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I'am working on a shorter proof of a theorem but to manage it I need to know if a lemma is true.

Conjecture: Given a manifold $M$ and an short exact sequence of vector bundles $$ 0 \rightarrow E' \rightarrow E \rightarrow E'' \rightarrow 0 $$ then $\Lambda^\cdot E' \otimes \Lambda^\cdot E'' \simeq \Lambda^\cdot E $, meaning by $\Lambda(\cdot)$ the vector bundle of external algebra. Is also a isomorphism of algebras?

Ivan Di Liberti
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1 Answers1

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This is not true unless the sequence splits. The only thing which is true in general is $\Lambda^{m}E'\otimes\Lambda^{n}E"\cong\Lambda^{m+n}E$ where $m,n$ are ranks of $E',E"$. Just as an illustration, let $m=n=1$. Then $\Lambda^. E'=\mathcal{O}\oplus E'$ and similarly for $E"$. So, the left side in your isomorphism is $\mathcal{O}\oplus (E'\oplus E")\oplus E'\otimes E"$, while the right side is $\mathcal{O}\oplus E\oplus \Lambda^2 E$, the last term as I said is just $E'\otimes E"$. In general these are not even isomorphic as vector bundles, unless $E\cong E'\oplus E"$.

Mohan
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  • In case the vector bundles are smooth over a smooth manifold, every s.e.s. splits. The smooth structure in the argument given here is crucial (to pick some Riemannian metric). Isn't the result true for topological vector bundles over a topological manifold? – Elías Guisado Villalgordo Jul 21 '23 at 09:37