# A topological consequence of Riemann-Roch in the almost complex case

This question originated from a conversation with Dmitry that took place here

Is there a complex structure on the 6-sphere?

The Hirzebruch-Riemann-Roch formula expresses the Euler characteristic of a holomorphic vector bundle on a compact complex manifold $M$ in terms of the Chern classes of the bundle and of the manifold. On the other hand, for complex manifolds the $\bar\partial$ operator is a differential (i.e. it squares to zero) and hence, the complex of sheaves of smooth $(p,q)$ forms for a fixed $p$ equipped with the $\bar\partial$ differential is a resolution of $\Omega^p$; this is a complex of soft sheaves and so, by taking global sections we can compute the cohomology of $\Omega^p$. Moreover, since the de Rham complex resolves the constant sheaf, the alternating sum of the Euler characteristics of $\Omega^p$'s is the Euler characteristic of $M$.

For an arbitrary pseudo-complex manifold the only part of the above that makes sense is the "right hand side" of the Riemann-Roch formula (i.e. the one involving Chern classes) and the (topological) Euler characteristic of the manifold itself. So it seems natural to ask whether the relation between the two that is true in the complex case remains true in the almost complex case. In other words, is it true that for a compact almost complex manifold $M$ of dimension $2n$ we have $$\chi(M)=\sum_{p=0}^n (-1)^p\sum_{i=0}^{n\choose{p}}\mathrm{ch}_{n-i}(\Omega^p)\frac{T_i}{i!}$$ where $\chi$ is the topological Euler characteristic, $\Omega^p$ is the $p$-th complex exterior power of the cotangent bundle (i.e., the complex dual of the tangent bundle), $\mathrm{ch}$ is the Chern character and $T_i$ is the $i$-th Todd polynomial of $M$?

In general, are there topological consequences of the existence of the Dolbeault resolution that would be difficult to prove (or, more ambitiously, would fail) for arbitrary pseudo-complex manifolds?

• A naive question: doesn't this follow from the Atiyah-Singer Index theorem? – Paul Siegel Apr 28 '10 at 20:37
• Paul -- briefly: maybe it does but I don't see how. The problem is that for example that when we deduce the Riemann-Roch formula from AS we have tensor a bundle by the Dolbeault resolution to get a elliptic complex. This does not make sense in the almost complex case since the $\bar\partial$ is not a differential. – algori Apr 28 '10 at 22:09

I believe that the displayed equation is valid for almost complex manifolds. This is closely related to a computation I talked about here.

Let $r_1$, $r_2$, ..., $r_n$ be the chern roots of the tangent bundle. Then $\sum (-1)^p \mathrm{ch}(\Omega^p) = \prod (1-e^{-r_i})$. Let $\mathrm{Td}$ denote the total Todd class, so $\mathrm{Td} := \sum T_i/i! = \prod \frac{r_i}{(1-e^{-r_i})}$. The quantity you are interested in is $$\int \mathrm{Td} \prod (1-e^{-r_i}) = \int \prod r_i.$$ In other words, the top chern class of the holomorphic tangent bundle.

So the question is "On an almost complex manifold, is it still true that the top chern class of the holomorphic tangent bundle is $\chi(M)$?" I believe the answer is yes. Take a generic smooth section $\sigma$ of the tangent bundle and integrate it to get a flow. I believe that the fixed points of that flow will precisely be, with multiplicity, the intersections of $\sigma$ with the zero section. So we are done by the Lefschetz fixed point theorem.

The reason I keep saying "I think" and "I believe" is that I don't spend much time working with nonintegrable complex structures, so I can easily believe that I missed some subtlety.

• David, thanks! Yes, the top Chern class of any complex bundle is the Euler class of the realization of that bundle, so I think this works. – algori Jan 22 '10 at 4:02
• Characteristic classes are topological, so all we need from the almost complex structure is a splitting of TM tensor C; we have an isomorphism over R between the tangent bundle and the holomorphic tangent bundle, with isomorphism given by v -> v + Jv. The top Chern class is the Euler class [Morita, Geometry of characteristic classes, Proposition 5.43: e(E) = c_n(E) for E an n-dim complex vector bundle]. Thus c_n(T_holo M) = e(TM) = chi(M). – Tom Church Jan 22 '10 at 4:58
In general, are there topological consequences of the existence of the Dolbeault resolution that would be difficult to prove (or, more ambitiously, would fail) for arbitrary pseudo-complex manifolds?

Since an almost complex manifold has a tangent bundle like that of a complex manifold, the place to measure the difference is not, I think, in things involving characteristic classes of bundles and indices of elliptic operators. David's answer illustrates this, and I'll say something more philosophical.

The topological restrictions imposed by integrability are stark in real dimension 4. Almost complex structures on 4-manifolds $X$ are cheap: all you need for existence is a candidate $c$ for the first Chern class, which should satisfy $w_2=c \mod 2$ and $c^2[X]=2\chi+3\sigma$ (where $\sigma$ is the signature, and the second equation rewrites $p_1=c_1^2-2c_2$). Integrable complex structures are hard to come by. So as to avoid recourse to Kaehler methods, let's say that $b_1$ should be odd. Complex surfaces of this kind are still not completely classified, but they have been hunted down to a few specific topological "locations", one of them being $\pi_1=\mathbb{Z}$ and $H^2$ negative-definite (Class VII surfaces).

