Symmetric power of a virtual representation, and decomposition of tensor power

abstract-algebrak-theoryrepresentation-theory

Given any representation of $G$ on $V$, one can define symmetric power and exterior power as usual. Then the symmetric algebra $S^\bullet (V)=\bigoplus_{k\geq 0}S^k(V)$ and exterior algebra $\Lambda^\bullet(V)=\bigoplus_{k\geq 0}\Lambda^k(V)$ are multiplicative: $$S^\bullet(V\oplus W)=S^\bullet(V)\otimes S^\bullet(W);\quad\Lambda^\bullet(V\oplus W)=\Lambda^\bullet(V)\otimes\Lambda^\bullet(W).$$

It is possible to extend this to virtual representations, i.e. elements of the representation ring $R(G)$. To keep track of the degree, we consider $S_t(-):R(G)\to R(G)[\![t]\!]$ by $S_t(V-W)=S_t(V)/S_t(W)$ and similarly for $\Lambda_t(-)$. By comparing coefficients of $t^k$ in $S_t(V)\cdot S_t(-V)=1$, one is led to expressions for $S^k(-V)$ in terms of in terms of $S^j(V)$. For example,
\begin{align*}
S^1(-V)&=-V=-\Lambda^1(V)\\
S^2(-V)&=-S^2(V)+S^1(V)^2=\Lambda^2(V)\\
S^3(-V)&=-S^3(V)+2S^1(V)S^2(V)-S^1(V)^3=-\Lambda^3(V)
\end{align*}
One can do the same calculations for $\Lambda^k(-V)$ and obtain similar result. My question is, how to see that $S^k(-V)=(-1)^k\Lambda^k(V)$ and $\Lambda^k(-V)=(-1)^kS^k(V)$ in general? In particular, it seems that there should be a general formula for decomposing $V^{\otimes k}$ as products of symmetric and exterior powers, which I am not quite sure how to derive. These all must be very general since once can do the same for vector bundles $K$-theory, so a reference to this calculation is also welcomed! Thanks.

Best Answer

One way to think about "negative" representations is by super-representations, i.e., by considering $(\mathbb{Z}/2\mathbb{Z})$-graded vector spaces. We require that the $G$ action preserve grading. Fixing notation, let's write $V=V_0\oplus V_1$ for the two components. The quantum dimension is $\operatorname{qdim}V=\dim V_0-\dim V_1$, which is the Hilbert polynomial evaluated at $-1$, or the supertrace of the identity map, and similarly there's a notion of characters by $\chi_V(g)=\operatorname{tr}(g|V_0)-\operatorname{tr}(g|V_1)$.

In the category of graded vector spaces, the correct permutation generator (the "braiding") is $\tau_{V,W}:V\otimes W\to W\otimes V$ defined by $v_i\otimes w_j\mapsto (-1)^{ij}w_j\otimes v_i$ for $v_i\in V_i$ and $w_j\in W_j$. The interesting thing about this is that if $V=V_0$ then $\tau_{V,V}$ is just $v\otimes w\mapsto w\otimes v$, but if $V=V_1$ it's $v\otimes w\mapsto -w\otimes v$. What this means is that if $S^k(V)\subseteq V^{\otimes k}$ denotes the image of the symmetric projector defined using $\tau_{V,V}$, then

  1. if $V=V_0$, then $S^k(V)=S^k(V_0)\oplus 0$, where $S^k(V_0)$ denotes the usual symmetric power; and
  2. if $V=V_1$, then $S^{2k}(V)=\Lambda^{2k}(V_1)\oplus 0$ and $S^{2k+1}(V)=0\oplus\Lambda^{2k+1}(V_1)$, where $\Lambda^k(V_1)$ denotes the usual exterior power.

Writing $-V$ for the opposite grading $V_1\oplus V_0$, then the second observation can be written as $S^k(V) = (-1)^k(\Lambda^{k}(V_1)\oplus 0)$ when $V=V_1$.

Similarly, defining $\Lambda^k(V)$ using $-\tau_{V,V}$, we have that

  1. if $V=V_0$, then $\Lambda^k(V)=\Lambda^k(V_0)\oplus 0$; and
  2. if $V=V_1$, then $\Lambda^k(V)=(-1)^k(S^k(V_1)\oplus 0)$.

Thus, if $V=V_0$, we have \begin{align*} S^k(-V)&=(-1)^k(\Lambda^k(V_0)\oplus 0)=(-1)^k\Lambda^k(V) \\ \Lambda^k(-V)&=(-1)^k(S^k(V_0)\oplus 0)=(-1)^kS^k(V). \end{align*}

What remains would be to relate this to the representation ring, where representations with $V_0=V_1$ would be sent to $0$ I believe. In any case, I think it's nice that there's a way to get the identity to hold with concrete vector spaces like this.

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