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

  • 1
    $\begingroup$ The columns of a unitary matrix are orthonormal. In particular they have norm $1$. $\endgroup$ Apr 1, 2016 at 13:33
  • $\begingroup$ Does this prove it is closed? How so? $\endgroup$
    – thinker
    Apr 1, 2016 at 14:50
  • $\begingroup$ No, that proves it's bounded. $\endgroup$ Apr 1, 2016 at 14:51
  • $\begingroup$ oh yes, with upper limit $1$. So if the elements in a matrix are bounded, the matrix as a whole is bounded? And then I guess I can use the fact that $U(n)$ is a closed set from the below answer $\endgroup$
    – thinker
    Apr 1, 2016 at 14:53

1 Answer 1


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.

  • $\begingroup$ 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 $\endgroup$
    – thinker
    Apr 1, 2016 at 14:51
  • $\begingroup$ The inverse image of a singleton set $\{1\}$ (which is also closed) under a continuous map is closed. $\endgroup$ Feb 15, 2017 at 10:10
  • $\begingroup$ 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, $||U||= 1$. $\endgroup$ Feb 15, 2017 at 10:12
  • $\begingroup$ @filipos this holds for finite $n$, right? if we take $n\to\infty$ limit, then $U(n)$ becomes non compact, right? $\endgroup$
    – QGravity
    Jan 19 at 19:54
  • 1
    $\begingroup$ @QGravity Correct. Indeed, it is precisely for finite $n$ that "closed and bounded" implies "compact". $\endgroup$
    – filipos
    May 18 at 19:26

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