13
$\begingroup$

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

  1. Is this identity known already?
  2. 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 $g_1 =3,\, g_2=2,\,g_3=1$.

$\endgroup$
8
$\begingroup$

Here is how to prove it with more standard methods. First of all, let me restate your identity:

Definition. Let $\mathbb{N}=\left\{ 0,1,2,\ldots\right\} $. A partition shall mean an integer partition, i.e., a weakly decreasing finite list of positive integers. If $\lambda$ is a partition and $i$ is a positive integer, then $m_{i}\left( \lambda\right) $ shall mean the number of times that $i$ appears as entry of $\lambda$. (For example, $m_{3}\left( \left( 4,3,3,1\right) \right) =2$ and $m_{2}\left( \left( 4,3,3,1\right) \right) =0$.) The size $\left\vert \lambda\right\vert $ of a partition is defined to be the sum of all entries of $\lambda$. If $n\in\mathbb{N}$, then a partition of $n$ means a partition of size $n$. We write "$\lambda\vdash n$" for "$\lambda$ is a partition of $n$".

Definition. We let $T_n\left(x\right)$ denote the Chebyshev polynomials of the first kind, which can be defined (e.g.) by the recurrence $T_0\left(x\right) = 1$ and $T_1\left(x\right) = x$ and $T_{n+1}\left(x\right) = 2x T_n\left(x\right) - T_{n-1}\left(x\right)$. We let $U_n\left(x\right)$ denote the Chebyshev polynomials of the second kind, which can be defined (e.g.) by the recurrence $U_0\left(x\right) = 1$ and $U_1\left(x\right) = 2x$ and $U_{n+1}\left(x\right) = 2x U_n\left(x\right) - U_{n-1}\left(x\right)$.

Theorem 1. For any $n\in\mathbb{N}$, we have \begin{equation} U_{n}\left( x\right) =\sum_{\lambda=\left( \lambda_{1},\lambda_{2} ,\ldots,\lambda_{k}\right) \vdash n}\left( \prod_{i=1}^{\infty}\dfrac {1}{i^{m_{i}\left( \lambda\right) }m_{i}\left( \lambda\right) !}\right) \cdot\prod_{j=1}^{k}\left( 2T_{\lambda_{j}}\left( x\right) \right) . \end{equation}

To prove this, I will use two well-known generating-function identities for Chebyshev polynomials, both of which appear on the Wikipedia: \begin{equation} \sum_{n=0}^{\infty}T_{n}\left( x\right) t^{n}=\dfrac{1-tx}{1-2tx+t^{2}} \label{darij1.eq.T-gen} \tag{1} \end{equation} and \begin{equation} \sum_{n=0}^{\infty}U_{n}\left( x\right) t^{n}=\dfrac{1}{1-2tx+t^{2} }. \label{darij1.eq.U-gen} \tag{2} \end{equation} These are identities in the ring $\left( \mathbb{Q}\left[ x\right] \right) \left[ \left[ t\right] \right] $ of formal power series in the variable $t$ over the polynomial ring $\mathbb{Q}\left[ x\right] $. Both identities can easily be derived from the above recurrent definitions of $T_n\left(x\right)$ and $U_n\left(x\right)$.

Now, subtracting the equality $\underbrace{T_{0}\left( x\right) } _{=1}\underbrace{t^{0}}_{=1}=1$ from the identity \eqref{darij1.eq.T-gen}, we obtain \begin{equation} \sum_{n=1}^{\infty}T_{n}\left( x\right) t^{n}=\dfrac{1-tx}{1-2tx+t^{2} }-1=t\cdot\dfrac{x-t}{1-2tx+t^{2}}. \end{equation} Dividing both sides of this by $t$, we obtain \begin{equation} \sum_{n=1}^{\infty}T_{n}\left( x\right) t^{n-1} = \dfrac{x-t}{1-2tx+t^{2}}. \end{equation} Integrating both sides of this equality over $t$, we find \begin{align} \sum_{n=1}^{\infty}T_{n}\left( x\right) \dfrac{t^{n}}{n} & =\int\dfrac {x-t}{1-2tx+t^{2}}dt\nonumber\\ & =\dfrac{1}{2}\log\dfrac{1}{1-2tx+t^{2}} \label{darij1.eq.T-ge2} \tag{3} \end{align} (as you can easily check by differentiation). (Note that this identity also appears on the Wikipedia, under the guise of $\sum_{n=1}^{\infty}T_{n}\left( x\right) \dfrac{t^{n}}{n}=\log\dfrac{1}{\sqrt{1-2tx+t^{2}}}$, apparently because someone finds square roots simpler than division by $2$.)

