Suppose we have a torus action on a compact oriented manifold M. Assume the action has isolated fixed points. Why is it that the equivariant Euler class of the normal bundle at the fixed point (i.e. the tangent space at that point) is (upto a sign) the product of the weights of the action of the lie algebra of the Torus on the tangent space at that point? (Maybe this is obvious but I don't see it).


Let me answer the question, under a different interpretation of the word "why".

First, what is the cohomology product [X] [Y] about, where X,Y are cycles and [X],[Y] their Poincar\'e duals? You could say it measures the inability to extricate X from Y.

Now, let's try to extricate the origin from itself, where [pt] denotes the class of the origin in a vector space V. Pretty easy (if dim V > 0). So we'd say [pt]^2 = 0.

Equivariantly, though, we may have trouble; there may be no nearby T-fixed points. This is exactly the question of whether there are any 0 weights in the action of T on V.

So [pt]^2 should be some multiple of [pt], that vanishes iff there's a 0 weight. Moreover it should be homogeneous of degree = codim(pt in V). Plus it should behave well under shrinking T. These come pretty close to forcing it to be the product of the weights.

  • $\begingroup$ This is very nice. Can this reasoning be extended to a complete proof? $\endgroup$ Feb 1 '10 at 23:18

I'm interpreting your question as "In what sense is the equivariant Euler class of a torus representation, thought of as a vector bundle on the point, just the product of the weights of the representation?" Because Euler classes multiply under direct sum, it's enough to answer this question for a 1-d representation.

Now, it's a general principle that if a space is nice in some sense, there will be a bijections between complex line bundles on that space and $H^2(X;\mathbb{Z})$ via first Chern class. One way to think of this correspondence is that a line bundle is defined by an element of $H^1(X;\mathcal{O}^*)$ where $\mathcal{O}^*$ is the nonvanishing elements of whatever sheaf of functions is relevant. As long as the sheaf cohomology of $\mathcal{O}$ is boring, the boundary map in the long exact sequence for the exponential sequence $\mathbb{Z}\to \mathcal{O}\to\mathcal{O}^*$ induces this isomorphism.

So, now what does this mean equivariantly? Well, remember that equivariant cohomology is the cohomology of a space. In the case of a point, it is the cohomology of the classifying space $BT$.

On the other hand, a line bundle on $BT$ is the same thing as a 1-dimensional representation of $T$; you use the standard associated bundle construction.

So, by the argument above, we get an identification between characters of $T$ and $H^2_T(pt;\mathbb{Z})$. This is being used implicitly when you make a statement like "the equivariant Euler class of the normal bundle at the fixed point (i.e. the tangent space at that point) is (upto a sign) the product of the weights of the action of the lie algebra of the Torus on the tangent space at that point," but having fixed this isomorphism, the statement above becomes tautological.


So you are saying that, we need to compute the equivariant Euler class of a complex line bundle over a point and since the equivariant first chern class is the same as this, the exponential sheaf sequence will help us. But, I am a little confused about this equivariant chern class business. I mean, I would be most grateful if you can make the correspondence between the characters of T and H^2 (pt) a little more explicit.

  • $\begingroup$ If you can edit your original question, perhaps it might be better to move the text of this comment into the main body of the question $\endgroup$
    – Yemon Choi
    Feb 1 '10 at 6:40
  • $\begingroup$ For some reason I can't seem to edit it. $\endgroup$
    – Vamsi
    Feb 1 '10 at 7:01
  • $\begingroup$ It might be because you have "insufficient reputation" on the system. I'll copy and paste it in for you, if that's all right. $\endgroup$
    – Yemon Choi
    Feb 1 '10 at 7:02
  • $\begingroup$ Insufficient reputation isn't the problem; an asker can always comment on his or her questions, and on answers to his or her questions. The problem is that MO doesn't recognize the user who posted this answer (3688) as the same user who posted the question (3686). It might help if you register; you could ask a moderator. $\endgroup$ Feb 1 '10 at 9:13
  • $\begingroup$ I merged all the Vamsi's. $\endgroup$
    – Ben Webster
    Feb 1 '10 at 17:31

Hi, perhaps this can make more explicit the correspondence above:

Let $G$ be a $\ell$-dimensional algebraic torus. Denote by $\Xi(G)\simeq \mathbb Z^\ell$ the character group of $G$ consisting of all continuous group homomorphisms $G\to \mathbb C^*$. Take any $\chi \in \Xi(G)$, it defines a one-dimensional complex representation of $G$ with space $\mathbb C_\chi$. We can associate a complex line bundle $L_\chi$ over $\mathbb B_G$ by:

$ L_\chi := (\mathbb E_G \times_G \mathbb C_\chi) \to \mathbb B_G. $

Denote by $c(\chi):= c_1(L_\chi)\in H^2(\mathbb B_G)$ its first Chern class. Let $\operatorname{Sym}_\mathbb Z^*(\Xi_G)$ be the symmetric algebra of the group $\Xi(G)$. It is a polynomial ring on $\ell$ generators of degree $1$, and the map $c\colon \chi\mapsto c(\chi)$ extends to a ring isomorphism:

$ c\colon \operatorname{Sym}_\mathbb Z^*(\Xi_G) \stackrel{\sim}\to H^*(\mathbb B_G) $

which doubles degrees. It is called the characteristic homomorphism.

Now, at an isolated fixed point under the action with characters $\chi_1,\dots, \chi_n$ one has:

$(\mathbb E_G \times_G T_X) = L_{\chi_1}\oplus\dots\oplus L_{\chi_n}.$

The equivariant euler class is just the top ordinary Chern class $c_{\operatorname{top}} (\mathbb E_G \times_G T_X)$, and you can conclude by Whitney sum formula.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.