Questions tagged [super-algebra]

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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/...
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5 votes
0 answers
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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 ...
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8 votes
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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 ...
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1 vote
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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 ...
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7 votes
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206 views

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 ...
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3 votes
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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 ...
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6 votes
1 answer
152 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 ...
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3 votes
2 answers
176 views

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 $\...
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10 votes
1 answer
534 views

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-...
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4 votes
0 answers
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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, ...
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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 ...
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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....
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1 vote
0 answers
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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: ...
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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 ...
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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 $I$,...
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4 votes
1 answer
505 views

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 ...
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1 vote
1 answer
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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 \...
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2 votes
0 answers
67 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 ...
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10 votes
1 answer
358 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 ...
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6 votes
1 answer
276 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: ...
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3 votes
1 answer
209 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 ...
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1 vote
1 answer
132 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(...
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4 votes
1 answer
328 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 ...
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11 votes
0 answers
601 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 ...
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3 votes
2 answers
268 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 ...
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1 vote
0 answers
166 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 ...
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4 votes
0 answers
110 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 $k$] ...
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4 votes
1 answer
146 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 ...
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3 votes
1 answer
529 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 ...
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4 votes
0 answers
179 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 ...
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4 votes
0 answers
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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 [...
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8 votes
2 answers
877 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 \...
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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 ...
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8 votes
3 answers
940 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 ...
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7 votes
1 answer
485 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 ...
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