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I would like to know if the odd Betti numbers of a projective bundle P(E) for some vector bundle E over say a compact complex smooth algebraic variety B are zero just as in the case for ordinary projective spaces over Spec(k), or more generally how to generalize standard calculations of the cohomology of projective space to projective bundles.

edited Jun 17 2011 at 20:44

Yemon Choi
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asked Jun 17 2011 at 20:03

DZN
135●5

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In Griffiths-Harris you find a description for the cohomology of a projective bundle. In particular, if h^i(B) is nonzero then h^i(P(E)) is also nonzero. – Remke Kloosterman Jun 17 2011 at 20:13

In a reasonable cohomology theory where one can define Chern classes, one always has this relation between the cohom. of $P(E)$ and of $X$ (see e.g. Grothendieck's paper on Chern classes). For singular cohom. one can apply Kunneth formula. – shenghao Jun 18 2011 at 14:34

In Grothendieck's Chern classes paper, "this specific" property you asked below is built into the axiom A1 (see p.5), and for singular cohom. he said this is well-known (see top of p.9). I don't know a precise reference, but I think it must be in some standard alg. top. book, maybe Bott-Tu? Anyway you may prove it using Leray to the map $f:P(E)\to B,$ which degenerates at $E_2$ by, for instance, Deligne's weight argument. Along the way you may need proper base change in topology. – shenghao Jun 19 2011 at 10:27

Sasha · Accepted Answer · 2011-06-18 04:22:10Z UTC

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If $E$ is of rank $r$ then $H^i(P_B(E)) = \sum_{t = 0}^{r-1} H^{i-2t}(B)$ (where the summands with negative $i - 2t$ are omitted). So $H^{odd}(P_B(E)) = 0$ if and only if $H^{odd}(B) = 0$.

answered Jun 18 2011 at 4:22

Sasha
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	Thanks. Any sources? – DZN Jun 18 2011 at 5:44
	Does anyone know a source for this specific result (I don't see this specific result in GH or Grothendieck's paper on Chern classes)? – DZN Jun 18 2011 at 23:13
	Any textbook. Griffiths-Harris, Fulton, etc. – Sasha Jun 19 2011 at 17:28

odd betti numbers of a projective bundle

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