I'm reading characteristic classes form the book Differential forms in Algebraic Topology by Bott and Tu. The Chern classes are defined as follows:

$E\xrightarrow{\rho} M$ is a vector bundle and $E_p$ be the fiber over $p$. Then $P(E)$, the projectivization of $E$ is a vector bundle with fiber $P(E_p)\colon=\{\text{1-dim subspaces of $E_p$}\}$ over $\ell_p\in P(E)$. It's then discussed that the first Chern class $x$ of the dual of the universal subbundle over $P(E)$ restricted to a fiber is the first Chern class of the fiber. Therefore the $1,x,\ldots,x^{n-1}$ are global classes on $P(E)$, whose restriction to each fiber $P(E)$ freely generates the cohomology of the fiber and hence by Leray-Hirsch $H^*(P(E))$ is a free module over $H^*(M)$ with basis $\{1,x,\ldots,x^{n-1}\}$ and then the Chern classes are the coefficients in the expression for $x^n$.

Also by Leray-Hirsch, $H^*(P(E))\cong H^*(M)\otimes H^*(Fiber)$. The fiber $P(E_p)$ has cohomology $\frac{\mathbb{R}[x]}{x^n}$, as $E_p$ is a vector space of complex dimension $n$. My question is, why $x^n\neq 0$.

Note: for the trivial bundle $M\times V$, $P(E)=M\times P(V)$ and by Künneth formula $H^*(P(E))\cong H^*(M)\otimes H^*(P(V))$ and then $x^n=0$ as $H^*(P(V))=\frac{\mathbb{R}[x]}{x^n}$. What is the difference in the general case and the trivial bundle?

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    $\begingroup$ The short answer is that Leray-Hirsch doesn't say (much) about the product structure on cohomology. It might help to work out a sufficiently generic example, eg take $E_n\to Gr_n(\mathbb C^{n+1})$ the tautological bundle over the Grassmanian; then $P(E_n)$ is the set of flags $L\subset E\subset \mathbb C^{n+1}$ , and $x$ is dual to the (real) codimension $2$ submanifold where $L\subset V$ for some codimension $1$ complex subspace $V$. Then $x^n$ is dual to the submanifold where $L$ is some fixed line, which is the inclusion of $Gr_{n-1}(\mathbb C^n)$ and thus not zero in homology. $\endgroup$ Jan 27, 2020 at 22:22
  • $\begingroup$ (I initially considered the general Grassmannian $Gr_k(\mathbb C^n)$ for general $k\neq n$; for $k +1 = n$, this is of course just $\mathbb{CP}^n$, with the tautological bundle the orthogonal complement of the tautological line bundle, so that its total Chern class is $(1+x)^{-1}$ as predicted by this calculation.) $\endgroup$ Jan 27, 2020 at 22:40
  • $\begingroup$ Basically what you are saying that for trivial bundle, because of the Kunneth formula, the chern class is 0. In other words the Chern class is the first obstruction to the trivialization of the bundle. $\endgroup$
    – user43326
    Jan 29, 2020 at 19:25
  • $\begingroup$ I realized that Leray-Hirsch is about the additive structure, thanks @BertramArnold . $\endgroup$ Feb 21, 2020 at 19:04

1 Answer 1


Bertram already mentioned this in the comments but I thought I'd write an answer for completness's sake.

The Leray-Hirsch theorem says that $H^{*}P(E)\cong H^{*}M\otimes H^{*}(Fiber)\ \ $ $\textit{as $H^{*}M$ modules}$.

So if $x$ is the first chern class of the tautological line bundle over $P(E)$, there's no reason to expect $x^{n}=0$, even if its pullback to the fiber satisfies that equation. Because if the fiber "twists around," there's no reason to expect that the behavior of a class on one fiber is indicative of its behavior globally.

On the other hand, when the bundle is trivial, the Kunneth formula (in this setting) tells you that $H^{*}P(E)\cong H^{*}M\otimes H^{*}(Fiber)\ \ $ $\textit{as rings}$.

Here's another fun example to illustrate the difference, where you can even draw pictures! You'll have to use cohomology over $\mathbb{Z}/2$ and you'll be calculating Stiefel-Whitney classes instead of chern classes, but the procedure is exactly the same: Leray-Hirsch, read off coefficients from a basis. Form $E\rightarrow S^1$ by taking the Mobius strip as a vector bundle over $S^1$ plus a copy of the trivial line bundle. Then $P(E)$ will be a familiar non-orientable surface (Klein bottle) whose cohomology ring you can compute from a delta-decomposition. Comparing it to the cohomology ring of the torus (which would be $P(E)$ for the trivial rank 2 bundle) where every positive degree class squares to zero, you can see exactly how the "twist" in the Klein bottle allows one of those classes to no longer square to zero!


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