LemmaStatement: AI-adaptedProof: AI-generatedPipeline-generatedjudge pass (gpt-6-luna)verified 2026-10-22 (gpt-6.1-sol)

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Two-generator SL2 word traces reduce to three coordinates

Statement

Let $A, B \in S L_{2} (C)$ and put $a = tr A$ , $b = tr B$ , $c = tr (A B)$ . Every word matrix in $A^{\pm 1}, B^{\pm 1}$ is a linear combination of $I, A, B, A B$ with coefficients in $Z [a, b, c]$ , computed by finite word induction. Consequently its trace is a polynomial in those three traces. If $a = b = x$ and $c = y$ , the coefficients specialize to $Z [x, y]$ . The assertion includes scalar, unipotent, reducible and irreducible pairs and uses no choice axiom.

Facts & Assumptions

Given: Two determinant-one complex $2 \times 2$ matrices, their indicated traces and any finite word in the matrices and their inverses.

[F1]

For a $2 \times 2$ matrix $T$ , the entrywise identity $T^{2} - (tr T) T + (\det T) I = 0$ follows by multiplying its four entries.

Proof

1.1F1givenconstructalgebra

Apply [F1] to $A$ and $B$ . Since both determinants are one, $A^{- 1} = a I - A, B^{- 1} = b I - B, A^{2} = a A - I, B^{2} = b B - I .$ 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 $\det (A + B) - \det A - \det B = a b - c$ . Hence $A B + B A = a B + b A + (c - a b) I,$ an identity valid without any genericity or independence of the four matrices.

2.1step 1.1constructalgebra

The span of $I, A, B, A B$ over $Z [a, b, c]$ is stable under right multiplication by $A$ and $B$ . Besides the products already displayed in step 1.1, the needed two are $A B A = c A + B - b I, A B B = b A B - A .$ For the first multiply the polarized identity by $A$ on the left and use $A^{2} = a A - I$ ; the second is $A B^{2} = A (b B - I)$ . The other products are $I A = A$ , $I B = B$ , $A A = a A - I$ , $A B = A B$ , $B A = - A B + a B + b A + (c - a b) I$ , $B B = b B - I$ . This is the complete multiplication table needed for one-letter extension of a word.

3.1step 1.1step 2.1constructinductionalgebra∎

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 $tr I = 2$ , $tr A = a$ , $tr B = b$ , $tr A B = c$ , 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.

Used by

Figure-eight character curve and longitude trace Lemma

Dependency tree · 0 levels

Nothing. This result depends on no other item in the library.

Sources

Explicit rank-two Cayley–Hamilton and polarized matrix identities (standard reference, not scraped)
Cooper, Culler, Gillet, Long and Shalen, Plane Curves Associated to Character Varieties of 3-Manifolds (standard reference, not scraped)