Character varieties and the A-polynomial

Let $F\subset \mathrm{\Omega }$ be a field extension with $\mathrm{\Omega }$ algebraic over $F$, let $F_0$ be an intermediate field, and let ${\varphi }_0:F_0\to {\mathrm{\Omega }}^{\mathrm{\prime }}$ be an embedding into an algebraically closed field ${\mathrm{\Omega }}^{\mathrm{\prime }}$. Order all extensions $\varphi :E\to {\mathrm{\Omega }}^{\mathrm{\prime }}$ with $F_0\subseteq E\subseteq \mathrm{\Omega }$ by extension. A chain has the union embedding, so Zorn's lemma from [F3] gives a maximal one. If $E\mathrm{\ne }\mathrm{\Omega }$, choose $\alpha \in \mathrm{\Omega }\setminus E$; it is algebraic over $E$. Apply $\varphi$ to its minimal polynomial and choose a root $\beta \in {\mathrm{\Omega }}^{\mathrm{\prime }}$. The transported minimal polynomial is irreducible over $\varphi (E)$ because $\varphi$ is an isomorphism onto its image, so sending $\alpha$ to $\beta$ extends $\varphi$ to an embedding of $E(\alpha )$, contradicting maximality. Hence $\varphi$ extends to $\mathrm{\Omega }$. If ${\varphi }_0$ maps $F$ onto $F$ and both source and target are algebraic closures of $F$, this extension is onto: its image is algebraically closed and contains $F$, so it contains every root of every polynomial over $F$, hence all of $\mathrm{\Omega }$.

Use [F2] to view $\bar{\mathbb{Q}}$ as a subfield of $\mathbb{C}$. By [F1], choose a transcendence basis $B$ of $\mathbb{C}$ over $\bar{\mathbb{Q}}$. Then $\mathbb{C}$ is algebraic over $\bar{\mathbb{Q}}(B)$ and, being algebraically closed, is an algebraic closure of that field.

If $z$ is algebraic over $\mathbb{Q}$ but $z\mathrm{\notin }\mathbb{Q}$, its irreducible minimal polynomial has degree at least two and has distinct roots because its derivative is nonzero and relatively prime to it. Choose another root $z^{\mathrm{\prime }}\in \bar{\mathbb{Q}}$. By step 1.1, the embedding $\mathbb{Q}(z)\to \bar{\mathbb{Q}}$ sending $z$ to $z^{\mathrm{\prime }}$ extends to an automorphism of $\bar{\mathbb{Q}}$ over $\mathbb{Q}$. Extend it coefficientwise to $\bar{\mathbb{Q}}(B)$ while fixing $B$, then apply step 1.1 to extend it to an automorphism of $\mathbb{C}$. This automorphism moves $z$.

If $z$ is transcendental over $\mathbb{Q}$, it is transcendental over the algebraic extension $\bar{\mathbb{Q}}$ by transitivity in [F2]. Extend $\{z\}$ to a transcendence basis $B$ over $\bar{\mathbb{Q}}$ using [F1]. The assignment fixing $\bar{\mathbb{Q}}$ and every element of $B\setminus \{z\}$ while sending $z$ to $z+1$ defines an automorphism of $\bar{\mathbb{Q}}(B)$; its inverse sends $z$ to $z-1$. By step 1.1 it extends to an automorphism of $\mathbb{C}$, and it moves $z$.

Every complex number outside $\mathbb{Q}$ is either algebraic nonrational or transcendental over $\mathbb{Q}$, and steps 2.1, 2.2 exhibit an automorphism fixing $\mathbb{Q}$ that moves it. Every element of $\mathbb{Q}$ is fixed by definition. Thus the common fixed field is exactly $\mathbb{Q}$.

If $A$ is scalar, choose an eigenvector of $B$ using [F1]; its span is preserved by $A$ and $B$.

If $A$ is not scalar, choose an eigenvalue $\lambda$ of $A$. Its eigenspace $E_{\lambda }$ is nonzero and has dimension one, since a two-dimensional eigenspace would force $A=\lambda I$. For $v\in E_{\lambda }$, commutation gives $A(Bv)=B(Av)=\lambda Bv$, so $Bv\in E_{\lambda }$; hence this line is invariant under both matrices.

