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Questions tagged [super-linear-algebra]

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Chern character of a super-connection (Heat kernels and Dirac operators)

Let $A$ be a super-connection on a super-bundle $E\to M$, then the differential form \begin{equation} \mathrm{ch}(A)=\mathrm{Str}(e^{-A^2}) \end{equation} is called the chern character of $A$ on page ...
Filippo'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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7 votes
2 answers
666 views

Super mixed Hodge structures?

It's common in subjects that have some version of the "yoga of weights" that you have a functor called "Tate twist" and that the most natural version of it seems like it should be ...
Ben Webster's user avatar
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6 votes
2 answers
372 views

Inner products on super vector spaces

Let $V=V^0\oplus V^1$ be a super vector space (https://en.wikipedia.org/wiki/Super_vector_space) Is there a special definition of an inner product on $V$ other than just an inner product on the ...
BinAcker's user avatar
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6 votes
1 answer
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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
7 votes
1 answer
265 views

Chain rule for the superderivative

A one dimensional complex supermanifold $X$ is locally described by an ordinary complex coordinate $z$ and an anticommuting coordinate $\theta$, $\theta^2 = 0$. The superderivative is the square root ...
user's user avatar
  • 749
2 votes
1 answer
152 views

Notation on supergeometry — parity

I know that given a manifold $M$ and its corresponding tangent bundle $TM$ we can call $\Pi TM$ the space of forms parametrized (via charts) by $\{x_i\}_{i=1,\dotsc,n}$ and its corresponding cotangent ...
Eggon Viana's user avatar
9 votes
2 answers
316 views

Schur Weyl duality for the supergroup $\text{GL}(m|n)$

Let $G$ be the supergroup $\text{GL}(m|n)$. It has a tautological representation $V= \mathbb{C}^{m|n}$. For every natural number $d$ we have a natural map $$\Phi_d:\mathbb{C} S_d\to \text{End}_G(V^{\...
Ehud Meir's user avatar
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12 votes
1 answer
810 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-...
Tim Campion's user avatar
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2 votes
0 answers
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$\mathbb{Z}_2$ 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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3 votes
2 answers
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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
5 votes
1 answer
202 views

Alternating elements in free graded-commutative algebras

It is classical that every alternating polynomial is (uniquely) the product of a symmetric polynomial with the Vandermonde polynomial, in particular the alternating polynomials are a free rank-one ...
Matthias Wendt's user avatar
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
1 vote
1 answer
225 views

Contraction of graded vector fields on de Rham complex

Given a commutative algebra $A$ smooth over a field $k$ of characteristic zero, the module of K\"ahler differentials $\Omega^{1}$ is projective of finite rank and so the sum of all wedge powers $\...
dhagbert's user avatar
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0 answers
798 views

In what sense does the Berezinian generalize the determinant?

One way of defining the determinant of a endomorphism of a vector space $\varphi:V \to V$ is by using the action of $End(V)$ on the underlying $\mathbb{Z}$-graded vector space of the exterior algebra $...
David Carchedi's user avatar
9 votes
1 answer
736 views

Strange boundary-like map on tensor algebra: what is its kernel?

Let $k$ be a commutative ring and $L$ a $k$-module. The tensor algebra $\otimes L$ is $\mathbb{Z}$-graded and $\mathbb{Z}_2$-graded (an element of $L^{\otimes n}$ has degree $n$ and $\mathbb{Z}_2$-...
darij grinberg's user avatar
15 votes
3 answers
855 views

Are supervector spaces the representations of a Hopf algebra?

Supervector spaces look a lot like the category of representations of $\mathbb{Z}/2\mathbb{Z}$ - the even part corresponds to the copies of the trivial representation and the odd part corresponds to ...
Qiaochu Yuan's user avatar