# Exterior power of a tensor product

Given 2 vector bundles $E$ and $F$ of ranks $r_1, r_2$, we can define $k$'th exterior power $\wedge^k (E \otimes F)$.

Is there some simple way to decompose this into tensor products of various exterior powers of individual bundles?

I am interested in the case when $F$ corresponds to the twisted line bundles $\mathcal{O}(k)$.

-
As far as I know, there is no general decomposition but it can be fibered into $\wedge^i E \otimes \wedge^{k-i} F$. –  only Oct 12 '12 at 16:07

In the special case when $F$ has rank $1$, the canonical homomorphism $$(E\otimes F)^{\otimes k}\to (\wedge^k E)\otimes (F^{\otimes k})$$ induces a homomorphism $$\wedge^k (E\otimes F)\to (\wedge^k E)\otimes (F^{\otimes k}).$$ This is because for all $u\in E$ and all $v_1, v_2\in F$, we have $(u\otimes v_1)\wedge (u\otimes v_2)=0$ (locally $v_i=a_ie$ with $a_i$ scalar and $e$ a basis, so the exterior product vanishes locally, hence vanishes globally). Now again locally (when $F$ is the trivial line the bundle), this homomorphism is clearly an isomorphism. So it is an isomorphism globally: $$\wedge^k (E\otimes F)\simeq (\wedge^k E)\otimes (F^{\otimes k}).$$

-
Thanks. If $F$ (rank > 1) can be resolved in terms of line bundles, one can say something about the general case as well. –  Amit Oct 13 '12 at 5:27
@Amit: sorry, I don't see how. –  user18119 Oct 13 '12 at 12:37
You won't get a formula like the one you gave for the case of line bundle. If $0 \rightarrow N_1 \rightarrow \oplus L_i \rightarrow F \rightarrow 0$ be a resolution of $F$, then you can get $\wedge^k (E \otimes F)$ as a quotient of $\wedge^k (E \otimes \oplus L_i)$ and terms coming from the filtration of $\wedge^k (E \otimes \oplus L_i)$ and various exterior products involving $N$. Try writing down the case for $k = 2$. For higher $k$, one can use induction. –  Amit Oct 13 '12 at 13:36

If $E$ and $F$ are arbitrary, decomposing $\bigwedge^k(E\otimes F)$ in terms of $\bigwedge^i E$ and $\bigwedge^j F$ for $i,j\leqslant k$ is quite subtle, and involves $\lambda$-rings (in fact, is inspiration for them). See this blog post for more details. Essentially, one looks at the action of the product $S_k\times S_k$ of symmetric groups on the polynomial ring $\mathbb Z[X_,\dots,X_k,Y_1,\dots,Y_k]$. The theory of symmetric polynomials tells us that for the elementary symmetric polynomials $E_1,\dots,E_k$ in the $X_i$ and $F_1,\dots,F_k$ in th $Y_i$, one has $$\mathbb Z[X_1,\dots,X_k,Y_1,\dots,Y_k]^{S_k\times S_k} = \mathbb Z[E_1,\dots,E_k,F_1,\dots,F_k] \text{.}$$ Since the formal power series
$$\prod_{i,j\geqslant 1} (1+X_i Y_jT)$$ is invariant under permutations of the $X_i$ and $Y_j$, we have $$\prod_{i,j\geqslant 1} (1+X_i Y_jT) = \sum_{k\geqslant 0} P_k(E_1,\dots,E_k,F_1,\dots,F_k) T^k \text{.}$$ It turns out that $$\textstyle\bigwedge^k(E\otimes F) = P_k(E,\textstyle\bigwedge^2 E,\ldots,\textstyle\bigwedge^k E, F, \textstyle\bigwedge^2 F,\ldots,\textstyle\bigwedge^k F)$$ where we interpret $V+W$ as $V\oplus W$ and $V\cdot W$ as $V\otimes W$.

All of this as a natural framework in $K$-theory. Briefly, if $X$ is a ringed space, the group $K_0(X)$ is the free abelian group generated by locally free sheaves on $X$, modulo the relation $[\mathscr E]+[\mathscr F]=[\mathscr G]$ whenever there is an exact sequence $$0 \to \mathscr E \to \mathscr G \to \mathscr F \to 0$$ The operation $[\mathscr E]\cdot [\mathscr F]=[\mathscr E\otimes \mathscr F]$ gives $K_0(X)$ the structure of a commutative ring. Even better, $K_0(X)$ is a $\lambda$-ring, with operations $\lambda^k:K_0(X) \to K_0(X)$ induced by $\lambda^k[\mathscr E] = [\bigwedge^k \mathscr E]$. The $\lambda$-ring structure on $K_0(X)$ features prominantly in the Grothendieck-Riemann-Roch theorem, a far-reaching generalization of the usual Riemann-Roch theorem for line bundles on compact connected Riemann surfaces.

-