Let $\mathcal{A} = \{V : \mathbb{U}_n \rightarrow \mathbb{C}\}$ where $\mathbb{U}_n$ is the group of the complex $n$-th roots of the unity. This group naturally acts on $\mathcal{A}$: for any $a \in \mathbb{U}_n$, and any $V \in \mathcal{A}$ $$ \begin{equation*} (a \star V) : b \mapsto V(a^{-1}b) \end{equation*} $$ I define $Fix(V) = \{a \in \mathbb{U}_n,~ a \star V = V\}$ the symmetry group of $V$.

My question is: can we define an equivariant function $\lambda : \mathcal{A} \rightarrow \mathbb{C}$ which measures the "asymmetry" of $V$, i.e., a function such that for all $a,V$ $$ \begin{align*} \lambda(a \star V) &= a \lambda(V) \\ \lambda(V) = 0 &\Leftrightarrow Fix(V) = \{1\} \end{align*} $$

Is it related to some known problem ? Satisfing the first condition is ok, for instance $$ \begin{equation*} \lambda_0(V) = \sum_{z \in \mathbb{U}_n} z V(z) \end{equation*} $$ but the second condition is not ok. Indeed, it suffices to take $V$ as the indicator function of, say, the union of a triangle and a square (in a suitable $\mathbb{U}_n$). The sum above is zero but this $V$ has a trivial symmetry group.

Thank you.

Let $\mathcal{A} = \{V : \mathbb{U}_n \rightarrow \mathbb{C}\}$ where $\mathbb{U}_n$ is the group of the complex $n$-th roots of the unity. This group naturally acts on $\mathcal{A}$: for any $a \in \mathbb{U}_n$, and any $V \in \mathcal{A}$ $$ \begin{equation*} (a \star V) : b \mapsto V(a^{-1}b) \end{equation*} $$ I define $Fix(V) = \{a \in \mathbb{U}_n,~ a \star V = V\}$ the symmetry group of $V$.

My question is: can we define an equivariant function $\lambda : \mathcal{A} \rightarrow \mathbb{C}$ which measures the "asymmetry" of $V$, i.e., a function such that for all $a,V$ $$ \begin{align*} \lambda(a \star V) &= a \lambda(V) \\ \lambda(V) = 0 &\Leftrightarrow Fix(V) \neq \{1\} \end{align*} $$

Is it related to some known problem ? Satisfying the first condition is ok, for instance $$ \begin{equation*} \lambda_0(V) = \sum_{z \in \mathbb{U}_n} z V(z) \end{equation*} $$ but the second condition is not ok. Indeed, it suffices to take $V$ as the indicator function of, say, the union of a triangle and a square (in a suitable $\mathbb{U}_n$). The sum above is zero but this $V$ has a trivial symmetry group.

Thank you.

Let $\mathcal{A} = \{V : \mathbb{U}_n \rightarrow \mathbb{C}\}$ where $\mathbb{U}_n$ is the group of the complex $n$-th roots of the unity. This group naturally acts on $\mathcal{A}$: for any $a \in \mathbb{U}_n$, and any $V \in \mathcal{A}$ $$ \begin{equation*} (a \star V) : b \mapsto V(a^{-1}b) \end{equation*} $$ I define $Fix(V) = \{a \in \mathbb{U}_n,~ a \star V = V\}$ the symmetry group of $V$.

My question is: can we define an equivariant function $\lambda : \mathcal{A} \rightarrow \mathbb{C}$ which measures the "asymmetry" of $V$, i.e., a function such that for all $a,V$ $$ \begin{align*} \lambda(a \star V) &= a \lambda(V) \\ \lambda(V) = 0 &\Leftrightarrow Fix(V) = \{1\} \end{align*} $$

Is it related to some known problem ?

Thank you.

Let $\mathcal{A} = \{V : \mathbb{U}_n \rightarrow \mathbb{C}\}$ where $\mathbb{U}_n$ is the group of the complex $n$-th roots of the unity. This group naturally acts on $\mathcal{A}$: for any $a \in \mathbb{U}_n$, and any $V \in \mathcal{A}$ $$ \begin{equation*} (a \star V) : b \mapsto V(a^{-1}b) \end{equation*} $$ I define $Fix(V) = \{a \in \mathbb{U}_n,~ a \star V = V\}$ the symmetry group of $V$.

My question is: can we define an equivariant function $\lambda : \mathcal{A} \rightarrow \mathbb{C}$ which measures the "asymmetry" of $V$, i.e., a function such that for all $a,V$ $$ \begin{align*} \lambda(a \star V) &= a \lambda(V) \\ \lambda(V) = 0 &\Leftrightarrow Fix(V) = \{1\} \end{align*} $$

Is it related to some known problem ? Satisfing the first condition is ok, for instance $$ \begin{equation*} \lambda_0(V) = \sum_{z \in \mathbb{U}_n} z V(z) \end{equation*} $$ but the second condition is not ok. Indeed, it suffices to take $V$ as the indicator function of, say, the union of a triangle and a square (in a suitable $\mathbb{U}_n$). The sum above is zero but this $V$ has a trivial symmetry group.

Thank you.