The type-A Soergel Hom formula

We first prove the stronger flagged-source form with $M\in F_{\mathrm{\Delta }}$ and $N$ a Soergel summand. For $M={\mathrm{\Delta }}_y(d)$, evaluation at its generator identifies $\mathrm{Hom}(M,N)$ with a shift of ${\mathrm{\Gamma }}_{y}N$. By [F2] this is free, and the lower-layer image theorem gives rank ${\sum }_e(N:{\mathrm{\nabla }}_y(e))v^{d-e}$. Thus the claimed formula holds for a single standard source, including arbitrary shifts and finite sums.

Let $0\to M_1\to M\to M_2\to 0$ be a short exact sequence from the canonical flag, with all three sources $\mathrm{\Delta }$-flagged and with their multiplicities additive. For a word sum $B$, the sequence $0\to \mathrm{Hom}(M_2,B)\to \mathrm{Hom}(M,B)\to \mathrm{Hom}(M_1,B)$ is always left exact. The special Hom formula [F1] computes all three modules as free, with graded rank polynomials satisfying the same additive identity. Hence their Hilbert series also satisfy that identity, and left exactness forces surjectivity in each degree: the kernel has the known dimension and the source then has precisely enough remaining dimension to fill the last term. Thus the sequence is short exact. If $N$ is a summand of $B$ with projection idempotent $e$, this sequence splits into its $e$ and $1-e$ parts by postcomposition. Exactness therefore holds with $N$ in place of $B$, including its final surjectivity; no Hom-exactness for arbitrary flagged targets has been assumed.

Induct on the finite length of the $\mathrm{\Delta }$ flag of $M$. Peel off the maximal-support first layer $M_1$, a finite sum of shifted standards; the quotient $M_2$ has a shorter flag. Step 1.1 proves the formula for $M_1$, and induction proves it for $M_2$. The short exact Hom sequence of step 1.2 splits as graded right modules, because its last term is free: lift homogeneous basis elements to obtain a homogeneous right-linear section. Thus $\mathrm{Hom}(M,N)$ is free and its rank is the sum of the two multiplicity formulas, giving exactly the displayed expression. This proves the flagged-source form for every summand target.

The dual flagged-target form is proved with the same argument: for a single $\mathrm{\nabla }$ target use the second image equality of [F2] and the free quotient ${\mathrm{\Gamma }}^{y}M$; for a flag extension use the dual special Hom formula [F1] to prove surjectivity for word sources degree by degree, then pass to a source summand by precomposition with its idempotent. Induction on the target flag and a homogeneous splitting of the resulting free Hom sequence give freeness and the same rank sum. In particular [F3] makes step 2.1 applicable to every pair of Soergel objects in the Statement.

Expanding the characters of [F4] in their orthonormal standard basis gives $⟨h_{\mathrm{\Delta }}(M),h_{\mathrm{\nabla }}(N)⟩={\sum }_{x,d,e}(M:{\mathrm{\Delta }}_x(d))(N:{\mathrm{\nabla }}_x(e))v^{d-e}$ by bilinearity, which is the established graded rank. For the unit the single graph layer has $d=e=0$; a bimodule endomorphism of $R$ is multiplication by its image of $1$, so ${\mathrm{End}}_{R\text{-}R}^{\bullet }(R)=R$, freely generated by the identity in degree zero. This proves all assertions.