Computation of the A-polynomial of the trefoil knot

Question

What is the A-polynomial of the trefoil knot? Show the steps.

I would like to compute the A-polynomial of the trefoil knot, whose fundamental group is given by: $$\langle x, y | x^2 = y^3 \rangle$$ But we have to figure out how to express the longitude $\lambda$ and meridian $\mu$ in terms of $x$ and $y$, and for reasons I do not understand it becomes something like $x = \lambda\mu\lambda$ and $\mu=y$ or something alike.

Can someone help? Thank you.

Answer 1

I followed the proof by D Cooper et al. in https://www.google.com/url?sa=t&source=web&rct=j&url=http://citeseerx.ist.psu.edu/viewdoc/download%3Fdoi%3D10.1.1.116.3012%26rep%3Drep1%26type%3Dpdf&ved=2ahUKEwiHq_uQoYDnAhUZfnAKHR00AWcQFjACegQIBBAB&usg=AOvVaw1m_3127P1umJezolXyi7YW

For the torus knot $K =K(p, q) $, and $X=S^3\backslash K$, the authors start from the fundamental group $\pi_1(X) = \langle x, y| x^p = y^q \rangle $ and note that the meridian $\mathfrak m$ is actually the generator of the homology $H_1(X) $, with linking number $+1$ with $K$. Note that $x = \mathfrak m^q$ and $y=\mathfrak m^p$ inside $H_1(X)$, so since $mp+nq=1$ for some integers $m$ and $n$, which is guaranteed by Euclidean algorithm, we can choose an abelian representation such that $\mathfrak m = x^n y^m$. But $\mathfrak l$ has linking number $0$ with $K$, so we choose the loop $\mathfrak l = x^p \mathfrak m^{-pq} $, since in this case the linking number is $q(p) - pq=0$.

Suppose $\mathfrak l, \mathfrak m$ get sent to $L,M$ respectively in the representation and $\lambda, \mu$ are the eigenvalues.

Then there is a non-abelian representation $\rho: \pi_1(X)\rightarrow SL_2(\mathbb C) $ such that $\rho(x^{2p}) = \rho(y^{2q}) = I_2$ and $LM^{pq} = \rho( x^p ) = \rho(x^p) ^{-1} $ has eigenvalues $\pm 1$.

We consider specifically the eigenvalue of $\rho (x^p)$ being $-1$. Now since $L,M$ commute, the eigenvalues of $LM^{pq}$ is the product $\lambda\mu^{pq} $. But note that $LM^{pq} = \rho(x^p) $, which has an eigenvalue of $-1$.

Thus $\lambda\mu^{pq} +1$ divides the A-polynomial.

Answer 2

That presentation of the fundamental group of the trefoil complement is explained in Hatcher's Algebraic topology. A torus knot is one that lies on a torus that splits $S^3$ into two solid tori, hence the complement of a torus knot can be formed from taking two solid tori and gluing them together along an annulus; then the fundamental group can be calculated with the van Kampen theorem. The two generators correspond to the core circles of each solid torus.

Fixing the basepoint at the boundary of this annulus, then there is a natural choice of meridian (take a small circle around the knot). There is also a longitude from a boundary component of the annulus, but to get a $0$-framed longitude you'll have to modify it with some number of meridians.

I've calculated a $\mu$ and $\lambda$ a few times (quoting Is every loop in a 3-manifold homotopic to some loop on its boundary?, which unfortunately doesn't show how I got it): \begin{align*} \mu &= y^{-1}x \\ \lambda &= \mu^{-5}xy. \end{align*}

If I remember correctly, I used the fundamental group of the trefoil coming from the Wirtinger presentation ($\langle a,b\mid aba=bab\rangle$, after eliminating a generator), since each generator is a meridian, and for a given meridian it is not too difficult to come up with a corresponding longitude. Then I matched it up with the $\langle x,y\mid x^2=y^3\rangle$ presentation. Something I find to be helpful to remember is that $x^2$ generates the center, as does $ababab$.

Answer 3

I am writing this answer because I was curious about how to compute A-polynomials, so I'm just going to leave some notes here. And a disclaimer: there's no guarantee what I'm doing is correct, but it did work for the trefoil.

The Wirtinger presentation of the standard $3$-crossing diagram of a trefoil is $$\langle a,b,c \mid cb=ac,ba=cb,ac=ba\rangle.$$ This reduces to $\langle a,b\mid aba=bab\rangle$. One of the two checkerboard surfaces is a Seifert surface, so taking its boundary curve, starting from the $a$ meridian, gives $\lambda=cb^{-1}ac^{-1}ba^{-1}$. After some conjugation by $a$, and substitution of $c=a^{-1}ba$, we get an equivalent longitude $\lambda=a^{-4}ba^{2}b$.

The A-polynomial is from the character variety of $SL_2(\mathbb{C})$ representations of $\pi_1(S^3-K)$. The set $R=\hom(\pi_1(S^3-K),SL_2(\mathbb{C}))$, which then you take a quotient by conjugation to get the character variety $X$, and then you restrict it to the peripheral subgroup $\langle \lambda,\mu\rangle$. From here, you project to $(\mathbb{C}^*)^2$ by putting it into $(L,M)$ coordinates, which are the eigenvalues of the two matrices for $\lambda$ and $\mu$. Then you remove the $L-1$ factor, remove multiplicities, and normalize.

One fact about $SL_2(\mathbb{C})$ is that the eigenvalues come in inverse pairs, so the trace of a matrix is the sum of an eigenvalue and its inverse. Also, the trace of the matrices is invariant under conjugation.

We construct a system of equations. Let $A,B\in SL_2(\mathbb{C})$ be indeterminate matrices and $t_\lambda,t_\mu\in\mathbb{C}$ be inderminates representing the traces of $\lambda$ and $\mu$ through the representation. \begin{align*} t_\mu&=\operatorname{trace}(A)\\ t_\lambda&=\operatorname{trace}(A^{-4}BA^2B)\\ 1 &= \det(A)\\ 1 &= \det(B). \end{align*} These are all polynomial equations in the entries of $A$ and $B$ and in $t_\lambda$ and $t_\mu$. If you calculate a Groebner basis for this system of equations with respect to $t_\lambda$ and $t_\mu$, while eliminating all the variables for the entries of $A$ and $B$ (this corresponds to restricting to the peripheral subgroup), then the system reduces to a single polynomial $$(t_\lambda - 2)(-2+ t_\lambda + 9t_\mu^2 - 6t_\mu^4 + t_\mu^6).$$ Substituting in $t_\lambda = L+L^{-1}$ and $t_\mu = M+M^{-1}$, this polynomial becomes $$L^{-1}M^{-6}(L-1)^2(L+M^6)(L^{-1}+M^6).$$ Choosing "the" eigenvalue of $\lambda$ is not unique, since it has $L$ and $L^{-1}$, so the last two terms correspond to the same curve in a sense (I'm not sure exactly what is going on here). Then $L^{-1}+M^6$ is an A-polynomial of the trefoil. After renormalizing by a factor of $L$, this is the A-polynomial listed in Hoste's table on KnotInfo.