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.

  • $\begingroup$ See math notation guide. I replaced the image with text. $\endgroup$ – 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.


per your request i'll add more details:

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.


Per your request, a prove that every matrix is of that form:

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.

  • $\begingroup$ I have tried working out the inverse of A but I doubt this will be of much use. I am trying to figure out a way to get from one to the other. A specific solution will be appreciated. $\endgroup$ – greg Apr 25 '15 at 0:40
  • $\begingroup$ Added details. If my answer is usefull please mark it as one and upvote it. $\endgroup$ – user234238 Apr 25 '15 at 0:48
  • $\begingroup$ why is $M \in SU(2)$ described as the matrix you have given? How is this obtained? $\endgroup$ – greg Apr 25 '15 at 0:57
  • $\begingroup$ added. Also please hit "accept this answer" when you think I helped you enough $\endgroup$ – user234238 Apr 25 '15 at 1:05

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