -(-1)^{|x_i||x_j|} x_jx_i)$$ for some index set $I$,...
Bubaya's user avatar
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5 votes
1 answer
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Definition of orthosymplectic supergroups

I found two versions of definitions of orthosymplectic supergroups. It seems that they are not equivalent. I don't know which version of the definition is standard. The first version of the ...
Jianrong Li's user avatar
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1 vote
1 answer
409 views

Super version of Poisson brackets of tensor products

Let $A$ be a Poisson super algebra ($A$ is a super algebra and $A$ satisfies super Jacobi identity, super commutativity, super Leibniz rule). Super version of the product of two tensor products is \...
Jianrong Li's user avatar
  • 6,201
2 votes
0 answers
74 views

$\mathbb{Z}_2$ graded analog of row operations for supermatrices

I'm working on some research involving supermatrices, and I was wondering if there was a $\mathbb{Z}_2$ graded analog of row operations for supermatrices. It seems to me that it makes sense to have ...
Alec Rhea's user avatar
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10 votes
1 answer
413 views

Super-plethysm?

Let $U$ be a representation of $S_m$ and $V$ a representation of $S_n$. Then the representation $\operatorname{Ind}_{S_m\wr S_n}^{S_{mn}}(U^{\otimes{n}}\otimes V)$ has a nice interpretation in terms ...
Nicholas Proudfoot's user avatar
6 votes
1 answer
301 views

Is the (super-)symmetric power of the exterior algebra free?

Let $V$ be a vector space over $k$ of dimension $m$. (I'm only interested in the case $k=\mathbb{Q}$.) Let $R:=\Lambda^*V$ be the exterior algebra. It carries the structure of a supercommutative ring: ...
Nicolas Malebranche's user avatar
3 votes
2 answers
270 views

Supercommutator of exterior multiplication operators and their adjoints

Let $\mathfrak{h}$ be a complex Hilbert space and consider Grassmann algebra $\mathcal{F}=\bigwedge\mathfrak{h}$ with its induced inner product. For $\omega\in\mathcal{F}$ we also consider the ...
Robert Rauch's user avatar
1 vote
1 answer
147 views

Non-graded representations over Lie superalgebra $\mathfrak{gl}(m,n)$

I have the following questions: Let $m,n$ be positive integers. Consider representations over the general linear Lie super-algebra $\mathfrak{gl}(m,n)$. Namely, modules over the associative algebra $U(...
GuNa's user avatar
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4 votes
1 answer
370 views

Quasicoherent sheaves on superschemes

I am interested in learning about super algebraic geometry (some objects I am studying seem to be naturally superstacks, at least in some sense). What would be the best reference for the subject? I am ...
Denis Nardin's user avatar
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11 votes
0 answers
668 views

What is the role of fiber functor in Deligne's theorem on Tannakian categories?

The theorem states that, for a field $k$ of characteristic 0, any $k$-linear tensor category with $End(1)=k$ satisfying a condition that each object is annihilated by a Schur functor, is equivalent to ...
Anonymous's user avatar
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3 votes
2 answers
323 views

How well is the classification of low-dimensional semisimple Hopf superalgebras (or braided Hopf algebras) understood?

As far as I know, low-dimensional semisimple Hopf algebras are classified (along with non-semisimple ones) up to dimension 60, with the first example of a semisimple Hopf algebra not coming from a ...
Manuel Bärenz's user avatar
1 vote
0 answers
178 views

Supertrace on Weyl algebra

Consider Weyl algebra, i.e. the algebra of $x^i$ and $p_i=\frac{\partial}{\partial x^i}$, its elements are differential operators $F(x,p)$. Weyl algebra is $\mathbb{Z}_2$ graded, hence one ask if ...
John's user avatar
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0 answers
113 views

Tensor categories with integer rank

I wonder the state of the following conjecture in "Deformation theory, Kontsevich, Soibelman": Conjecture 3.3.5. Rigid [abelian symmetric] tensor categories [over an algebraically closed field $k$] ...
Mostafa - Free Palestine's user avatar
4 votes
1 answer
178 views

Does every equivalence class in a Brauer-Wall group have a (graded) division algebra?

It is known that each equivalence class in a Brauer group has a division algebra (or, in other words, every central simple algebra is isomorphic to $\mathrm{Mat}(D)$ for some division algebra $D$). Is ...
user50311's user avatar
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3 votes
1 answer
586 views

A good reference for learning about super-differentiation & super-integration?

I've looked at a couple of books for basic information for super-differentiation & super-integration - Rogers Supermanifolds, and Khrennikovs Superanalysis. Unfortunately both books lack a clear ...
Mozibur Ullah's user avatar
4 votes
0 answers
218 views

FRT construction in the case of super algebras

I'm looking on papers which are talking about the super quantum algebra osp(2|1). I want to understand how one applies the FRT construction in the case of osp(2|1). Of course there is a super ...
kieffer's user avatar
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0 answers
215 views

Super group GL(m,m) and Koszul (deRham) complex. (Is there brigde from super-math to usual-math ?)

Consider vector space with coordinates x1, ... xn. Consider polynomial deRham complex (also known as Koszul complex) which is generated by xi and dx_i. As an algebra it is just $C[x_i]\otimes \Lambda [...
Alexander Chervov's user avatar
8 votes
2 answers
976 views

How can I write down a point in the Berezinian of a super vector space?

A vector space $V$ of dimension $n$ has an associated determinant line $Det(V)$. An element of $Det(V)$ is represented as a (formal limear combination) of expresstions of the form $v_1 \wedge v_2 \...
André Henriques's user avatar
3 votes
1 answer
1k views

Witten Index, letter partition function and superconformal representations.

Except in a few papers I have seen so little written about this that I am not sure I can even frame this question properly. I would like to know of expository references and explanations on the ...
Anirbit's user avatar
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8 votes
3 answers
1k views

Derivations of C(X)? or Why Must Supermanifolds be Smooth?

What are the derivations of the algebra of continuous functions on a topological manifold? A supermanifold is a locally ringed space (X,O) whose underlying space is a smooth manifold X, and whose ...
Chris Schommer-Pries's user avatar
7 votes
1 answer
498 views

Does some version of U_q(gl(1|1)) have a basis like Lusztig's basis for \dot{U(sl_2)}?

There's a non-unital algebra $\dot{U}$ formed from $U_q (sl_2)$ by including a system of mutually orthogonal idempotents $1_n$, indexed by the weight lattice. You can think of this as a category with ...
Sammy Black's user avatar
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