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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 underlying vector space (forgetting the grading)?

  • Does the supertrace have a role in the induced inner product on the space $End(V)$ just like the trace has a role in the non graded case?

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  • $\begingroup$ I would be interested in a reference answering this question! $\endgroup$
    – Jonas
    Commented Sep 29, 2022 at 8:54

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I am missing well written references for this (sign) sensitive topic! But here is what I learned from a few talks I have attended:

An inner product on the super vector space $V = V^0 \oplus V^1$ should be a morphism $B:V\otimes V \to \mathbb{K}$ of super vector spaces. The standard convention is to have $\mathbb{K} = \mathbb{K}^{1|0}$ to be purely even. (But I have seen people use $\mathbb{K}^{0|1}$ or $\mathbb{K}^{1|1}$ instead.)

Then additionally one can require $B$ to be an even morphism, meaning that it vanishes on odd elements of $V\otimes V$, implying that homogenous vectors of opposite parity are orthogonal to each other.

Now $B$ is called symmetric if $B \circ c = B$ where $c:V\otimes V \overset{\sim}{\to} V\otimes V$ is the braiding which on homogenous elements $v,w$ is given by $v\otimes w \mapsto (-1)^{|v||w|}w\otimes v$.

(edited:) Assume now $V$ is finite dimensional. Then $\mathrm{str}:\mathrm{End}(V) \to \mathbb{K}$ can be defined to be the unique linear map vanishing on superommutators with normalisation $\mathrm{str}(\mathrm{id}) = \mathrm{sdim} V = \dim V^0 - \dim V^1$.

Then \begin{align*} \mathrm{End}(V)\otimes \mathrm{End}(V) &\to \mathbb{K}\\ A\otimes B &\mapsto \mathrm{str}(AB) \end{align*} defines an even, symmetric, non-degenerate bilinear form.

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  • $\begingroup$ Perhaps I am confused. Doesn't your symmetry condition mean that the restriction of B to the odd part of V is a (non-super) anti-symmetric form? So a symplectic form if non-degenerate. Then it would not have the basis you claim, but a different one. $\endgroup$
    – Chris Schommer-Pries
    Commented Sep 30, 2022 at 14:05
  • $\begingroup$ No, you are right it should be a (non-super) symplectic basis of $V^1$. I will edit the answer, thanks. $\endgroup$
    – Jonas
    Commented Oct 2, 2022 at 9:32
  • $\begingroup$ @ChrisSchommer-Pries I removed the choice of basis and the presentation of the trace in coordinates. I believe one should actually use a 'difference' of two scalar products on $V^0$ and $V^1$ to do this (then negative definite on $V^1$), but this seems arbitrary to me. $\endgroup$
    – Jonas
    Commented Oct 3, 2022 at 13:39
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In addition to the other answer, a very good reference on that matter is the book of Yuri Ivanovich Manin "Gauge Field Theory and Complex Geometry", specifically Chapter 3 "Introduction to superalgebra", even more specifically Section 5 "Scalar products" of that chapter.

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