Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $\mathcal{O}(k)$.

share|improve this question
    
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

2 Answers 2

up vote 1 down vote accepted

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: $$ \wedge^k (E\otimes F)\simeq (\wedge^k E)\otimes (F^{\otimes k}).$$

share|improve this answer
    
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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.