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Question. Suppose that $X$ is a transitive $G$-set of size greater than $1$ and let $\pi$ be the associated permutation representation with the character $\chi$. Show that some element $g \in G$ has no fixed points.

The start of the solution goes: Suppose all $g \in G$ has fixed points, so $\chi(g) \geq 1$. Then define $\chi_2=\chi-\chi_{\mathbb{I}}$, so $\chi_2(g) \geq 0$ for all $g \in G$. $\chi_2$ is also a character and $\langle \mathbb{I},\chi_2 \rangle=0$ since the action is transitive. Now $$0=\langle \mathbb{I},\chi_2 \rangle=\frac{1}{|G|}\sum_{g \in G}\chi_2(g) >0$$ because all terms are non-negative and the one for $g=1$ is $\dim \chi_2 >0$.

I cannot see why $\chi_2$ is also a character.

I cannot see why $\langle \mathbb{I},\chi_2 \rangle=0$ because the action is transitive.

Nor can I see why $\dim \chi_2 >0$.

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Notice that $[\chi,\mathbb{I}]=k$ is equal to number of the orbits.. Since our action is transitive $k=1$.

$$\chi=\mathbb{I}+n_2\phi_2+ n_3\phi_3...$$ where $\phi_i$ are irreducible character of $G$. Form this, it is easy to see that why $\chi_2$ is an character.

Notice also that if $\dim(\chi_2)=0$ then $\chi_2=0\implies \chi=\mathbb{I}$ hence $X$ has only one elements which is a contradiction.

EDIT: $$\chi=\mathbb{I}+n_2\phi_2+... n_k\phi$$

where $n_i$ are nonnegative integers and $\phi_i$ are irreducible characters. $$\chi_2=\chi-\mathbb{I}=n_2\phi_2+...+n_k\phi_k$$

$$\dim(\chi_2)=\chi_2(e)=n_2\phi_2(e)+...n_k\phi_k(e)$$

if $0=\dim(\chi_2)$ then we must have $n_i=0$ for $i\geq 2$ which means $\chi=\mathbb{I}\implies$ $G$ fixes every elemets of $X$ which is not the case as $|X|>1$ and the action is transitive.

Notice that if a function is a linear combination of irreducible characters with non-negative integer coefficient (and at least one of them is positive) then the function is a character. Thus, $\chi_2$ is an character.

Edit $2$: Having $p:G\to Gl_n(\mathbb C)$ is equivalent that you have an $n$ dimensional $\mathbb C G$ module $V$ and vice versa.

Existence of $\phi_i$ (irreducable character) means that we have irreducable $\mathbb C G$ module $V_i$ and corresponding represantation $p_i:G\to Gl_{n_i}(\mathbb C)$.

A natural question is that can we find representation for $V_i\oplus V_j$ ?

\begin{pmatrix} p_i & \huge 0 \\ \huge 0 & p_j \\ \end{pmatrix} You can see that above is the representatin of $V_i\oplus V_j$. And the corresponding character is exactly $Tr(p_i)+Tr(p_j)=\phi_i+\phi_j$." By similiar reasoning $n\phi_i+m\phi_j$ is the character of the module

$$V_i\oplus ...V_i\oplus V_j\oplus....\oplus V_j$$

Where $V_i$ is summed $n$ times and $V_j$ is summed by $m$ times. (Of course $m,n$ are nonnegative integer and at least one of them is positive)

By above argument, you can see that linear combinations of characters by non-integers (at least one of them is positive) is a character.

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  • $\begingroup$ It is not easy to see that $\chi_2$ is a character. Could you also justify $\dim(\chi_2)=0$ then "$\chi_2=0\implies \chi=\mathbb{I}$" please $\endgroup$ – Permian Apr 28 '15 at 10:59
  • $\begingroup$ @sandstone: I try to give some explanation. I hope it is clear for you now. $\endgroup$ – mesel Apr 28 '15 at 16:41
  • $\begingroup$ Thanks, theres still one thing. "Notice that if a function is a linear combination of irreducible characters with non-negative integer coefficient (and at least one of them is positive) then the function is a character. " I cannot see why this holds $\endgroup$ – Permian Apr 28 '15 at 16:56
  • $\begingroup$ There are many equivalent definations of character, from some of them, it is very claer. You should say yours for me to explain this. $\endgroup$ – mesel Apr 28 '15 at 19:56
  • $\begingroup$ My definition is that the character of a representation $\rho : G \rightarrow GL_n(\mathbb{C})$ the list of traces of the conjugacy classes $\endgroup$ – Permian Apr 30 '15 at 8:45

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