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
- 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
- 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
- if $V=V_0$, then $\Lambda^k(V)=\Lambda^k(V_0)\oplus 0$; and
- 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.
Best Answer
I'll assume you meant $\mathbb{N}$ everywhere. Then the set of such homomorphisms is pretty limited. For every $n \geq 1$ we have $f(n) = f(1)$:
$$f(n) = f(1 + \ldots + 1) = f(1) \cup \ldots \cup f(1) = f(1)$$
Since by definition $f(0) = \emptyset$ we are only left with one degree of freedom by setting $f(1)$ -- depending on whether you see restricting $f(1)$ included in your homomorphism definition.
The above equality can also be seen in a different, more general light. Every semiring homomorphism $f$ gives rise to a monoid homomorphism $$f_\mathrm{Mon}: (S, +) \to (S', \oplus)$$ wrt. the additive operation. This is just because a semiring "contains" a monoid. The additive monoid here is $(\mathbb{N}, +)$, which is generated by $1\in\mathbb{N}$. This in turn means that $f_\mathrm{Mon}$ is already fully determined by its image of $0$ and $1$. Since we have $f = f_\mathrm{Mon}$, we can conclude the same for $f$.
In fact you could have made similar considerations for the multiplicative monoid to conclude that $f$ is determined by its image on $0$, $1$ and primes.