Spectral projection of an eigenvalue associated to a generator of Hecke algebras In his paper "Hecke Algebras of type $A_n$ (Inv. Math. 1988, EUDML link) and subfactors", in section 2 "Orthogonal representations...", Wenzl takes the usual third relation of a standard generator of Hecke algebras, namely $g_i^2=(q-1)g_i+q$, where $q$ is a complex number that is neither $0$ nor $-1$, and says that $g_i$ can only have at most two eigenvalues. Understandable given that the minimal polynomial of $g_i$ must divide $x^2-(q-1)x-q$. So the possible eigenvalues must be $q,-1$. Then proceeds to say that the spectral projection belonging to $-1$ is $e_i= \frac{q-g_i}{q+1}$. Can someone explain to me what he means by spectral projection and how he comes up with the formula for $e_i$?
 A: Since $e_i$ kills any eigenvectors with eigenvalue $q$ (clear) and squares to itself (due to the quadratic relation for $g_i$ and normalisation), it is a projector onto the $-1$-eigenspace of $g_i$. For the usual (reducible) representation of $H_n(q)$ on $V^{\otimes n}$, where $g_i$ becomes the simple transposition $s_i$ as $q\to 1$, the element $e_i$ acts as the antisymmetriser in the $i$ and $i+1$st factors of $V^{\otimes n}$. So one may think of $e_i$ as a $q$-antisymmetriser.
For comparison, the spectral projector onto the $q$-eigenspace of $g_i$ would be $-(1+g_i)/(q+1)$. Note that these two projectors orthogonal (again due to the quadratic relation) and add to the identity. One can think of this as the corresponding $q$-symmetriser.
(If $g_i$ acts on $V^{\otimes n}$ and $V \cong \mathbb{C}^2$, then the left- and right-hand sides of Wenzl's relation (H1) both vanish, and the $H_n(q)$-representation on $V^{\otimes n}$ factors through a representation of the Temperley--Lieb algebra, with generators $e_i$ up to normalisation depending on conventions.)
I hope this helps.
PS. Nowadays the Hecke generators are usually denoted by $T_i$ rather than $g_i$. The notation $e_i$ for (something proportional to) $q-T_i$ is standard.
