Questions tagged [super-linear-algebra]
The super-linear-algebra tag has no usage guidance.
17 questions
1
vote
1
answer
90
views
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 ...
1
vote
0
answers
50
views
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/...
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 ...
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 ...
6
votes
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 ...
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 ...
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 ...
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^{\...
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-...
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 ...
3
votes
2
answers
269
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 ...
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 ...
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 ...
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 $\...
5
votes
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 $...
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 $\mathbb{Z}_2$-...
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 ...