A topological space is said to be connected if it cannot be written as $X=X_1\cup X_2$, where $X_1,X_2$ are both open and $X_1\cap X_2=\emptyset$. Otherwise, X is called disconnected.

Is it wrong to write $$O(3)=SO(3)\cup O(3)^{det=-1}$$ where $O(3)^{det=-1}$ are those elements of $O(3)$ with determinant $=-1$.

EDIT : 1. Should the disconnectedness be understood in terms of non-existence of a continuous path in the manifold that interpolates between these two sets?

  1. Is there a way to show that, starting from identity, by a continuous change of group parameters, along a certain path in the manifold, I can reach all elements of $SO(3)$ but not those of $O(3)^{det=-1}$.

3 Answers 3


Here is a very easy mathematical proof that $O(n,\mathbb{R})$ i.e set of all $n\times n$ real orthogonal matrices is disconnected.

Standard Result 1: $ f: M_n(\mathbb{R})\to \mathbb{R}$ given by $f(A)=\det A$ is continuous map.

Standard Result 2: $f:A\to B$ continuous map, if $A$ is connected or $X\subseteq A$ is connected then $f(A)$ is connected and also $f(X)$ is connected in $B$

Now consider $O(n,\mathbb{R})\subseteq M_n(\mathbb{R})$ and suppose it is connected set. so by $\det$ map from $M_n(\mathbb{R})\to \mathbb{R}$ image set must be connected in $\mathbb{R}$, but what is the image set under $\det$ map of $O(n,\mathbb{R})$? it is just $\{+1,-1\}$ which is a disconnected set so we get a contradiction.

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    $\begingroup$ +1 great, succinct answer. You probably already know this but may be worth emphasising: one needs to be wary of one thing: one either (1) needs to use the result that $O(n,\,\mathbb{R})\subset M_n(\mathbb{R})$ is a topological embedding so that $O(n,\,\mathbb{R})$ gets the relative topology (so that standard result 1 survives the restriction of its domain from $M_n(\mathbb{R})$ to $O(n,\,\mathbb{R})$ or (2) prove result 1 from first principles for $O(n,\,\mathbb{R})$. Not all matrix Lie groups are topological embeddings in $M_n(\mathbb{R})$ (although $\det$ is still cts for the examples ... $\endgroup$ Oct 6, 2014 at 1:26
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    $\begingroup$ ...I know of: e.g. the irrational slope one parameter subgroup of the torues). It's not a problem here, as there is only one possible topology a compact, simple Lie group like $O(n,\,\mathbb{R})$ can have and still be a Lie group: see my account of a famous result be van der Waerden here for example. $\endgroup$ Oct 6, 2014 at 1:26
  • $\begingroup$ @WetSavannaAnimalakaRodVance Thanks,thanks for the info :-) $\endgroup$
    – Marso
    Oct 6, 2014 at 6:56

No, it is not wrong to write that, you're spot on the mark; therefore, your conclusion is right. Your example has an interesting generalisation beyond $O(3)$ and indeed beyone Lie groups as the following is true for all topological groups. For a topological group $\mathfrak{G}$, the "identity component" $\mathfrak{G}_\mathrm{id}$ (i.e. the connected component of the group with the identity in it) is always:

  1. Clopen (closed and open at once), therefore, by the connectedness argument, the whole group $\mathfrak{G}$ is connected if and only if $\mathfrak{G}_\mathrm{id}$ is a proper subset of $\mathfrak{G}$;
  2. A normal subgroup of $\mathfrak{G}$, therefore the connected components of $\mathfrak{G}$ are precisely $\mathfrak{G}_\mathrm{id}$ and its cosets.

See the discussion as Theorem 9.31 on my website here of this situation. Mine is a retelling of this slick little proof that I found a long time ago in

Sagle, A. A. and Walde, R. E., “Introduction to Lie Groups and Lie Algebras“, Academic Press, New York, 1973. §3.3

This is widely known, BTW, but it's a little pearl of a proof, showing how powerful the connectedness argument is. I still enjoy reading it, just like I still like listening to "Walk Like an Egyptian" (for me, they come from about the same time of life!)