A basis beginning with a nonzero vector in the common invariant line makes both matrices upper triangular.

The invariant tensor dimension. In the tensor power $V^{\otimes d}$ use the standard orthonormal tensor basis with letters $+$ and $-$. Let $H=\mathrm{diag}(1,-1)$, let $E(-)=+$, $E(+)=0$, and let $F(+)=-$, $F(-)=0$; on tensor powers each acts by the sum of its actions in each slot. Then $E^{\ast }=F$ and $[E,F]=H$ by direct multiplication in each slot, with distinct-slot terms commuting. Invariance under the diagonal matrices forces weight zero, and invariance under the upper unipotents forces $Ew=0$. Conversely, for weight-zero $w$ with $Ew=0$, the identity $\mathrm{\parallel }Fw{\mathrm{\parallel }}^2-\mathrm{\parallel }Ew{\mathrm{\parallel }}^2=⟨Hw,w⟩$ gives $Fw=0$. Both unipotent one-parameter groups then fix $w$ by their finite exponential series. They generate $S{L}_2(\mathbb{C})$: Gaussian elimination handles a matrix with nonzero upper-left entry; a zero entry is first made nonzero by an elementary row operation, and diagonal matrices are also products of elementary upper and lower matrices. Thus these are exactly the invariant tensors. Odd $d$ gives zero invariants, since the central element $-I$ acts by $-1$.

The group hypothesis for the quotient passage is proved. If $U$ is a closed normal unipotent subgroup of $S{L}_2$, its action on the nonzero rational module $V$ has a nonzero fixed vector by [F1]. The subspace $V^U$ is $S{L}_2$-stable: for $v\in V^U$, $g\in S{L}_2$ and $u\in U$, normality gives $u(gv)=g(g^{-1}ug)v=gv$. The standard representation is irreducible; a nonzero vector can be carried to any other by a determinant-one change of basis, obtained by completing each vector to a basis with determinant one. Hence $V^U=V$. The standard representation is faithful, so $U$ is trivial. By [F1] $S{L}_2(\mathbb{C})$ is reductive in precisely the convention used by [F2].

A finite rank computation. Put $d=2n$. The weight-zero space has dimension $\left(\genfrac{}{}{0px}{}{2n}{n}\right)$ and weight-two space has dimension $\left(\genfrac{}{}{0px}{}{2n}{n-1}\right)$, with the latter zero when $n=0$. The adjoint of $E:W_0\to W_2$ is $F:W_2\to W_0$. This $F$ is injective: if $Fw=0$ for a weight-two vector, the commutator identity gives $-\mathrm{\parallel }Ew{\mathrm{\parallel }}^2=2\mathrm{\parallel }w{\mathrm{\parallel }}^2$, so $w=0$. Finite-dimensional orthogonal-complement linear algebra therefore makes $E:W_0\to W_2$ surjective. Step 1.1 gives invariant dimension $\left(\genfrac{}{}{0px}{}{2n}{n}\right)-\left(\genfrac{}{}{0px}{}{2n}{n-1}\right)$. This computation includes the one-dimensional empty tensor power.

Actual epsilon tensors form a basis. For each noncrossing matching of the ordered $2n$ positions, insert the invariant tensor $+\otimes ---\otimes +$ in each matched pair, smaller position first, and permute tensor slots to their specified positions. All these tensors are invariant because the determinant form and its inverse are preserved by $S{L}_2$. Order tensor words lexicographically with $+>-$. The leading word has $+$ at every left endpoint and $-$ at every right endpoint. It is a Dyck word: each initial segment has at least as many left endpoints as right endpoints. Noncrossing matchings correspond bijectively to these words by closing the most recent unclosed endpoint. Distinct matchings therefore have distinct leading words, so their tensors are independent: in any nonzero linear relation choose the largest leading word among its nonzero coefficients. The number of Dyck words is $\left(\genfrac{}{}{0px}{}{2n}{n}\right)-\left(\genfrac{}{}{0px}{}{2n}{n-1}\right)$, by reflecting the initial segment through the first negative partial sum in each bad balanced word. The dimension computation in step 2.1 shows these actual tensors form a basis of the entire invariant space. The same holds for invariant multilinear forms, using the equivariant identification $V\to V^{\ast }$ given by the nondegenerate determinant form. For $n=0$ the matching is empty and gives the scalar $1$.

