Let $V$ be a complex vector space. Let $G$ be a finite group and let $$\rho: G \rightarrow GL(V)$$ be a representation of $G$.

Let $V^*$ denote the dual space of $V$, and let $\mathcal{O}(V)$ denote the algebra of functions $F: V \rightarrow \mathbb{C}$ generated by the elements of $V^*$. Elements of $\mathcal{O}(V)$ are called regular functions.

The dual representation $\rho^*: G \rightarrow GL(V^*)$ is given by $$(\rho^{*}(g)f)(v) = f(\rho(g)^{-1}v), \forall g \in G, f \in V^*, v \in V$$

Define the ring of invariant functions to be $$\mathcal{O}(V)^G = {\{f \in \mathcal{O}(V) : gf = f \hspace{2mm} \forall g \in G}\}$$ where $gf$ is just shorthand for $\rho^{*}(g)f$.

Now, when we restrict to $V^{*} \subset \mathcal{O}(V)$, where $V^{*}$ has basis $x_1,...,x_n$, the regular functions are polynomials in the $x_i$ and the action of $G$ is given by $$gp(x_1,...,x_n) = p(gx_1,...,gx_n)$$ for some polynomial $p$.

$(1)$ - I don't understand this last bit? What is the distinction between $V^{*}$ and $\mathcal{O}(V)$? Is $V^{*}$ just the set of $\mathbb{C}$-linear sums of $x_1,...,x_n$? There is no multiplication between the $x_i$ defined? And an element of $\mathcal{O}(V)$ is of the form $p(x_1,...,x_n)$ for any polynomial $p$ with complex coefficients(as multiplication is defined here)?

$(2)$ - What does restrict to $V^{*}$ mean? Does it mean to restrict the action of $G$ to $V^{*}$? I thought the action was only defined on $V^{*}$ to begin with?

$(3)$ - Also, when something like $\mathbb{C}[V]$ is written in this context, what does this mean? What does $G$ acting on $\mathbb{C}[V]$ mean?


1 Answer 1


For (1) you are completely correct.

For (2), the idea is that the statement is meant to define the action on all of $\mathcal{O}(V)$, so the part about restriction to $V^*$ seems to be an error (since you are correct that the action was originally just defined on $V^*$).

For (3), $\mathbb{C}[V]$ refers to the polynomial ring in $n$ variables where $n$ is the dimension of $V$, with the variables identified with a basis of $V$. But writing is like that means we don't need to pick a basis, which can have some advantages.
The main idea is that any representation on $V$ can be extended to $\mathbb{C}[V]$ similarly to what was done for $\mathcal{O}(V)$.

  • $\begingroup$ The action was defined on $V^{*}$, so how would the group act on $\mathcal{O}(V)$? Also, how exactly is this representaion extended to $\mathbb{C}[V]$? $\endgroup$
    – the man
    Mar 31, 2019 at 1:12

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