2
$\begingroup$

I am trying to read the proof of the Riemann-Roch-Hirzebruch theorem using the index theorem (Heat Kernel and Dirac operators, [BGV]), but I do not understand the few last steps (theorem 4.11, page 148).

Let $M$ be a compact Kähler manifold of even dimension, $W$ a Hermitian and holomorphic vector bundle over $M$ and $E := \Lambda (T^{0,1}M)^{*}\otimes W $ ; $E$ is a Clifford module and we have a Clifford connection $ \nabla ^{E}$ on it.

Propositon 3.43 (page 117) says that the curvature $R^E$ of this Clifford connection decomposes under the isomorphism $End(E) = C(M)\otimes End_{C(M)}(E)$ ($C(M)$ being the Clifford bundle) as follows :

$$ R^E = R^S + R^{E/S},$$

where $R^S \in \Omega ^2(M,C(M))$ and $R^{E/S} \in \Omega (M, End_{C(M)}(E)).$

In our case, for $E := \Lambda (T^{0,1}M)^{*}\otimes W $, they obtain

$$ R^{E/S} = \frac {1} {2} Tr_{T^{1,0}M}( R^+ ) + R^W,$$

where $R^+$ is the curvature of the bundle $T^{1,0}M$ and $R^W $ the one of the canonical connection of $W$, but I do not see how that works.

If you have any idea or any other references that would be nice. Also if you want more details about the set up just tell me. Thanks.

$\endgroup$

1 Answer 1

0
$\begingroup$

I think the top of page 148 computes the case where $W$ is trivial, whose second term $\frac{1}{2}\sum_k(Rw_k,\bar{w}_k)$ is exactly $\frac{1}{2}\text{Tr}_{T^{1,0}M}(R^+)$. The general case then follows.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.