For $F =\mathbb{R}$ and $F=\mathbb{C}$ , let $O_n (F)=${ $A\in M_{n\times n}(F)| AA^t =I$}. Is $O_n(\mathbb{R}) $ compact? Is $O_n(\mathbb{R}$) connected? Is $O_n(\mathbb{C})$ compact?
This question is from my topology assignment.
I have been following Wayne Patty's book and understand the subject definitions and theorems.
But particularly in questions related to compactness and connectedness of the matrix I am unable to prove or contradict the given assertions.
For all the three parts I assumed what is given and tried to prove or contradict the definitions.
I also want to discuss what statergy should be used/ some subtle points that need to be kept in mind to prove or disprove the questions related to matrices in topology? I have been trying some qualifying exam questions and there also I got struck on question in topology involving matrices.
Thanks for any guidence that you provide?
Here's a general fact: if $G$ is a topological group and $H$ is a subgroup, then:
(i) if $H$ and $G/H$ are connected, then $G$ is connected.
(ii) if $H$ and $G/H$ are compact, then $G$ is compact.
Now, we have that ${\rm O}_n(\Bbb R)$ acts transitively on the sphere $S^{n-1}$ and the stabilizer of a point is isomorphic to ${\rm O}_{n-1}(\Bbb R)$, so the orbit-stabilizer theorem gives that ${\rm O}_n(\Bbb R)/{\rm O}_{n-1}(\Bbb R) \cong S^{n-1}$. Since spheres are compact and ${\rm O}_1(\Bbb R)$ is compact, then ${\rm O}_n(\Bbb R)$ is compact for all $n \geq 1$ (by induction).
The similar argument for connectedness fails because each matrix in ${\rm O}_n(\Bbb R)$ has determinant equal to $1$ or $-1$, and this gives two connected components for ${\rm O}_n(\Bbb R)$. But the same idea as above shows that ${\rm SO}_n(\Bbb R)/{\rm SO}_{n-1}(\Bbb R)\cong S^{n-1}$, so since spheres are connected and compact and ${\rm SO}_2(\Bbb R) \cong S^1$ is connected and compact, we have that ${\rm SO}_n(\Bbb R)$ is connected and compact for $n\geq 1$ (again by induction, the case $n=1$ trivial for unrelated reasons).
Same argument shows that the unitary group ${\rm U}(n)$ is compact and connected for $n\geq 1$, by playing with $S^{2n-1}$ instead.
As for ${\rm O}_n(\Bbb C)$, the situation changes, as the hypersurface in $\Bbb C^n$ given by the equation $z_1^2+\cdots + z_n^2=1$ is not compact (it is not bounded; try to find an unbounded subgroup of ${\rm O}_n(\Bbb C)$). And ${\rm O}_n(\Bbb C)$ is not connected either, as $\det$ gives two connected components (with positive det and negative det).
The general fact is a particular instance of a more general fact. Let $X$ be a topological space with an equivalence relation $\sim$. Then:
(i) if $\sim$ has connected equivalence classes and $X/_\sim$ is connected, then $X$ is connected.
(ii) if $\sim$ has compact equivalence classes, the quotient projection is an open map, and $X/_\sim$ is compact, then $X$ is compact.
It is a nice exercise in topology (don't think about matrices, they are a distraction at this point) to prove this and apply this to the case where $X=G$ and $\sim$ is congruence mod $H$.