The Jones polynomial can be computed from the representation theory of $\mathcal{U}\_q(\mathfrak{sl}(2))$. The Alexander polynomial has an analogous description in terms of the representation theory of the superalgebra $\mathcal{U}\_q(\mathfrak{gl}(1|1))$ (see, for example, Sartori <http://arxiv.org/abs/1308.2047>). 


From this perspective, is there a nice explanation of the fact that $V\_K|\_{q = i} = \Delta\_K|\_{t = -1} = \det K$? Here $K$ is a knot, $V\_K(q)$ denotes the Jones polynomial of $K$, $\Delta\_K(t)$ denotes the Alexander polynomial of $K$ ($t$ and $q$ are related by $t = q^2$), and $\det K$ is the knot determinant.



Why, in terms of quantum groups, does the knot determinant appear as an evaluation of both the Jones and Alexander polynomials?
You haven't specified what sorts of explanations you would find "nice," so the following collection of thoughts might not be helpful, but here goes (or it might be helpful, but contain only things you already know). [By the way, this answer only attempts to "explain" why $V\_K(i) = \Delta(-1)$ -- it does not address the additional question of why they both yield the knot determinant. That's a great question; I guess one "explanation" is simply that the invariants are determined by their skein relation.]


Firstly, a reference I like, in addition to the one you mention by Sartori, [is this paper by Kauffman and Saleur](https://scholar.google.com/scholar?q=Free%20fermions%20and%20the%20Alexander-Conway%20polynomial&btnG=&hl=en&as_sdt=0%2C31). In that paper, many (though not all) of the theorems and computations are for the general Lie superalgebra $\mathfrak{gl}(m|n)$, where $m,n \in \mathbb{Z} >= 0$ are not necessarily equal. The Jones and Alexander cases are each specializations of this general case, with $(m,n) = (2,0)$ or $(1,1)$, respectively, so therefore a general strategy for a certain kind of "explanation" would be: observe that for certain specializations of the quantum parameter $q$, the generally-defined $\mathfrak{gl}(m|n)$ quantum link invariants, and even much of the structure of the quantum group $U\_q(\mathfrak{gl}(m|n))$, depends only on $m+n$.


Here are two specific instantiations of the strategy. For the first, see page 299 (= page 7) of the paper I linked to. There the authors point out that for $q = i$, the quantum $R$ matrices for $U\_q(\mathfrak{gl}(m|n))$ and $U\_q(\mathfrak{gl}(m+n))$ agree. This is the essential ingredient for showing the relevant polynomial invariants agree at $q=i$, but to be complete one would want to go through the details of how those invariants are defined, in terms of the $R$ matrix.


The second is along those lines: the first link invariant defined in the Kauffman and Saleur paper has this skein relation (see p. 302):


$$ q^{n-m}P^+ - q^{-(n-m)}P^- = (q-q^{-1})P\_{||} $$


(hopefully my notation is clear enough). However, one can see, by passing from a given knot to a split (1,1) tangle, say, that this invariant $P$ is forced to be zero for all knots if the quantum dimension $D\_q$ of $U\_q(\mathfrak{gl}(m|n))$ is zero. However, one can define another invariant $P':= P/D\_q,$ which will be non-zero. The quantum dimension is $$ D\_q = \frac{q^{n-m} - q^{-(n-m)}}{q-q^{-1}},$$ which is zero for all $q$ if $n=m$. Anyway, I haven't done it carefully myself, but I think it should be fairly straightforward to go from the above skein relation to the skein relation satisfied by $P'$ at $q = i$, and see that this only depends on $m+n$. (A final step would be to relate these normalizations of the invariants to those in the literature).


Let me know if I've said anything confusing or wrong, and if this helps in any way; or if you were imagining something different in terms of an explanation.


