Fixed-embedding BM self-intersection and Euler units

By [F1], $F$ is smooth and $T_{x}F=(T_{x}X{)}^T$. The quotient $T_{x}X\mathrm{/}{T}_{x}F$ therefore has no invariant vector. On an affine open of a connected fixed component the base ring has trivial $T$-action, so the coaction on the finite-rank normal bundle is linear over that ring. Each of finitely many local module generators has finite character support; the character-coefficient projections are orthogonal idempotents summing to the identity. Their images are locally free direct summands. These decompositions agree on overlaps because the projections are defined by the group coaction. The zero-weight image has zero fiber at every point by $T_{x}F=(T_{x}X{)}^T$, so ${\nu }_F$ is the finite direct sum of nontrivial character subbundles. Thus $i$ is a closed regular embedding with the stated complex normal bundle.

The complex normal orientation gives the refined Gysin map in the normalized Borel–Moore grading. The Thom class is normalized to restrict to the positive complex-oriented generator on every normal fiber, so $i^{!}[X{]}_T=[F{]}_T$. A $K$-invariant tubular neighborhood of each fixed component identifies the embedding locally in the Borel construction with the zero section of its normal bundle. By [F3] and the Thom class in [F4], this zero-section operation satisfies $i^{!}{i}_{\ast }\beta =e_T({\nu }_F)\cap \beta$; the degree is lowered by twice the complex codimension $c$. The result is independent of the chosen tubular neighborhood because the refined exchange and orientation comparisons are canonical.

Now let $F_0$ be a connected component of proper $F$. It is a compact smooth complex manifold, and $T$ acts trivially on it. The Borel normal bundle is a complex vector bundle over $BT\times F_0$; [F5] puts its base in the Thom theorem's scope. By [F6], $H_T^{\ast }(F_0;\mathbb{Q})=H^{\ast }(BT;\mathbb{Q}){\otimes }_{\mathbb{Q}}{H}^{\ast }(F_0;\mathbb{Q})$. The weight subbundles from step 1.1 have ranks $m_{\chi }$ constant on $F_0$. Restricting the Borel normal bundle to $BT\times \{x\}$, for any $x\in F_0$, gives ${⨁}_{\chi \mathrm{\ne }1}{L}_{\chi }^{\oplus m_{\chi }}$. By [F4] and the zero-section definition of Euler class, the restriction of $e_T({\nu }_F)$ is $w_0={\prod }_{\chi \mathrm{\ne }1}{u}_{\chi }^{m_{\chi }}$. Each $u_{\chi }$ is a generator of $S_T$, so $w_0$ is invertible after localization.

Under the Künneth decomposition of step 2.2, write $e_T({\nu }_F)=w_0+\eta$, where $\eta \in J_0{\otimes }_{\mathbb{Q}}{H}^{\ast }(BT;\mathbb{Q})$; restriction to a point detects the degree-zero $F_0$ component $w_0$. By [F5], $J_0^N=0$ for some finite $N$, so $z=w_0^{-1}\eta$ has $z^N=0$ in the localized ring. Therefore $e_T({\nu }_F{)}^{-1}=w_0^{-1}{\sum }_{a=0}^{N-1}(-z{)}^a$ is a finite inverse. If the normal rank is zero, the empty product is $w_0=1$, $\eta =0$, and the same formula gives $e_T=1$. There are finitely many connected components because $F$ is finite type; the componentwise inverses give a unit on all of $F$. If $F=\mathrm{\varnothing }$, the Euler-unit clause is vacuous.