数学代写|MATH11144 Representation theory

(5 pts) Let $G$ be a finite group. Show that the function
$$
\begin{aligned}
\mathbb{C}[G] \times \mathbb{C}[G] & \longrightarrow \mathbb{C} \
\left(f_1, f_2\right) & \longmapsto\left\langle f_1, f_2\right\rangle=\frac{1}{|G|} \sum_{g \in G} f_1(g) \overline{f_2(g)}
\end{aligned}
$$
defines an inner product on $\mathbb{C}[G]$.
Solution: We have
$$
\begin{aligned}
\left\langle c_1 f_1+c_2 f_2, f_3\right\rangle & =\frac{1}{|G|} \sum_{g \in G}\left(c_1 f_1+c_2 f_2\right)(g) \overline{f_3(g)} \
& =\frac{1}{|G|} \sum_{g \in G}\left(c_1 f_1(g) \overline{f_3(g)}+c_2 f_2(g) \overline{f_3(g)}\right) \
& =\frac{1}{|G|} \sum_{g \in G} c_1 f_1(g) \overline{f_3(g)}+\frac{1}{|G|} \sum_{g \in G} c_2 f_2(g) \overline{f_3(g)} \
& =c_1 \frac{1}{|G|} \sum_{g \in G} f_1(g) \overline{f_3(g)}+c_2 \frac{1}{|G|} \sum_{g \in G} f_2(g) \overline{f_3(g)} \
& =c_1\left\langle f_1, f_3\right\rangle+c_2\left\langle f_2, f_3\right\rangle
\end{aligned}
$$
Also
$$
\begin{aligned}
& \overline{\left\langle f_1, f_2\right\rangle}=\overline{\frac{1}{|G|} \sum_{g \in G} f_1(g) \overline{f_2(g)}} \
& \frac{1}{|G|} \sum_{g \in G} \overline{f_1(g) \overline{f_2(g)}} \
& \frac{1}{|G|} \sum_{g \in G} \overline{f_1(g)} \overline{\overline{f_2(g)}} \
& \frac{1}{|G|} \sum_{g \in G} \overline{f_1(g)} f_2(g)=\left\langle f_2, f_1\right\rangle \
&
\end{aligned}
$$
Lastly, we have
$$
\langle f, f\rangle=\frac{1}{|G|} \sum_{g \in G} f(g) \overline{f(g)}=\frac{1}{|G|} \sum_{g \in G}|f(g)|^2 .
$$
This sum is greater than or equal to zero and is zero if and only if $f(g)=0$ for all $g \in G$.

Instead of taking the trace of $\phi_g$ to define the character, one might try to do the same by taking the determinant of $\phi_g$ instead. This problem shows that this is not as useful since such a character would tell us nothing about non-abelian simple groups (and these are important).
For $\phi$ a representation of a finite group $G$, define a function
$$
\begin{aligned}
\operatorname{det} \phi: G & \longrightarrow \mathbb{C}^{\times} \
g & \longmapsto(\operatorname{det} \phi)(g)=\operatorname{det}\left(\phi_g\right)
\end{aligned}
$$
(a) (4 pts) Show det $\phi$ is a representatation and hence it’s a character since characters are same as representations for one-dimensional representations.

(b) (5 pts) Show that if $G$ is a non-abelian simple group, then det $\phi$ is the trivial character.
Solution:
(a) Follows from the multiplicativity of the determinant.
(b) Since $\operatorname{det} \phi$ is a homomorphism, its kernel is a normal subgroup of $G$. Since $G$ is simple, it must be that $\operatorname{ker}(\operatorname{det} \phi)={e}$ or $\operatorname{ker}(\operatorname{det} \phi)=G$. If the former is true, then $\operatorname{det} \phi$ is injective, and its image is a subgroup of $\mathbb{C}^{\times}$, which is abelian. This $G$ would be isomorphic to an abelian group, but this cannot by by assumption. So it must be that $\operatorname{ker}(\operatorname{det} \phi)=G$, in which case $\operatorname{det} \phi$ is the trivial homomorphism.

