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


[Is there a complex structure on the 6-sphere?](https://mathoverflow.net/questions/1973/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 topological consequence of Riemann-Roch in the almost complex case
I believe that the displayed equation is valid for almost complex manifolds. This is closely related to a computation I talked about [here](https://mathoverflow.net/questions/10630/why-do-todd-classes-appear-in-grothendieck-riemann-roch-formula).


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.