Polarize invariant polynomials. Separate an invariant polynomial into its homogeneous degrees; the action is linear, so each part is invariant. For a homogeneous degree-$d$ polynomial on $W=M_2(\mathbb{C}{)}^r$, its polarization is the symmetric multilinear form $T(w_1,\dots ,w_d)=\frac{1}{d!}{\sum }_{A\subseteq \{1,\dots ,d\}}(-1{)}^{d-\mathrm{\mid }A\mathrm{\mid }}f({\sum }_{j\in A}{w}_j)\mathrm{.}$ Expansion of each monomial shows that terms missing any argument cancel and the surviving terms use each argument once. Hence it is multilinear and $T(w,\dots ,w)=f(w)$. The formula also proves invariance. Expand each slot according to the finite direct sum of $r$ matrix spaces; each resulting form is an invariant multilinear form on $d$ copies of $\mathrm{End}(V)$. The determinant identification $\mathrm{End}(V)=V\otimes V^{\ast }\cong V\otimes V$ turns it into the invariant $2d$-tensor problem of step 3.1. Consequently it is a linear combination of pairwise epsilon contractions of the two legs of each matrix. Degree zero is simply a scalar and needs no polarization.

Every contraction is a product of word traces. Write $J=\left(\begin{array}{cc}0& 1\\ -1& 0\end{array}\right)$. The tensor of a matrix $A$ in $V\otimes V$ is $AJ$: indeed $(gA{g}^{-1})J=g(AJ)g^T$ because $gJ{g}^T=J$. Contract tensor legs by $J^{-1}$. Regard a complete pairing as a graph with one two-legged vertex for each matrix and one edge for each contraction. It is a disjoint union of cycles, allowing loops and doubled edges. Traverse a cycle and sum its repeated indices. Passing a vertex in its given leg order contributes $(AJ)J^{-1}=A$; passing it in the reverse order contributes $(AJ{)}^T{J}^{-1}=-J{A}^T{J}^{-1}=-\mathrm{adj}(A)$. Reversing an edge pairing merely changes a sign because $(J^{-1}{)}^T=-J^{-1}$. Closing the repeated indices is exactly a trace. Thus each cycle is, up to a sign, a trace of a word in the matrices and their adjugates, and different cycles multiply. Finally $\mathrm{adj}(A)=(\mathrm{tr}A)I-A$ is the explicit $2\times 2$ entry identity. Expanding each adjugate in every word expresses the contraction as a polynomial in ordinary word traces, with $\mathrm{tr}I=2$ for an empty word. Together with polarization, this proves the ambient matrix-polynomial assertion.

Pass to the actual reduced representation equations. For a closed invariant reduced $Z\subseteq M_2^r$, apply [F2] and step 1.2 to its inclusion. The invariant coordinate algebra of $Z$ is a quotient of $\mathbb{C}[M_2^r{]}^{S{L}_2}$. Every ambient invariant is a polynomial in word traces by step 5.1, so their restrictions generate the quotient. A finitely presented group representation variety is such a $Z$: determinants equal one and relators equal the identity are finite polynomial equations after inverse letters are replaced by adjugates. The vanishing ideal of their point set is radical and invariant, whether or not the displayed raw equation ideal is radical. Thus [F2] applies to this actual reduced variety and yields the asserted invariant trace generation. On determinant-one matrices the adjugates used along the way are precisely inverse matrices, so the traces are traces of group words. Every polynomial calculation and pairing list was finite; full AC enters only in [F2]. No classification of individual orbits or rational descent of generic representations has been substituted for this calculation.

