# Show that the set of all $n \times n$ orthogonal matrices, $O(n)$, is a compact subset of $\mbox{GL} (n,\mathbb R)$

I have only concept in topology, metric space, and functional analysis. How do I tackle this? Also I want to know that is the set connected?

• Please add the source if you have collected it from some question paper of entrance examination. Most probably it is a question of NBHM but I have forgotten the year. – Dutta Dec 16 '13 at 14:27

Hints:

• $\mathcal O(n)$ is the continuous inverse image of a closed set…

• There exists a rather obvious bound for every element in $\mathcal O(n)$ and thus this set is bounded…

Now Heine-Borel and we're done.

Added: Thanks to the comments by Berci I think some confusion may happen here, since the orthogonal group is not the inverse image of $\,\{1,-1\}\,$ under the determinant map in the whole $\,GL(n,\Bbb R)\,$ (as there are matrices with determinant $\,\pm 1\,$ which are not orthogonal, of course).

So let us try the following approach: the map $\,T:GL(n,\Bbb R)\to GL(n,\Bbb R)\,$ defined by $\,T(A):=AA^t\,$ is continuous since every entry in the matrix $\,AA^t\,$ is a polynomial on the entries of $\,A\,$ ,and now we can see that

$$\mathcal O(n)=T^{-1}(\{I\})$$

and, of course, $\,\{I\}\,$ is a closed set, so $\,\mathcal O(n)\,$ is closed.

• Antonio,Is the mapping $A\mapsto AA^t$. I can not understand why the set is bounded.can you explain me plz? – Andy Mar 30 '13 at 14:38
• If $A\in O(n)$ then $\|A\|=1$. – Berci Mar 30 '13 at 15:29
• @AnindyaGhatak: If $\,A=(a_{ij})\in\mathcal O(n)\,$ , then$$AA^t=1\implies \sum_{k=1}^na_{ik}^2=1\;,\;\;\forall\;i=1,2,\ldots,n$$ – DonAntonio Mar 30 '13 at 15:45
• Oh, and the continuous map I was talking about is the determinant map: $$\det:GL(n,\Bbb R)\to\Bbb R$$Of course, the set $\,\{-1,1\}\,$ is closed in $\,\Bbb R\,$ ... – DonAntonio Mar 30 '13 at 15:47
• First @Berci, I think you may have meant $\,SO(n)\,$ , and second $$\mathcal O(n)=\det^{-1}\{-1,1\}\;,\;\;SO(n)=\det^{-1}\{1\}\,$$unless, as many other times, I misunderstood something here. – DonAntonio Mar 30 '13 at 16:14

In fact, you can say more:

Using the canonical action of $SO(n)$ on $\mathbb{S}^{n-1}$, you can show that $SO(n)$ is homeomorphic to $\mathbb{S}^{n-1}$. But you have the exact sequence $1 \to SO(n) \to O(n) \to \mathbb{Z}_2 \to 1$, so $[O(n):SO(n)]=2$ and $O(n)$ is homeomorphic to $\mathbb{S}^{n-1} \coprod \mathbb{S}^{n-1}$.