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I’m interested in the question for which $n$ the special orthogonal group is homeomorphic to the product

$$ \mathrm{SO}(n) \approx S^{n-1} \times \mathrm{SO}(n-1). $$

Allen Hatcher [1, p. 293 f.] claims (?) that this is true for $n \in \{ 2, 4, 8 \}$ and wrong for all other values (although I’m not sure what is meant by “twisted product”).

Does anyone have a reference where this is done in more detail? Maybe I just didn’t have the right keywords for a proper search.

[1] Algebraic Topology, downloadable at https://www.math.cornell.edu/~hatcher/AT/ATpage.html

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  • $\begingroup$ "I’m not sure what is meant by “twisted product”": as written in the book, look at example 4.55. There is a fibration $SO(n-1) \hookrightarrow SO(n) \twoheadrightarrow S^{n-1}$. $\endgroup$
    – Najib Idrissi
    Commented Oct 6, 2015 at 12:08
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    $\begingroup$ Sorry for the dumb question. What do you mean with $SO(n)\cong S^{n-1}\times SO(n-1)$? Should this be an Isomorphism, homeomorphism or something else? $\endgroup$
    – Oliver Straser
    Commented Oct 6, 2015 at 16:29

4 Answers 4

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I think this is not an answer but a precisation. Assume that the statement is true for some $n$. Then S^{n-1}\times SO(n-1)has a Lie group strcture and in particular a parallelizable tangent bundle. This implies that the tangent bundle of S^{n-1} is parallelizableand this is true if and only if n=2,4,8 (the original reference is //projecteuclid.org/euclid.bams/1183522319). Edit: As explained below (by Bertram Arnold), this implies that the sequence

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

splits, and this split give rise to a section in the frame bundle of $S^{n-1}$, i.e it gives a trivialization of the tangential bundle of $S^{n-1}$. This is true if and only if $n=2,4,8$ (the original reference is https://projecteuclid.org/euclid.bams/1183522319). Now what about the group structure?

a) If $n=2$ then $S^{1}\times SO(1)\cong S^{1}$

b) If $n=3$, then $SO(4)\cong S^{3}\times SO(3)$, in particular SO(4) is diffeomophic to a product of two Lie groups but it isn't a Lie groups isomorphism (see the comments below). (The Lie group structure on $S^{3}$ is the one give by the quaternions with norm $1$)

c) If $n=8$, then $SO(8)\cong S^{7}\times SO(7)$, but it is not longer a product of Lie group . $S^{7}$ may be viewed as the set of octonions with norm 1, it is in particular a quasi Lie group (or non associative Lie group).