To get that far, one uses nearly all the complex geometry one can think of. The story starts with Dolbeault and the degeneration of the Hodge to de Rham spectral sequence (which uses Serre duality), but it invokes many further arguments (see Barth et al., Compact complex surfaces). It's expected that Class VII surfaces contain non-separating 3-spheres; to prove this when $b_2=1$, A. Teleman carefully analysed compactified moduli spaces of stable rank 2 bundles.

Your question was inspired by Dmitri's quotation of Gromov, who asserted in the quoted passage from Spaces and Questions that "complex manifolds have not stood up to their fame!" In the case of non-Kaehler surfaces, he might be right; a great deal of work turns up only a handful of quirky specimens which it is hard to fall in love with.

• Thanks, Tim! What surprised me is this: an almost complex structures is something that comes from smooth topology; then one can impose the integrability condition and one gets the Dolbeault resolution; the fact that this doesn't seem to have global consequences is a bit surprising (to me). Anyway, it's good to know all this. – algori Jan 22 '10 at 22:52

I may be repeating what has been said, but I think the point is this. The index theory always works in the "almost" case because one can set up a 2-step elliptic complex with operator D + D^* where D is d-bar. Moreover I believe that, in real dimension 6 or more, there are no known obstructions to the existence of an integrable complex structure beyond those for an almost complex structure. The case of 4 real dimensions is special because we have Kodaira's classification of complex surfaces. (PS: I don't think it was really necessary to have so many math formulae above!)

• Simon -- thanks. The fact that there are no known obstructions in higher dimensions seems surprising to me. It would be a miracle if any almost complex manifold of dimension $\geq 6$ admitted a complex structure, wouldn't it? One possible place where these hypothetical obstructions may be hiding is the Dolbeault resolution: on the one hand it exists iff the structure is integrable and on the other it doesn't seem completely crazy to expect it to have topological implications. The example in the posting is the simplest explicit one (that's why one needs the formulae) that one can think of. – algori Jul 25 '10 at 23:24

I just wanted to point out how this question is related to (spin${}^c$) Dirac operators and their indicies since this was alluded to in the comments to the question.

Let $(M, g)$ be an $2n$-dimensional closed Riemannian manifold. Given a spin${}^c$ structure, one can form the complex spin${}^c$ bundles $\mathbb{S}^+_{\mathbb{C}}$ and $\mathbb{S}^-_{\mathbb{C}}$. For any hermitian vector bundle $E \to M$, there is a twisted spin${}^c$ Dirac operator $D^c_E : \Gamma(\mathbb{S}^+_{\mathbb{C}}\otimes E) \to \Gamma(\mathbb{S}^-_{\mathbb{C}}\otimes E)$ which has index

$$\operatorname{ind}(D^c_E) = \int_M\exp(c_1(L)/2)\operatorname{ch}(E)\hat{A}(TM)$$

where $L$ is the complex line bundle associated to the spin${}^c$ structure.

Suppose now that $M$ admits an almost complex structure $J$ and $g$ is hermitian. Then there is a canonical spin${}^c$ structure which has associated line bundle $L = \det_{\mathbb{C}}(TM)$, so $c_1(L) = c_1(M)$. Using the fact that $\exp(c_1(M)/2)\hat{A}(TM) = \operatorname{Td}(M)$, the index becomes

$$\operatorname{ind}(D^c_E) = \int_M \operatorname{ch}(E)\hat{A}(TM).$$

Moreover, $\mathbb{S}^+_{\mathbb{C}} \cong \bigwedge^{0,\text{even}}M$ and $\mathbb{S}^-_{\mathbb{C}} \cong \bigwedge^{0, \text{odd}}M$.

When $J$ is integrable and $E$ is holomorphic, $D^c_E = \sqrt{2}(\bar{\partial}_E + \bar{\partial}^*_E)$ and the above equality becomes the statement of the Hirzebruch-Riemann-Roch Theorem. In particular, if $E = \Omega^p$, then $D^c_p := D^c_{\Omega^p}$ has index $\operatorname{ind}(D^c_p) = \chi(M, \Omega^p)$.

In the general case, Hirzebruch defined $\chi^p(M) := \operatorname{ind}(D^c_p)$ and introduced the Hirzebruch $\chi_y$ genus $\chi_y(M) = \sum_{p=0}^n\chi^p(M)y^p$. With this notation, your question is whether or not the equality $\chi_{-1}(M) = \chi(M)$ holds. Using Chern roots $x_i$ of $TM$, one finds that

$$\chi_y(M) = \int_M\prod_{i=1}^n\frac{x_i(1+ye^{-x_i})}{1-e^{-x_i}}.$$

Setting $y = -1$, we conclude $\chi_{-1}(M) = \chi(M)$ exactly as in David E Speyer's answer.

See this note for more details on Hirzebruch's $\chi_y$ genus and its properties, as well as references for the claims made above.