Multiplying both sides of the equality \eqref{darij1.eq.T-ge2} by $2$, we obtain \begin{equation} 2\sum_{n=1}^{\infty}T_{n}\left( x\right) \dfrac{t^{n}}{n}=\log\dfrac {1}{1-2tx+t^{2}}. \end{equation} Hence, \begin{equation} \log\dfrac{1}{1-2tx+t^{2}}=2\sum_{n=1}^{\infty}T_{n}\left( x\right) \dfrac{t^{n}}{n}=\sum_{n=1}^{\infty}2T_{n}\left( x\right) \dfrac{t^{n}}{n}, \end{equation} so that \begin{equation} \dfrac{1}{1-2tx+t^{2}}=\exp\left( \sum_{n=1}^{\infty}2T_{n}\left( x\right) \dfrac{t^{n}}{n}\right) . \end{equation} Hence, \eqref{darij1.eq.U-gen} becomes \begin{equation} \sum_{n=0}^{\infty}U_{n}\left( x\right) t^{n}=\dfrac{1}{1-2tx+t^{2}} =\exp\left( \sum_{n=1}^{\infty}2T_{n}\left( x\right) \dfrac{t^{n}} {n}\right) . \label{darij1.eq.T-ge3} \tag{4} \end{equation}

Now, we recall one of the staple formulas of algebraic combinatorics (probably in EC or Wilf or similar sources):

Proposition 2. Let $R$ be a commutative $\mathbb{Q}$-algebra (for example, $\mathbb{Q}$ or $\mathbb{Q}\left[ x\right] $). Let $b_{1},b_{2},b_{3} ,\ldots\in R$ and $c_{0},c_{1},c_{2},\ldots\in R$ be such that \begin{equation} \sum_{n=0}^{\infty}c_{n}t^{n}=\exp\left( \sum_{n=1}^{\infty}b_{n}\dfrac {t^{n}}{n}\right) \end{equation} in the ring $R\left[ \left[ t\right] \right] $ of formal power series. Then, \begin{equation} c_{n}=\sum_{\lambda=\left( \lambda_{1},\lambda_{2},\ldots,\lambda_{k}\right) \vdash n}\left( \prod_{i=1}^{\infty}\dfrac{1}{i^{m_{i}\left( \lambda\right) }m_{i}\left( \lambda\right) !}\right) \cdot\prod_{j=1}^{k}b_{\lambda_{j}} \end{equation} for each $n\in\mathbb{N}$.

