What I know:
$SU(n)=${$A \in U(n): detA=1$}
where
$U(n)=${$n \times n$ matrices $A: AA^*=I=A^*A$} with elements in $\mathbb{C}$ and $A^*$ is the complex transpose of $A$
A topological group is a Hausdorff topological space with a continuous group operation with continuous inverse
My idea is to first show that $U(n)$ is compact, which would then imply the compactness of $SU(n)$ since any closed subset of a compact space is itself compact.
Heine Borel criterion of compactness: A subset $V \subset \mathbb{R^n}$ is compact $\iff$ V is closed and bounded
So how can I show that $U(n)$ is closed and bounded?
Perhaps we can find a function whose preimage is closed and is $U(n)$
Would very much appreciate your help. Thanks
Best Answer
You just defined $SU(n)$ as the preimage of 1 under $\det$, so it is necessarily closed.
You also defined $U(n)$ as the preimage of $I$ under the continuous function $A \mapsto A^*A$.
Also, the operator norm of an unitary matrix is always 1, so $U(n)$ is bounded.
Additional information:
By definition of continuity, the preimage of a closed set is closed. Also, points are closed in $\mathbb{C}$ and $\mathbb{C}^{n\times n}$.
The operator norm of a matrix is defined by $\max_{x \neq 0}\frac{|Ax|}{|x|}$. It is easy to see that it is a norm and that it equals 1 for unitary operators.
In any case, whatever norm you use, it should not be hard to see that $U(n)$ is bounded: All the entries in an unitary matrix have absolute value at most one.