The actual rank-two matrix-algebra argument. Let $A\subseteq M_2(\mathbb{C})$ be a unital subalgebra acting irreducibly. It contains a nonscalar element, since scalar matrices preserve every line. If some $T\in A$ has distinct eigenvalues, conjugate to an eigenbasis. Its two rank-one spectral projections belong to $A$, being degree-one polynomials in $T$, so $E_{11},E_{22}\in A$. Irreducibility forces an element $B\in A$ with $b_{12}\mathrm{\ne }0$, since otherwise every element preserves the second coordinate line; it also forces $C\in A$ with $c_{21}\mathrm{\ne }0$, since otherwise every element preserves the first coordinate line. Thus $E_{11}B{E}_{22}=b_{12}{E}_{12}$ and $E_{22}C{E}_{11}=c_{21}{E}_{21}$ put all four matrix units in $A$. If no element initially has distinct eigenvalues, take a nonscalar $T$ and subtract its repeated eigenvalue. The resulting nonzero nilpotent is conjugate, after scaling one basis vector, to $N=E_{12}$. Irreducibility again supplies $B$ with $b_{21}\mathrm{\ne }0$. Then $NB=\left(\begin{array}{cc}{b}_{21}& b_{22}\\ 0& 0\end{array}\right)$ has distinct eigenvalues $b_{21},0$ and reduces to the previous case. Hence every such algebra is $M_2(\mathbb{C})$. In particular an irreducible group image spans the full matrix algebra by [F1].

Irreducibility is detected by the trace character. Choose four group elements $g_i$ whose images under an irreducible $\rho$ are a basis of $M_2$; such a finite basis can be extracted from its spanning image in step 1.1. The trace pairing $⟨A,B⟩=\mathrm{tr}(AB)$ is nondegenerate: pairing with $E_{ij}$ reads the $(j,i)$ matrix entry. Therefore its Gram matrix $(\mathrm{tr}\rho (g_i{g}_j){)}_{i,j=1}^4$ is invertible. The equal character gives the identical Gram matrix for the four ${\rho }^{\mathrm{\prime }}(g_i)$. Those four matrices are independent, because a dependence would give a kernel vector of that Gram matrix. A reducible representation is triangular in an invariant-line basis and its image spans at most the three-dimensional upper-triangular algebra. Thus ${\rho }^{\mathrm{\prime }}$ cannot be reducible.

Trace equality defines an algebra isomorphism. Both images now span $M_2$. Define on finite linear combinations $\mathrm{\Phi }({\sum }_i{c}_i\rho (g_i))={\sum }_i{c}_i{\rho }^{\mathrm{\prime }}(g_i)\mathrm{.}$ This is well defined. Indeed if the source combination is zero, its proposed image pairs to zero with every ${\rho }^{\mathrm{\prime }}(h)$, since ${\sum }_i{c}_i\mathrm{tr}{\rho }^{\mathrm{\prime }}(g_{i}h)={\sum }_i{c}_i\mathrm{tr}\rho (g_{i}h)=0.$ The second image spans $M_2$ and its trace pairing is nondegenerate, so that proposed image is zero. The inverse construction with the representations exchanged proves bijectivity. Multiplication on the group image, then bilinearity, proves that $\mathrm{\Phi }$ is a unital algebra isomorphism and $\mathrm{\Phi }(\rho (g))={\rho }^{\mathrm{\prime }}(g)$.

The algebra isomorphism is explicitly inner. Put $Q_{ij}=\mathrm{\Phi }(E_{ij})$. The nonzero proper idempotent $Q_{11}$ has rank one, because $Q_{11}^2=Q_{11}$ and $Q_{22}=I-Q_{11}$ is also nonzero. Choose $v_1\mathrm{\ne }0$ in its image and put $v_2=Q_{21}{v}_1$. The identity $Q_{12}{Q}_{21}=Q_{11}$ implies $v_2\mathrm{\ne }0$ and $Q_{12}{v}_2=v_1$. Also $Q_{11}{v}_1=v_1$ and $Q_{11}{v}_2=0$, so $v_1,v_2$ are independent. The matrix-unit identities now show $Q_{ij}{v}_k={\delta }_{jk}{v}_i$. Hence, for the invertible matrix $P_0$ with these columns, $\mathrm{\Phi }(A)=P_{0}A{P}_0^{-1}$ for every $A$. Multiplying $P_0$ by a scalar whose square is $(\det P_0{)}^{-1}$ makes its determinant one without changing conjugation; a nonzero complex number has such a square root. Thus the asserted $P\in S{L}_2(\mathbb{C})$ exists. These selections involve only finitely many nonempty sets.