(4 pts) Show that $\chi_{\phi \oplus \psi}=\chi_\phi+\chi_\psi$
Solution: The matrix for each $(\phi \oplus \psi)g$ is a block matrix with blocks $\phi_g$ and $\psi_g$. The trace of a matrix like that is the sum of the traces of the blocks. Hence $\chi\phi+\chi_\psi$.

(5 $\mathrm{pts} / \mathrm{part})$
(a) Suppose $A$ is a matrix over $\mathbb{C}$ of finite order, i.e. $A^n=I$ for some positive integer $n$. Show that the eigenvalues $\lambda_i$ of $A$ are the $n$th roots of unity, namely they satisfy $\lambda_i^n=1$. (Hint: Use the result that $A$ is diagonalizable, which in turn follows from fact that for a representation of a finite group $G$, there exists a matrix $T$ such that $T \phi_g T^{-1}$ is diagonal for all $g \in G$. Then look up what diagonalizability has to do with eigenvalues.)
(b) Prove that for an irreducible representation of a finite group $G$,
$$
\chi_\phi(g)=\lambda_1+\cdots+\lambda_d
$$
where $\lambda_i$ are the eigenvalues of $\phi_g$ and $d$ is the dimension of $\phi$.
(c) Show that, if a complex number $\omega$ is a root of unity, then $\omega^{-1}=\bar{\omega}$.
(d) Prove that $\chi_\phi\left(g^{-1}\right)=\overline{\chi_\phi(g)}$.

Solution:
(a) If $A$ is diagonalizable, then it is a basic result of linear algebra that its diagonal entries are its (distinct) eigenvalues $\lambda_i$. Denote this diagonal matrix by $D$. Then
$$
D^n=\left(T A T^{-1}\right)^n=T A^n T^{-1}=T I T^{-1}=T T^{-1}=I
$$
But powers of a diagonal matrix are obtained by taking powers of its diagonal entries. Thus it follows that $\lambda_i^n=1$ as desired.
(b) Since $G$ is finite, $\phi_g$ has finite order. Hence, by the previous part, there exists a matrix $T$ such that $T \phi_g T^{-1}=D$, where $D$ is diagonal with eigenvalues $\lambda_i$ as the diagonal entries. Since characters take the same value on similar matrices, we have
$$
\chi_\phi(g)=\operatorname{Tr}\left(\phi_g\right)=\operatorname{Tr}\left(T \phi_g T^{-1}\right)=\operatorname{Tr}(D)=\lambda_1+\cdots+\lambda_d
$$
(c) A conjugate of a complex number $r e^{i \theta}$ is $r e^{-i \theta}$. If $\omega$ is a root of unity, then it has the form $e^{i \theta / n}$ for some $n$ (but the important thing is that $r=1$; if $r \neq 1$, then $\left(r e^{i \theta}\right)^n=r^n e^{i \theta n}$, and this number could not have size (modulus) 1). Then
$$
\omega \bar{\omega}=e^{i \theta / n} e^{-i \theta / n}=e^0=1
$$
So $\bar{\omega}$ is the inverse of $\omega$.
(d) Consider
$$
\chi_\phi\left(g^{-1}\right)=\operatorname{Tr}\left(\phi_{g^{-1}}\right)=\operatorname{Tr}\left(\phi_g^{-1}\right)
$$
Since trace is same on similar matrices, by part (a) we can replace $\phi_g$ by the diagonal matrix with diagonal entries the eigenvalues $\lambda_i$, and consequently we can replace $\phi_{g^{-1}}$ by the diagonal matrix whose entries are $\lambda_i^{-1}$ since to obtain the inverse of a diagonal matrix, you take the inverse of the diagonal entries. Thus
$$
\operatorname{Tr}\left(\phi_{g^{-1}}\right)=\sum_i \lambda_i^{-1}
$$

Since $\lambda_i$ are roots of unity, by part (b) we have
$$
\operatorname{Tr}\left(\phi_{g^{-1}}\right)=\sum_i \bar{\lambda}i $$ But this sum is precisely $\operatorname{Tr}\left(\overline{\phi_g}\right)$ (or rather the trace of the conjugate of the diagonal matrix replacing $\phi_g$ ). I.e. the sum is precisely $\overline{\chi\phi}$.