# Do these two sets of matrices form groups?

Stimulated by some Physics backgrounds, consider the following two sets of matrices.

Notations and definitions:Let $A,B$ be two complex $n\times n$ matrices, then $\left [ A,B \right ]\overset{\underset{\mathrm{def}}{}}{=}AB-BA$ and $e^A\overset{\underset{\mathrm{def}}{}}{=} \sum_{m=0}^{\infty }\frac{A^m}{m!}.$

Let $M_1,M_2,M_3$ be three $n\times n$ Hermitian matrices, and they satisfy $[M_1,M_2]=iM_3,[M_2,M_3]=iM_1,[M_3,M_1]=iM_2$ identities, where $i=\sqrt{-1}$.

Now define a set $X$ of unitary matrices:$X=\left \{ e^{i\alpha M_3}e^{i\beta M_2}e^{i\gamma M_3} :\alpha,\beta,\gamma \in \mathbb{R}\right \},$ and another set $Y$ of unitary matrices:$Y=\left \{ e^{i(\alpha M_1+\beta M_2+\gamma M_3)} :\alpha,\beta,\gamma \in \mathbb{R}\right \},$ where $i=\sqrt{-1}$ ( Note that the number indices of the Hermitian matrices $M$ are different in $X$ and $Y$ ).

And my questions are as follows:

(1) Is $X=Y$ ?

(2) If (1) is true, is the set $X$ a group ?

(3) If both (1) and (2) are true, is $X\cong SU(2)$ true ?

Supplements: For concreteness, let's take a look at the following simple example. Consider the Physically called spin-$\frac{1}{2}$ "Pauli" matrices $M_1=\frac{1}{2}\bigl(\begin{smallmatrix} 0& 1\\ 1&0 \end{smallmatrix}\bigr),M_2=\frac{1}{2}\bigl(\begin{smallmatrix} 0& -i\\ i&0 \end{smallmatrix}\bigr)$ and $M_3=\frac{1}{2}\bigl(\begin{smallmatrix} 1& 0\\ 0&-1 \end{smallmatrix}\bigr),$ and it's easy to find that $M_1^2+M_2^2+M_3^2=\frac{1}{2}(\frac{1}{2}+1)\mathbb{I}$, where $\mathbb{I}$ is a $2\times2$ identity matrix.

In the above example, direct calculation of matrices $e^{i\alpha M_3}e^{i\beta M_2}e^{i\gamma M_3}$ in $X$ shows that $X=SU(2)$ (then $X$ is a group), and it's also easy to verify that $Y\subseteq SU(2)$. So now the question is, is $SU(2)\subseteq Y$ too ?

Thanks in advance.

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In the definition of X there is $M_3$ twice. Is this an error or not? :) – Riccardo May 3 '13 at 21:06
I would imagine that any answer would depend heavily on $M_1, M_2$ and $M_3$, which you have not elaborated on. Could be wrong though. I don't immediately grasp what those commutation conditions say. – rschwieb May 3 '13 at 21:20
@ Ric Ped， Dear Ped, $M_3$ definitely appears twice, it's not an error. – Kai Li May 3 '13 at 21:33
@ Ric Ped, in fact, if you choose $M_1=\bigl(\begin{smallmatrix} 0 &1 \\ 1&0 \end{smallmatrix}\bigr),M_2=\bigl(\begin{smallmatrix} 0 &-i \\ i&0 \end{smallmatrix}\bigr),M_3=\bigl(\begin{smallmatrix} 1 &0 \\ 0&-1 \end{smallmatrix}\bigr)$, then $X$ is just the Euler's angles $(\alpha,\beta ,\gamma)$ representation of group $SU(2).$ – Kai Li May 3 '13 at 21:39
Those relations don't hold for the Euler's angles matrices, instead $[M_1,M_2]=2iM_3$ $[M_2,M_3]=2iM_1$ $[M_3,M_1]=2iM_2$. Also for those matrices $\sum M_i^2=3\mathbb{I}$, where you say it should be $\frac{3}{4}\mathbb{I}$. Are you missing a factor of $2$ somewhere? – Alexander Gruber May 3 '13 at 22:48

## 1 Answer

Comments to the question (v3):

(We will from now on assume that the $n\times n$ matrices $M_1$, $M_2$ and $M_3$ are linearly independent, and that $n\geq 2$.)

OP defines two sets:

$$\tag{1} X_n~:=~ \{ e^{i\alpha M_3}e^{i\beta M_2}e^{i\gamma M_3}\in {\rm Mat}_{n\times n}(\mathbb{C}) \mid \alpha,\beta,\gamma \in \mathbb{R} \},$$

$$\tag{2} Y_n~:=~ \{ e^{i(\alpha M_1+\beta M_2+\gamma M_3)} \in {\rm Mat}_{n\times n}(\mathbb{C}) \mid\alpha,\beta,\gamma \in \mathbb{R} \}.$$

The set $Y_n$ is the image $Y_n=\rho(SU(2))$ of an $n$-dimensional $SU(2)$ group representation $\rho:SU(2) \to GL(n,\mathbb{C})$, also known in physics as a spin $\frac{n-1}{2}$ representation. In particular, the set $Y$ is itself a group.

$$\tag{3} Y_n~\cong~\left\{ \begin{array}{rcl} SU(2)/ \mathbb{Z}_2 ~\cong~SO(3) &\text{for}&n&\text{odd}, \\ SU(2) &\text{for}&n&\text{even}.\end{array} \right.$$

The set $X_n\subseteq Y_n$ is a subset of $Y_n$; due, among other things, to the Baker-Campbell-Hausdorff formula.

In fact, the set $X_n$ is a generalized Euler angle realization of $SU(2)$, as mentioned by OP himself. One may check by inspection that the $X_n$ hit every element in $Y_n$, so that $X_n=Y_n$.

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@ Qmechanic, thanks for your wonderful comment. So do you mean that set $Y$ is always "larger" than set $X$ ? Is this conclusion also true for single spin-$\frac{1}{2}$ ? In spin-$\frac{1}{2}$ example, is $Y$ a group ? And is only the set $X$ the spin-rotation group for spin-$\frac{1}{2}$ while $Y$ not ? – Kai Li May 5 '13 at 17:33
@ Qmechanic, In spin-$\frac{1}{2}$ example, it's easy to show that $Y$ is a subset of group $SU(2)$(then a subset of $X$), if what you say "$X$ is a subset of $Y$" is simultaneously true, then we arrive at $X=Y=SU(2)$ for spin-$\frac{1}{2}$, right? – Kai Li May 6 '13 at 8:13
I updated the answer. – Qmechanic May 7 '13 at 15:20
@ Qmechanic, thank you very much. To conclude your explanations, for spin-$\frac{n-1}{2}$ , we have $X_n=Y_n\cong SO(3)$ for $n=odd$ and $X_n=Y_n\cong SU(2)$ for $n=even$ . So $X_n$ and $Y_n$ are in fact the same one called spin-rotation group for spin-$\frac{n-1}{2}$. Am I right ? – Kai Li May 7 '13 at 19:44
@ Qmechanic, and for a general angular momentum $M_1,M_2,M_3$ instead of spin , is $X_n$ still identical to $Y_n$ and $X_n$ or $Y_n$ still the rotation group associated with $M_1,M_2,M_3$ ? Thanks a lot. – Kai Li May 10 '13 at 11:01