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    $\begingroup$ I don't understand how it follows that $S^{n-1}$ is parallelizable if $S^{n-1}\times SO(n-1)$ is - don't you just get that the tangent bundle is stably trivial? For instance, $S^{n-1}\times\mathbb{R}$ is always parallelizable since it is diffeomorphic to $\mathbb{R}^n\setminus\{0\}$. $\endgroup$
    – Bertram Arnold
    Commented Oct 6, 2015 at 22:43
  • $\begingroup$ Isn't $S^{n-1}\times SO(n-1)$ always parallelizable? $\endgroup$
    – PVAL
    Commented Oct 6, 2015 at 23:32
  • $\begingroup$ @PVAL Yes - it is the sum of the pullback of the tangent bundle of $S^{n-1}$ and a trivial bundle, and the first becomes trivial when you add a trivial line bundle to it. user51223, in general you do not get a trivialization of $E/F$ from a trivialization of $E$ and $F$. $\endgroup$
    – Bertram Arnold
    Commented Oct 7, 2015 at 9:04
  • $\begingroup$ I became quite excited about c): does it mean that one can somehow express the group structure of $SO(8)$ via group structure of $SO(7)$ and the quasigroup structure of $S^7$, possibly using some kind of cocycle? $\endgroup$
    – მამუკა ჯიბლაძე
    Commented Oct 8, 2015 at 20:34
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    $\begingroup$ @მამუკაჯიბლაძე Identify $S^3$ with the unit quaternions. Then the map $S^3\times S^3\to SO(4), (v,w)\mapsto v\cdot w^{-1}$ is a two-fold covering. In this picture, the projection $SO(4)\to S^3$ sends $(v,w)$ to $vw^{-1}$, so it is not a group homomorphism. Also, the two Lie groups $SO(4)$ and $SO(3)\times S^3$ are not isomorphic, since the first has an outer automorphism of order 2 (orientation reversal) and the second doesn't. $\endgroup$
    – Bertram Arnold
    Commented Oct 9, 2015 at 12:00
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If this was the case, then $\pi_{n-2}(SO(n))\cong \pi_{n-2}(SO(n-1))$. The first group is in the stable range and hence one of $\mathbb{Z},\mathbb{Z}/2$ or 0; in particular every surjective endomorphism of it is an isomorphism. Now the orthonormal frame bundle of $S^{n-1}$ is a fiber sequence $SO(n-1)\to SO(n) \to S^{n-1}$ which gives a long exact sequence $$ \pi_{n-1}(SO(n))\to\pi_{n-1}(S^{n-1})\to \pi_{n-2}(SO(n-1))\to \pi_{n-2}(SO(n))\to\pi_{n-2}(S^{n-1}). $$ The last group is zero, so the third arrow is a surjection; by what was said before, it is an isomorphism. Hence the first map is surjective, and a preimage of the canonical generator gives a section of the frame bundle, in other words, a trivialization of the tangent bundle of $S^{n-1}$. But it is well-known (see user51223's answer) that this is only possible in the dimensions you have listed above.

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Another line of reasoning: If $SO(n)$ were homeomorphic to the product $S^{n-1}\times SO(n-1)$ then $S^{n-1}$ would be a retract (not deformation retract!) of $SO(n)$, hence $S^{n-1}$ would be an H-space since a retract of an H-space is an H-space, assuming the retract contains the identity element, which we can arrange here by a suitable choice of retraction. Then Adams' theorem restricts $n$ to $2,4,8$.

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This is another formulation of the celebrated Hopf invariant theorem of Adams. If the decomposition holds, then the fibration $$SO(n−1)\to SO(n)\to S^{n−1}$$ must admit a section $S^{n−1}\to SO(n)$ which in particular is nontrivial in homology. Now, compose this with the unstable J-homomorphism $SO(n)\to \Omega^n S^n$. The composition still is nontrivial in $\mathbb{Z}/2$-homology. The outcome then is a spherical class in $H_{n-1}(\Omega^nS^n;\mathbb{Z}/2)$ that is coming from the image of J-homomorphism. But, it is known that the only spherical classes in these spaces, coming from the image of J-homomorphism, are the Hopf invariant one elements, that is this can happen if and only if the adjoint map $S^{2n-1}\to S^n$ is detected by the mod 2 Hopf invariant. This is possible only in the above dimensions according to Adams.

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    $\begingroup$ This is probably a naive question, but is it always true that if $C \cong A \times B$, then any fibration $A \to C \to B$, but where the maps involved are not necessarily the canonical ones, always admits a section? $\endgroup$
    – Najib Idrissi
    Commented Oct 6, 2015 at 12:46
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    $\begingroup$ Nope, not true in general: Take a countable union of copies of $S^1$ and define a map $\cup S^1\rightarrow S^1$ by the double cover $S^1\rightarrow S^1$ on each summand. The fiber is countably many points, the total space is a product of countably many points with $S^1$, but there's no section. $\endgroup$
    – Achim Krause
    Commented Oct 6, 2015 at 13:17
  • $\begingroup$ @NajibIdrissi We known there is a fibration such as $SO(n-1)\to SO(n)\to S^{n-1}$. If you assume that the decomposition holds, then your input data are: the existence of the fibration and the decomposition of its total as the product of fibre and base, by which you to find a section easily. This situation is slightly different from what you have asked and the upcoming answers afterwards. $\endgroup$
    – user51223
    Commented Oct 7, 2015 at 6:03

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