4
$\begingroup$

Let us first precise the question : let $T$ be a torus, $\alpha : T \to \mathbb{C}$ be an irreducible character. I am interested in the $T$-equivariant Euler class of the ($T$-equivariant) bundle $\xi_\alpha:\mathbb{C}\to pt$ where $T$ acts on $\mathbb{C}$ via $\alpha$. $\alpha$ corresponds to an integral form $\alpha^\ast$ on the dual of the Lie algebra of $T$, and in their magnificent paper "The Moment Map and Equivariant Cohomology", Atiyah and Bott say that the equivariant Euler class of the bundle $\xi_\alpha$ is exactly $\alpha^\ast$ (seen as an equivariant cohomology class of $pt$).

My question then is : why is this true ?

$\endgroup$

2 Answers 2

5
$\begingroup$

Well, there's going to be some map $T^* \to H^2_T(pt)$ taking a weight $\lambda$ to the equivariant Euler class of the corresponding line bundle over a point. If you add weights, that tensors the line bundles, so adds the first Chern classes; hence this is an additive map between these two free abelian groups. Since it's natural it's going to be $Aut(T)$-equivariant. Already we see that Atiyah and Bott have to be right, up to scale.

$\endgroup$
3
  • $\begingroup$ I can only see the first sentence as a rephrasing of my question. Is the map you're talking about the identification between $H^\ast_T(pt)$ and $ST^\ast$ ? If yes, why must it map a weight to the corresponding line bundle over a point ? If not, what is it ? $\endgroup$ Jun 18, 2013 at 12:32
  • 2
    $\begingroup$ I see; your question is more about why the Cartan model that A-B are using for equivariant cohomology is set up well to admit this identification. Then what I'm saying is that we have two maps: $m: T^* \to H^2_T(pt)$ the identification you mention in the Cartan model, and $e: T^* \to H^2_T(pt)$ taking a weight to $c_1$ of the corresponding line bundle. So $m^{-1} \circ e$ is an endomorphism of $T^*$, but up to scale the only ones that commute with $Aut(T)$ are multiples of the identity. $\endgroup$ Jun 18, 2013 at 14:08
  • $\begingroup$ Okay, fine, now I understand. Thank you for this neat answer ! =D $\endgroup$ Jun 18, 2013 at 15:39
2
$\begingroup$

Having worked a bit on the question, I am now able to provide an answer. It is technical and not so elegant, so I would sill be glad to see better answers, but I think it is rather elementary. The idea is to compute the equivariant Thom class of the bundle, and to restrict it to the zero-section. Since the induced bundle over $BT$ admits a $\mathbb{C}^\ast$ action, it is orientable, so it admist a Thom class, and we will see that there is only one class in $H^2_T(\mathbb{C},\mathbb{C}^\ast)$, so it will have to be the Thom class.

So let us compute $H^2_T(\mathbb{C},\mathbb{C}^\ast)$. We will proced through the equivariant long exact sequence of the pair $(\mathbb{C},\mathbb{C}^\ast)$ : $$ \dots \to H^1_T(\mathbb{C}^\ast) \to H^2_T(\mathbb{C},\mathbb{C}^\ast) \stackrel{j^\ast}{\to} H^2_T(\mathbb{C}) \stackrel{i^\ast}{\to} H^2_T(\mathbb{C}^\ast) \to \cdots $$ Let us write $(\mu^i) $ for a basis of $T^\ast$, the dual of the Lie algebra of $T$. Now use any way you like to see that $H^1_T(\mathbb{C}^\ast)=0$ and $H^2_T(\mathbb{C}^\ast)=\bigoplus_i \mathbb{R}\mu^i / \alpha^*$. So $j^\ast$ is injective, so $H^2_T(\mathbb{C},\mathbb{C}^\ast) = \ker i^\ast$. Now $\mathbb{C}$ is ($T$-equivariantly) homotopic to a point, so $H^2_T(\mathbb{C}) = \bigoplus_i \mathbb{R}\mu^i$, and $i^\ast$ simply maps $\mu^i$ to its class modulo $\alpha^\ast$, so $H^2_T(\mathbb{C},\mathbb{C}^\ast) = \ker i^\ast = \mathbb{R}\alpha^\ast$ and we have what we wanted.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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