# Index of Dirac operator and Chern character of symmetric product twisting bundle

I am having trouble understanding a couple of lines of computation from Theorem 13.30 in Besse's Einstein Manifolds text We are twisting the spinor bundle (on Einstein 4-manifold) $$\Sigma$$ with an auxiliary bundle $$S^3\Sigma^-$$. We form the Dirac operator $$\mathscr{D}^D$$ formed by twisting the Levi-Civita connection on $$\Sigma$$ with a copy $$D$$ acting on $$S^3\Sigma^-$$ (and composing with the Clifford action acting trivially on $$S^3\Sigma$$). Besse evaluates the index of this operator using the APS theorem as an integral over Chern and Pontrjagin classes.

I understand that the $$(1-\frac{1}{24}p_1)$$ terms comes from the $$\widehat{A}$$-genus of the manifold, and further the signature theorem, $$\tau=\frac13 \int_M p_1(M)$$. However, I do not understand the evaluation of the Chern character of the twisting bundle as $$(4-10c_2)$$ and the subsequent evaluation in terms of Euler characteristic and signature.

Any insight would be greatly appreciated.

Edit: I am still not sure about the final calculation relating the index to the Euler character and signature, however, working backwards it seems we require $$c_2(\Sigma^-)=\frac12e(M)-\frac14p_1(M)$$ (or perhaps something a bit different if there are lower degree terms which could multiply with with the $$\frac{1}{24}p_1$$ term from the $$\widehat{A}-$$genus), where $$e(M)$$ is the Euler class of $$M$$.

Edit#2: Solved by Michael Albanese.

If $$V \to X$$ is a complex vector bundle of rank two, then $$c_1(S^3V) = 6c_1(V)$$ and $$c_2(S^3V) = 11c_1(V)^2 + 10c_2(V)$$.

Proof: By the splitting principle, there is a map $$p : Y \to X$$ such that $$p^*$$ is injective on integral cohomology and $$p^*V \cong L_1\oplus L_2$$, so $$p^*(S^3V) \cong S^3(p^*V) \cong S^3(L_1\oplus L_2)$$. In general, we have $$S^n(E_1\oplus E_2) \cong \bigoplus_{i+j=n} S^i(E_1)\otimes S^j(E_2)$$, so

\begin{align*} &\, S^3(L_1\oplus L_2)\\ \cong&\, S^3(L_1)\otimes S^0(L_2)\oplus S^2(L_1)\otimes S^1(L_2)\oplus S^1(L_1)\otimes S^2(L_2)\oplus S^0(L_1)\otimes S^3(L_2)\\ \cong&\, L_1^3\oplus L_1^2\otimes L_2\oplus L_1\otimes L_2^2\oplus L_2^3. \end{align*}

It follows that $$c_1(S^3(L_1\oplus L_2)) = 6c_1(L_1) + 6c_1(L_2) = 6c_1(L_1\oplus L_2)$$. So $$p^*c_1(S^3V) = c_1(p^*S^3V) = c_1(S^3(L_1\oplus L_2)) = 6c_1(L_1\oplus L_2) = 6c_1(p^*V) = p^*(6c_1(V)).$$ By the injectivity of $$p^*$$, we have $$c_1(S^3V) = 6c_1(V)$$.

Similarly, one can compute that

\begin{align*} c_2(S^3(L_1\oplus L_2)) &= 11c_1(L_1)^2 + 11c_1(L_2)^2 + 32c_1(L_1)c_1(L_2)\\ &= 11(c_1(L_1)+c_1(L_2))^2 + 10c_1(L_1)c_1(L_2)\\ &= 11c_1(L_1\oplus L_2)^2 + 10c_2(L_1\oplus L_2) \end{align*}

and hence $$c_2(S^3V) = 11c_1(V)^2 + 10c_2(V)$$. $$\quad\square$$

In this case, $$\Sigma^-$$ is an $$SU(2)$$ bundle and hence $$c_1(\Sigma^-) = 0$$. So we see that $$c_1(S^3\Sigma^-) = 0$$ and $$c_2(S^3\Sigma^-) = 10c_2(\Sigma^-)$$. Therefore

$$\operatorname{ch}(S^3\Sigma^-) = \operatorname{rank}(S^3\Sigma^-) + c_1(S^3\Sigma^-) + \frac{1}{2}(c_1(S^3\Sigma^-)^2 - 2c_2(S^3\Sigma^-)) = 4 - 10c_2(\Sigma^-).$$

To find $$c_2(\Sigma^-)$$, or $$c_2(\Sigma^+)$$, we can proceed as follows.

As $$\Sigma^{\pm}$$ is an $$SU(2)$$-bundle which is a lift of the $$SO(3)$$-bundle $$\Lambda^{\pm}$$, there is a relationship between $$c_2(\Sigma^{\pm})$$ and $$p_1(\Lambda^{\pm})$$, namely $$p_1(\Lambda^{\pm}) = -4c_2(\Sigma^{\pm})$$; see Appendix E of Instantons and Four-Manifolds by Freed and Uhlenbeck, also this question. So now we just need to know $$p_1(\Lambda^{\pm})$$, but this is given by $$\pm 2e(M) + p_1(M)$$; see Chapter $$6$$, Proposition $$5.4$$ of Metric Structures in Differential Geometry by Walschap. Therefore

$$c_2(\Sigma^{\pm}) = \mp\frac{1}{2}e(M) - \frac{1}{4}p_1(M).$$

Note, as $$M$$ is assumed to be spin, its signature is a multiple of $$16$$ by Rohklin's Theorem, so $$\frac{1}{4}p_1(M)$$ is an integral class. As the signature of $$M$$ is even, so is the Euler characteristic, and hence $$\frac{1}{2}e(M)$$ is also an integral class.

Finally, as $$c_2(\Sigma^-) = \frac{1}{2}e(M) - \frac{1}{4}p_1(M)$$, we see that

\begin{align*} \int_M(10c_2(\Sigma^-) - 4)\left(1 - \frac{1}{24}p_1(M)\right) &= \int_M 10c_2(\Sigma^-) + \frac{1}{6}p_1(M)\\ &= \int_M 5e(M) - \frac{5}{2}p_1(M) + \frac{1}{6}p_1(M)\\ &= \int_M 5e(M) - \frac{7}{3}p_1(M)\\ &= 5\chi(M) - 7\tau(M) \end{align*}

as claimed.

• Perfect, thank you. Although, I am unsure why symmetric product of line bundles reduces to the standard tensor product. Jan 31, 2020 at 23:16
• If $V$ is a one-dimensional vector space, then every tensor on $V$ is symmetric, so $\bigotimes^nV = S^nV$. Feb 1, 2020 at 1:06