TITLE: Atiyah Bott localisation applied to Euler characteristic
QUESTION [10 upvotes]: 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).

REPLY [0 votes]: 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.
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