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I'm having the following problem: let $T \subset G := SO(2k)$ be the maximal torus acting on $V := \mathbb{R}^{2k}$ by linear transformations on each $2$-dimensional component. Denote by $V_T := (V \times ET) / T$ the homotopy quotient, where $ET \to BT$ is the universal principal bundle associated with $T$. I would like to compute the Euler class of the (oriented) vector bundle (with fiber $V$) \begin{equation} \pi : V_T \to BT, \end{equation} in terms of the cohomology groups $H^*(BT)$.

So far, I have the following, mainly coming from Chern-Weil theory: we first consider the action of the whole group $G$ (we still denote by $\pi$ the vector bundle above for this action). The Chern-Weil homomorphism (which is an isomorphism here since $G$ is compact) provides an algebra isomorphism \begin{equation} \xi : S(g^*)^G \simeq H^*(BG), \end{equation} where $S(g^*)^G$ is the algebra of (say complex-valued if we look at cohomology with complex coefficients) $G$-invariant polynomial functions on the Lie algebra $g = so(2k)$. This isomorphism is given by applying polynomials on $g$ to the curvature $\Omega$ of any given connection form on the universal principal vector bundle $EG \to BG$, and one can show that the classes obtained this way are independent on the choices made. A particular element in $S(g^*)^G$ is called the Pfaffian $Pf$, and its image $Pf(\Omega)$ through $\xi$ is called an Euler form for $EG \to BG$. Now, choose a metric on the bundle $\pi$, and consider the associated principal bundle, that is the bundle $$\pi_F : F(V_G) := (F(V) \times EG) \to BG,$$ where $F(V)$ is the set of oriented orthonormal frames on V. Of course, $G$ acts freely on the total space $F(V_G)$, and one can see that it is also contractible, since it is homeomorphic to $(G \times EG) / G$. Therefore $\pi_F$ can be seen as the universal principal bundle of $G$, and the Euler form $Pf(\Omega)$ as an Euler form for $\pi_F$.

My questions are the following:

  1. How can I prove that this Euler form is indeed a representative of the Euler class of the vector bundle $\pi$ ?
  2. Is there a preferred curvature form for which the computation of $Pf(\Omega)$ would be simple ?
  3. Is there a more direct way (without passing through the Chern-Weil homomorphism) of computing the Euler class of $\pi$ ?
  4. Going back to the action of the torus $T$, how can I compute the Euler class of the vector bundle $\pi : V_T \to BT$ in terms of the cohomology groups $H^*(BT) \simeq \mathbb{C}[u_1,...,u_k]$ (I should find that it is equal to $u_1...u_k$, where $u_i$ are generators of degree $2$ of $H^*(BT)$) ?

My ultimate goal is to answer question 4, but it seems to me that understanding the above is necessary. Of course, if someone has a more straightforward way of treating question 4, it would be great.

I apologize in advance if I don't seem clear (I'm not a specialist of Chern-Weil theory).

Thanks a lot.

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I write this as an answer because it is too long to post it as a comment.

You need to be careful when using the term connection on a vector bundle over $BG$ since $BG$ is not finite dimensional. Technically, you have to work with finite dimensional approximations for $BG$.

In any case you can find the answer to your question in Chapter 8 of this book.

More precisely, question 1 is answered in Thm. 8.3.17 of Section 8.3.2.

Question 4 is answered in Proposition 8.3.14 in Sect. 8.3.1.

The answer to question 2 is that in general there is no nice curvature that makes the computation of the Pfaffian simpler. On the other hand I refer to Appendix B of this paper for a particularly convenient way of defining the Pfaffian.

The answer to question 3 is yes, but it requires a different definition for the Euler class in terms of the Thom class. See for example chapter 4 of these lecture notes.

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  • $\begingroup$ I cannot find the relation between my question 4 and proposition 8.3.14 of your book though. Could you be a bit more specific about how this gives the Euler class in terms of the generators of the cohomology of $BG$ ? $\endgroup$ – BrianT May 17 '18 at 20:01
  • $\begingroup$ If the bundle reduces to $BT^n$ it splits as a direct sum of oriented plane bundles $L_1, \dots, L_n$. and the Euler class of a sum is the product of the Euler classes of the splitting factors (this is prop 8.3.14 applied to $T^n=\underbrace{SO(2)\times\cdots\times SO(2)}_{n}$. The Euler classes of $L_i$ have degree $2$ and generate the cohomology ring of $BT^n$. $\endgroup$ – Liviu Nicolaescu May 17 '18 at 20:49
  • $\begingroup$ Oh great, thank you very much for your help. $\endgroup$ – BrianT May 17 '18 at 22:32
  • $\begingroup$ I'm sorry, I'm going to bother you one last time. You said that if the bundle reduces to $BT^n$, it splits as a direct sum of oriented plane bundles $L_1,...,L_n$. 1- Are we taking about the bundle $V_{T^n} \to BT^n$ (this is the one for which I'm trying to compute the Euler class) ? 2- Could you tell me what result tells me that it splits as $L_1 \oplus ... \oplus L_n$, and what would be each $L_i$ ? Thank you again for all your explanations. $\endgroup$ – BrianT May 17 '18 at 23:40
  • $\begingroup$ Also, is it clear that the cohomology ring of $BG $ is generated by the Euler classes of the $L_i$ ? $\endgroup$ – BrianT May 17 '18 at 23:47

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