# Chebyshev polynomials and trace of $A \in SL_2(\mathbb{C})$

Defining $$C_n(z) = \frac{z^m + z^{-m}}{2}$$, the Chebyshev polynomials are defined by

$$T_n(C_1(z)) = C_n(z)$$ and are given by $$T_1(z) = z, T_2(z) = 2z^2-1, T_3(z) = 4z^3-3z$$, etc. Since for $$z=e^{i\theta}$$ we have $$C_1(z) = \cos\theta$$, they also satisfy

$$T_n(\cos\theta) = \cos(n\theta)$$

thereby generalizing the double angle trig identity.

The notes I'm reading also claim $$T_n(\text{tr } A/2) = \text{tr } (A^n/2)$$ for $$A \in SL_2(\mathbb{C})$$. Why does this follow?

Attempt: if $$A= \begin{pmatrix} a&b \\ c &d \end{pmatrix}$$ then choosing $$z=\frac12(\sqrt{(a+d)^2-4}- (a+d))$$ implies $$C_1(z) = \text{tr }A/2$$, but then $$T_n(C_1(z)) = \frac{z^n+z^{-n}}{2}$$ does not simplify as far as I see to $$\text{tr} A^n/2$$.

• Note that the trace of $A$ is the sum of its eigenvalues and the product of its eigenvalues is by definition $1$. – WimC Dec 25 '18 at 18:09
• @WimC thanks, that resolves it – Dwagg Dec 25 '18 at 18:57

I found several sources with a proof, e.g., the paper by Francis Bonahon, Lemma $$8$$ on page $$9$$, using Cayley-Hamilton. Another interesting reference is the paper by Traina on trace polynomials for $$SL_2(\Bbb{C})$$.