Character varieties and the A-polynomial — Examples and Counterexamples

Since $N^2=0$, $U=\left(\begin{array}{cc}1& 1\\ 0& 1\end{array}\right)$, so $U\in S{L}_2(\mathbb{C})$, has only eigenvalue $1$, and is not $I_2$. Also $V=I_2\in S{L}_2(\mathbb{C})$. By [F2], the pair commutes and has common eigenline $\mathbb{C}(1,0)$ with eigenvalues $M=L=1$.

For $(p,q)=(1,0)$, the eigenvalue equation gives $M^p{L}^q=1^1{1}^0=1$, while $U^p{V}^q=U\mathrm{\ne }{I}_2$. This refutes the statement. The example is only a matrix pair and does not claim to extend to any knot-group representation.

Put $x=M+M^{-1}$. By [F1] and [F2], $L+L^{-1}=x^4-5x^2+2$. Since $x^2=M^2+2+M^{-2}$ and $x^4=M^4+4M^2+6+4M^{-2}+M^{-4}$, the right side is $M^4-M^2-2-M^{-2}+M^{-4}$. Multiplying by the unit $M^{4}L$ gives $Q(M,L)=M^4{L}^2+(-M^8+M^6+2M^4+M^2-1)L+M^4=0.$ Thus every eigenvalue pair from an irreducible representation lies on this explicit Laurent curve.

Irreducibility is checked by an exact factorization. Put $X=M^2$ and $B(X)=-X^4+X^3+2X^2+X-1$. As a quadratic in $L$, $Q=X^2{L}^2+B(X)L+X^2$ has discriminant $B(X{)}^2-4X^4=(X-1{)}^2(X+1{)}^2(X^2+X+1)(X^2-3X+1)\mathrm{.}$ This follows by factoring $B-2X^2=-(X-1{)}^2(X^2+X+1)$ and $B+2X^2=-(X+1{)}^2(X^2-3X+1)$. A root $r$ of $X^2-3X+1$ is nonzero, is simple, and is neither $1,-1$ nor a root of $X^2+X+1$: subtracting the two quadratics at a common root would give $4r=0$. Each square root $M$ of $r$ is consequently a simple zero of the discriminant after substitution $X=M^2$, since $2M\mathrm{\ne }0$. Thus the discriminant is not a square in $\mathbb{C}(M)$. The quadratic $Q$ is irreducible over $\mathbb{C}(M)$, and [F3] gives its irreducibility in the Laurent polynomial ring; its nonzero constant term excludes an inverted $L$ factor.

There are actual representations with infinitely many meridian values. Use the presentation $⟨u,v\mid vw=wu⟩$, $w=u{v}^{-1}{u}^{-1}v$, from [F4]. For $M\mathrm{\ne }0$, put $a=M^2+M^{-2}-3$, $F=t^2+at-a$ and $A=\left(\begin{array}{cc}M& 1\\ 0& M^{-1}\end{array}\right),B=\left(\begin{array}{cc}M& 0\\ t& M^{-1}\end{array}\right),W=A{B}^{-1}{A}^{-1}B\mathrm{.}$ Both matrices have determinant one. Multiplication gives $BW-WA=\left(\begin{array}{cc}0& -F\\ tF& 0\end{array}\right),$ so each root of $F=0$ defines an actual representation with $\rho (u)=A$, $\rho (v)=B$. Exclude $M=\pm 1$ and the finitely many solutions of $a=0$. The monic quadratic $F$ has a complex root $t$, and $F(0)=-a\mathrm{\ne }0$ makes $t\mathrm{\ne }0$. Write $d=M-M^{-1}\mathrm{\ne }0$. The only eigenlines of $A$ are $\mathbb{C}(1,0)$ and $\mathbb{C}(1,-d)$. The first is not $B$-invariant since $t\mathrm{\ne }0$; the second is $B$-invariant exactly when $t=-d^2$. Since $d^2=a+1$, substitution gives $F(-d^2)=1$, excluding that case. Thus the representation is irreducible. Commutation with $A$ makes the preferred longitude preserve its $M$-eigenline, producing an eigenvalue pair $(M,L)$. Infinitely many distinct $M$ remain, and their pairs satisfy $Q=0$ by step 1.1. No normal-form converse or image-completeness theorem is assumed.

The exact curve vanishing ideal follows by polynomial Euclid. Suppose a Laurent polynomial $P(M,L)$ vanishes on all those irreducible eigenvalue pairs. If $Q$ does not divide $P$ in $\mathbb{C}(M)[L^{\pm 1}]$, step 2.1 makes their gcd one. The finite Euclidean algorithm gives a Bezout identity there. Clearing its finitely many nonzero denominators in $M$, while retaining the already allowed Laurent powers of $L$, gives Laurent coefficients $U,V$ and a nonzero Laurent polynomial $h(M)$ with $UP+VQ=h(M)$. Step 2.2 would make $h(M)$ vanish at infinitely many nonzero meridian values, impossible for a nonzero one-variable Laurent polynomial. Therefore $Q$ divides $P$. Its primitive irreducible Laurent equation from step 2.1 and [F3] then divides $P$ in $\mathbb{C}[M^{\pm 1},L^{\pm 1}]$ itself. The converse is step 1.1. Hence the vanishing ideal of the full irreducible peripheral eigenvalue locus is exactly $(Q)$, and its reduced closure is the whole $Q$ curve. This includes any exceptional central or parabolic pairs, since they too satisfy step 1.1, while the dense family already forces this ideal.

Reducible representations contribute precisely $L=1$ by [F4]. Since $Q(M,1)=-M^8+M^6+4M^4+M^2-1$ is not identically zero, $Q$ and $L-1$ are distinct irreducible Laurent factors. The defining reduced union has ideal $(Q)\cap (L-1)=((L-1)Q)$ by [F3]. By the ordinary and enhanced definitions, the enhanced equation is $Q$ and the ordinary one is $(L-1)Q$, up to a Laurent unit. The displayed integer polynomials are primitive and square-free. This proves both formulas using the actual representation family and preferred longitude, rather than an assumed image-completeness premise.