1
$\begingroup$

Let $O(m)$ denote the group of orthogonal matrices under multiplication, and let $SO(m)$ be the special orthogonal group over $\mathbb{R}$. Let \begin{equation*} (O(k)\times O(n-k))\cap SO(n):=\left\{A=\begin{pmatrix} B & 0\\ 0 & C \end{pmatrix} \in \mathbb{R}^{n\times n} \mid B\in O( k) ,\ C\in O( n-k) ,\ \det( B)\det( C) =1\right\}. \end{equation*}

I want to prove/disprove $O(k)\times O(n-k)$, $k=1,\dots,n-1$ is closed in $SO(n)$.

I do not really know how.

Even for $k=1$, I am not too sure. For $k=1$, we have \begin{equation*} \left\{\begin{pmatrix} b & 0\\ 0 & C \end{pmatrix} \in \mathbb{R}^{n\times n} \mid b\in O( 1) ,\ C\in O( n-1) ,\ b\det( C) =1\right\} =H_{1} \cup H_{-1} , \end{equation*} where for $b=1,-1$ we let \begin{equation*} H_{b} :=\left\{\begin{pmatrix} b & 0\\ 0 & C \end{pmatrix} \in \mathbb{R}^{n\times n} \mid \ C\in O( n-1) ,\ \det( C) =b\right\} . \end{equation*} I want to say this is essentially the union of $SO(n-1)$ and thus closed, but I am not really comfortable with seeing $SO(n-1)$ as a subset of $SO(n)$ incorporating the topological consistency.

$\endgroup$
1
$\begingroup$

Since each orthogonal group is compact, $O(k)\times O(n-k)$ is compact and therefore a closed subset of $\mathbb{R}^{n\times n}$. And your set is a closed subset of this one, because it's the set of thos elements whose determinant is $1$. Since $\det$ is continuous and $\{1\}$ is closed, this is again a closed set.

$\endgroup$
  • $\begingroup$ I see. Just to check:$(O(k)\times O(n-k))\cap SO(n)=\det^{-1}(\{1\})$, i.e. it is the preimage of $\det:O(k)\times O(n-k)\to \{1\}$. Thus, $(O(k)\times O(n-k))\cap SO(n)$ is closed in $O(k)\times O(n-k)$. But $O(k)\times O(n-k)$ is closed in the whole space $\mathbb{R}^{n\times n}$. Thus, $(O(k)\times O(n-k))\cap SO(n)$ is closed in $\mathbb{R}^{n\times n}$ (cf., proofwiki.org/wiki/Closed_Set_in_Topological_Subspace). But $SO(n)$ is closed in $\mathbb{R}^{n\times n}$ and $(O(k)\times O(n-k))\cap SO(n)\subset SO(n)$. Thus $(O(k)\times O(n-k))\cap SO(n)$ is closed in $SO(n)$? $\endgroup$ – user41467 Feb 27 '18 at 9:14
  • $\begingroup$ (The question I am asking in the comment above is if the above was the argument you used, but there wasn't enough room to put my intention of the comment, which I am typing here.) $\endgroup$ – user41467 Feb 27 '18 at 9:22
  • 1
    $\begingroup$ @user41467 That's it, with this extra detail: $O(k)\times O(n-k)$ is closed because it's compact and it is compact because both $O(k)$ and $O(n-k)$ are compact. $\endgroup$ – José Carlos Santos Feb 27 '18 at 10:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.