# How to understand the Todd class?

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).

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### (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

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

is.

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).

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(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:

http://mathoverflow.net/questions/60478/hirzebruchs-motivation-of-the-todd-class/60481#60481