# A random question on the local family index theorem

A random question.

Say we have a complex vector bundle $E\to M$ that is isomorphic to a trivial bundle, and $d$ is a trivial connection on it. Let $M\to B$ be a fiber bundle with even dimensional closed connected oriented spin fibers. If we run the process of the local family index theorem, what happens to the kernel bundle (assume it exists) and what about the connection on it? Would the $k$-fold direct sum of the kernel bundle be trivial for some large $k$?

In the literature we see some secondary index theorems but I never know what is the (primary) family index theorem for trivial (or trivialized) vector bundle.

• Actually, what do you mean by "primary index theorem for flat vetcor bundles"? If you demand oriented and spin fibres, do you want to take the fibrewise Dirac operator? Then maybe you want the fibrewise tangent bundle to be flat over all of $M$. Otherwise, why should the kernel bundle be flat? – Sebastian Goette Apr 7 '16 at 17:54
• Hi, prinary means we are not talking about Cheeger Chern Simons class nor differential characters. We are just talking about Chern character form. For the second question, yes, we are taking the fiberwise Dirac operators. For the third question I dont want the vertical bundle to be flat. Acutally I am looking for some conditions that the kernel bundle to be at least stably trivial. – GRR Apr 8 '16 at 0:38

## 1 Answer

The $k$-fold direct sum of the kernel bundle $H\to B$ (assuming it exists) is trivial as a topological bundle for some $k$ if and only if $\mathrm{ch}(H)=0\in H^\bullet(B)$ because the Chern character is an isomorphism $K^0(B)\otimes_{\mathbb Z}\mathbb Q\to H^{\mathrm{even}}(B;\mathbb Q)$. The primary cohomological index theorem takes the form $$\mathrm{ch}(H)=\int_{M/B}\hat A(TX)\,\mathrm{ch}(E)=\mathrm{rk} E\int_{M/B}\hat A(TX)\;,$$ see e.g. the books by Lawson-Michelsohn or Berline-Getzler-Vergne.

On the other hand, if you put the natural connection $\nabla^H$ on $H$, then its $k$-fold direct sum can be trivial as a bundle with connection if and only if $(H,\nabla^H)$ itself is trivial as a bundle with connection. However, I don't know a form of the family index theorem that is strong enough to detect this. The best I know is a geometric (or, if you like, secondary) index theorem that computes the Cheeger-Simons Chern character $\widehat{\mathrm{ch}}(H)$. But I doubt that $\widehat{\mathrm{ch}}(H,\nabla^H)=0$ already implies triviality of $(H,\nabla^H)$ as a vector bundle with connection.

• Let me be more precise. I would like to find conditions on $E\to M$ such that the kernel bundle $H\to B$ is isomorphic to a trivial bundle (or its $k$-fold direct sum is so) while imposing any condition on the fiber bundle $M\to B$. Of course it should be easier to take $E\to M$ as a flat bundle or even a trivial bundle. As you said I don't find any index theorem that can detect the triviality of the kernel bundle. – GRR Apr 8 '16 at 4:23
• On the other hand, I don't intend to think that $\widehat{ch}(E, \nabla^E)=0$ implies the triviality of $E\to X$ (for any $E\to X$). Of course we know that the condition implies $E$ is torsion in ordinary $K$-theory. I would love to see a proof if your thought is true. – GRR Apr 8 '16 at 4:32