# Some questions about group representations

1. Nov 11, 2014

### ShayanJ

I was rethinking about some things I learnt but I came to things that seemed to be not firm enough in my mind.

1) When we want to find the unitary matrix that block-diagonalizes a certain matrix through a similarity transformation, we should find the eigenvectors of that matrix and stick them together to get a square matrix. But this process of sticking two column matrices together doesn't seem rigorous to me. Is there a rigorous way of doing this?
2)Consider the group of rotations around an axis. It has two irreducible representations, $e^{\pm i \varphi}$ on the vector space of complex numbers. But we also have the reducible representation of $\mathbb R ^2$ with Rotation matrices!
a) They both seem to be two dimensional. So sub-representations can have the same dimension of the bigger representation??? Seems strange!!!
b) For $e^{i\varphi}$, we write the vectors as $z=x+iy=\rho e^{i\alpha}$. Should we write the vectors differently when we consider $e^{-i\varphi}$?
c)can we say the group of rotations around an axis has a left-handed and a right-handed irreducible representation?

Thanks

2. Nov 13, 2014

### Matterwave

1) Why doesn't this method seem rigorous? By the way, I'm pretty sure (but sadly not 100% sure) that only Hermitian matrices are diagonalized by Unitary similarity transformations. But what about this method seems non-rigorous? What kind of a method do you have in mind that is "rigorous"?

2) I don't think I can answer this question with any authority, so I will stay away from this one. :D

3. Nov 13, 2014

### ShayanJ

I was thinking that there is no mathematical operation that let's you stick two column vectors to get a square matrix. But after seeing this page, the equivalent conditions part, I'm now good enough with it!
I have the answer to the a part. I was wrong that the $e^{\pm i \varphi}$ representations are 2 dimensional. They're one dimensional because they associate 1 by 1 matrices to each element of the group. I also think c was not a good question so I can abandon it.
Now only 2-b remains! which it seems to me for both representations $e^{\pm i \varphi}$, we're using the same vector space ($\mathbb C$) to define the automorphisms on so I think the vectors are the same and only the automorphisms are different. But as you put it, I'm sadly not 100% sure!

4. Nov 13, 2014

### Matterwave

I'm a little confused by why the group of rotations has "two representations $e^{\pm i\varphi}$", it seems to me that the + case is a counter-clock-wise rotation, and the - case is a clock-wise rotation, so they are not representing the same thing...

5. Nov 13, 2014

### ShayanJ

Yeah, they're different in the sense that they are different irreducible representations of the same group. This is actually what we can expect from the beginning. Because rotations around an axis can be done with either of the two possible orientations and no one of them has priority and so if we had only one irreducible representation for rotations around an axis, we were missing this symmetry between the two orientations.

6. Nov 13, 2014

### ShayanJ

I found something interesting:
$\left( \begin{array}{cc} \cos\varphi \ \ \ -\sin\varphi \\ \sin\varphi \ \ \ \cos\varphi \end{array}\right) \left( \begin{array}{cc} x \\ y \end{array}\right)=\left( \begin{array}{cc} x' \\ y' \end{array}\right) \Rightarrow \\ \frac{1}{2\sqrt{2}}\left( \begin{array}{cc} 1 \ \ \ i \\ 1 \ -i \end{array}\right) \left( \begin{array}{cc}\cos\varphi \ \ \ -\sin\varphi \\ \sin\varphi \ \ \ \cos\varphi \end{array}\right) \left( \begin{array}{cc} 1 \ \ \ 1 \\ -i \ \ \ i \end{array}\right) \left( \begin{array}{cc} 1 \ \ \ i \\ 1 \ -i \end{array}\right) \left( \begin{array}{cc} x \\ y \end{array}\right)=\frac{1}{\sqrt 2}\left( \begin{array}{cc} 1 \ \ \ i \\ 1 \ -i \end{array}\right) \left( \begin{array}{cc} x' \\ y'\end{array}\right) \Rightarrow \\ \left( \begin{array}{cc} e^{i\varphi} \ \ \ 0 \\ 0 \ \ \ e^{-i\varphi} \end{array}\right)\left( \begin{array}{cc} x+iy \\ x-iy \end{array}\right)=\left( \begin{array}{cc} x'+iy' \\ x'-iy' \end{array}\right)$
Which means for $e^{i\varphi}$ the vectors are $x+iy$ and for $e^{-i\varphi}$, they're $x-iy$.

7. Nov 13, 2014

### Fredrik

Staff Emeritus
You don't need an operation. If a,b,c,d are real numbers, then $\begin{pmatrix}a & b\\ c & d\end{pmatrix}$ is a square matrix, period. The fact that $\begin{pmatrix}a\\ c\end{pmatrix}$ and $\begin{pmatrix}b\\ d\end{pmatrix}$ are 2×1 matrices doesn't change that.

But the sets $\{x+iy|x,y\in\mathbb R\}$ and $\{x-iy|x,y\in\mathbb R\}$ are the same. (They're both $\mathbb C$).

Last edited: Nov 13, 2014
8. Nov 13, 2014

### Fredrik

Staff Emeritus
You started out with a set of 2×2 complex matrices. Their components may be real, but you're using the multiplication operation defined on the set of 2×2 complex matrices, so all your matrices should be thought of as complex. They correspond to linear operators on $\mathbb C^2$. This is a 2-dimensional vector space over $\mathbb C$.

Then you found two invariant subspaces, each of them 1-dimensional. ($\mathbb C$ is a 1-dimensional vector space over $\mathbb C$), and you used them to define irreducible representations. So you went from 2 dimensions to 1.

No need. They are arbitrary elements of $\mathbb C$.

I don't think that terminology is standard, but I suppose you can call them whatever you want.