# $t$-analogue of the symmetric power of an additive character over $\Bbb{F}_q^*$

Let $G$ be a finite group and let $f: G \longrightarrow \Bbb{C}$ be any complex-valued function. For integers $k, n \geq 0$, an indeterminant $t$, and $x \in G$ let $f_k(x) := f \big( x^k \big)$ and define the $t$-analogue of $n$-th symmetric power $\text{sym}^n_t f$ by the recursion

$$\text{sym}^n_t f \, = \, {1 \over n} \, \sum_{k=1}^n \, t ^{k-1} \, f_k \cdot \text{sym}^{n-k}_tf$$

If $f$ happens to be the character of a finite dimensional complex representation $V$ of $G$ then upon specialising $t$ to the values $1$ and $-1$ the $t$-analogue $\text{sym}^n_t f$ will be respectively the character of the $n$-th symmetric power and the $n$-th exterior power of the representation $V$; when $t = 0$ it is ${1 \over {n!}}$ times the character of the $n$-th tensor product of $V$. It is easy to see that the recursive formula defining $\text{sym}^n_t f$ is in fact the Laplace expansion of the determinant of a certain $n \times n$ matrix, an example of which, in the case of $n= 4$, makes the pattern evident:

$$\det \begin{pmatrix} {1 \over 4} f_1 & {1 \over 4} f_2 & {1 \over 4}f_3 & {1 \over 4} f_4 \\ -t & {1 \over 3} f_1 & {1 \over 3} f_2 & {1 \over 3} f_3 \\ 0 & -t & {1 \over 2} f_1 & {1 \over 2} f_2 \\ 0 & 0 & -t & f_1 \end{pmatrix}$$

Consider now the case of a non-trivial additive character $\psi$ of the finite field $\Bbb{F}_q$ --- that is to say a map $\psi: \Bbb{F}_q \longrightarrow \Bbb{C}^*$ such that $\psi(0) = 1$ and $\psi(x+y) = \psi(x) \, \psi(y)$ for all $x,y \in \Bbb{F}_q$. We may restrict $\psi$ to the multiplicative group $\Bbb{F}_q^*$ consisting of all non-zero field elements and over this group form $\text{sym}^n_t \psi$. For example when $n=2$ and $t=1$ this is the function who's value at $x \in \Bbb{F}_q^*$ is

$${1 \over 2} \psi \big(2x\big) + {1 \over 2}\psi \big( x^2 \big)$$

Do these $t$-analogues of the symmetric powers have any number-theoretic significance or meaning --- for example in terms of the various symbols of multiplicative characters or in regard to the evaluation of Gauss sums taken over the field ?

regards, A. Leverkühn

In view of the defining recursion, when $t=1$, the symmetric power $\text{sym}^n \psi$ has an elegant expression --- namely as the coefficient of $s^n$ in the exponent of "logarithmic" expansion

$$\begin{array}{ll} &{\displaystyle \exp \Big( s \, \psi_1 \ + \ {s^2 \over 2} \, \psi_2 \ + \ {s^3 \over 3} \, \psi_3 \ + \ \cdots \Big)} \ &= \\ &{\displaystyle \exp \big( s \, \psi_1\big) \cdot \exp \big( {s^2 \over 2} \, \psi_2 \big) \cdot \exp \big({s^3 \over 3} \, \psi_3\big) \cdots } \ &= \end{array}$$

In the caes of the real field $\Bbb{R}$ and the additive character $\psi(x) = \exp (-x)$ we may expand the factor

$$\exp \Big( {s^k \over k} \, \psi_k(x) \Big) \ = \sum_{r \geq 0} \ {s^{kr} \over {n! \, k^r}} \, \psi\big( {\scriptstyle -} \, rx^k \big)$$

If this expansion is formally recycled into the previous infinite product we may conclude that $\text{sym}^n \psi$ is the coefficient of $s^n$ in

$$\sum_{ \text{p} } {s^{\, p_1} \cdot s^{\ p_2} \cdot s^{\, p_3} \cdots \, s^{p_d} \over { {\scriptstyle (p_1)! \,(p_2)! \,(p_3)! \, \cdots \, (p_d)!} \cdot {\scriptstyle 2^{p_2} \, 3^{p_3} \, \cdots \, d^{p_d} }}} \, \psi\big( {\scriptstyle -} \text{p} \big)$$

where the sum is taken over all polynomials $\text{p} = p_1x + p_2x^2 + \dots + p_dx^d$ with non-negative integer coefficients and with zero constant term. Alternatively we may encode such a polynomial $\text{p}$ as a composition ${\bf p} = \big(p_1, \dots, p_d\big) \,$ in which case the desired coefficient of $s^n$ is

$$\sum_{ {\bf \text{p}} \, \vdash n } {2^{-p_2} \, \cdots \, d^{-p_d} \over { (p_1)! \, \cdots \, (p_d)!} } \, \psi \big( {\scriptstyle -} \text{p} \big)$$

where the sum is taken over all compositions ${\bf p} = \big(p_1, \dots, p_d\big) \,$ whose tally $p_1 + 2p_2 + 3p_3 + \dots + dp_d$ is exactly $n$.

yours, in crank-heit

A. Leverkühn

• I'm confused: is this an answer to your quesion? Apr 6, 2015 at 22:17

Not sure if this helps, but the determinantal expression for $\text{sym}^n_t f$ is in fact a quantum determinant --- albeit of a scalar-valued matrix and not a quantum matrix (whose entries must satisfy the Faddeev-Reshetikhin-Takhtadzhyan relations). For example, when $n=4$ we have:

$$\det \, \begin{pmatrix} {1 \over 4}f_1 & {1 \over 4}f_2 & {1 \over 4}f_3 & {1 \over 4}f_4 \\ -t & {1 \over 3}f_1 & {1 \over 3}f_2 & {1 \over 3}f_3 \\ 0 & -t & {1 \over 2}f_1 & {1 \over 2}f_2 \\ 0 & 0 & -t & f_1\end{pmatrix} \ = \ \text{det}_t \, \begin{pmatrix} {1 \over 4}f_1 & {1 \over 4}f_2 & {1 \over 4}f_3 & {1 \over 4}f_4 \\ -1 & {1 \over 3}f_1 & {1 \over 3}f_2 & {1 \over 3}f_3 \\ 0 & -1 & {1 \over 2}f_1 & {1 \over 2}f_2 \\ 0 & 0 & -1 & f_1\end{pmatrix}$$

yours, Ines