Lie Group Homomorphism from $U(n)$ into $SO(2)$

Question

The task was to find an onto Lie Group Homomorphism from $U(n)$ onto $SO(2)$ and I know how to do this.

My prof then said that the crux was that the task said onto, if it would have said into you could have just used the identity map. I can't get my head around this statement.

I mean $id(AB)=id(A)id(B)=AB$ with $A,B \in U(n)$ is obviously not in $SO(2)$, is it?

Answer 1

No, that is not a map from $SU(n)$ into $SO(2,\mathbb R)$. What your teacher meant was the map $M\mapsto e_{SO(2,\mathbb R)}$.

But consider the map $\det\colon U(n)\longrightarrow\mathbb C\setminus\{0\}$, and the map$$\begin{array}{rccc}\psi\colon&\mathbb C\setminus\{0\}&\longrightarrow&GL(2,\mathbb R)\\&a+bi&\mapsto&\begin{bmatrix}a&-b\\b&a\end{bmatrix}.\end{array}$$Then $\psi\circ\det$ is a group homomorphism and its image is precisely $SO(2,\mathbb R)$, since\begin{align}\det\bigl(U(n)\bigr)&=\{z\in\mathbb C\mid\lvert z\rvert=1\}\\&=\bigl\{\cos(\theta)+i\sin(\theta)\mid\theta\in\mathbb R\bigr\}.\end{align}