# Why $O(n)$ has exactly two connected components? [duplicate]

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It is well know that $SO(n)$ is connected and $O(n)$ has two connected components: $O^+(n)=\{A\in O(n):\det A=+1\}$ and $O^-(n)=\{A\in O(n):\det A=-1\}$.

In what book can I find this property?

## marked as duplicate by Dietrich Burde, Shailesh, Leucippus, TastyRomeo, C. FalconJan 6 '17 at 1:11

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• I would think most books that introduce $O(n)$ and $SO(n)$ proves it. – Arthur Jan 5 '17 at 20:42
• You can also find it here at MSE. Start here and follow the references. Nice answers, as detailed as in a book, are also here. Actually, $O(n)$ and $GL_n(\mathbb{R})$ are homotopy equivalent, and have the same number of connected components then. – Dietrich Burde Jan 5 '17 at 20:50

## Why $O(n)$ has exactly two connected components?

Suppose $A \in O(n)$. Then, by definition, $A^tA=\mathbb{I}$, where $A^t$ represents the transpose of the matrix $A$ and $\mathbb{I}$ represents the identity. Taking the determinant of this equation, we get

\begin{align*} \text{det}(A^t)\text{det}(A) &=1 \\ \Rightarrow (\text{det}(A))^2 &= 1 \\ \Rightarrow \text{det}(A) &= \pm 1 \tag{1}\\ \end{align*}

Therefore, we see that $$O(n)=O^-(n)\ \coprod \ O^+(n) \tag{2}$$ where $\coprod$ stands for disjoint union.

Now we know that the determinant is a continuous map. Therefore, the pre-image of an open set in $\mathbb{R}$ must be open in $O(n)$. Consider two disjoint open intervals $B_\epsilon(-1)\equiv(-1-\epsilon,-1+\epsilon)$ and $B_\epsilon(+1)\equiv(1-\epsilon,1+\epsilon)$ for some small $\epsilon>0$.

By continuity of the determinant, $\text{det}^{-1}(B_\epsilon(-1))$ and $\text{det}^{-1}(B_\epsilon(+1))$ are open in $O(n)$. But observe that, we have precisely:

$$\text{det}^{-1}(B_\epsilon(-1))=O^-(n) \tag{3}$$ $$\text{det}^{-1}(B_\epsilon(+1))=O^+(n) \tag{4}$$

Therefore, (2), (3) and (4) prove that $O(n)$ is not connected and that there are at least two connected components of $O(n)$. In order to prove that $O(n)$ has exactly two connected components, one further needs to prove that $O^\pm (n)$ are connected, in fact arcwise-connected.

## In what book can I find this property?

I would recommend Matrices: Theory and Applications by Denis Serre. Read Chapter 10 where he discusses all this in great detail. Moreover, the book is very beginner-friendly I guess. If not, it is very readable nonetheless.

• Your proof has helped me a lot. Many thanks! – FUUNK1000 Jan 5 '17 at 22:39