I am having trouble understanding a couple of lines of computation from Theorem 13.30 in Besse's Einstein Manifolds text

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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.


Your first question can be answered by using the splitting principle.

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.

  • $\begingroup$ Perfect, thank you. Although, I am unsure why symmetric product of line bundles reduces to the standard tensor product. $\endgroup$ – Guest123412341234 Jan 31 '20 at 23:16
  • 4
    $\begingroup$ If $V$ is a one-dimensional vector space, then every tensor on $V$ is symmetric, so $\bigotimes^nV = S^nV$. $\endgroup$ – Michael Albanese Feb 1 '20 at 1:06

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