11

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?

Rodrigo de Azevedo
  • 1
Andy
  • 2,246
  • 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. – Supriyo Dec 16 '13 at 14:27

2 Answers2

19

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.

DonAntonio
  • 211,718
  • 17
  • 136
  • 287
  • 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
  • 2
    If $A\in O(n)$ then $|A|=1$. – Berci Mar 30 '13 at 15:29
  • 2
    @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
  • @DonAntonio ,thanks – Andy Mar 30 '13 at 15:54
  • Anytime,@AnindyaGhatak.... – DonAntonio Mar 30 '13 at 16:02
  • But.. $O(n)\ne\det^{-1}({-1,1})$, it's rather something like $SL(n)$. – Berci Mar 30 '13 at 16:12
  • 1
    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
  • 1
    For example $\pmatrix{1&1\0&1}$ has determinant $1$, though it's not orthogonal. By definition, $SL(n)=\det^{-1}(1)$. – Berci Mar 30 '13 at 17:43
  • Oh, I see the confusion: sorry. Of course, it must be that $,A^t=A^{-1},$, or the matrix's rows are orthogonal vector in $,\Bbb R^n,$ . There is where all happens. – DonAntonio Mar 30 '13 at 18:24
  • Thanks @Berci for noting that. I added some stuff to my answer that I hope will clean this mess. – DonAntonio Mar 30 '13 at 18:43
2

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 $\mathbb{S}^{n-1} \coprod \mathbb{S}^{n-1}$.

Seirios
  • 33,157