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{$\mathrm{SU}(2)$}

Its commutator is one-dimensional, thus a single dot in the Dynkin diagram.

These are the rotations of a sphere. They can be represented by 2x2 complex matrices, or 1x1 quaternions.

Start with a circle group - rotations around an axis (say, z). In three-dimensional space we add rotations around two other axes (say, x and y). Then composing rotations should yield opposite effects depending on the order.

{$\mathrm{U}(2)$}

Given a matrix:

{$\begin{pmatrix} a_{11}+ib_{11} & a_{12}+ib_{12} \\ a_{21}+ib_{21} & a_{22}+ib_{22} \end{pmatrix} $}

Suppose the determinant is 1. Then the inverse is:

{$\begin{pmatrix} a_{22}+ib_{22} & -a_{12}-ib_{12} \\ -a_{21}-ib_{21} & a_{11}+ib_{11} \end{pmatrix} $}

And the conjugate transpose is:

{$\begin{pmatrix} a_{11}-ib_{11} & a_{21}-ib_{21} \\ a_{12}-ib_{12} & a_{22}-ib_{22} \end{pmatrix} $}

Setting the latter two equal, we have the equations:

{$a_{11}-ib_{11}=a_{22}+ib_{22}$}

{$-a_{21}-ib_{21}=a_{12}-ib_{12}$}

Thus the matrix M is:

{$\begin{pmatrix} a_{11}+ib_{11} & -a_{12}+ib_{12} \\ a_{12}+ib_{12} & a_{11}-ib_{11} \end{pmatrix} $}

In other words, we have that M is:

{$\begin{pmatrix} a_{11}+ib_{11} & i(b_{12}-ia_{12}) \\ i(b_{12}+ia_{12}) & a_{11}-ib_{11} \end{pmatrix} $}

In terms of complex numbers, M is:

{$\begin{pmatrix} x & i\overline{y} \\ iy & \overline{x} \end{pmatrix} $}

Alternatively, M is:

{$\begin{pmatrix} x & 0 \\ 0 & \overline{x} \end{pmatrix} + i \begin{pmatrix} 0 & \overline{y} \\ y & 0 \end{pmatrix} $}

So {$i$} can be understood as signifying the permutation of the rows, which is to say, a switch of axes, a rotation of 90 degrees.

{$\mathrm{SU}(2)$}

Note that we have an additional equation that the {$\textrm{det}(M)=1$}, so that {$x \overline{x} + y \overline{y} = 1$}. And {$x$} and {$y$} are complex numbers. Thus there are three independent real variables.