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Trace functions are regular and conjugation invariant
Statement
Assume full AC. For every word in the generators of a finitely presented group , the function is a regular function on and is invariant under simultaneous conjugation.
Statement references
Facts & Assumptions
Given: Full AC, a finite presentation of , its affine representation variety, and a word in the generators and their inverses.
For the given finite presentation, impose determinant-one equations and the matrix-entry relator equations in the finite generator matrix space. By the inverse formula [F2], these are polynomial equations for every finitely presented group, not just a knot group.
If , then , as verified by direct multiplication using .
Proof
The equations of [F1] cut out exactly the representation tuples, since a generator assignment satisfies them precisely when it respects every presented relator. Evaluate by multiplying its generator matrices and the adjugates for inverse letters from [F2]. Each entry of the resulting matrix is a polynomial in the representation coordinates, so its trace is regular.
For any , conjugation sends the word matrix to . The identity proves invariance.
Thus each word trace defines an element of the conjugation-invariant coordinate ring of the representation variety.
Depends on
Used by
Dependency tree · two levels
8 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.
Sources
- Léo Bénard, François Florens, and Adrien Rodau, A slope invariant and the A-polynomial of knots, §2.1 (standard reference, not scraped)
- Stephan Tillmann, Varieties associated to 3-manifolds, §§4.2–4.4 (standard reference, not scraped)