What is the determinant of the R-matrix defining the colored Jones polynomial? Let $V_n$ be the $(n+1)$-dimensional irreducible representation of $\mathcal U = \mathcal{U}_q(\mathfrak{sl}_2)$, and let $\mathbf R \in \mathcal{U} \widehat \otimes \mathcal{U}$ be the universal $R$-matrix. The $R$-matrix $R_n$ defining the $n$th colored Jones polynomial is the matrix of the linear map given by the action of $\mathbf R$ on $V_n \otimes V_n$.
Actually, this is not quite true. The partial quantum trace $\operatorname{ptr}_q^r (\tau R) : V_n \to V_n$ of the braiding induced by $R_n$ is not the identity map, but the identity map times a certain scalar $\theta_n$, so the corresponding link invariant depends on the framing of the link. To remove this dependence we can instead use $\tau R_n/\theta_n$.
Is there a straightforward way to compute $\det(\tau R_n/\theta_n)$, or equivalently to compute $\det R_n$? Perhaps you could figure out the eigenvalues of $R_n$ via Jones-Wenzl projectors, but this computation seems somewhat tricky and I'm curious if anyone has already worked it out.
 A: There is a trick that makes it easy to compute
$$
\det R_n = 1.
$$
It is similarly well-known that $\theta_n = q^{n^2/2}$ (up to some minus signs), so
$$
\det \theta_n^{-1} \tau R_n = q^{-n^4/2} (-1)^{n(n+1)/2}.
$$
This is probably known, but I'll include an argument below in case it helps someone else.
We can write
$$ \mathbf{R} = q^{H \otimes H/2} \exp_q\left(\frac{E \otimes F}{(q -q^{-1})^2}\right) = q^{H \otimes H/2} \sum_{k = 0}^\infty \frac{q^{\binom{k}{2}}}{[k]_q! (q - q^{-1})^{2k}} (E \otimes F)^k.$$
$[k]_q!$ is a quantum factorial, but its value will not affect our answer.
The action of $\mathbf{R}$ makes sense because $E$ and $F$ both act nilpotently on $V_n$ and we can pick an eigenbasis for $K = q^H$.
In fact, because $E \otimes F$ is nilpotent, the determinant of the $q$-exponential is $1$.
Because $(E \otimes F)^{n+1}$ acts by $0$ on $V_n \otimes V_n$, we can replace the exponential with
$$ \exp_q^{\le n+1}(z) = \sum_{k=0}^{n+1} \frac{q^{\binom{k}{2}}}{[k]_q!} z^k. $$
Choosing a basis of $V_n \otimes V_n$ in which $E \otimes F$ is strictly upper-triangular, we see that
$$
\exp_q\left(\frac{E \otimes F}{(q -q^{-1})^2}\right)
= 1 + p(E \otimes F)
$$
where $p$ is some polynomial.
In particular, the action of $p$ is strictly-upper triangular, so $$\det(1 + p(E \otimes F)) = 1.$$
The determinant of $q^{H \otimes H/2}$ is easy to compute.
Pick a weight basis $\{v_k, k = 0, \dots, n\}$ with $K \cdot v_k = q^{n-2k}$.
Then
$$
(H \otimes H) \cdot v_k \otimes v_l = (n - 2k)(n-2l)
$$
and since
$$
\sum_{k, l = 0}^{n} (n - 2k)(n-2l) = 0
$$
we see that
$$
\det\left(q^{H \otimes H/2}\right) = 1.
$$