Proof of Proposition 2. An infinite sequence $\left( k_{1},k_{2} ,k_{3},\ldots\right) \in\mathbb{N}^{\infty}$ of nonnegative integers will be called a weak composition if all but finitely many $i\geq1$ satisfy $k_{i}=0$. There is a bijection \begin{align} \left\{ \text{partitions}\right\} & \rightarrow\left\{ \text{weak compositions}\right\} ,\nonumber\\ \lambda & \mapsto\left( m_{1}\left( \lambda\right) ,m_{2}\left( \lambda\right) ,m_{3}\left( \lambda\right) ,\ldots\right) \label{darij1.pf.p2.1} \tag{5} \end{align} (since any partition $\lambda$ is uniquely determined by the numbers $m_{1}\left( \lambda\right) ,m_{2}\left( \lambda\right) ,m_{3}\left( \lambda\right) ,\ldots$ which record how often each positive integer appears in $\lambda$). We notice that any partition $\lambda$ satisfies \begin{equation} 1m_{1}\left( \lambda\right) +2m_{2}\left( \lambda\right) +3m_{3}\left( \lambda\right) +\cdots=\left\vert \lambda\right\vert \label{darij1.pf.p2.2} \tag{6} \end{equation} (because $\left\vert \lambda\right\vert $ is the sum of all entries of $\lambda$, while $1m_{1}\left( \lambda\right) +2m_{2}\left( \lambda\right) +3m_{3}\left( \lambda\right) +\cdots$ is what becomes of this sum after equal addends are bunched together). Moreover, any partition $\lambda=\left( \lambda_{1},\lambda_{2},\ldots,\lambda_{k}\right) $ satisfies \begin{equation} \prod_{i=1}^{\infty}b_{i}^{m_{i}\left( \lambda\right) }=\prod_{j=1} ^{k}b_{\lambda_{j}} \label{darij1.pf.p2.3} \tag{7} \end{equation} (for a similar reason: the product $\prod_{i=1}^{\infty}b_{i}^{m_{i}\left( \lambda\right) }$ is what will become of the product $\prod_{j=1} ^{k}b_{\lambda_{j}}$ if you bunch factors corresponding to equal entries of $\lambda$ together).

We have the following product rule (i.e., analogue of the distributivity law) for infinite products of infinite sums: If $\left( a_{i,k}\right) _{i\geq1\text{ and }k\geq0}$ is a family of elements of $R\left[ \left[ t\right] \right] $ satisfying $a_{i,0}=1$ for each $i\geq1$, then \begin{equation} \prod_{i=1}^{\infty}\sum_{k=0}^{\infty}a_{i,k}=\sum_{\substack{\left( k_{1},k_{2},k_{3},\ldots\right) \text{ is a}\\\text{weak composition}} }\prod_{i=1}^{\infty}a_{i,k_{i}}, \label{darij1.pf.p2.prodrule} \tag{8} \end{equation} provided that everything formally converges (i.e., for each given $N\in\mathbb{N}$, all but finitely many pairs $\left( i,k\right) \in\left\{ 1,2,3,\ldots\right\} ^{2}$ satisfy $t^N \mid a_{i,k}$ in $R\left[\left[t\right]\right]$).