The reducible character is its diagonal one. In a basis beginning with a common invariant-line vector, write $\rho (g)=\left(\begin{array}{cc}\chi (g)& c(g)\\ 0& \chi (g{)}^{-1}\end{array}\right)\mathrm{.}$ The product law makes $\chi$ a multiplicative map to ${\mathbb{C}}^{\ast }$, and the trace of every group word is $\chi (g)+\chi (g{)}^{-1}$, exactly the trace of the diagonal representation. Conjugation by $\left(\begin{array}{cc}0& 1\\ -1& 0\end{array}\right)\in S{L}_2$ exchanges its two diagonal characters. This proves the semisimplification assertions, not conjugacy of every triangular representative to its diagonal part. Steps 1.1–4.1 prove the irreducible conclusion. All spans, bases, roots and matrix operations used are finite, so no choice axiom is introduced.

Only the specified scalar signs can occur. The element $h$ commutes with $x$ and $y$, since it is a power of each. A nonscalar two-by-two complex matrix either has two distinct eigenvalues or has a repeated eigenvalue and a unique one-dimensional eigenspace. In the first case every commuting matrix preserves each eigenspace; in the second it preserves the kernel of the nonzero nilpotent part, which is that unique line. Thus a nonscalar $\rho (h)$ would provide a line invariant under both generator matrices, contradicting irreducibility. Consequently $\rho (h)=ϵI$, and determinant one forces $ϵ=\pm 1$. If $b=2$ and $ϵ=1$, the equation $Y^2=I$ for $Y=\rho (y)$ makes $Y$ diagonalizable because $(z-1)(z+1)$ has distinct roots. Its determinant one forces its two eigenvalues to be equal, so $Y=I$ or $-I$. An eigenline of $\rho (x)$ would then be invariant under both generators. Hence only $ϵ=-1$ is possible for $b=2$. In every case $\rho (\lambda )=ϵ\rho (\mu {)}^{-ab}$, so every common peripheral eigenvalue pair satisfies $L=ϵM^{-ab}$, including nonsemisimple meridian matrices.

Choose noncentral roots with noncentral meridian powers. Fix a permitted sign. For a positive integer $n\ge 2$ choose ${\zeta }_n=\exp (2\pi i\mathrm{/}n)$ when $ϵ=1$ and ${\zeta }_n=\exp (\pi i\mathrm{/}n)$ when $ϵ=-1$. The first case is used only for $n\ge 3$, since $ϵ=1$ is permitted only when $a,b>2$. Set $\xi ={\zeta }_a$ and $\eta ={\zeta }_b$. They satisfy ${\xi }^a={\eta }^b=ϵ$ and neither is $\pm 1$. Also ${\xi }^{\beta }\mathrm{\ne }{\xi }^{-\beta }$ and ${\eta }^{\alpha }\mathrm{\ne }{\eta }^{-\alpha }$. Indeed the Bezout identity gives $\gcd (\beta ,a)=\gcd (\alpha ,b)=1$. In the positive-sign case equality of the two powers would imply $a\mid 2\beta$ or $b\mid 2\alpha$, impossible for the relevant integer greater than two. In the negative-sign case the roots have orders $2a,2b$; equality would imply $a\mid \beta$ or $b\mid \alpha$, also impossible. These explicit roots require no choice family.

An actual irreducible matrix family. Put $s=\eta +{\eta }^{-1}$ and, for $v\in \mathbb{C}$, define $X=\left(\begin{array}{cc}\xi & 0\\ 0& {\xi }^{-1}\end{array}\right),Y_v=\left(\begin{array}{cc}v& 1\\ v(s-v)-1& s-v\end{array}\right)\mathrm{.}$ The second matrix has determinant one and trace $s$, hence distinct eigenvalues $\eta ,{\eta }^{-1}$. Therefore $X^a=Y_v^b=ϵI$, and these are actual representations of the displayed presentation for every $v$. Since $X$ has distinct eigenvalues, a common invariant line must be one of its coordinate lines. The upper-right entry of $Y_v$ is $1$, so the second coordinate line is never invariant. The first is invariant exactly when $v(s-v)-1=0$, whose two roots are $\eta ,{\eta }^{-1}$. Thus the representation is irreducible for every $v$ other than these two values.

