24
$\begingroup$

My question is about how to think about the Todd class.

Usually this is presented via Grothendieck Riemann Roch (GRR): if $X$ is a smooth projective scheme over a field $\mathbf{C}$, the chern character does not commute with integration/proper pushforward: $\require{AMScd}$ \begin{CD} K_0(X) @>\int >> K_0(\text{pt})=\mathbf{Z}\\ @V \text{ch} V V @VV \text{ch} V\\ \text{H}^*(X) @>\int >> \text{H}^*(\text{pt})=k \end{CD} and the Todd class $\text{Td}(\mathcal{T}_X)\in \text{H}^*(X)$ is the correction term to make the above commute. You can then do a computation to explicitly write down what $\text{Td}$ is in terms of chern roots.

I will take this story as the definition of the Todd class. However, on its own the content of the above is just: there exists a correction factor, and here is an explicit formula. It doesn't do a good job of explaining what this correction factor really is, or why it should show up in GRR. Ideally there would be a couple more examples where the Todd class shows up, or ways of understanding the Todd class, so that you could understand the correction factor as "that thing that shows up in $\ldots$". Notice that the quality of this answer improves the further away these examples are from GRR. For instance, if the folk result about the Todd class being an Euler class had been written down and made precise, it would be a good answer.

My question: in what other contexts (fundamentally different from GRR) does the Todd class appear?

Improve this question
$\endgroup$
5
  • 1
    $\begingroup$ mathematik.ur.de/hoyois/papers/grr.pdf might be useful: in 5.2, they invoked a result which says that Todd classes appear in comparing the HKR isomorphism and pushforwards. $\endgroup$
    – Z. M
    Commented Apr 19, 2022 at 11:12
  • 1
    $\begingroup$ I don't know enough to say precisely how it relates to your question, but some version of this has a nice interpretation which follows from the Duflo isomorphism where one look at the shifted tangent sheaf of $X$ as a Lie algebra in an appropriate symmetric monoidal category and where the square root of the Todd class is then nothing but the Duflo element. See e.g. the introduction of arxiv.org/abs/1711.09402 $\endgroup$
    – Adrien
    Commented Apr 19, 2022 at 14:07
  • 5
    $\begingroup$ This is basically the same argument as in the usual proof of GRR, but it lends to generalization: by the splitting principle, let's reduce to the universal case X = CP^infty, so we're trying to understand the relationship between the KU-Chern class and the usual Chern class in cohomology. These are given by the complex orientations of KU and rational cohomology, respectively, and the Todd class can be viewed as the ratio between these two complex orientations. $\endgroup$
    – skd
    Commented Apr 19, 2022 at 14:39
  • 3
    $\begingroup$ This lets you generalize GRR: if you have two complex oriented cohomology theories A, B, let R = A smash B. Then you can compare the composites f: A^*(X) -> A^*(pt) -> R^*(pt) and g: A^*(X) -> R^*(X) -> R^*(pt). For a class z in A^*(X), one would then have f(z) = g(z * u), where u is the ratio (in R^*(X)) between the complex orientations of A and B. (When A = KU and B = rational cohomology, this gives you GRR.) $\endgroup$
    – skd
    Commented Apr 19, 2022 at 14:39
  • 1
    $\begingroup$ For another usage of the Todd class: The Todd class is used widely in the theory of circle actions on symplectic manifolds (i.e. take a compatible almost complex structure and takes its Todd class.) The point is that it "localises " over the fixed point set, and this often leads to something useful. For the localisation result see "manifolds and modular forms" Friedrich Hirzebruch; Thomas Berger; Rainer Jung. It is not GRR or even HRR since there is no integrable complex structure a priori. $\endgroup$
    – Nick L
    Commented Apr 20, 2022 at 8:50

1 Answer 1

Reset to default
14
$\begingroup$

To my knowledge, currently the best "motivation" for the Todd class comes from the so called "orientation theory" and the formal group laws associated to "oriented" cohomology theories. As far as I know, these ideas go back to Quillen's work on complex cobordism, although current formulation is mainly due to Panin-Smirnov's works (here, here and here). I suggest to read this exposition, which briefly summarizes the main ideas.

Let me summarize the explanation without going into every detail. We say that a cohomology theory $H\colon \mathrm{Var}_k\to \mathrm{Rings}$ is oriented if it has a theory of Chern classes with all usual properties one might expect (invariance under pullback, Withney sum and projective bundle formula, nilpotence...). (I am cheating here, people define orientations differently, but there is a theorem stating that both things, orientations and reasonable Chern classes, are in fact equivalent under mild conditions). It is worth noticing two facts about this definition:

  1. the cohomology is not assumed to be graded (so that the $K$-theory is such a cohomology)

  2. For $L,L'\to X$ two line bundles $c_1(L\otimes L')$ need not to be equal to $c_1(L) + c_1(L')$. Such cohomologies are called additive. For example, $K$-theory satisfies that $c_1(L)=1-[L^*]$ so that $c_1(L\otimes L')=c_1(L) + c_1(L')-c_1(L)\cdot c_1(L')$, which is said to be multiplicative (the name comes from the multiplicative formal group law, where the set of coordinates is taken with respect to 1. That is to say, (1-x)(1-y)=(1-x-y+xy)). The target of the Chern character is always and additive cohomology.