We have \begin{align} \sum_{n=0}^{\infty}c_{n}t^{n} & =\exp\left( \sum_{n=1}^{\infty}b_{n} \dfrac{t^{n}}{n}\right) =\prod_{n=1}^{\infty}\underbrace{\exp\left( b_{n}\dfrac{t^{n}}{n}\right) }_{\substack{=\sum_{k=0}^{\infty}\dfrac{1} {k!}\left( b_{n}\dfrac{t^{n}}{n}\right) ^{k}\\\text{(since }\exp z=\sum_{k=0}^{\infty}\dfrac{1}{k!}z^{k}\text{)}}}\nonumber\\ & \qquad \left(\text{since $\exp\left(\sum_{i\in I} a_i\right) = \prod_{i\in I} \exp a_i$ for any family $\left(a_i\right)_{i\in I}$}\right) \nonumber\\ & =\prod_{n=1}^{\infty}\sum_{k=0}^{\infty}\dfrac{1}{k!}\left( b_{n} \dfrac{t^{n}}{n}\right) ^{k}=\prod_{i=1}^{\infty}\sum_{k=0}^{\infty}\dfrac {1}{k!}\underbrace{\left( b_{i}\dfrac{t^{i}}{i}\right) ^{k}}_{=\dfrac {b_{i}^{k}t^{ik}}{i^{k}}}\nonumber\\ & \qquad\left( \text{here, we have renamed the index }n\text{ as }i\right) \nonumber\\ & =\prod_{i=1}^{\infty}\sum_{k=0}^{\infty}\dfrac{1}{k!}\cdot\dfrac{b_{i} ^{k}t^{ik}}{i^{k}}=\prod_{i=1}^{\infty}\sum_{k=0}^{\infty}\dfrac{b_{i} ^{k}t^{ik}}{i^{k}k!}\nonumber\\ & =\sum_{\substack{\left( k_{1},k_{2},k_{3},\ldots\right) \text{ is a}\\\text{weak composition}}}\prod_{i=1}^{\infty}\dfrac{b_{i}^{k_{i}} t^{ik_{i}}}{i^{k_{i}}k_{i}!}\nonumber\\ & \qquad\left( \text{by the product rule \eqref{darij1.pf.p2.prodrule}, applied to }a_{i,k}=\dfrac{b_{i}^{k}t^{ik}}{i^{k}k!}\right) \nonumber\\ & =\sum_{\lambda\text{ is a partition}}\underbrace{\prod_{i=1}^{\infty} \dfrac{b_{i}^{m_{i}\left( \lambda\right) }t^{im_{i}\left( \lambda\right) }}{i^{m_{i}\left( \lambda\right) }m_{i}\left( \lambda\right) !}}_{=\left( \prod_{i=1}^{\infty}\dfrac{1}{i^{m_{i}\left( \lambda\right) }m_{i}\left( \lambda\right) !}\right) \left( \prod_{i=1}^{\infty}b_{i}^{m_{i}\left( \lambda\right) }\right) \left( \prod_{i=1}^{\infty}t^{im_{i}\left( \lambda\right) }\right) }\nonumber\\ & \qquad\left( \begin{array} [c]{c} \text{here, we have substituted }\left( m_{1}\left( \lambda\right) ,m_{2}\left( \lambda\right) ,m_{3}\left( \lambda\right) ,\ldots\right) \\ \text{for }\left( k_{1},k_{2},k_{3},\ldots\right) \text{ in the sum, due to the bijection \eqref{darij1.pf.p2.1}} \end{array} \right) \nonumber\\ & =\sum_{\lambda\text{ is a partition}}\left( \prod_{i=1}^{\infty}\dfrac {1}{i^{m_{i}\left( \lambda\right) }m_{i}\left( \lambda\right) !}\right) \left( \prod_{i=1}^{\infty}b_{i}^{m_{i}\left( \lambda\right) }\right) \underbrace{\left( \prod_{i=1}^{\infty}t^{im_{i}\left( \lambda\right) }\right) }_{\substack{=t^{1m_{1}\left( \lambda\right) +2m_{2}\left( \lambda\right) +3m_{3}\left( \lambda\right) +\cdots}\\=t^{\left\vert \lambda\right\vert }\\\text{(by \eqref{darij1.pf.p2.2})}}}\nonumber\\ & =\sum_{\lambda\text{ is a partition}}\left( \prod_{i=1}^{\infty}\dfrac {1}{i^{m_{i}\left( \lambda\right) }m_{i}\left( \lambda\right) !}\right) \left( \prod_{i=1}^{\infty}b_{i}^{m_{i}\left( \lambda\right) }\right) t^{\left\vert \lambda\right\vert }. \label{darij1.pf.p2.6} \tag{9} \end{align}