The meridian trace varies with a nonzero linear coefficient. Polynomial division modulo $(z-\eta )(z-{\eta }^{-1})$ gives $Y_v^{\alpha }=c{Y}_v+dI$ for constants $c,d$ independent of $v$, including negative $\alpha$ because both eigenvalues are nonzero. Explicitly $c=({\eta }^{\alpha }-{\eta }^{-\alpha })\mathrm{/}(\eta -{\eta }^{-1})\mathrm{\ne }0$ and $d={\eta }^{\alpha }-c\eta$. With $r={\xi }^{\beta }$, one obtains $\mathrm{tr}(Y_v^{\alpha }{X}^{\beta })=c(r-r^{-1})v+c{r}^{-1}s+d(r+r^{-1})\mathrm{.}$ The coefficient is nonzero by step 1.2. For each $M\mathrm{\ne }0$ there is consequently exactly one $v$ giving meridian trace $M+M^{-1}$. The two forbidden $v$ values of step 2.1 exclude at most four meridian values, since each equation $M+M^{-1}=q$ is a quadratic after multiplication by $M$. Discard also $M=\pm 1$. For every remaining $M$ the meridian has distinct eigenvalues $M,M^{-1}$, so an eigenline with eigenvalue $M$ exists and the longitude, being $ϵ{\mu }^{-ab}$, acts there by $L=ϵM^{-ab}$. This gives the claimed cofinite graph family for each permitted sign.

Exact ideals, including all exceptional representations. Evaluation $L\mapsto ϵM^{-ab}$ defines a surjection $\mathbb{C}[M^{\pm 1},L^{\pm 1}]\to \mathbb{C}[M^{\pm 1}]$. Its kernel is $(L{M}^{ab}-ϵ)$: multiplying by a monomial reduces any expression to a polynomial in $L$, and division by the monic $L-ϵM^{-ab}$ over $\mathbb{C}[M^{\pm 1}]$ leaves exactly its evaluated constant remainder. A Laurent polynomial vanishing on the cofinite graph family of step 3.1 therefore has zero evaluated remainder, since a nonzero one-variable Laurent polynomial has only finitely many nonzero roots. Its ideal is precisely this graph ideal. Step 1.1 puts every irreducible eigenvalue pair, including all exceptional meridian values, in the permitted graphs, and step 3.1 supplies a dense family in each. The ideal of the entire locus and of its closure is consequently the intersection of these graph ideals. For $b=2$ the single sign gives $(L{M}^{2a}+1)$. For $b>2$ put $q=L{M}^{ab}$; the two ideals $(q-1)$ and $(q+1)$ are comaximal since their difference is $2$, an invertible scalar. Their intersection equals their product: writing $1=A+B$ with $A$ in one and $B$ in the other expresses an element of the intersection as a sum of two product elements. Thus their intersection is $(q^2-1)=(M^{2ab}{L}^2-1)$. Each graph ideal is prime because its quotient is the Laurent domain just displayed; their intersection is radical. This proves the exact reduced closure assertions by finite algebra with no assumption about completeness of an unspecified image.

Apply [F1] to $A$ and $B$. Since both determinants are one, $A^{-1}=aI-A,B^{-1}=bI-B,A^2=aA-I,B^2=bB-I\mathrm{.}$ Thus it suffices to reduce positive words. Apply [F1] also to $A+B$ and subtract the identities for $A,B$. The entrywise determinant polarization is $\det (A+B)-\det A-\det B=ab-c$. Hence $AB+BA=aB+bA+(c-ab)I,$ an identity valid without any genericity or independence of the four matrices.

