Borel localization for compact torus actions

For any invariant pair $Z\subseteq Y$, use the relative Borel notation of [F6]. Put $X_T=ET{\times }_{T}M$ and $A_T=ET{\times }_T(M\setminus F)$. The fixed set is closed by [F1], so $M\setminus F$ is invariant and open; its Borel construction is open because the quotient map by the diagonal $T$-action is open. The pair long exact sequence [F5] has segment

By [F3], after localization both outside terms vanish in every degree, and the sequence remains exact. Hence the relative-to-absolute map

is an isomorphism.

Let $U\subset M$ be the invariant open tubular disk of radius $ϵ$ around $F$ from [F1], and choose $0<\rho <ϵ$. The Borel opens $ET{\times }_{T}U$ and $A_T$ cover $X_T$. Their intersection is $ET{\times }_T(U\setminus F)$, so excision [F5] identifies

The equivariant tubular diffeomorphism identifies this pair with the Borel open disk bundle pair $(D({\nu }_{F,T}),D({\nu }_{F,T})\setminus F_T)$, where $F_T=BT\times F$. Let $S_{\rho }({\nu }_{F,T})$ be the sphere bundle of radius $\rho$, which lies inside the open disk. On the punctured disk the radial formula $v\mapsto ((1-t)\parallel v\parallel +t\rho )v\mathrm{/}\parallel v\parallel$, for $0\le t\le 1$, stays inside the disk because both endpoint radii are less than $ϵ$; it gives a continuous equivariant deformation retraction onto $S_{\rho }({\nu }_{F,T})$. Thus the inclusion of pairs $(D({\nu }_{F,T}),S_{\rho }({\nu }_{F,T}))\to (D({\nu }_{F,T}),D({\nu }_{F,T})\setminus F_T)$ induces a cohomology isomorphism: in the long exact pair sequences the absolute disk terms agree and the sphere-to-punctured-disk terms are isomorphisms by [F5]. Let ${\bar{D}}_{\rho }({\nu }_{F,T})$ be the closed subdisk bundle of radius $\rho$. The inclusion of pairs $({\bar{D}}_{\rho },S_{\rho })\hookrightarrow (D({\nu }_{F,T}),S_{\rho })$ is a homotopy equivalence: the equivariant radial clamp $R_t(0)=0$, $R_t(v)=((1-t)\parallel v\parallel +t\min \{\parallel v\parallel ,\rho \})v\mathrm{/}\parallel v\parallel$ for $v\mathrm{\ne }0$, stays inside the open disk and fixes ${\bar{D}}_{\rho }$, including $S_{\rho }$. Rescaling identifies $({\bar{D}}_{\rho },S_{\rho })$ with the unit closed disk/sphere pair in the definition of the Thom isomorphism. Hence [F6] identifies these relative groups componentwise with $H_T^{\ast -c_j}(F_j;k)$, where $F_j$ ranges over the finitely many components and $c_j={\mathrm{rank}}_{\mathbb{R}}{\nu }_{F_j}$. Composing this identification with relative-to-absolute defines the Gysin map $i_{\ast }^T$. Pulling its tubular Thom representative back to the zero section gives

on every component, by the zero-section definition of the Euler class and naturality in [F6].

Let $\gamma \in {\mathcal{S}}_T(k{)}^{-1}{H}_T^n(F;k)$. The finite component decomposition writes $\gamma =({\gamma }_j)$. By [F4], $e_j=e_T({\nu }_{F_j})$ has a localized inverse on every component. Define ${\beta }_j=e_j^{-1}{\gamma }_j$, which has degree $n-c_j$; the componentwise Thom isomorphism in step 2.1 combines the ${\beta }_j$ into a relative class, and its Gysin image satisfies $i^{\ast }{i}_{\ast }^T(\beta )=\gamma$. Therefore $i^{\ast }$ is surjective after localization.

Suppose $\alpha \in {\mathcal{S}}_T(k{)}^{-1}{H}_T^n(M;k)$ satisfies $i^{\ast }\alpha =0$. Step 1.1 writes $\alpha =i_{\ast }^T(\beta )$ for a relative class $\beta =({\beta }_j)$. Step 2.1 gives $0=i^{\ast }\alpha =(e_j{\beta }_j{)}_j$; since every $e_j$ is a unit by [F4], each ${\beta }_j=0$, so $\alpha =0$. This proves injectivity. If $F=\mathrm{\varnothing }$, step 1.1 and [F3] give zero localized cohomology on both sides. If $T$ is trivial, then $F=M$, ${\mathcal{S}}_T(k)=\{1\}$, the normal bundle has rank zero, and restriction is the identity. Empty $M$ is included in these cases. Hence the map in the Statement is an isomorphism.