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# Index of a differential operator between trivial bundlesoveraparallelizablemanifold.

Let $M$ be a closed parallelizable manifold and $D: \Gamma(E) \to \Gamma(F)$ an elliptic differential operator between trivial vector bundles $E,F \to M$. The Atiyah Singer index theorem implies that the index of $D$ is zero. Is there a way to prove this with less machinery?

By the way, this question is a cross-post from math.SE.

EDIT:

Actually I think I made a mistake in my reasoning, which was that the symbol class $[\sigma_D] \in K(TM)$ is zero (I actually didn't need triviality of $TM$). Viewing $K$-theory as sequences of bundles the symbol class is $$0 \to \pi^* E \stackrel{\sigma_D}{\to} \pi^* F\to 0$$ where $\pi: TM \to M$. Now if $TM^+$ is the one-point compactification of $TM$ then the isomorphism $K(TM) \to \tilde K(TM^+)$ is given by extending the sequence to $TM^+$. I thought that the extension would have to involve trivial bundles as well, from which it will follow that $\sigma_D = 0$ since for a compact space any sequences involving trivial bundles is zero in $\tilde K$. But now I think this extension need not involve trivial bundles: $K(\mathbb R^2) \simeq \tilde K(S^2) = \mathbb Z$. But every bundle over $\mathbb R^2$ is trivial so my argument would give $K(\mathbb R^2) = 0$.

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# Index of a differential operator between trivial bundles over a parallelizable manifold

Let $M$ be a closed parallelizable manifold and $D: \Gamma(E) \to \Gamma(F)$ an elliptic differential operator between trivial vector bundles $E,F \to M$. The Atiyah Singer index theorem implies that the index of $D$ is zero. Is there a way to prove this with less machinery?

By the way, this question is a cross-post from math.SE.