From a comment on [this question](https://mathoverflow.net/questions/111770/cycling-through-the-zeta-garden-zeta-functions-for-graphs-cycle-index-polynomi?noredirect=1#comment550560_111770):



> 
> @draks, there is a connection between the Chebyshev polynomials and the Faber polynomials (a.k.a. Shur polynomials), which 'invert" the cyclic partition polynomials (a.k.a. the refined partition polynomials of the first kind--signed OEIS [A036039](http://oeis.org/A036039)) to isolate the indeterminates in their definition.
> 
> 
> 


Interesting. Do you have something where I can read on this?



> 
> @Draks, see [oeis.org/A263916](http://oeis.org/A263916) and [oeis.org/A127672](http://oeis.org/A127672) along with the Damianou and Damianou and Evripidou links. 
> 
> 
> 


The links don't help. Can anyone explain the connection?



Connection between the Chebyshev polynomials and the Faber polynomials
The characteristic polynomial and the traces of the powers of a square matrix $A$ are related by


$$\operatorname{det}(I-x \; A)=\exp\left(-\sum\_{m\geq 1} \frac{\operatorname{tr}(A^m) \; x^m}{m} \right)$$


$$=1 + \sum\_{k \ge 1} P\_k(b\_1,..,b\_k) \; x^k / k! = 1 + \sum\_{k \geq 1} d\_k \; x^k$$


where $b\_k=-\operatorname{tr}(A^k)$ for $k>0$ and $P\_m(b\_1,..,b\_m)$ are the cycle index partition polynomials of OEIS [A036039](http://oeis.org/A036039) for the symmetric groups (see the Lang link there for a compilation). Presented umbrally,


$$\operatorname{det}(I - x \; A)=\exp[\ln(1 - b.\;x)]=e^{P. \; x}=1/(1-d.\;x) \; .$$


The Faber polynomials $F\_m$ of [A263916](http://oeis.org/A263916) can, in general, invert the the determinant polynomial, or the partition polynomials $P\_n$, to generate the power sums, or traces:


$$F\_m[P\_1(b\_1),P\_2(b\_1,b\_2)/2!,..,P\_m(b\_1,..,b\_m)/m!] = F\_m(d\_1,d\_2,..,d\_m) = -b\_m = tr(A^m) \; .$$


Damianou points out that the adjacency matrix $A\_n$ for the simple Lie algebra $B\_n$ has the characteristic polynomial


$$det(xI - A\_n) = 2\; T\_n(x/2)= a\_n(x)\; ,$$


the Chebyshev polynomials given by [A127672](http://oeis.org/A127672), so with


$$ det(I - x\;A\_n) = x^n \; a\_n(1/x) = 2 x^n \; T(1/(2x)) = 1 + \sum\_{k = 1}^{n} d\_k \; x^k \; ,$$


$$F\_m(d\_1,d\_2,..,d\_m) = -b\_m = tr(A^m) \;,$$


unwrapping the determinant relation.


An example is afforded by the adjacency matrix for for the simple Lie algebra $B\_3$, given by Damianou:


$$A\_3=\begin{bmatrix}
0 & 1 & 0\\ 
1 & 0 & 2\\ 
0 & 1 & 0
\end{bmatrix} \; .$$ 


The characteristic polynomial $\operatorname{det}(xI-A\_3)=a\_3(x) = 2T\_3(x/2)= x^3 -3x=$ has the zeroes $x\_1=\sqrt{3},x\_2=-\sqrt{3},x\_3=0$, so $\operatorname{tr}(A\_3^k)= x\_1^k + x\_2^k +x\_3^k=3^{k/2} + (-1)^k 3^{k/2}$ (aerated [A025192](http://oeis.org/A025192)), which vanishes for odd $k$ and equals $2\; 3^{k/2}$ for even $k$. Then $P\_k(b\_1,..,b\_k)$ vanishes except for $P\_2(0,-6) = -6$, and $\operatorname{det}(I-uA\_3)= 1-3u^2=u^3a\_3(1/u) = 2x^3T\_3(1/2x)) $, so only $d\_2 = -3$ is non-vanishing and $F\_k[0,-3,0,..,0]= -b\_k = \operatorname{tr}(A\_3^k)$ for $k>0$.


