Index of a differential operator between trivial bundles. - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-18T17:12:16Z http://mathoverflow.net/feeds/question/112546 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/112546/index-of-a-differential-operator-between-trivial-bundles Index of a differential operator between trivial bundles. Eric O. Korman 2012-11-16T02:32:29Z 2012-11-16T19:21:02Z <p>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?</p> <p>By the way, this question is a cross-post from math.SE.</p> <p>EDIT:</p> <p>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 $K(\mathbb R^2) = 0$.</p> http://mathoverflow.net/questions/112546/index-of-a-differential-operator-between-trivial-bundles/112609#112609 Answer by Johannes Ebert for Index of a differential operator between trivial bundles. Johannes Ebert 2012-11-16T19:21:02Z 2012-11-16T19:21:02Z <p>The result is wrong; the case of a point as base manifold creates counterexamples. Here is a less trivial construction in dimension $2$:</p> <p>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).</p> <p>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 $2$.</p> <p>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 $2$.</p> <p>I do not see how to produce an order $1$ operator of index $1$, though.</p> <p>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.</p>