Exercise 1964Topology › Topology of Mn(ℝ)L1Added 4 Nov 2026

Compactness of orthogonal and orthogonal groups

#compactness#O_n#SO_n#boundedness

Statement

Show that $O_n(\mathbb{R})$ and $SO_n(\mathbb{R})=\{A\in O_n:\det A=1\}$ are compact subsets of $M_n(\mathbb{R})$ .

Hint

$O_n$ is closed (previous exercise) and bounded: every entry of an orthogonal matrix satisfies $|a_{ij}|\le1$ .

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Solution

Closed. $O_n=\Phi^{-1}(\{0\})$ with $\Phi(A)=A^TA-I_n$ , continuous: closed.

Bounded. For $A\in O_n$ and any row $i$ : $\|(\text{row }i)\|_2=1$ , so $|a_{ij}|\le1$ . Hence $\|A\|_\infty\le1$ : bounded.

Conclusion. $O_n$ is closed and bounded, hence compact (Heine–Borel in $\mathbb{R}^{n^2}$ ). $SO_n$ is closed (as $O_n\cap SL_n$ ) and bounded, hence compact.

Compactness of orthogonal and orthogonal groups

Statement

Solution

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