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Let $O(n)$ be the isometry group of an inner product space on $\mathbb{R}^n$ with the induced norm $||\cdot||$ and induced metric $d$. As a curiosity, is it possible to show that $O(n)$ has exactly two connected components and to characterize these, without using determinants?

One possible approach could be based on the Cartan-Dieudonne Theorem: we could try to show that isometries requiring an at most even vs. at most odd number of reflections across hyperplanes form two disjoint classes. There's a constructive proof of the theorem that doesn't use determinants here by Aragon-Gonzalez et al. Though I'm rather lost how to exactly carry out this procedure.

user
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  • You could use another homomorphism $O(n)\rightarrow \mathbb{R}^{\times}$, different from the determinant. For example, all determinant-functions $\Delta$, with $\Delta(I_n)\neq 1$. I suppose you would not like this, but it is different from "the" determinant. – Dietrich Burde Nov 15 '17 at 19:36
  • @DietrichBurde well if you could show how to construct $\Delta$ without using $\det$ in its definition, that'd be a perfect answer! – user Nov 15 '17 at 20:27
  • Just take axioms $2$ and $3$ only, at this question, ignoring $1$. Then you obtain $\Delta$, by deleting the word "determinant" everywhere. It is then no longer unique, of course. But you can enforce this again, by saying $\Delta(I_n)=3$, for example. Then it is still different from "the" determinant you want to exclude. Still, I fear, you will not be happy. – Dietrich Burde Nov 15 '17 at 20:41
  • Well we know we're using $\det|_E$ with $E = End(V) \setminus {id(\cdot)}$, because the theorem you linked proves that those properties uniquely characterize $\det|_E$, which is equivalent to using $\det$ in a piecewise definition of $\Delta$. – user Nov 15 '17 at 21:25
  • Well, no, I don't think so, but perhaps this doesn't matter for the question. It is not clear to me what without exactly means. Is it literally, or should we avoid all kinds of determinant-like constructions? This seems pointless, or as you said, just a "curiosity". – Dietrich Burde Nov 15 '17 at 23:23
  • By without, I mean avoiding $\det$, all its equivalent characterizations, and all other objects defined using these, while proceeding with the decomposition. Succeeding with the decomposition this way will necessarily give rise to a characterization of $\det$. Well yes, it's just a curiosity. – user Nov 16 '17 at 01:53
  • Do you allow an arbitrary metric? If $d$ is the discrete metric ($d(x,x)=0$, $d(x,y)=1$ otherwise), every bijection is an isometry. – celtschk Nov 16 '17 at 02:44
  • @celtschk Of course, thanks; I edited the question to clarify that $d$ should be obtained from a norm. – user Nov 16 '17 at 08:09
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    You certainly want not only a norm but a norm coming from an inner product; the isometry groups of other norms are genuinely different groups, with different numbers of connected components. – Qiaochu Yuan Nov 16 '17 at 08:41
  • @QiaochuYuan thanks, I fixed that in my question, that's a very important point. – user Nov 16 '17 at 10:34

1 Answers1

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Here's one approach. By the spectral theorem, every orthogonal matrix $X$ has an orthonormal basis of eigenvectors $v_i$ over $\mathbb{C}$. By considering the action of $X$ on pairs $\{ v_i, \overline{v_i} \}$ you can show that this implies that $X$ is conjugate, in $O(n)$, to a block matrix whose blocks are $2 \times 2$ matrices in $O(2)$ or, when $n$ is odd, a $1 \times 1$ block with entries $\pm 1 \in O(1)$.

Now $O(1) = \{ \pm 1 \}$ certainly two connected components. It's also not hard to see that $O(2)$ has two connected components given by the rotations and the reflections. Now we have a natural invariant of a block matrix as above to consider, which is the number of reflections that appear in it. You can show that a direct sum of two reflections is conjugate to a direct sum of two rotations, and consequently when $n \ge 3$ there are two cases:

  • An even number of reflections. In this case $X$ is conjugate to a direct sum of rotations and hence is itself a rotation, so lies in the connected component of the identity: said another way, it lies in the image of the exponential map $\mathfrak{o}(n) \to O(n)$.
  • An odd number of reflections. In this case $X$ is conjugate to a direct sum of rotations and a single reflection, so is connected to every other reflection (because we can multiply $X$ by a suitable rotation to connect it to a diagonal matrix with an odd number of $-1$s, and all of these can be connected to each other). Hence the connected component of the identity has exactly one other coset as desired.

Edit (9/11/19): Mea culpa; this argument is incomplete. It doesn't show that the reflections are actually disconnected from the rotations, just that there are at most two connected components. We can argue that the image of the exponential map generates the connected component of the identity, then try to show that the product of two rotations is another rotation, but I don't see a clean way to do this off the top of my head.

Here's another approach. The idea is to use inductively the action of $O(n)$ on the sphere $S^{n-1}$, which produces a sequence of fiber bundles

$$O(n-1) \to O(n) \to S^{n-1}.$$

Associated to each of these is a long exact sequence in homotopy

$$\dots \to \pi_1(S^{n-1}) \to \pi_0(O(n-1)) \to \pi_0(O(n)) \to \pi_0(S^{n-1})$$

which, using the fact that $S^{n-1}$ is simply connected for $n \ge 3$, implies that the map $\pi_0(O(n-1)) \to \pi_0(O(n))$ is an isomorphism for $n \ge 3$. Thus, as above, it suffices to check the number of connected components of $O(1)$ and $O(2)$ to get the result in general. (Usually these long exact sequences are used to compute the fundamental group and higher homotopy groups of the $O(n)$.)

Qiaochu Yuan
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