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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 ?

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2 Answers 2

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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.

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  • $\begingroup$ 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 ? $\endgroup$
    – Samuel Tinguely
    Commented Jun 18, 2013 at 12:32
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    $\begingroup$ 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. $\endgroup$
    – Allen Knutson
    Commented Jun 18, 2013 at 14:08
  • $\begingroup$ Okay, fine, now I understand. Thank you for this neat answer ! =D $\endgroup$
    – Samuel Tinguely
    Commented Jun 18, 2013 at 15:39
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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.

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