Clarification of $G$-orbits acting on a finite set $X$ and subspaces of $\mathbb{C}[X].$

Question

Suppose that $G$ acts on a finite set $X$, and define $\mathbb{C}[X]=\{f\mid f:X\to \mathbb{C}\}$.

I'm asked to show that every $G$-orbit in $X$ defines an invariant subspace in $\mathbb{C}[X]$, where the group representation of $G$ on $\mathbb{C}[X]$ is given by $(g\cdot f)(x)=f(g^{-1}x)$ for all $g\in G$.

However, I'm having difficulty understanding where each object lives, per se, and what the invariant subspaces of $\mathbb{C}[X]$ actually are.

If I'm not mistaken, the $G$-orbits should be of the form $\mathcal{O}_x=\{f\in\mathbb{C}[X]\mid f(x)=x\}$, but I'm finding the representation given and the set $\mathbb{C}[X]$ to be abstract and difficult to understand.

This is homework, so I would ultimately like to prove the claim, but I would like some guidance on how to approach this.

Answer 1

You can understand a function $f \in \mathbb{C}[X]$ as an assignment of a complex number to each element of $X$, in other words just writing a complex number next to each point of $X$. The $G$-action on $\mathbb{C}[X]$ must take some assignment and produce a new one. It makes sense that given an action of $G$ on $X$, there is a nice way to induce an action on the assignments $\mathbb{C}[X]$. An invariant function $f$ will be an assignment that is unchanged under the action of any $g \in G$, and so it makes sense that this is related to the orbits of $X$ under the $G$-action.

Take for example $X$ being four points, and $G = C_3$ (cyclic of order 3), with the action on $X$ joining three points into a cycle, while not touching the fourth. A generator $g$ of $G$ will act on an assignment by permuting the complex numbers attached to the cycle, while leaving the complex number attached to the isolated fourth point alone. (But check the direction of this cycle: it goes the opposite direction to what you think!) If you want a function which is invariant under the whole of $G$, you need an assignment that doesn't change no matter what element of $G$ you hit it with, so it will have to assign the same number to each element of the cycle. The number assigned to the fourth point doesn't interact with the others, and so it can be varied independently.

So in this example it seems intuitive that the space of invariant functions is two-dimensional, since you have two complex numbers which may be varied independently. Does this picture give a good starting point on how to approach the question?

Answer 2

Imagine $X$ to be a set of students in a class. Instead of complex numbers, for the sake of ease in understanding, take the set of numbers $\{0,1,2,\ldots,\}$. A function on $X$ to this set of numbers might be thought of scores awarded to the students in an exam.

Assume the students exchange the answerscripts among themselves (this is the action on $X$). Now the scores awarded is a different function. An orbit on $X$ is a subset of the set of students whose answer scripts get distributed among themselves.

Now a second professor aware of the paper swapping going on, awards scores to a student as the sum of the scores awarded by the first professor, the sum being taken over the 'orbit' of students.

Can you see that this second professors score function is invariant?