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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

\begin{equation} \text{sym}^n_t f \, = \, {1 \over n} \, \sum_{k=1}^n \, t ^{k-1} \, f_k \cdot \text{sym}^{n-k}f \end{equation}

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:

\begin{equation} \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} \end{equation}

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

\begin{equation} {1 \over 2} \psi \big(2x\big) + {1 \over 2}\psi \big( x^2 \big) \end{equation}

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

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

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

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:

\begin{equation} \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} \end{equation}

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

\begin{equation} {1 \over 2} \psi \big(2x\big) + {1 \over 2}\psi \big( x^2 \big) \end{equation}

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

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

\begin{equation} \text{sym}^n f \, = \, {1 \over n} \, \sum_{k=1}^n \, t ^{k-1} \, f_k \cdot \text{sym}^{n-k}f \end{equation}

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:

\begin{equation} \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} \end{equation}

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

\begin{equation} {1 \over 2} \psi \big(2x\big) + {1 \over 2}\psi \big( x^2 \big) \end{equation}

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

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

\begin{equation} \text{sym}^n_t f \, = \, {1 \over n} \, \sum_{k=1}^n \, t ^{k-1} \, f_k \cdot \text{sym}^{n-k}f \end{equation}

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:

\begin{equation} \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} \end{equation}

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

\begin{equation} {1 \over 2} \psi \big(2x\big) + {1 \over 2}\psi \big( x^2 \big) \end{equation}

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

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

\begin{equation} \text{sym}^n f \, = \, {1 \over n} \, \sum_{k=1}^n \, t ^{k-1} \, f_k \cdot \text{sym}^{n-k}f \end{equation}

If $f$ happens to be the character of a finite dimensional complex representation $V$ of $G$ then upon specialising $t$ to the values $1$, $-1$, and $0$ the $t$-analogue $\text{sym}^n_t f$ will be respectively the character of the $n$-th symmetric power, the $n$-th exterior power, and the $n$-th tensor power of the representation $V$; in the later case you should divide by $n!$. 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:

\begin{equation} \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} \end{equation}

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

\begin{equation} {1 \over 2} \psi \big(2x\big) + {1 \over 2}\psi \big( x^2 \big) \end{equation}

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

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

\begin{equation} \text{sym}^n f \, = \, {1 \over n} \, \sum_{k=1}^n \, t ^{k-1} \, f_k \cdot \text{sym}^{n-k}f \end{equation}

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:

\begin{equation} \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} \end{equation}

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

\begin{equation} {1 \over 2} \psi \big(2x\big) + {1 \over 2}\psi \big( x^2 \big) \end{equation}

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