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. – User3568 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. – User3568 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.

There is also a nice description in terms of Schur functor and Young diagrams.

A Young diagram $\lambda$ is a picture made by a finite set of cells, left-aligned in rows, such that the length of the rows decreases going down. Any Young diagram can be transposed (exchanging rows and columns) to obtain another Young diagram $\lambda'$.

To any Young diagram you can associate a functor $S_\lambda$, called Schur functor, which is an endofunctor of the category of finite vector spaces over a fixed field. The way how the Schur functor is constructed starting from the Young diagram is a rather complicated and you can find the details in this page of ncatlab.

To make an example, the Schur functor associated to a diagram made by only one row with $n$ cells is the functor sending any vector space $V$ to the vector space of symmetric $n$-powers $Sym^n(V)$, while the transposed diagram gives the exterior $n$-powers $\Lambda^n(V)$.

The product of two Schur functors can be decomposed into the linear combination of other Schur functors thanks to the Littlewood–Richardson rule.

The application of this rule in the case of the $n$th exterior power of a tensor product yields the following formula:

$\Lambda^n(V \otimes W) = \bigoplus (S_\lambda(V) \otimes S_{\lambda'}(W))$

where $\lambda$ in the direct sum runs over al the Young diagrams with $n$ cells and at most dim($V$) rows and dim($W$) columns.

You can find a reference for this formula in Fulton, Harris, Representation Theory, exercise 6.11.

[edit: this is an expaned comment based on the hint given in this mathoverflow discussion: mathoverflow.net/questions/126219/exterior-and-symmetric-powers-of-external-tensor-products-of-representations]

• Though this may answer the question it is good to provide the essence of the link in the answer – Shailesh May 8 '16 at 10:03
• Ok, I expand my answer. Even though, ifI have to be honest, I don't know much about them too, since I am still working on it :-) – Ramac May 8 '16 at 13:00