The local intersection form on Hom(B_y, B_x B_s)

The setting. Fix $x\in S_n$, a simple reflection $s$ with $xs>x$, and $y<xs$ in the Bruhat order; assume $\mathrm{S}(z)$ for all $z<xs$. Choose reduced-word embeddings $B_y^R\subseteq \text{\hspace{0.17em}⁣}\text{\hspace{0.17em}⁣}⨁\mathrm{B}\mathrm{S}(\underset{‾}{y})$ and $B_x^R\subseteq \text{\hspace{0.17em}⁣}\text{\hspace{0.17em}⁣}⨁\mathrm{B}\mathrm{S}(\underset{‾}{x})$, and equip $B_y^R$, $B_x^R$ with the positively normalized non-degenerate invariant forms obtained by restriction, so that the bottom Lefschetz values are $+1$; equip $M:=B_x^R{B}_s^R$ with the induced invariant form of The trace, the invariant intersection form on Bott-Samelson bimodules and its reduction. Under $\mathrm{S}(y)$ and $\mathrm{S}(x)$ these forms exist and are unique up to positive scalars, and the choices enter only through positive factors (Invariant forms on an indecomposable Soergel bimodule are unique up to scalar, non-degenerate, and agree with the restricted Bott-Samelson form).

The local intersection form. Let $E_y:={\mathrm{Hom}}^0(B_y^R,M)$ be the finite-dimensional real space of degree-zero bimodule maps. For $f,g\in E_y$ define the local intersection form $(f,g{)}_y^{x,s}:=g^{\ast }\circ f\in {\mathrm{End}}^0(B_y^R)=\mathbb{R},$ where $g^{\ast }:M\to B_y^R$ is the adjoint of $g$ with respect to the two non-degenerate forms, uniquely characterized by $⟨g(b),m{⟩}_M=⟨b,g^{\ast }(m){⟩}_{B_y}$ for all $b\in B_y^R$, $m\in M$. Here the dual is the graded right $S$-module dual $DB:={\mathrm{Hom}}_{-S}^{\bullet }(B,S)$, with $(r\phi )(b)=\phi (rb)$ and $(\phi r)(b)=\phi (b)r$. If ${\mathrm{\Phi }}_B:B\to DB$ is the perfect-form isomorphism ${\mathrm{\Phi }}_B(b)(b^{\mathrm{\prime }})=⟨b,b^{\mathrm{\prime }}{⟩}_B$, then $g^{\ast }={\mathrm{\Phi }}_{B_y}^{-1}\circ Dg\circ {\mathrm{\Phi }}_M,$ where $Dg:DM\to D{B}_y$ is precomposition by $g$. Right bilinearity and left invariance make ${\mathrm{\Phi }}_B$ and $Dg$ degree-zero bimodule maps, so this formula proves the existence, uniqueness and degree of the adjoint. The adjoint construction identifies ${\mathrm{Hom}}^0(B_y^R,M)$ with ${\mathrm{Hom}}^0(M,B_y^R)$.

Basic properties. The form $(-,-{)}_y^{x,s}$ is symmetric: the adjoint of $g^{\ast }f$ is $f^{\ast }g$, and a scalar endomorphism of $B_y^R$ is self-adjoint for its invariant form. It is $\mathbb{R}$-bilinear, and it changes only by a positive scalar when the chosen normalized forms or reduced-word embeddings are varied: scaling the form on $M$ by $a>0$ and the form on $B_y^R$ by $b>0$ scales $(-,-{)}_y^{x,s}$ by $a\mathrm{/}b$, and the invariant-form uniqueness lemma shows that any two choices compare by positive scalars. Define ${\mathrm{S}}^{\pm }(y,x,s)$ to mean that $(-,-{)}_y^{x,s}$ is definite of sign $(-1{)}^{(\mathrm{\ell }(x)+1-\mathrm{\ell }(y))\mathrm{/}2}$, respectively its negative, with the parity convention of Primitive subspaces, hard Lefschetz and the Hodge-Riemann signs: the dictionary of hypotheses S, hL and HR: when the numerator is odd the corresponding primitive space is zero and the sign condition is not used.

Status. The definition uses degree-zero Hom throughout; no statement identifies the rank of the full graded Hom module with a Kazhdan-Lusztig multiplication coefficient. The dimension of $E_y$ in terms of the multiplication formula, and the equivalence of non-degeneracy of these forms with the actual summand multiplicities of $M$, are proved in the detection theorem later on this page; they are not part of this definition.

• The adjoint $g^{\ast }$ exists and is unique because the forms involved are non-degenerate and invariant: an invariant form on a finite right-free bimodule is the same as an isomorphism to the dual bimodule, and composing $g$ with the two form-isomorphisms gives a degree-zero bimodule map $M\to B_y^R$.
• The normalizations are chosen so that the bottom Lefschetz value is $+1$; by the uniqueness lemma the comparison between any two such normalizations is by a positive scalar, so the sign of the local intersection form, hence the statement ${\mathrm{S}}^{\pm }(y,x,s)$, is well defined independently of the choices.