# Are (finite dimensional?) inner product spaces also super vector spaces?

I recently read a little about super vector spaces and naturally I have a question and I don't know if I'm correct or if there's something in the definition that I don't quite understand.

My understanding is that a super vector space is a $$\mathbb{Z}_2$$-graded vector space, that is a vector space $$V$$ such that $$V=V_0\oplus V_1$$, $$0,1\in \mathbb{Z}_2$$ and $$\forall x \in V_i$$ we denote the parity of $$x$$ by $$\left|x\right|=i$$ (so $$x\in V_0$$ has parity $$0$$ and $$x\in V_1$$ has parity $$1$$).

Also, given a finite dimensional inner product space $$(V,\langle\cdot,\cdot \rangle)$$, and a subspace $$F$$ we can construct the subspace $$F^{\perp}=\left\lbrace u \in V, \langle u,v \rangle =0, v\in F\right\rbrace,$$ the subspace orthogonal to $$F$$. Furthermore, one can readily show that $$V=F\oplus F^\perp$$. (Not sure if the same can be said about infinite dimensional?). So call $$F=V_0$$ and $$F^\perp=V_1$$ then we can make $$(V,\langle\cdot,\cdot\rangle)$$ into a super vector space by fixing a subspace $$F$$, and viewing $$V$$ as the direct sum of $$F$$ and $$F^\perp$$ and saying that if $$x\in F, \left|x\right|=0$$ and if $$x\in F^\perp$$ then $$\left|x\right|=1$$.

So is it correct to say that any finite dimensional inner product space can be made into a super vector space ?

Converting my comments into an answer: any vector space $$V$$ can be made into a super vector space in many different ways, corresponding to any direct sum decomposition $$V \cong V_0 \oplus V_1$$. This is extra structure in general so it doesn't make sense to say that $$V$$ "is" a super vector space this way, only that it "can be made" a super vector space this way.
There are two canonical such decompositions, namely $$V_0 = V$$ (concentrated in even degree) or $$V_1 = V$$ (concentrated in odd degree). The even one is distinguished because that construction is symmetric monoidal.