The span of $I,A,B,AB$ over $\mathbb{Z}[a,b,c]$ is stable under right multiplication by $A$ and $B$. Besides the products already displayed in step 1.1, the needed two are $ABA=cA+B-bI,ABB=bAB-A\mathrm{.}$ For the first multiply the polarized identity by $A$ on the left and use $A^2=aA-I$; the second is $A{B}^2=A(bB-I)$. The other products are $IA=A$, $IB=B$, $AA=aA-I$, $AB=AB$, $BA=-AB+aB+bA+(c-ab)I$, $BB=bB-I$. This is the complete multiplication table needed for one-letter extension of a word.

Start the empty word at $I$ and apply the table of step 2.1 once per positive letter. Replace each inverse letter by its two-term expression from step 1.1, or use linearity to extend that same table to inverse letters. Induction gives a finite coefficient expression for every word. Taking traces uses only $\mathrm{tr}I=2$, $\mathrm{tr}A=a$, $\mathrm{tr}B=b$, $\mathrm{tr}AB=c$, proving the trace assertion. Substituting $a=b=x,c=y$ gives the stated equal-trace specialization. No matrix-basis independence, diagonalization or infinite choice of word expressions is required.

Find a noncentral semisimple word and a word meeting both eigenspaces. Some group matrix has trace different from $2,-2$. Otherwise, if all matrices were scalar the representation would be reducible; choose a nonscalar $D=ϵI+N$ with $ϵ=\pm 1$, $N\mathrm{\ne }0$ and $N^2=0$ by [F1]. For every group matrix $C$ and integer $m\ge 0$, $\mathrm{tr}(D^{m}C)={ϵ}^m(\mathrm{tr}C+m{ϵ}^{-1}\mathrm{tr}(NC))\mathrm{.}$ All these traces belong to $\{2,-2\}$ by the supposition. In characteristic zero a nonconstant affine function on the nonnegative integers takes infinitely many values, so $\mathrm{tr}(NC)=0$ for every $C$. In a basis over $F$ with $N=\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right)$ this forces the lower-left entry of every $C$ to be zero. Its first coordinate line would be invariant, a contradiction. Thus choose a word $A$ with trace different from $\pm 2$, and adjoin one root $\xi$ of its characteristic polynomial. Its two distinct eigenvalues are $\xi ,{\xi }^{-1}$, so it is diagonalizable already over $F(\xi )$. In that eigenbasis some group word $B$ has both off-diagonal entries nonzero. Indeed the elements preserving the first and second eigenlines form two proper subgroups of $\mathrm{\Gamma }$ by absolute irreducibility. A group cannot be the union of two proper subgroups: for elements in their respective differences, their product belongs to neither. Choose $B$ outside that union. These are finitely many existential selections, not an infinite choice principle.

Normalize and recover every generator by trace equations. In the chosen $A$ eigenbasis write $B_{12}=u\mathrm{\ne }0$ and conjugate by $\mathrm{diag}(u^{-1},1)$ to make this entry one. The lower-left entry becomes the nonzero product $q=B_{12}{B}_{21}$. Put $a=\mathrm{tr}A$, $b=\mathrm{tr}B$, $c=\mathrm{tr}(AB)$. In the normalized basis $A=\left(\begin{array}{cc}\xi & 0\\ 0& {\xi }^{-1}\end{array}\right),B=\left(\begin{array}{cc}{d}_1& 1\\ q& d_2\end{array}\right),d_1=\frac{c-{\xi }^{-1}b}{\xi -{\xi }^{-1}},d_2=b-d_1,q=d_1{d}_2-1\mathrm{\ne }0.$ All these entries lie in $K^{\mathrm{\prime }}=K(\xi )$. For each generator $C$ put $e=\mathrm{tr}C$, $f=\mathrm{tr}(AC)$, $g=\mathrm{tr}(BC)$ and $h=\mathrm{tr}(ABC)$, which belong to $K$ by its definition. Its diagonal entries are $c_{11}=\frac{f-{\xi }^{-1}e}{\xi -{\xi }^{-1}},c_{22}=e-c_{11}\mathrm{.}$ Its off-diagonal entries are uniquely solved from $c_{21}+q{c}_{12}=g-d_1{c}_{11}-d_2{c}_{22},\xi c_{21}+{\xi }^{-1}q{c}_{12}=h-\xi d_1{c}_{11}-{\xi }^{-1}{d}_2{c}_{22}\mathrm{.}$ The determinant of this two-by-two linear system is $q({\xi }^{-1}-\xi )\mathrm{\ne }0$. Thus every normalized generator matrix has entries in $K^{\mathrm{\prime }}$. These are the actual conjugated original matrices in $F(\xi )$, not merely formal matrices with the correct traces: the normalization was performed first and the displayed linear equations uniquely recover their entries. Therefore their determinant-one equations and every group relation hold in $K^{\mathrm{\prime }}$, by its injection into $F(\xi )$. The conjugated representation is an actual model over $K^{\mathrm{\prime }}$, with no new relator-existence assumption.