In your example, $\mathfrak{G}=O(3)$, the identity component is the smallest Lie group containing $\exp(\mathfrak{g})$, where $\mathfrak{g}=\operatorname{Lie}(\mathfrak{G}) = \mathfrak{so}(3)$; indeed, in this compact case, $\mathfrak{G}_\mathrm{id}=\exp(\mathfrak{g})$ is the whole connected component (in noncompact groups, e.g. $SL(2,\mathbb{C})$, the identity component can be strictly bigger than $\exp(\mathfrak{g})$). $\mathfrak{so}(3)$ is of course the Lie algebra of skew-symmetric, real $3\times 3$ matrices and, since $\det(\exp(H)) = \exp(\mathrm{tr}(H))=1$ for any such matrix $H$, we see that $SO(3)$ is the whole of $\mathfrak{G}_\mathrm{id}$. It is a normal subgroup of $O(3)$, and the group of cosets is simply $O(3)/SO(3)\cong\{+1,\,-1\}\cong\mathbb{Z}_2$, which is another way of writing your decomposition.

Answer to edit 1: For manifolds, which are locally Euclidean (locally homeomorphic to $\mathbb{R}^N$) path connectedness and connectedness are the same notion. Moreover, path connectedness always implies connectedness (to prove this, assume otherwise and let $\alpha,\,\beta\in\mathbb{X}$ belong to separate connected components $\mathbb{U},\,\mathbb{U}^\sim$ linked by path $\sigma:[0,\,1]\to\mathbb{X}$ where $\mathbb{X}$ is the topological space in question and $\mathbb{X}=\mathbb{U}\bigcup\mathbb{U}^\sim$. By assumption (path connectedness), $\sigma$ is continuous when $[0,\,1]$ has its wonted topology. But $\mathbb{U},\,\mathbb{U}^\sim$ are disjoint, therefore so is $\mathbb{V}=\sigma([0,\,1])\bigcap\mathbb{X}$, so the inverse image $\sigma^{-1}(\mathbb{V})$, namely $[0,\,1]$, must also be the union of disjoint open sets, contradicting the known connectedness of $[0,\,1]$). However, not all connected topological spaces are path connected (look up the weird "topologist's Sine Curve" as a counterexample).

Answer to Edit 2. We have actually already done this above, because every matrix in $SO(3)$ can be written as a $\exp(H)$, where $H\in\mathfrak{so}(3)$, so you can take your path to be $\sigma:[0,\,1]\to SO(3);\;\sigma(\tau) = e^{\tau\,H}$. To prove that every $SO(3)$ matrix can be written in this was from first principles, simply witness that $\gamma\in SO(3)$ is normal, i.e. commutes with its Hermitian transpose and thus always has a diagonalisation with orthonormal eigenvectors, so $\exists U\ni\,\gamma=U\,\Lambda\,U^\dagger$, where $\Lambda$ is the diagonal matrix of eigenvalues, none of which are nought ($SO(3)$ is a group). Thus, you can always define $\log\gamma = H = U\,\log\Lambda\,U^\dagger$, and you are done.

Note that this does not work for $O(3)$, because now $H$, even though definable as $H = \log \gamma$, $H$ now has diagonal elements if $\gamma\not\in SO(3)$, so $H\not\in\mathfrak{so}(3)$ and you can't find a path through $O(3)$'s charts, because $\mathfrak{so}(3)$ is the Lie algebra of $O(3)$ as well.

  • $\begingroup$ The question I asked is wrong as pointed out by Red Act, I think. I will modify the question. $\endgroup$
    – SRS
    Oct 5, 2014 at 5:38
  • $\begingroup$ @Roopam I've modified my answer $\endgroup$ Oct 5, 2014 at 6:27

A topological space is connected if it cannot be written as $X=X_1\cup X_2$, where $X_1,X_2$ are both open and $X_1\cap X_2=\emptyset$. But you've just written $O(3)$ as a union, and

$$SO(3)\cap O(3)^{det=-1} =\emptyset\ ,$$

so $O(3)$ is not a connected topological space, due to that overlooked word "cannot" in the definition.

  • $\begingroup$ You are right. I had something else in mind. I will modify the post. $\endgroup$
    – SRS
    Oct 5, 2014 at 5:28

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