Putting an inner product on superspace

Question

I was wondering if there is a natural way of explicitly defining an inner product on superspace.

In this paper, for a point x = $x_1,x_2,...,x_{n+1},\eta_1,\xi_1,\eta_2,\xi_2,...,\eta_m,\xi_m \in \mathbb{R}^{n+1|2m}$ an inner product is defined as (x,x) $= x_1^2+x_2^2+...+x_{n+1}^2 + 2\eta_1\xi_1+2\eta_2\xi_2+...+2\eta_m\xi_m$.

This is confusing to me as if we apply the anticommutativity property and swap two odd numbers in the coordinates of a point we can get two different results for the inner product (e.g. consider the product of $(0,1)=-(1,0)$ with $(1,0)$ in $\mathbb{R}^{0|2}$), so it doesn't seem well defined. We also get the strange result that some points are simultaneously orthogonal and non orthogonal which is even weirder.

What I would like help with is:

1. Am I misunderstanding the above inner product from the paper I cited somehow? It seems strange how something with weird properties would be considered an inner product, but I'm not sure what I'm missing.
2. If not, what is a way of defining an inner product on superspace?

Answer 1

I believe that the first question has been addressed in the comments. To answer your second question:

On an ordinary vector space $V$, you want an inner product to be a symmetric non-degenerate bilinear form $\text{Sym}^2(V)\to \mathbb{R}$. The whole 'super' construction arises from replacing ordinary commutativity by graded commutativity. Thus, you want the inner product to be graded symmetric. Put in formulas, let $V$ be a super vector space, and denote for parity-homogenous elements $v,w\in V$ their parities by $p(v)$ and $p(w)$, respectively. Any inner product should satisfy $$ g(v,w) = (-1)^{p(v)p(w)} g(w,v)$$ (extended to non-homogenous elements by $\mathbb R$-bilinearity). In particular, if both $v$ and $w$ are odd ($p(v)=p(w)=1$), it should hold $g(v,w)=-g(w,v)$, such that on the odd part, $g$ looks like a symplectic form.

The mathematical construction is as follows: The category $\mathbf{sVect}_\mathbb R$ of super vector spaces constitutes an abelian category with the commutativity isomorphism $$\sigma: V\otimes W\xrightarrow{\sim} W\otimes V,\quad v\otimes w\mapsto (-1)^{p(v)p(w)} w\otimes v, $$ extended to non-homogenous elements by linearity. This tells you what the appropriate interpretations of e.g. $\text{Sym}^\bullet(V)$ and $\wedge^\bullet(V)$ are. In particular, given a super vector space $V$, an inner product $g$ is just a non-degenerate element of $\text{Sym}^2(V)^\vee$.

This second point of view also generalises straight-forwardly to the case where you replace super vector spaces by free modules over a supercommutative ring $R$ (e.g., if you extend the ring of functions by odd generators, i.e., introducing 'Grassmann numbers'); in which case you simply replace the tensor product by the graded tensor product over $R$: The commutativity isomorphism in $\mathbf{freeMod}_R$ is given by $$\sigma:V\otimes_R W\xrightarrow{\sim} W\otimes_R V,\quad v\otimes_R w \mapsto (-1)^{p(v)p(w)}w\otimes_R v.$$ This is what is done in the paper: Consider $\mathbb R^{0|2}$ with (odd!) basis $(e,f)$, and furnish it with additional odd numbers $\xi,\eta$ (strictly speaking, we consider $R^{0|2}$, where $R=\wedge[\eta,\xi]$). Thus, products of the form $\xi e$ or $\eta f$ have even parity. We can define an inner product by $$ 1 = g(e,f) = -g(f,e),\quad g(e,e)=g(f,f)=0$$ (note that it is graded symmetric), such that $$ \begin{aligned} g(\xi e + \eta f, \xi e + \eta f) &= g(\xi e, \eta f) + g(\eta f, \xi e)\\ &= -\xi\eta g(e,f) - \eta\xi g(f,e) \\ &= -\xi\eta + \eta\xi = 2\eta\xi,\end{aligned}$$ which illustrates exactly what happens in the paper you linked.