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Let $X$ be a space good enough to have a fundamental class, and $E$ a complex vector bundle on $X$. Let $P$ be some polynomial expression, and say I want to evaluate $P(c_i(E)) \cap [X]$.

Is there some naturally associated space, $\mathbf{P}(X, E)$ such that (e.g.) the topological Euler characteristic $\chi(\mathbf{P}(X, E)) = P(c_i(E)) \cap [X]$?

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Take $X = S^2$, and $E$ any line bundle. I can't think of what the space would be that computes the degree. If you have a section, then you're in much better shape. – Will Sawin Mar 8 '13 at 2:25
I agree with Will. In the definition of the Chern classes you've already pushed most of the bundley love magic into spacey love magic via the use of the associated projective bundle. It seems unlikely you can push all of the bundley love magic into a space together with its tangent bundle (which is basically what you're saying: when is evaluating some complicated characteristic class the same as evaluating the Euler class of some tangent bundle of a space?) – Dylan Wilson Mar 8 '13 at 4:54
@Dylan Wilson: +1 on your comment which, as far as I know, is the first occurrance of the term "love magic" in a serious mathematical context. And now I beg you to excuse me, I have to get back to some differential equation love magic! – drbobmeister Mar 8 '13 at 5:53

Let $B = Gr_{\dim E}(\infty)$ denote the classifying space for $\dim E$-bundles. Your $P(c_i(E))$ defines a class on $B$, which (if $P$ has integer coefficients, which you better've meant it to!) can pretty obviously be represented by an actual cycle, $\mathcal P \subseteq B$.

Assume the classifying map $c : X \to B$ is transverse to $\mathcal P$. Then $c^{-1}(\mathcal P)$ is the space you're looking for. Too bad it's finite!

Not quite. I want $c$ to be not just transverse to $\mathcal P$, but to meet it positively (unless you're okay with $\chi$ of a "negative point" being $-1$). Something like, $X$ should be algebraic, and the classifying map should be algebraic, some condition like $E$ being globally generated.

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The cycle might be very ugly, in particular not smooth, so what do you mean by transverse? It also seems like you're basically saying "The Euler characteristic of the Poincaré dual of the cohomology class I want is the same as evaluating the cohomology class on the manifold." But in general it's not true that this happens: for example, on $\mathbb{C}P^n$ I might want to evaluate $C_1^n$ on $[\mathbb{C}P^n]$. I'll get $1$. On the other hand, the Poincaré dual of $c_1^n$ is $[\mathbb{C}P^n]$, which has Euler characteristic $n$. – Dylan Wilson Mar 8 '13 at 13:43

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