Let $H_n$ be the Hecke algebra of GL(n), i.e., the algebra over $\mathbb{Q}(q)$ with generators $T_1, \ldots, T_{n-1}$ which satisfy the braid relations and also $T^2 = (q-1) T + q$.

Recall the Markov trace: it is defined to be the unique family of maps $M_n:H_n \to \mathbb{Q}(a,q)$ satisfying the following.

Note $H_{n-1}$ naturally includes inside $H_n$. Then for $h \in H_{n-1}$, we require:

$$M_n(h) = \left(\frac{1-a^2}{1-q}\right)\cdot M_{n-1}(h)$$ $$M_n(T_{n-1} h) = M_{n-1}(h)$$

This determines the trace up to a normalization, we take

$$M_1(1 \in H_1) = \frac{1-a^2}{1-q}$$

Since the Hecke algebra satisfies the braid relations, the braid group maps to it. For a braid $\beta$ on $n$ strands with writhe $w$, and its braid closure $\overline{\beta}$ one puts $$P(\overline{\beta}) = \left(\frac{a}{q^{1/2}} \right)^{w-n} M_n(\beta)$$

Then $P$ is famously a knot invariant: the HOMFLY polynomial.

The above inductive description is a bit mysterious, though subsequent work has shed some algebraic and geometric light on the issue.

I have recently encountered a formula of a seemingly different flavor for the $a=0$ specialization of the above. To state it, recall that $H_n$ has a basis $T_w$ where $w \in S_n$ are the elements of the symmetric group, and $T_w$ is by definition $T_{i_1} \ldots T_{i_k}$ where $w = s_{i_1} \ldots s_{i_k}$ is a reduced expression. We will write $$\mathrm{Coeff}_{T_{id}} : H_n \to \mathbb{Q}(q)$$ for the function extracting the coefficient of $T_{id}$ in the above basis.

Let $\Delta^2$ be the full twist in the braid group, or its image in the Hecke algebra. Then:

$$ M_n(h) \bigg|_{a=0} = \frac{\mathrm{Coeff}_{T_{id}}(h \Delta^2)}{q^{n \choose 2} (q-1)^n}$$

(The denominator is, not accidentally, the number of points over $\mathbb{F}_q$ in the Borel subgroup.)

Is this formula known? In particular, does it follow from one of the above-linked items?

More importantly:

Does the above formula have a generalization to other coefficients of $a$ ?

And...

What of Hecke algebras of other types?

For the curious: the only proof I know of this formula takes a rather long detour through the subject of Legendrian knots. This proceeds in broad outlines as follows. A Legendrian knot serves as a boundary condition for Lagrangians; the corresponding Fukaya category is a natural invariant of the knot. When the knot is a (positive) braid closure, it turns out that the moduli space of rank one objects in this category is the fibre over the trivial flag of the map from the open Bott-Samelson of the braid times a full twist. Counting points via the Hecke algebra gives the right side of the formula. On the other hand, the objects in this category admit certain canonical filtrations; this determines a stratification on the moduli space. (I believe, but have not checked, that it is an analogue of the Deodhar stratification of the Richardson varieties.) The strata are determined by certain diagrams and each contribute some $q^a (q-1)^b$ points over $\mathbb{F}_q$. But the same diagrams with the same contributions are known to sum to the appropriate coefficient of the HOMFLY polynomial. This gives the left hand side.