TITLE: Curious identity between the two kinds of Chebyshev polynomials
QUESTION [14 upvotes]: I have found, by accident, an identity that relates a sum of Chebyshev polynomials of the first kind to a Chebyshev polynomial of the second kind. It goes as follows:
Given an integer partition of $n$, let $g_a$ be the number of times $a$ appears in said partition.${}^1$ Then the following identity holds for all $n \in\mathbb{N}$:
$$
U_n(x) = \sum_{\substack{n_i>0\\ \sum_i n_i = n}} \frac{1}{\prod_{a\in \{n_i\}} g_a!} \prod_{i=1} \frac{2}{n_i} T_{n_i}(x)\,.
$$
The sum is over all integer partitions of $n$, the product is on all $n_i$'s in the partition, with repetitions.
I have a very roundabout way to prove this identity (I'll skip the details). The left hand side is obtained by contracting two symmetric traceless tensors of $SO(4)$. That is, letting $|x|=|y|=1$ and $x,y\in \mathbb{R}^4$ then
$$
(x^{i_1}\cdots x^{i_n} - \mathrm{traces}) (y_{i_1}\cdots y_{i_n} - \mathrm{traces}) \propto U_n(x\cdot y)\,.
$$
The right hand side instead comes from the same contraction but in spinor notation. Namely we let
$$
\mathrm{x} = \left(\begin{matrix}x_3-x_4 & x_1 - i x_2 \\ x_1 + i x_2 & -x_3-x_4\end{matrix}\right)\,,\quad \bar{\mathrm{x}} = \epsilon\, \mathrm{x}\, \epsilon^T\,,
$$
and $\mathrm{y}$ in a similar way ($\epsilon$ is the Levi Civita tensor). Then we introduce two dimensional spinors $\eta,\tilde{\eta}$ let $\partial_{\eta^\alpha}\eta^\beta = \delta_\alpha^\beta$ (similar for $\tilde{\eta}$) and finally
$$
(\partial_\eta \mathrm{x} \partial_{\tilde{\eta}})^n (\eta \mathrm{y}\tilde{\eta})^n \sim  \sum_{\substack{n_i>0\\ \sum_i n_i = n}} \# \prod_{i} \mathrm{tr}\,((\mathrm{x}\bar{\mathrm{y}})^{n_i})\,.
$$
The sum over partitions comes from a combinatoric argument. Then it's a simple exercise to show that $\mathrm{tr}\,((\mathrm{x}\bar{\mathrm{y}})^n) \propto T_n(x\cdot y)$.
My questions are

Is this identity known already?
If not, could you come up with some more direct argument to prove it?


$\;{}^1$ For example, $(1,1,1,2,2,3)$ is an integer partition of $n=10$ with $g_1 =3,\, g_2=2,\,g_3=1$.

REPLY [5 votes]: I think I can sketch a shorter proof.
Let $z_j = x_j+x_j^{-1}$, and let $p_m$ and $h_m$ denote the power-sum and complete homogeneous symmetric polynomial.
Then (see e.g p.3 in this preprint)
$$
2 T_m(z_j/2) = p_m(x_j,x_j^{-1})
\text{ and }
U_m(z_j/2) = h_m(x_j,x_j^{-1})
$$
Now, we can use the Newton identities, to express $h_m$
in terms of the power-sum symmetric functions.
This gives a relation between the $U_m$ and the $T_m$.
Looking at your formula, it is very similar to the Newton identity.