The fun thing is that a given cohomology theory $H$ may have different orientations, in other words, different theories of Chern classes. However two orientations can be related concretely: if $H$ is oriented, with some first Chern classes $c_1$, and we have a new orientation, with a new first Chern classes $c_1^{\mathrm{new}}$, then it is easy to check that $$ c_1^{\mathrm{new}}(L)=F(c_1(L))\cdot c_1(L) $$ where $F(t)\in H(pt)[[t]]$ is an invertible series and with coefficients in the cohomology ring of a point. This $F$ is the (inverse) Todd series associated to the new orientation and can be applied, through the usual multiplicative extension, to any virtual vector bundle.

Let $\varphi \colon H\to H'$ be a morphism of cohomology theories (i.e., natural transformation between contravariant functor of rings) it is also easy to check that the orientation on $H$ provides an orientation on $H'$ or, in other words, new Chern classes on $H'$ and therefore, an associated (inverse) Todd series.

Main example: Consider the Chern character $\mathrm{ch}\colon K\to H_{\mathbb{Q}}$, which is a morphism of cohomologies where $H_{\mathbb{Q}}$ is additive. Let $c_1^K$ denote the Chern classes with values in $K$-theory, to avoid confussion. Let us check the associated (inverse) Todd series to $\mathrm{ch}$: $$ \mathrm{ch}(c^K_1(L))=\mathrm{ch}(1-[L^*]))=1-e^{c_1(L^*)}=1-e^{-c_1(L)}=\frac{1-e^{-c_1(L)}}{c_1(L)}\cdot c_1(L) $$ Hence, its associated series is $\frac{1-e^{-t}}{t}$, which is precisely the inverse of the classical Todd series.

Now, to concretely answer your question, here you have:

Other examples: The universal map from cobordsim to any oriented cohomology theory $\Omega\to H$ also has its associated (inverse) Todd series, but it is trivial (the series is $F(t)=1$). However, the associated Riemann-Roch type theorem is not at all trivial, is the so-called Conner-Floyd theorem.

On any oriented cohomology theory $H$ you can consider a new theory of Chern classes: $c_1^\mathrm{new}(L):=-c_1(L^*)$, its associated (inverse) Todd series is $F(t)=1$ if $H$ is additive, but it is the formal inverse otherwise.

-----0-----

In the aformentioned exposition you have these kind of arguments with details and "without cheating" (orientations are defined properly, and not just as a "theory of Chern classes"). The Riemann-Roch theorem in this context is as corollary of the universal property of $K$-theory (which is the universal multiplicative oriented cohomology theory).

Improve this answer
$\endgroup$
7
  • 2
    $\begingroup$ That's nice! Do you know if there is a version of this game for the map $\Omega\to K$? Or, say, for the Miller character $TMF\to K$? $\endgroup$
    – მამუკა ჯიბლაძე
    Commented May 5, 2022 at 11:33
  • $\begingroup$ Yes, you can play this game for the natural map $\Omega \to K$. The $K$-theory is an example of oriented cohomology theory, so it is cointained in the first "other" example mentioned. The associated Todd series is trivial. This fact is contained in the so called Conner-Floy theorem, whose statement makes emphasis in the fact that it induces an isomorphism between $K$ and a quotient of $\Omega$ rather than commuting with Gysin maps. For the other question: I have not studied neither TMF nor the Miller character. Does TMF have Chern classes/Gysin maps? Is Miller character a morphism of rings? $\endgroup$
    – Tintin
    Commented May 5, 2022 at 12:43
  • 1
    $\begingroup$ Well I surely know less than you. I suspect that what Miller denotes by $E\ell^*$ is in fact a complex-oriented version of TMF (the latter, I believe, is not). With this version, I believe both answers must be yes: I think the Miller character is produced from the Witten genus by standard procedure (although I am not sure as I don't know much about it) $\endgroup$
    – მამუკა ჯიბლაძე
    Commented May 5, 2022 at 13:23
  • 1
    $\begingroup$ Is there a projective bundle theorem in TFM? If so, then it is certainly reasonable to expect that the same game works. The associated Todd series appearing will come from computing the Miller character of the first Chern class of the tautological bundles in TFM-cohomology. In other words, the image through the Miller character of $c_1^{TFM}(\mathcal{O}_{\mathbb{P}^n}(-1))$ will be a polynomial of degree $n$ in $c_1^K(\mathcal{O}_{\mathbb{P}^n}(-1))=1-\mathcal{O}_{\mathbb{P}^n}(1)$. These will be the coefficients of the associated (inverse) Todd series up to degree $n$. $\endgroup$
    – Tintin
    Commented May 8, 2022 at 16:52
  • 1
    $\begingroup$ This is very interesting consideration, but I don't even know how to formulate the projective bundle theorem in TMF, let alone whether it is true. Is there such thing in $\Omega$? $\endgroup$
    – მამუკა ჯიბლაძე
    Commented May 8, 2022 at 17:03

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .