Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I am reading the article "K-Theory and Elliptic Operators"(http://arxiv.org/abs/math/0504555), which is about Atiyah-Singer index theorem. In page 14 the article discussed the Thom isomorphism: $$\psi:H^{k}(X)\rightarrow H^{n+k}_{c}(E)$$ and $$\phi:K(X)\rightarrow K(E)$$ with $\psi: x \rightarrow \pi^{*}x* \lambda_{E}$ and $\phi:x \rightarrow \pi^{*}x\cup \mu$. Greg further defined a correction factor $\mu(E)$ such that $$\psi(\mu(E)\cup \operatorname{ch}(x))=\operatorname{ch}(\phi(x))$$ He analyzed $\mu(E)$ by splitting principle and give the expression $$\mu(E)\cup e(E_{\mathbb{R}})=\operatorname{ch}\left(\sum^{n}_{i=0}(-1)^{i}\wedge^{i}(E)\right)=\operatorname{ch}\left(\prod^{n}_{i=1}(1-L_{i})\right)=\prod^{n}_{i=1}(1-e^{x_{i}})$$

He argued that since $e(E_{\mathbb{R}})=c_{n}E=\prod x_{i}$ we can conclude $$\mu(E)=\prod^{n}_{i=1} \frac{1-e^{x_{i}}}{x_{i}}$$ He then define the Todd class $$\operatorname{td}(E)=\prod^{n}_{i=1}\frac{x_{i}}{1-e^{-x_{i}}}$$ such that we have $$\mu(E)=(-1)^{n}\operatorname{td}(\overline{E})^{-1}$$

My questions are:

  1. Is the step from $\mu(E)\cup e(E_{\mathbb{R}})$ to $\mu(E)$ justified? I feel uncertain about this as I have only seen it somewhere in Milnor & Stasheff's appendix, I do not know if this is the cap product or some other operation. Normally cup product made $H^{*}_{c}(E)$ to be a ring instead of a field. I think I need to clarify details in here.

  2. Why we define the Todd class in terms of the relationship $$\operatorname{td}(E)=\prod^{n}_{i=1}\frac{x_{i}}{1-e^{-x_{i}}}$$ instead of just using the result for $\mu(E)$? Is there any deeper motivation for it? On the other hand, why cannot we simply define $\mu(E)$ by the relation $\mu(E)\cup \psi(\operatorname{ch}(x))=\operatorname{ch}(\phi(x))$? I am not sure what the calculation of this will be but I feel it should be easier than the calculation in previous definition.

I did take a look at the wikipedia article and found the two definitions are mostly the same. I think maybe with time I can understand Todd class better. It is not mentioned in the book Characteristic classes (at least the part I covered) so I feel I do not really understand it (I hope I can understand it enough to understand the Atiyah-Singer index theorem).

share|improve this question

2 Answers 2

up vote 4 down vote accepted

(Re: 2)

AFAIK, the Todd class is slightly more convenient (than $\mu$) in various forms of (Grothendieck-Hirzebruch-)Riemann-Roch theorem. For example, if $f\colon X\to Y$ is a map of (compact stably almost complex) manifolds, the diagram

$$\begin{array}{ccc} K(X) & \stackrel{ch}{\longrightarrow} & H(X;\mathbb Q)\\ \downarrow{f_*} && \downarrow{f_*}\\ K(Y) & \stackrel{ch}{\longrightarrow} & H(Y;\mathbb Q) \end{array}$$

is not commutative, but the diagram

$$\begin{array}{ccc} K(X) & \stackrel{td(X)\cdot ch}{\longrightarrow} & H(X;\mathbb Q)\\ \downarrow{f_*} && \downarrow{f_*}\\ K(Y) & \stackrel{td(Y)\cdot ch}{\longrightarrow} & H(Y;\mathbb Q) \end{array}$$


By the way, it also explains the role, the Todd class plays in the Atiyah-Singer index theorem: $\int_M ch([\sigma(D)])\cdot td(M)$ of the RHS is nothing else but $\int_M[\sigma(D)]$ (where $\int_M$ is the direct image under the projection $M\to pt$ in cohomology/K-theory), so Atiyah-Singer boils down to just $\operatorname{ind} D=\int_M[\sigma(D)]$ (of course, when one actually tries to apply A-S, the traditional form is more convenient).

share|improve this answer
(Somewhat related: mathoverflow.net/questions/10630/…) –  Grigory M Jun 10 '11 at 9:06

I just wish to answer that a nice source regarding this can be found in here:


very ashramed that did not found this before.

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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