I've been learning about induced representations recently and I've come across something which I'm very confused about;

For any $G$-set $X$, the number of orbits is equal to $(1_G, \chi_{\mathbb{C}[X]})$ where $1_G$ denotes the character of the trivial representation and $\chi_{\mathbb{C}[X]}$ denotes the character of the permutation representation defined below;

$\mathbb{C}[X]$ is the natural permutation representation with basis ${\{e_x : x \in X}\}$ and the action of $g$ being $ge_{x} = e_{gx}$.

I have no idea why this is true. I know that $Ind_{H}^{G} \mathbb{C} \simeq \mathbb{C}[G/H]$ and the explanation I've seen goes like this;

If $X$ is transitive, then $(1_G, \chi_{\mathbb{C}[X]}) = 1$, which can be shown through Frobenius Reciprocity. If $X$ isn't transitive, then $X = \bigcup_{i=1}^{n} X_i$ where the $X_i$ are transitive. Then, $(1_G, \chi_{\mathbb{C}[X]}) = (1_G, \chi_{\mathbb{C}[X_1] \oplus \cdots \oplus \mathbb{C}[X_n]})= (1_G, \chi_{\mathbb{C}[X_1]}) + ... + (1_G, \chi_{\mathbb{C}[X_n]}) = n$.

I mean, I understand every line of the proof and can follow it, but why does the value $(1_G, \chi_{\mathbb{C}[X]})$ give you the number of orbits of a $G$-set? What does this value have to do with orbits of $G$-sets?


2 Answers 2


I really like Eric Wofsey's answer (+1), which was posted as I was writing this. However, this is a different solution, so I'll post it too.

My solution follows from the orbit-stabilizer theorem.

Let's take a closer look at what $\newcommand\CC{\mathbb{C}}\chi_{\CC[X]}$ actually is. $$\chi_{\CC[X]}(g) = \newcommand\tr{\operatorname{tr}}\tr(\pi_g) = \sum_{x\in X} \langle e_x,ge_x\rangle = \sum_{x\in X} [gx=x], $$ i.e. $\chi_{\CC[X]}(g)$ counts the number of elements of $X$ fixed by $G$.

Then $$(1_G,\chi_{\CC[X]}) = \frac{1}{|G|}\sum_{g\in G} \chi_{\CC[X]}(g) =\frac{1}{|G|}\sum_{g\in G}\sum_{x\in X} [gx=x],$$ i.e. the inner product is the average number of elements of $X$ fixed. Rearranging the sums, we have $$(1_G,\chi_{\CC[X]})=\frac{1}{|G|}\sum_{x\in X} \sum_{g\in G}[gx=x] = \frac{1}{|G|} \sum_{x\in X} |\operatorname{Stab}(x)|.$$ Now pull the cardinality of the group in, and recognize the term in the summation as the index of the stabilizer $$(1_G,\chi_{\CC[X]})=\sum_{x\in X} \frac{|\newcommand\Stab{\operatorname{Stab}}\Stab(x)|}{|G|}=\sum_{x\in X}\frac{1}{[G:\Stab(x)]},$$ but the index of the stabilizer equals the size of the orbit, so we have $$(1_G,\chi_{\CC[X]})=\sum_{x\in X} \frac{1}{|Gx|} =\sum_{\text{orbits}} 1 =\#\{\text{orbits}\} $$

A note on notation

For a statement $P$, I use the notation $$[P]:=\begin{cases} 1 & \text{ $P$ is true} \\ 0 & \text{otherwise}.\end{cases}$$


Also this result is essentially just Burnside's lemma (modulo my comment on the character counting the number of fixed points).


The inner product $(1_G,\chi_{\mathbb{C}[X]})$ is just the dimension of the space of intertwining maps from the trivial representation $\mathbb{C}$ to $\mathbb{C}[X]$. A linear map $T:\mathbb{C}\to\mathbb{C}[X]$ is determined by $T(1)$, and is intertwining iff $T(1)$ is fixed by every element of $G$. So, $(1_G,\chi_{\mathbb{C}[X]})$ is just the dimension of the space of vectors in $\mathbb{C}[X]$ which are fixed by every element of $G$.

Now, when is an element $\sum c_xe_x\in\mathbb{C}[X]$ fixed by every element of $G$? Exactly when the coefficients $c_x$ are constant on each orbit (since the action of $G$ just permutes these coefficients transitively within each orbit). Letting $X_1,\dots,X_n$ be the orbits, this means that the vectors $v_i=\sum_{x\in X_i} e_x$ are a basis for the space of vectors fixed by $G$, and in particular that space has dimension $n$.

  • $\begingroup$ What is the relation between the dimension of the space of sign representation and orbits? Can we say the dimension is the number of orbits containing an even number of elements or something similar? $\endgroup$
    – khashayar
    Dec 22, 2022 at 7:32
  • $\begingroup$ I'm not sure what you mean by "the space of sign representation". $\endgroup$ Dec 22, 2022 at 14:21
  • $\begingroup$ The number of times that the sign representation occurs in the decomposition into irreducible representations, or multiplicity with which the sign representation occurs. This number is equal to the inner product $(\chi,\psi)$, where $\chi$ and $\psi$ are the characters of our permutation representation and the sign representation, respectively. $\endgroup$
    – khashayar
    Dec 22, 2022 at 22:44
  • $\begingroup$ What do you mean by the sign representation? Are you assuming $G=S_n$? $\endgroup$ Dec 22, 2022 at 22:59
  • 1
    $\begingroup$ A general group does not have a unique such "sign representation". I would suggest you post a new question and spell out in detail a precise definition of what you want to ask. $\endgroup$ Dec 23, 2022 at 14:40

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .