# $n$th symmetric power of a superspace

Given a vector space $$V$$, we consider the (trivial) associated even superspace $$V\oplus 0$$ and odd superspace $$0\oplus V$$. For any (super) vector space $$W$$ we define the $$n$$th symmetric power as $$\operatorname{Sym}^n(W) = W^{\otimes n}/\langle v_1\otimes v_2\otimes\cdots\otimes v_n - \sigma(v_1\otimes v_2\otimes\cdots\otimes v_n),\ \sigma\in S_n\rangle$$ for $$S_n$$ the symmetric group of $$n$$ symbols. In already two sources I read it is mentioned without proof that for any usual vector space V we get \begin{align} \operatorname{Sym}^n(V\oplus 0) &= \operatorname{Sym}^n(V)\tag{1}\\ \operatorname{Sym}^n(0\oplus V) &= \wedge^n(V).\tag{2} \end{align} Nevertheless, I cannot see why this holds; in fact the result is counter-intuitive to me, given how tensor products of direct sums decompose. In the simplest case, where $$n=2$$, I believe that $$(V\oplus 0)^{\otimes 2} \cong V_1\oplus 0$$ for $$V_1 = (V\otimes V)\oplus (V\otimes 0)\oplus (0\otimes V)$$ while in the other case we have $$(0\oplus V)^{\otimes 2} \cong 0 \oplus V_2$$ for $$V_2 = (0\otimes V)\oplus (V\otimes 0)\oplus (V\otimes V)$$. Now how these quotient by the ideal of the symmetric elements to lead to $$(1)$$ and $$(2)$$ is something I cannot see at all, because $$V_1\cong V_2$$. I am probably wrong on something, but I ignore what.

Edit

I am aware of the fact that $$(1)$$ and $$(2)$$ follow from the following, more general result $$\operatorname{Sym}^n(A\oplus B) = \bigoplus_{k=0}^n \left(\operatorname {Sym}^{k}(A)\otimes \bigwedge^{n-k}B \right) \tag{3}$$ but this is not something I am able to prove either.

• Are you aware that $0 \otimes W = 0$? I suggest you prove these results using the universal property of the symmetric power Commented Apr 10 at 13:26
• Also this should be $\mathrm{Sym}^{n-k} (B)$. You can try to first see $A$ and $B$ as two sub vector spaces of another space $C$ Commented Apr 10 at 13:33
• The (2) is false Commented Apr 10 at 15:10
• @julio_es_sui_glace, if you believe that $(2)$ is false, please provide a reasoning as an answer, since that is stated in two articles, namely here arxiv.org/abs/1905.08047 and here arxiv.org/pdf/math/0402057.pdf. Note that these statements are made in the context of supergeometry. Commented Apr 10 at 15:20
• Just using the dimension of each space you can ave the result Commented Apr 10 at 15:30

The reason of this whole misunderstanding is that the action of the symmetric group on the odd subspace $$V_1$$ of a supervector space $$V$$ is equal to minus its action on the even subspace $$V_0$$. This is due to the fact that the category $$\tt\text{SuperVect}_\mathbb{K}$$ of supervector spaces over $$\mathbb{K}$$ is a monoidal category under the tensor product $$\otimes$$ and that, in the convention used in both papers, the permutation of components of a tensor is defined through the braiding, given for any two supervector spaces $$V$$ and $$W$$ as a map \begin{align} V\otimes W&\to W\otimes V\\ v\otimes w&\mapsto (-1)^{|v||w|}w\otimes v \end{align} for $$|x|$$ the grade of $$x$$. Consequently and put simply, the action of the symmetric group $$S_n$$ is itself graded. $$(1)$$ and $$(2)$$ follow from this fact, and one could alternatively see these as definitions of the symmetric power of "purely" even and "purely" odd superspaces.
From this graded action of the symmetric group, one derives $$(3)$$, although not for any vector spaces $$A$$ and $$B$$, but rather for $$A = V_0$$ and $$B = V_1$$ being the even, respectively odd subspaces composing a superspace $$V = V_0\oplus V_1$$. That is, properly speaking one should say that given $$V_0$$ and $$V_1$$ as the respectively even and odd subspaces of a supervector space $$V= V_0\oplus V_1$$, it follows from the chosen braiding that $$\operatorname{Sym}^n(V) = \operatorname{Sym}^n(V_0\oplus V_1) = \bigoplus_{k=0}^n \left(\operatorname {Sym}^{k}(V_0)\otimes \bigwedge^{n-k}V_1 \right)$$ because by construction $$\operatorname{Sym}^k(V_1) = \bigwedge^k(V_1)$$.