# The special unitary group SU(2) is homeomorphic to the 3-sphere

The special unitary group is defined as $$\mathrm{SU}(2) = \{A\in M_{2\times 2}(\mathbb{C}) \mid A(\overline{A})^T=I_{2\times2}\}.$$ Show that this is homeomorphic to the $$3$$-sphere $$\mathbb{S}^3 = \{(a,b)\in \mathbb{C}^2\mid |a|^2+|b|^2=1\}.$$

I know that for a map to be a homeomorphism the function needs to be continuous, function is bijective and the inverse of this function is continuous. However, I am not sure how to show that two things are homeomorphic.

• See math notation guide. I replaced the image with text. – user147263 Apr 25 '15 at 3:29

Two topological spaces $X,Y$ are said to be homeomorphic if there is a function $f:\:X\to Y$ which is a homeomorphism.

So all you got to do is find one. Usually there is a natural way to define such function, and that is the case right here : think of a general matrix in $SU2$ - how does it look? how many parameters, or values, define it? where would you send it then?

If needed i'll give a specific solution.

p.s

every $M\in SU(2)$ can be desribed as $\begin{bmatrix} a & -\overline b \\ b & \overline a \end{bmatrix}$, and every $a,b$ describes a unique $M$. therefore send $f(M)=(a,b)$. this is clearly a bijection (the inverse is $(a,b)\mapsto \begin{bmatrix} a & -\overline b \\ b & \overline a \end{bmatrix}$ , and I'll leave it to you to try to show that is a homeomorphism. Again, if required, i'll assist.
Let $M=\begin{bmatrix} a & c \\ b & d \end{bmatrix}\in U$. we have $M\overline M^t$= \begin{bmatrix} aa^*+cc^* & ab^*+cd^* \\ ba^*+dc^* & bb^*+dd^* \end{bmatrix}, and we must have $M\overline M^t=I$. the solution for these 4 equations is $c=-\overline b$ and $d=\overline a$, please check yourself.
• why is $M \in SU(2)$ described as the matrix you have given? How is this obtained? – greg Apr 25 '15 at 0:57