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Suppose that $K$ is a non-trivial, alternating knot.

Is it possible that $\det K = 1$ where $\det K=\Delta_K(-1)$?

Using knotinfo, I checked that all non-trivial alternating knots with crossing less than or equal to 12 have non-trivial determinant.

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1 Answer

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If $K$ is alternating and $\det K=1$, then $K$ is the unknot.

Let $D$ be an alternating diagram of the knot $K$, and let $G$ be the checkerboard graph (see picture) of $D$.

The determinant of $K$ is the number of spanning trees of $G$. One way to see this is via Kauffman's state sum expansion for the Alexander polynomial. If $\det K=1$, then $G$ has only one spanning tree, and hence $G$ is a tree (Edit: possibly with loops). If the checkerboard graph is a tree with loops, then it can be transformed into the standard diagram of the unknot using only Reidemeister one moves.

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