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Let $z$ satisfying the equation $z^3=1$ be a generator of the cyclic group $\mathbb{Z}_3= \{ 1 , z,z^2 \}$. You are given that $\rho : \mathbb{Z}_3 \to GL(\mathbb{C}^2)$ defined by $$\rho(z) = \begin{bmatrix} -1 & 1 \\ -1 & 0 \end{bmatrix}$$ is a representation of the group $\mathbb{Z}_3$ in the vector space $\mathbb{C}^2$.

(a) Find an inner product on the complex vector space $\mathbb{C}^2$ which is $\rho$-invariant: $$\langle \rho(g)u,\rho(g)v \rangle = \langle u,v \rangle$$ for all $g\in\mathbb{Z}_3, u,v\in\mathbb{C}^2$.

(b) Describe all $\rho$-invariant inner products on the vector space $\mathbb{C}^2$ explicitly, in terms of the coordinates of vectors.

I've done (a) but how do I do (b)?

2 Answers 2

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$\def\p#1{\left\langle#1\right\rangle}\def\C{\mathbb C}\def\Mat{\operatorname{Mat}}$Let $\p{\cdot,\cdot}$ an inner product on $\C^2$, then there is a positive definite $A \in \Mat_2(\C)$ such that $\p{x,y} = x^tA\bar y$ for all $x,y\in \C^2$. $\p{\cdot,\cdot}$ is $\rho$ invariant, iff for any $x,y$: $$ x^tA\bar y = \p{x,y} = \p{\rho(z)x,\rho(z)y} = x^t\rho(z)^tA\rho(z)\bar y $$ that is $A = \rho(z)^tA\rho(z)$. We have \begin{align*} \rho(z)^tA \rho(z) &= \begin{bmatrix} -1 & -1 \\ 1 & 0 \end{bmatrix} \begin{bmatrix} a & b \\ c & d \end{bmatrix} \begin{bmatrix} -1 & 1 \\ -1 & 0 \end{bmatrix}\\ &= \begin{bmatrix} -a-c & -b-d \\ a & b\end{bmatrix} \begin{bmatrix} -1 & 1 \\ -1 & 0 \end{bmatrix}\\ &= \begin{bmatrix} a+b+c+d & -a-c \\ -a-b & a\end{bmatrix} \end{align*} That is, we must have \begin{align*} a &= a+b+c+d\\ -a-c &= b\\ -a-b &= c\\ a &= d \end{align*} which is equivalent to \begin{align*} b + c + d &= 0\\ a + b + c &= 0\\ a + b + c &= 0\\ a - d &= 0 \end{align*} So we must have $a = d$ and $b+c = -a$. As $A$ is hermitian $b = \bar c$, and by definiteness $a > 0$, $ad - bc = a^2 - |b|^2 > 0$. $b+ c = b + \bar b = 2\Re b$. That is $a = -2\Re b$. We need therefore $\Re b < 0$ and $$ a^2 = 4\Re^2 b > |b|^2 = \Re^2 b + \Im^2 b \iff |\Im b| < -\sqrt 3|\Re b| $$

So the $\rho$-invariant inner products are given by the matrices $$ \begin{bmatrix} 2\lambda & -\lambda + \mu i \\ -\lambda - \mu i & 2\lambda \end{bmatrix}, \qquad \lambda > 0, |\mu| < \sqrt 3 \lambda $$

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    So if $\boldsymbol{x} = \begin{bmatrix} x_1 \\ y_1 \end{bmatrix}$ , $\boldsymbol{y} = \begin{bmatrix} y_1 \\ y_2 \end{bmatrix}$ we express the inner products in terms of coordinates by working out $\boldsymbol{x} A \boldsymbol{ \bar{y} }$? For (a), I have $$\langle u , v \rangle _{\text{invariant}} = \frac{1}{|G|} \sum_{g\in G} \langle \rho(g) u, \rho(g) v \rangle$$ where $\langle \cdot , \cdot \rangle$ is the standard inner product on $\mathbb{C}^2$: $$\langle u , v \rangle = u_1 \bar{v}_1 + u_2 \bar{v}_2$$ but how do you plug a general g\in G in $\rho$ when it is only defined using $z$?2012-11-24
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    Note that $G$ is generated by $z$, and for each $n\in\mathbb N$ we have $\p{\rho(z^n)x, \rho(z^n)y} = \p{\rho(z)^nx,\rho(z)^ny} = \p{x,y}$ as $\rho$ is a homomorphism.2012-11-24
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    At the top of your proof above, you said $\langle \cdot , \cdot \rangle$ is $\rho$ -invariant $\iff \langle x ,y \rangle = \langle \rho (z) x , \rho(z)y \rangle$ instead of $\langle \rho (g) x , \rho(g)y \rangle$ (for any $g\in G$) - is that OK? Also, at the bottom you said $$\Im ^2 b < 3\Re^2 b \iff | \Im b | < - \sqrt{3} | \Re b |.$$ Why do we need the minus sign and why do we need the modulus signs since $\Re b, \Im b \in\mathbb{R}$? Also by $\boldsymbol{x}^t A \boldsymbol{\bar{y}}$ do you mean the $\textit{conjugate}$ transpose of $x$?2012-11-24
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    In fact I don't think that second 'if and only if' is true?2012-11-24
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I think that second $\iff$ in martini's solution should be (no minus sign): $$\text{Im}(b)^2 < 3 \text{Re}(b)^2 \iff |\text{Im}(b)| < \sqrt{3}|\text{Re}(b)|$$ which then gives the matrices as $$\begin{bmatrix} -2\lambda & \lambda + \mu i \\ \lambda - \mu i & -2\lambda \end{bmatrix}$$ for $\lambda <0, |\mu|< \sqrt{3}|\lambda|$ ($\lambda = \text{Re}(b), \mu = \text{Im}(b)$).

These are the same set of matrices.