Character localization in Borel equivariant cohomology

The rational character formula and the unique integer vector $\mathbf{n}$ are [F2]. To pass to real coefficients, let $j:\mathbb{Q}\hookrightarrow \mathbb{R}$ and, for any space $Y$, send a rational singular cochain $\phi$ to $j\circ \phi$. By [F4], $\delta (j\circ \phi )=j\circ \phi \circ \mathrm{\partial }=j\circ \delta \phi$, so this additive map sends cocycles to cocycles and coboundaries to coboundaries and induces $j_{\ast }:H^2(Y;\mathbb{Q})\to H^2(Y;\mathbb{R})$. It commutes with pullback, since postcomposition by $j$ commutes with precomposition by the induced singular chain map. Using the coefficient convention in [F3] and $\mathbb{Z}\to \mathbb{Q}\to \mathbb{R}=\mathbb{Z}\to \mathbb{R}$, it sends $c_1(L;\mathbb{Q})$ to $c_1(L;\mathbb{R})$. The coordinate associated lines in [F2] are the tautological lines whose real coefficient images define $u_i$ in [F1]. Applying $j_{\ast }$ to the rational identity therefore gives $c_1(L_{{\chi }_{\mathbf{n}}};\mathbb{R})={\sum }_i{n}_i{u}_i$. Uniqueness of $\mathbf{n}$ is the character classification in [F2]; the polynomial generators are independent in both coefficient fields.

The empty product is $1$, concatenation of two finite character lists multiplies their products, so $1\in {\mathcal{S}}_T(k)$ and ${\mathcal{S}}_T(k)$ is multiplicatively closed. For each nontrivial character $\mathbf{n}\mathrm{\ne }0$, step 1.1 gives the nonzero linear polynomial ${\sum }_i{n}_i{u}_i$; a polynomial ring over a field is an integral domain, so every finite product of these classes remains nonzero. If $r=0$, the generating list is empty and the set is $\{1\}$. Hence $0\mathrm{\notin }{\mathcal{S}}_T(k)$ in all cases.

The fraction relation is reflexive using multiplier $1$, symmetric by negating its witness equation, and transitive: if $a(s^{\mathrm{\prime }}n-s{n}^{\mathrm{\prime }})=0$ and $b(s^{\mathrm{\prime }\mathrm{\prime }}{n}^{\mathrm{\prime }}-s^{\mathrm{\prime }}{n}^{\mathrm{\prime }\mathrm{\prime }})=0$, then $ab{s}^{\mathrm{\prime }}(s^{\mathrm{\prime }\mathrm{\prime }}n-s{n}^{\mathrm{\prime }\mathrm{\prime }})=0$, with $ab{s}^{\mathrm{\prime }}\in {\mathcal{S}}_T(k)$. Thus it is an equivalence relation. If $n\mathrm{/}s\sim n^{\mathrm{\prime }}\mathrm{/}{s}^{\mathrm{\prime }}$ with witness $a$ and $m\mathrm{/}t\sim m^{\mathrm{\prime }}\mathrm{/}{t}^{\mathrm{\prime }}$ with witness $b$, then

Multiplying by $ab$ makes the right side zero, proving the displayed addition is independent of representatives. For the scalar action, if $n\mathrm{/}s\sim n^{\mathrm{\prime }}\mathrm{/}{s}^{\mathrm{\prime }}$ with witness $a$, then the cross-multiplied numerator difference between $(c\mathrm{/}q)(n\mathrm{/}s)$ and $(c\mathrm{/}q)(n^{\mathrm{\prime }}\mathrm{/}{s}^{\mathrm{\prime }})$ is $cq(s^{\mathrm{\prime }}n-s{n}^{\mathrm{\prime }})$, killed by $a$. If $c\mathrm{/}q\sim c^{\mathrm{\prime }}\mathrm{/}{q}^{\mathrm{\prime }}$ with witness $b$, the corresponding difference between $(c\mathrm{/}q)(n\mathrm{/}s)$ and $(c^{\mathrm{\prime }}\mathrm{/}{q}^{\mathrm{\prime }})(n\mathrm{/}s)$ is $sn(q^{\mathrm{\prime }}c-q{c}^{\mathrm{\prime }})$, killed by $b$. Thus the scalar action is independent of both representatives. These formulas define the stated localized module, and every $s\in {\mathcal{S}}_T(k)$ acts invertibly with inverse $1\mathrm{/}s$.

For a continuous left $T$-action on $X$, the projection $ET{\times }_{T}X\to BT$ makes singular cohomology a module over $A_T(k)$ by pullback and cup product; pulling back first Chern classes defines the character actions used in the localization. For empty $X$ the Borel space is empty and its cohomology module is zero, and for $r=0$ the localization changes no module because ${\mathcal{S}}_T(k)=\{1\}$. This verifies the constructions in the Definition.