A topological consequence of Riemann-Roch in the almost complex case - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T17:14:58Z http://mathoverflow.net/feeds/question/12601 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/12601/a-topological-consequence-of-riemann-roch-in-the-almost-complex-case A topological consequence of Riemann-Roch in the almost complex case algori 2010-01-22T03:12:23Z 2010-07-25T21:37:49Z <p>This question originated from a conversation with Dmitry that took place here</p> <p><a href="http://mathoverflow.net/questions/1973/is-there-a-complex-structure-on-the-6-sphere" rel="nofollow">http://mathoverflow.net/questions/1973/is-there-a-complex-structure-on-the-6-sphere</a></p> <p>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 $M$.</p> <p>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 $M$?</p> <p>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?</p> http://mathoverflow.net/questions/12601/a-topological-consequence-of-riemann-roch-in-the-almost-complex-case/12604#12604 Answer by David Speyer for A topological consequence of Riemann-Roch in the almost complex case David Speyer 2010-01-22T03:49:25Z 2010-01-22T03:49:25Z <p>I believe that the displayed equation is valid for almost complex manifolds. This is closely related to a computation I talked about <a href="http://mathoverflow.net/questions/10630/why-do-todd-classes-appear-in-grothendieck-riemann-roch-formula" rel="nofollow">here</a>.</p> <p>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.</p> <p>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.</p> <p>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.</p> http://mathoverflow.net/questions/12601/a-topological-consequence-of-riemann-roch-in-the-almost-complex-case/12681#12681 Answer by Tim Perutz for A topological consequence of Riemann-Roch in the almost complex case Tim Perutz 2010-01-22T19:25:00Z 2010-01-22T19:25:00Z <blockquote> 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? </blockquote> <p>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.</p> <p>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). </p> <p>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., <i>Compact complex surfaces</i>). 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. </p> <p>Your question was inspired by Dmitri's quotation of Gromov, who asserted in the quoted passage from <i>Spaces and Questions</i> 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.</p> http://mathoverflow.net/questions/12601/a-topological-consequence-of-riemann-roch-in-the-almost-complex-case/33330#33330 Answer by Simon Salamon for A topological consequence of Riemann-Roch in the almost complex case Simon Salamon 2010-07-25T21:37:49Z 2010-07-25T21:37:49Z <p>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 <em>2-step</em> elliptic complex with operator D + D^* where D is d-bar. Moreover I believe that, in real dimension 6 or more, there are <em>no</em> known obstructions to the existence of an <em>integrable</em> 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!) </p>