# Are there good product rules on the $k$-sphere?

I have heard sometimes that the only dimensions $k$ for which there exists a "good" smooth product $P:S^k\times S^k\to S^k$ are $k = 0,1,3,7$ (the above products corresponding to $\mathbb Z_2, U(1)\subset\mathbb C$, the product of unit quaternions and of unit Cayley numbers).

I would like to ask for references about such a result.

More precisely, I am interested in finding out how is it that one can define "good" so that the above is true (with proofs, preferably).

## 4 Answers

I guess that what you want to say is that for $k=0, 1,3, 7$ the sphere $S^k$ is a H-space (as in the comment).

Also, Lie groups are a word to look at for having good'' products (better than the above, this works for $S^k$ with $k= 0,1,3$ not $7$, see the comment below).

See Theorem 2.16 of the following nice book.

• $S^7$ is not a Lie group, did you mean H-space? en.wikipedia.org/wiki/H-space – Gjergji Zaimi Jul 23 '10 at 12:47
• Yes, sorry, I will correct that. – rpotrie Jul 23 '10 at 12:53
• Right, the H-space version of the result is extremely convincing. The product law $P$ in the question is "good" if it is (a) continuous, and (b) has an element $1$ such that $a1 = 1a = a$. Or even such that $a \mapsto 1a$ and $a \mapsto a1$ are homotopic to the identity. That's not asking for a whole lot, but it already restricts you to the four examples, up to homotopy. – Greg Kuperberg Jul 23 '10 at 18:24
• dear rpotrie, thanks for the answer and the link. I didn't want to have necessarily Lie groups, but I guess some two-sided identity is quite natural. Why is $S^7$ not a Lie group? – Mircea Feb 4 '11 at 14:09
• Hi, S7 is not a Lie group because the octonions (which are the ones which induce the $H$-space structure) lack of associativity. – rpotrie Feb 7 '11 at 17:36

$S^0$, $S^1$, and $S^3$ have well-known smooth group structures. These can be obtained from well-known bilinear multiplications in $\mathbb R$, $\mathbb R^2$, and $\mathbb R^4$. $S^7$ does not have a smooth group structure; this fact can be obtained from the theory of Lie groups and their Lie algebras. Stronger statement: $S^7$ does not have a continuous group structure; this can be deduced from the previous fact using the deep solution of a Hilbert problem. But $S^7$ does have an $H$-space structure, by which I mean a continuous multiplication law with a two-sided identity. This can be obtained from a (non-associative) bilinear multiplication on $\mathbb R^8$, so it can even be chosen in such a way that multiplication is smooth and multiplication by a fixed element on either side has a smooth inverse.

Outside these dimensions, $S^{n-1}$ has no $H$-space structure. For $n>1$ odd, this is easy using cup products in $S^{n-1}\times S^{n-1}$. For $n$ not a power of $2$, it can be done using Steenrod operations in mod $2$ cohomology. The general case uses $K$-theory.

A corollary is that real division algebras (even in a nonassociative sense) are impossible in dimensions other than $1$, $2$, $4$, and $8$. Another corollary is that no spheres other than those are parallelizable.

The question of how many linearly independent tangent vector fields the $(n-1)$-sphere admits was also settled (by Adams) using more subtle $K$-theory arguments.

One condition we might like for a product on a manifold to be nice is that it admits a 2-sided identity element. Another condition we might like is that left multiplication is nonsingular near the identity. In other words, left multiplication is infinitesimally injective, i.e., the differential of left multiplication takes any nonzero tangent vector at the identity to a nonvanishing vector field. This condition is equivalent to the manifold being parallelizable, and Kervaire, Bott, and Milnor showed that the only parallelizable spheres have dimension 0, 1, 3, and 7.

You can find proofs in most books on (topological) K-theory.

'Good' in the sense of being a lie group. I think this is proven using K-theory. Other people here will have a lot more to say about that.

(EDIT: I'd just post this as a comment, but not enough reputation.)

• I don't think you mean to say field. See rpotrie's answer above. – Mike Skirvin Jul 23 '10 at 12:39