Are there good product rules on the $k$-sphere? - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T21:14:50Z http://mathoverflow.net/feeds/question/33084 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/33084/are-there-good-product-rules-on-the-k-sphere Are there good product rules on the $k$-sphere? Mircea 2010-07-23T12:10:49Z 2010-10-01T00:54:33Z <p>I have heard sometimes that the only dimensions $k$ for which there exists a "<em>good</em>" 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).</p> <p>I would like to ask for references about such a result. </p> <p>More precisely, I am interested in finding out how is it that one can define "<em>good</em>" so that the above is true (with proofs, preferably).</p> http://mathoverflow.net/questions/33084/are-there-good-product-rules-on-the-k-sphere/33085#33085 Answer by Eivind Dahl for Are there good product rules on the $k$-sphere? Eivind Dahl 2010-07-23T12:29:36Z 2010-07-23T17:24:47Z <p>'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.</p> <p>(EDIT: I'd just post this as a comment, but not enough reputation.)</p> http://mathoverflow.net/questions/33084/are-there-good-product-rules-on-the-k-sphere/33086#33086 Answer by rpotrie for Are there good product rules on the $k$-sphere? rpotrie 2010-07-23T12:32:34Z 2010-07-23T12:55:38Z <p>I guess that what you want to say is that for $k=0, 1,3, 7$ the sphere $S^k$ is a <em>H-space</em> (as in the comment).</p> <p>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).</p> <p>See Theorem 2.16 of the following <a href="http://www.math.cornell.edu/~hatcher/VBKT/VBdoublepage.pdf" rel="nofollow">nice book</a>. </p> http://mathoverflow.net/questions/33084/are-there-good-product-rules-on-the-k-sphere/33094#33094 Answer by S. Carnahan for Are there good product rules on the $k$-sphere? S. Carnahan 2010-07-23T13:52:21Z 2010-07-23T13:52:21Z <p>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.</p> <p>You can find proofs in most books on (topological) K-theory.</p> http://mathoverflow.net/questions/33084/are-there-good-product-rules-on-the-k-sphere/33119#33119 Answer by Tom Goodwillie for Are there good product rules on the $k$-sphere? Tom Goodwillie 2010-07-23T18:09:37Z 2010-10-01T00:54:33Z <p>$S^0$, $S^1$, 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.</p> <p>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. </p> <p>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.</p> <p>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. </p>