In the paper "Self-duality in Four-dimensional Riemannian Geometry" (1978), Atiyah, Hitchin and Singer present a proof that the space of self-dual irreducible Yang-Mills connections is a Hausdorff manifold, and if it is not the empty set, then the dimension is given by $$p_1(\text{Ad}(P))-\frac{1}{2}\dim G(\chi(M)-\tau(M))$$ Where $\chi(M)$ is the Euler characteristic and $\tau(M)$ the signature.

EDIT: It turns out the original paper contained an error/typo. It should in fact be $$2p_1(\text{Ad}(P))-\frac{1}{2}\dim G(\chi(M)-\tau(M))$$ End of edit.

Although I would love to be able to understand the full paper, I am not in a position to be able to do so yet, I am only trying to understand the computation of this dimension, because I am interested in some applications of the Atiyah-Singer index theorem.
To compute this dimension, the following is utilised in the paper: Let $D:\Gamma(V_-\otimes E)\to\Gamma(V_+\otimes E)$ be the Dirac operator for a spinor bundle with values in some auxiliary bundle $E$. By the index theorem, $$\text{ind}(D)=\int_M\text{ch}(E)\widehat{A}(M)$$ In dimension four, we have $\widehat{A}(M)=1-\frac{1}{24}p_1(M)$ (but where is this used?). For the proof, we take $E=V_-\otimes\text{Ad}(P)$. Then $\text{ch}(E)=\text{ch}(\text{Ad}(P))\text{ch}(V_-)$. So far, so good. I lose track in the following computation: $$\text{ind}(D)=\int_M\text{ch}(\text{Ad}(P))\text{ch}(V_-)\widehat{A}(M)\\ \color{red}{=p_1(\text{Ad}(P))+\dim G(\text{ind}(D'))}=\\ p_1(\text{Ad}(P))-\frac{1}{2}\dim G(\chi-\tau)$$ Where $D':\Gamma(V_+\otimes V_-)\to\Gamma(V_-\otimes V_-)$. I have been trying to find a result that explains the red coloured part of the equation, because this step seems completely non-trivial, and in spite of that, it is not elaborated upon within the paper at all, and I am not able to find any sources that explain this step. In Index of Dirac operator and Chern character of symmetric product twisting bundle the accepted answer seems to give an answer that goes some way towards explaining how this result is obtained, in a very particular case. However, I am not very experienced in this area and I don't know how to generalise the result to an arbitrary principal $G$-bundle. I am looking for an explanation of the above, whether someone is able to provide their own response or a reference. Either one would be greatly appreciated.


1 Answer 1


Hopefully I remember this well. My adviser explained this computation to me I don't even want to think how many years ago.

The deformation complex of the SD equation is $\DeclareMathOperator{\Ad}{Ad}$

$$L=d_A^-\oplus d_A^*:\Omega^1\big(\, \Ad(P)\,\big)\to\Omega^2_-\big(\; \Ad(P)\;\big)\oplus \Omega^0\big(\;\Ad(P)\;\big). $$

The dimension of the moduli space of self-dual connections is the index of this operator. $\DeclareMathOperator{\ind}{ind}$ $\DeclareMathOperator{\ch}{ch}$ $\DeclareMathOperator{\hA}{\widehat{A}}$This operator is obtained by twisting with $\Ad(P)$ the operator

$$ D=d^-+d^*:\Omega^1(M)\to \Omega^2_-(M)\oplus \Omega^0(M) $$

This is the operator $D: \Gamma(V_+\oplus V_-)\to \Gamma(V_-\oplus V_-)$ in the paper you mentioned.

The Atiyah-Singer index theory shows that $\ind L$ is

$$\ind L= \int_M \big[\; \ch(\Ad(P)) \hA(X)\ch(V_-)\;\big]_4, $$

where $[--]_4$ denotes the degree $4$ part of a non-homogeneous differential form.

We deduce

$$\ch(\Ad(P))=\dim G +\ch_2(\Ad(P))+\cdots = \dim G+p_1(\Ad(P))+\cdots, $$

$$\ind L= \int_M \big(\; p_1(\Ad(P))+(\dim G)\rho_D\;\big) $$

where the degree $4$ from $\rho_D= [\hA(X)\ch(V_-)]_4$ is the index density of $D$ appearing in the Atiyah-Singer index theorem $$ \ind D=\int_M \rho_D. $$


$$ \ind L=\int_M p_1(\Ad(P))+\dim G\ind D= \int_M p_1(\Ad(P))+\dim G(b_1 -b_2^--b_0). $$

Now express $(b_1-b_2^--b_0)$ in terms of the signature $\tau=b_2^+-b_2^-$ and the Euler characteristic $\chi=2b_0-2b_1+b_2^++b_2^-$.

  • $\begingroup$ Trying to go through your explanation in detail - I have one small question. You immediately write $\text{ch}(\text{Ad}(P))=\dim G+\text{ch}_2(\text{Ad}(P))+\dots$. Is it valid to immediately omit the first Chern character? Since we're taking the product with another term that contains degree $2$ forms, could it not combine with terms there to add a term to the index density? The only candidate for that term would be $\text{ch}(\text{Ad}(P))\text{ch}_1(V_-)$. I guess that this term has to vanish somehow? $\endgroup$ Commented Dec 16, 2020 at 14:38
  • 2
    $\begingroup$ $Ad(P)$ is a real vector bundle $c_1(Ad(P)\otimes \mathbb{C})=0$. $\endgroup$ Commented Dec 16, 2020 at 14:43
  • $\begingroup$ Final remark incoming. According to what I computed, we get the $\dim G$ term, and $\text{rank}V_-(-2c_2(\text{Ad}(P))=-4c_2(\text{Ad}(P))=p_1(\text{Ad}(P))$, so I am not sure if the mistake is on my end, but I thought I would point out the discrepancy between what I got and what you've posted (namely the sign of the Pontrjagin class and possibly some minor details in the Chern character of the adjoint bundle). $\endgroup$ Commented Dec 16, 2020 at 17:00
  • $\begingroup$ You do not need to know what $ch(V_-)$ is. (The computation of $ch(V_-)$ is tricky. The clincher is that $D=d_-+d^*$ and you know what is the index $D$. $\endgroup$ Commented Dec 16, 2020 at 17:55
  • $\begingroup$ Although you've already helped me more than I could have hoped for, I made an edit to my original post explaining what I am talking about in more detail, because I think I did not make it clear (also the computation in my comment forgot about a factor of $1/2$). If you wouldn't mind, could you explain where the difference in our results comes from? Because your result seems to be off by a sign on the Pontrjagin class, whereas my result is off by a factor of $2$. $\endgroup$ Commented Dec 16, 2020 at 19:11

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.