The quadratic swap is explicit conjugation. If $K^{\mathrm{\prime }}=K$ there is nothing to prove. Otherwise its automorphism $\tau$ exchanges $\xi$ and ${\xi }^{-1}$ and fixes every word trace. The formulas of step 2.1 give $\tau (d_1)=d_2$, $\tau (q)=q$. Put $P=\left(\begin{array}{cc}0& 1\\ q& 0\end{array}\right),P^{-1}=\left(\begin{array}{cc}0& q^{-1}\\ 1& 0\end{array}\right)\mathrm{.}$ Direct multiplication gives $PA{P}^{-1}=\tau A$ and $PB{P}^{-1}=\tau B$. For each generator $C$, both $PC{P}^{-1}$ and $\tau C$ have traces $e,f,g,h$ when paired respectively with $I,\tau A,\tau B,(\tau A)(\tau B)$: conjugation preserves trace, and $\tau$ fixes the four word traces. The diagonal/off-diagonal recovery in step 2.1 applies to this swapped pair as well; its determinant is still nonzero. Uniqueness therefore gives $PC{P}^{-1}=\tau C$ for every generator, hence for all group words. This proves simultaneous conjugacy over $K^{\mathrm{\prime }}$ itself, without an appeal to pointwise complex-character classification.

Projective descent for compatible invariant constructions. Under the hypotheses on $z$, applying $\tau$ to its construction yields the point constructed from $\tau \rho$, while conjugation invariance and step 3.1 identify that point with $z$. Hence $\tau z=z$. Choose one nonzero homogeneous coordinate $z_j$ and normalize it to $1$. Equality of projective points implies that each ratio $z_i\mathrm{/}{z}_j$ is fixed by $\tau$. Every element of the quadratic field has the unique form $u+v\xi$ with $u,v\in K$; being fixed gives $v(\xi -{\xi }^{-1})=0$, so $v=0$. Thus every normalized coordinate lies in $K$, proving that $z$ is a $K$-point. In the trivial-extension case this is immediate. The argument uses only finite matrix and field operations; it neither asserts that an arbitrary merely pointwise invariant assignment is rational nor supplies any analytic or chain-complex prerequisite for constructing $z$.

Finite field normal forms with all transformations retained. Over $F$, Gaussian elimination is an actual finite algorithm: choose the first nonzero entry of the remaining rectangle in the fixed row/column order, swap its row and column into the next pivot position, divide the pivot row by that nonzero entry and subtract its multiples to clear its column and row. Each operation has an explicit inverse and removes one row and column from the remaining problem. It ends after at most the smaller matrix size, with a block identity and a zero rectangle. This proves that its number of pivots is the rank, since invertible operations preserve the column-span dimension and $\mathrm{diag}(I_r,0)$ has rank $r$; the same normal form also gives row-span dimension $r$; no real-only rank-minor theorem is invoked. Apply this to $C$, recording its changes of bases in $F^n$ and $F^t$. Its kernel has basis the last $n-\gamma$ domain coordinates. The identity $CB=0$ places every column of the transformed $B$ in those coordinates. Apply the same algorithm to this $(n-\gamma )$-row matrix, using only row operations within $\ker C$ and column operations on $F^s$. Its first $\beta$ coordinates are now the image of $B$, and its remaining $h=n-\gamma -\beta$ coordinates represent the quotient homology. Thus the entire chain pair has a split block form over $F$, not just separately computed nullities.