A quadratic generic character model specializes on a dense principal open

A dominant source and an actual generic representation. Let $R$ be the representation set and $\pi :R\to R\mathrm{/}\text{\hspace{0.17em}⁣}\mathrm{/}S{L}_2$ its quotient. For this arbitrary finitely presented group, finite trace coordinates follow directly from [F1] and [F2]: express each of a finite list of invariant algebra generators as a polynomial in finitely many word traces and take the union of those finite word lists. They generate the same coordinate algebra and hence give its closed affine model. The closed set ${\pi }^{-1}(X)$ has finitely many irreducible components $Z_i$ by [F2]. The closures of their images cover $X$, because $\pi$ is onto. A nonempty irreducible set cannot be the union of finitely many proper closed subsets: split the union into its first member and the rest and repeat the defining irreducibility condition. Hence one $Z$ has dense image in $X$. Pullback injects $A$ into $\mathbb{C}[Z]$, since a function vanishing on that dense image vanishes on $X$. Put $F=\mathrm{Frac}\mathbb{C}[Z]$, so $K\subset F$. The coordinate generator matrices over $F$ satisfy determinant and relator identities by [F1]; thus they give an actual representation $\rho :G\to S{L}_2(F)$, whose word traces are the corresponding elements of $K$. No section of the quotient has been asserted.

A trace certificate for absolute irreducibility. Choose an irreducible complex character of $X$ and a representative using [F2]. The construction in [F3] supplies two words $a_0,b_0$ whose matrices have $a=\mathrm{tr}(a_0)$ different from $\pm 2$ and whose normalized second matrix has both off-diagonal entries nonzero. Write $b=\mathrm{tr}(b_0)$, $c=\mathrm{tr}(a_0{b}_0)$ and $D=a^2-4$. In a basis diagonalizing the first word with eigenvalues $\xi ,{\xi }^{-1}$, the product of the off-diagonal entries of the second is $q=\frac{abc-c^2-b^2-a^2+4}{a^2-4}\mathrm{.}$ Indeed its diagonal entries are $(c-{\xi }^{-1}b)\mathrm{/}(\xi -{\xi }^{-1})$ and their complementary entry in trace $b$; their product minus one is the displayed expression by $\xi +{\xi }^{-1}=a$. Thus $D$ and the numerator of $q$ are nonzero regular functions on $X$. They remain nonzero in $F$. Over any field extension diagonalizing the first word, the nonzero product $q$ makes both off-diagonal entries of the second word nonzero. A common invariant line would have to be one of the two eigenspaces of the first, neither preserved by the second. This proves absolute irreducibility of the actual generic $\rho$, and also gives the same certificate at every complex character where these two regular functions do not vanish.

Recover the model over at most a quadratic extension of $K$. Adjoin one root $\xi$ of $T^2-aT+1$ to $F$ and put $L=K(\xi )\subset F(\xi )$. Apply the explicit normalization and recovery equations of [F3] to $\rho$ and these two words. Their coefficients are word traces in $K$, so they give generator matrices in $S{L}_2(L)$. These are the actual conjugated matrices over $F(\xi )$, since the four trace equations for each generator have nonzero determinant $q({\xi }^{-1}-\xi )$. Their determinant, relator and word-trace identities therefore hold in $L$, which injects into $F(\xi )$. Denote this representation by $\sigma$. Absolute irreducibility follows from the same two-word certificate. If $L\mathrm{\ne }K$, the explicit matrix $P=\left(\begin{array}{cc}0& 1\\ q& 0\end{array}\right)$ conjugates every $\sigma (g)$ to $\tau \sigma (g)$: the recovery equations and the proof of [F3] apply unchanged with the larger coefficient field $K$. No assertion that $K$ equals its prime-field trace subfield is needed.

Spread the actual matrices and specialize every point of an open set. If $L=K$, express the finitely many generator entries and the required nonzero elements and their inverses as fractions in $A$ and invert the product of their nonzero denominators. Also invert $D$ and the numerator of $q$. This gives $B=A_f$. If $[L:K]=2$, each entry has a unique form $u+v\xi$ with $u,v\in K$. Clearing their finitely many coefficient denominators gives $B=A_f[\xi ]\subset L$. The monic equation reduces every polynomial to degree at most one, and independence of $1,\xi$ over $K$ proves that this ring is exactly the displayed free rank-two quotient. For any additional nonzero $z\in L$, first clear its coefficient denominators. Its norm $z\tau (z)\in K^{\ast }$ is nonzero; after clearing and inverting its numerator, $z^{-1}=\tau (z)\mathrm{/}(z\tau (z))$ belongs to $B$. Thus any finite further localization of $B$ can be achieved by enlarging $f$ in $A$. All matrix identities remain identities in $B$ because it is a subring of $L$. At a point $\chi \in D_X(f)$, in the quadratic case choose either complex root of $T^2-a(\chi )T+1$; evaluation gives a homomorphism $B\to \mathbb{C}$. In the degree-one case evaluate directly. The specialized determinant and relator identities give an actual representation. Its finite trace-coordinate list equals $\chi$, since those identities already hold in $B$; [F1] then gives equality of every word trace. The nonzero $D,q$ certificate proves irreducibility. The same argument works for every root and every point in this principal open, rather than merely for a constructible dense set.

Projective descent with no quotient shortcut. In the quadratic case the assumed compatibility and step 2.1 give $\tau z=z$ for the constructed projective point. Normalize one nonzero homogeneous coordinate to one. Each other coordinate is fixed by $\tau$; writing it as $u+v\xi$, with $u,v\in K$, gives $v(\xi -{\xi }^{-1})=0$, and hence $v=0$. The normalized coordinates lie in $K$. In the degree-one case this is automatic. They define a rational map on $X$ by clearing their finitely many denominators; one normalized coordinate is one, so they have no simultaneous zero there. Step 3.1 permits simultaneous clearing of any prescribed additional matrix or kernel-chart denominators, so later actual specializations may be checked on one smaller dense principal open. This proves all family and descent assertions, with no claim that a merely pointwise invariant assignment was rational before a construction was given.