Now, let $n\in\mathbb{N}$. Comparing coefficients of $t^{n}$ on both sides of the equality \eqref{darij1.pf.p2.6}, we obtain \begin{align*} c_{n} & =\underbrace{\sum_{\substack{\lambda\text{ is a partition;} \\\left\vert \lambda\right\vert =n}}}_{\substack{=\sum_{\lambda\vdash n}\\\text{(since the partitions of }n\\\text{are precisely the partitions }\lambda\\\text{with }\left\vert \lambda\right\vert =n\text{)}}}\left( \prod_{i=1}^{\infty}\dfrac{1}{i^{m_{i}\left( \lambda\right) }m_{i}\left( \lambda\right) !}\right) \prod_{i=1}^{\infty}b_{i}^{m_{i}\left( \lambda\right) }\\ & =\sum_{\lambda\vdash n}\left( \prod_{i=1}^{\infty}\dfrac{1}{i^{m_{i}\left( \lambda\right) }m_{i}\left( \lambda\right) !}\right) \prod_{i=1}^{\infty }b_{i}^{m_{i}\left( \lambda\right) }\\ & =\sum_{\lambda=\left( \lambda_{1},\lambda_{2},\ldots,\lambda_{k}\right) \vdash n}\left( \prod_{i=1}^{\infty}\dfrac{1}{i^{m_{i}\left( \lambda\right) }m_{i}\left( \lambda\right) !}\right) \underbrace{\left( \prod _{i=1}^{\infty}b_{i}^{m_{i}\left( \lambda\right) }\right) } _{\substack{=\prod_{j=1}^{k}b_{\lambda_{j}}\\\text{(by \eqref{darij1.pf.p2.3})}}}\\ & =\sum_{\lambda=\left( \lambda_{1},\lambda_{2},\ldots,\lambda_{k}\right) \vdash n}\left( \prod_{i=1}^{\infty}\dfrac{1}{i^{m_{i}\left( \lambda\right) }m_{i}\left( \lambda\right) !}\right) \prod_{j=1}^{k}b_{\lambda_{j}}. \end{align*} This proves Proposition 2. $\blacksquare$

Proof of Theorem 1. Recall the identity \eqref{darij1.eq.T-ge3}. Thus, Proposition 2 (applied to $R=\mathbb{Q}\left[ x\right] $ and $c_{n} =U_{n}\left( x\right) $ and $b_{n}=2T_{n}\left( x\right) $) yields that \begin{equation} U_{n}\left( x\right) =\sum_{\lambda=\left( \lambda_{1},\lambda_{2} ,\ldots,\lambda_{k}\right) \vdash n}\left( \prod_{i=1}^{\infty}\dfrac {1}{i^{m_{i}\left( \lambda\right) }m_{i}\left( \lambda\right) !}\right) \cdot\prod_{j=1}^{k}\left( 2T_{\lambda_{j}}\left( x\right) \right) \end{equation} for each $n\in\mathbb{N}$. This proves Theorem 1. $\blacksquare$

$\endgroup$
3
  • 1
    $\begingroup$ After your proof I realized that this could be trivially generalized to general Gegenbauer polynomials by letting $U_n(x)\to G_n^{(\alpha)}(x)$ and $2T_{\lambda_j}(x) \to 2\alpha T_{\lambda_j}(x) $. Furthermore, for $\alpha = (d-2)/2$, $d\in\mathbb{N}$ I can prove it in my way too. $\endgroup$ – MannyC Feb 15 '20 at 8:23
  • $\begingroup$ Why only for the canonical scaling dimension $\alpha$? There is no reason to be afraid of fractional free fields. $\endgroup$ – Abdelmalek Abdesselam Feb 15 '20 at 23:38
  • $\begingroup$ I was about to show the details of my proof in a self-answer but I did not have time today. Essentially I want to express the left hand side as a contraction of $d$ dimensional, rank $n$ symmetric traceless tensors and the right hand side as traces of products of matrices belonging to the Clifford algebra in $d$ dimensions. While I believe these concepts can be generalized to any $d$, I have no rigorous way of doing so. $\endgroup$ – MannyC Feb 16 '20 at 0:12
5
$\begingroup$

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.

$\endgroup$
2
  • 1
    $\begingroup$ +1. I agree this is probably the best way to understand the identity. Another relevant reference is the article "Miraculous cancellations for quantum SL2" by Francis Bonahon, arxiv.org/abs/1708.07617 $\endgroup$ – Abdelmalek Abdesselam Feb 15 '20 at 23:33
  • 2
    $\begingroup$ Ah -- I actually got my proof by reverse engineering a symmetric functions argument, but I wasn't aware of you proving the necessary formulas in your preprint! $\endgroup$ – darij grinberg Feb 16 '20 at 0:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.