The Group of Special Unitary Matrices
We’ve been talking about complex vector spaces with an inner product. Last time we discussed linear transformations that preserved the default inner product,
$$ \langle M\cdot x , M \cdot x \rangle = \langle x , x \rangle, $$
where
\begin{equation}\label{unitary} M^{\dagger}\cdot M = \mathbb{1}.\end{equation}
Unitary Matrices
The condition in \eqref{unitary} defines what we call a unitary matrix. The most general $2\times2$ unitary matrix is given by,
\begin{equation}\label{genericU} U^{\dagger} = U^{-1} \Rightarrow \left(\begin{array}{cc} \alpha & \beta \\ -e^{i\theta}\bar{\beta} & e^{i\theta}\bar{\alpha} \end{array}\right),\end{equation}
where
$$|\alpha|^{2} + |\beta|^{2} = 1.$$
Here’s a quick exercise to test yourself:
Use \eqref{genericU} to verify by explicit computation both that $U^{\dagger}\cdot U = \mathbb{1}$ and $\det U = e^{i\theta}$.
It might be illuminating to make contact with the space of real operators by considering the case where
$$\theta = 0,\quad \alpha = \cos\phi,\quad \beta = \sin\phi.$$
This gives the matrix operator $O$ where
\begin{equation}\label{ortho} O = \left(\begin{array}{cc} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{array}\right), \end{equation}
which is the familiar rotation matrix in two dimensions.
So indeed, \eqref{genericU} can be thought of a rotation in $\mathbb{C}^{2}$, whose restriction to real parameters restricts to a familiar rotation.
To further our intuition, let's consider the other real case,
$$\theta = \frac{\pi}{2}, \quad \alpha = \cos\phi, \quad \beta = \sin\phi.$$
In this case we find,
\begin{equation}\label{ortho2} O_{(-1)} = \left(\begin{array}{cc} \cos\theta & \sin\theta \\ \sin\theta & -\cos\theta \end{array}\right).\end{equation}
Here’s another quick exercise to test yourself:
Show by explicit computation that $\det O = 1 , \det O_{(-1)} = -1$, and use \eqref{ortho2} verify expicitly that $O^{\dagger} = O^{T} = O^{-1}$.
The difference between $O$ and $O_{(-1)}$ is that the latter involves a reflection,
$$ (x,y) \rightarrow (x,-y).$$
The determinant, $\det O_{(-1)} = -1 $ is what signals this reflection. While $\theta \rightarrow 0$ sends $O \rightarrow \mathbb{1}$,
$$\lim_{\theta \rightarrow 0} O_{(-1)} = \left(\begin{array}{cc} 1 & 0 \\ 0 & -1 \end{array}\right).$$
Similarly, for an unitary transformation where $\det U = e^{i\theta}$, we find an inhomogeneous phase rotation:
$$ \left(\begin{array}{c} z_{1} \\ z_{2} \end{array}\right) \rightarrow \left(\begin{array}{c} z_{1} \\ e^{i\theta}z_{2} \end{array}\right)$$
which preserves the default inner product
$$\langle z,z \rangle = |z_{1}|^{2} + |z_{2}|^{2}.$$
The set of all such unitary matrices that act on $\mathbb{C}^{N}$ is denoted $\mathsf{U}_{N}$.
Notice that any product of unitary matrices in a given dimension is also unitary. To see this, consider, say
$$ M = U_{1} \cdot U_{2} \cdot U{3},$$
where the $U_{i}$ are unitary, i.e. $U_{i}^{\dagger} = U_{i}^{-1}$. We can then compute $M^{\dagger}$,
$$ M^{\dagger} = \left(U_{1} \cdot U_{2} \cdot U{3}\right)^{\dagger} = U_{3}^{\dagger} \cdot U_{2}^{\dagger} \cdot U{1}^{\dagger},$$
where we've used the reversed multiplicative property of the transpose operation.
$$M^{\dagger}\cdot M = U_{3}^{\dagger} \cdot U_{2}^{\dagger} \cdot U{1}^{\dagger} \cdot U_{1} \cdot U_{2} \cdot U{3},$$
which, since $U_{i}^{\dagger} = U_{i}^{-1}$, means that
$$M^{\dagger}\cdot M = \mathbb{1}.$$
It's not hard to see from this that the matrix product among any unitary matrices is itself a unitary matrix. The inverse of any such matrix exists by definition, and of course $\mathbb{1}$ is unitary. Therefore, $\mathsf{U}_{N}$ forms a \textbf{group} under the matrix multiplication operation. And that group precisely reflects the symmetries associated to the default inner product on $\mathbb{C}^{N}$.
Special Unitary Matrices
Unitary matrices with unit determinant - $\det U = 1$ - are special. As we saw with quite explicitly with \eqref{ortho2}, unit determinant means you are simply connected with the identity. No reflections. No phase rotations.
Because of the determinant theorem, the subgroup of $\mathsf{U}_{N}$ that has unit determinant closes under matrix multiplication. This subset is special, since it is simply connected to the identity in a manner akin to \eqref{ortho}. As such, we refer to these matrices as the special unitary matrices or $\mathsf{SU}_{N}$.
Here’s yet another quick exercise to test yourself:
Show that $\mathsf{SU}_N$ is a subgroup of $\mathsf{U}_N$.
The default inner product is used all the time in physics. In quantum field theory, it is used to model the rotational invariance of wavefunctions for particles like the electron. It's also used to compute other, ``internal'' properties such as quark flavor and color. By considering what effectively amounts to a nontrivial $\mathbb{C}^{N}$ bundle over spacetime, $\mathsf{SU}_{N}$ describes the local symmetries of Yang-Mills theory, and is directly related to some of the force carrying particles we observe in particle colliders.
We shall have much to say about $\mathsf{SU}_{N}$ in the future, particularly when $N=2$.
TL;DR
Often, we work with vector spaces because they're easy to compute with, but that doesn't make their mathematics any less rich. Linear transformations of complex vector spaces are prototyped by matrices. Those matrices that preserve the default inner product we discussed last time are called Unitary. The set of all those unitary matrices simply connected to the identity matrix are called $\mathsf{SU}_{N}$, have the structure of a group, and are responsible for our understanding of nuclear physics, electron spin and more.
