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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

thinker
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1 Answers1

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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.

filipos
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  • Can I just ask for some more information? Why is the pre-image of 1 necessarily closed? Also, what is the operator norm of an unitary matrix? Thanks – thinker Apr 01 '16 at 14:51
  • The inverse image of a singleton set ${1}$ (which is also closed) under a continuous map is closed. – Shubham Namdeo Feb 15 '17 at 10:10
  • By definition of operator norm, $||U|| = \text{sup}{ ||U(x)||: ||x|| =1}$. But $||U(x)||^2= \langle U(x), U(x)\rangle= \langle x, x\rangle = ||x||^2.$ Therefore, $||U||= 1$. – Shubham Namdeo Feb 15 '17 at 10:12
  • @filipos this holds for finite $n$, right? if we take $n\to\infty$ limit, then $U(n)$ becomes non compact, right? – QGravity Jan 19 '22 at 19:54
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    @QGravity Correct. Indeed, it is precisely for finite $n$ that "closed and bounded" implies "compact". – filipos May 18 '22 at 19:26