# Why is $O(n;k)$ not connected, and has four connected components? [closed]

Why is $O(n;k)$ not connected and has four connected components when $nk\ge 1$? Here $O(n;k) =\{A\in GL(n+k,\mathbb{R}) \mid A^{T}GA=G\}$

where $G=\begin{pmatrix} 1&&&&&\\ &\ddots& & & &\\ &&1&& &\\ && &-1& &\\ && & &\ddots &\\ && & & &-1 \end{pmatrix}$,

with $n$ instances of $1$, and $k$ instances of $-1$.

## closed as off-topic by Todd Trimble♦Apr 17 '18 at 10:33

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• where n and k are non-zero. If k=0, get an orthogonal group O(n) with two connected components. – user25309 Apr 16 '18 at 12:53
• This was a reasonable question with an interesting and useful answer, you could have migrated it to MathSE instead of closing it. – YCor Apr 16 '18 at 13:11
• Ah, here's the deal: it can't be migrated (blocked from asking). – Todd Trimble Apr 17 '18 at 10:35
• @j.c. thanks for flagging! I suggest to pursue the discussion on meta meta.mathoverflow.net/questions/3699/… – YCor Apr 17 '18 at 10:55
• Since $O(n,k)$ as defined is closed under transposes, the group $O(n,k)\cap O(n+k)=O(n)\times O(k)$ is a maximal compact subgroup of $O(n,k)$. Therefore, the number of connected components of $O(n,k)$ is the same as that of $O(n)\times O(k)$, and this is $2\times 2=4$. – Venkataramana Apr 17 '18 at 11:00

To see that there are at least four connected components, write $A$ blockwise: $$A=\begin{pmatrix} B & C \\ D & E \end{pmatrix},\qquad B\in M_n, E\in M_k.$$ Then $A^TGA=G$ decomposes as three identities, among which $$B^TB=I_n+C^TC,\qquad E^TE=I_k+D^TD.$$ The matrix $B^TB$ is thus larger (in the sense of symmetric matrices) than $I_n$, hence positive definite. There follows that $B$ is always non-singular. The same is true for the block $E$. Now form the map $$f:O(n;k)\rightarrow\{\pm1\}^2,\qquad f(A)=({\rm sgn}(\det B),{\rm sgn}(\det E)).$$ From above, this is a continuous function (the determinants don't vanish). When $nk\ge 1$, it is obviously onto (consider diagonal elements of the group). Hence $O(n;k)$ has at least as many connected components as the target $\{\pm\}^2$, that is four.
To see that there are exactly four connected components, you have to prove that $O(n;k)$ is stable under the polar decomposition. Next prove that $O(n;k)\cap SPD_{n+k}$ is homeomorphic (through the exponential map) to a vector space, and check that $O(n;k)\cap O(n+k)\sim O(n)\times O(k)$. Since $O(n)$ and $O(k)$ have two connected components, you are done.
Remark. The map $f$ defined above is a group homomorphism !