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.


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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    $\begingroup$ I don't think you need $E$ or $F$ to be trivial: if $M$ is parallelizable then the Todd class is $0$, and that's enough to kill the index. $\endgroup$ Nov 16, 2012 at 5:24
  • $\begingroup$ @Paul: that is not true; the Todd class is $1$, and there is enough room for nontrivial index. $\endgroup$ Nov 16, 2012 at 9:22
  • $\begingroup$ How does this follow from Atiyah-Singer? I am aware of the statement that a differential operator on an odd-dimensional manifold has zero index (without triviality of the bundle). $\endgroup$ Nov 16, 2012 at 9:28
  • $\begingroup$ I looked it up in Atiyah-Singer IoEO III. There is Proposition 2.17, which has the effect that the index of a geometric operator (associated with a reduction of the sructure group of the tangent bundle) only depends on $TM$ and the bundles (and it then zero in your situation). I section 9, they prove that any pseudodifferential operator acting on trivial bundles of rank at most half the dimension of $M$ has trivial index (this is false if the rank becomes larger). $\endgroup$ Nov 16, 2012 at 9:59
  • $\begingroup$ @Johannes, Paul: Now I think I was actually wrong-- I edited the question with my reasoning and will probably delete it soon. $\endgroup$ Nov 16, 2012 at 14:10

1 Answer 1


The result is wrong; the case of a point as base manifold creates counterexamples. Here is a less trivial construction in dimension $2$:

Let $M$ be a manifold and $V \to M$ be any vector bundle. There is an elliptic differential operator $D$ of order $2$ on $V$, which is self-adjoint and has thus index $0$: take a connection $\nabla$ on $V$ and put $D=\nabla^{\ast} \nabla$ (this is a Laplace type operator).

Now let $M= T^2$ and let $W \to T^2 $ be a holomorphic line bundle of degree $1$. By Riemann-Roch, the operator $\bar{\partial}_W$ has index $1$; and it goes from sections of $W$ to sections of $W$, since the canonical line bundle of a torus is trivial. Therefore one can form the composite $P:=(\bar{\partial}_W)^2$, and $P$ has index $2$.

Now let $V$ be a complex vector bundle such that $V \oplus W$ is trivial; with the operator $D$ constructed above. Consider the operator $D \oplus P$; this is an order $2$ elliptic operator on the trivial vector bundle over a parallelizable manifold and has index $2$.

I do not see how to produce an order $1$ operator of index $1$, though.

The vanishing theorems in Aityah-Singer, IoEO III, are quite optimal. My construction does not work in odd dimensions; and it is clear that the resulting trivial vector bundle has dimension at least $2$. If the dimension of the trivial vector bundle is too small, each (pseudo)differerential operator will have index $0$, as proven by Atiyah-Singer.


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