# Multiplications on Spheres (Jie Wu)

The n-dimensional sphere Sn is the set of unit vectors in the (n+1)-dimensional Euclidean space Rn+1. As a (topological) space, Sn is regarded as a subspace of Rn+1. For instance, S0={-1,1} consists of two points, S1 is the circle and S2 is the usual sphere.

200 years ago people developed group theory in mathematics. Roughly speaking, a group means a set in which there is a binary operation (so-called multiplication) with the properties 1) there is an identity element, that is, there is an element, denoted by 1, such that 1x=x1=x; 2) this binary operation is associative, that is, (ab)c=a(bc) and 3) the inverse exists, that is, for each element x, there is another element, denoted by x-1, such that xx-1=x-1x=1.

A topological group G means a (topological) space G with a continuous multiplication on G such that G is group under this multiplication and the function which sends x to x-1 is continuous. For instance, the space of orthogonal matrices is a topological group, where the multiplication is given by the matrix multiplication. The Euclidean space is a topological group under the vector addition.

An H-space (Hopf space) means a space X with a continuous multiplication on X such that this multiplication has the identity element. In other words, for an H-space, the multiplication need NOT be associative and we do not require that the inverse exists too. A topological group is certainly an H-space, but conversely it may not be true in general.

### Question 1. Is Sn a topological group?

When n=0,1,3, the answer is YES. The multiplication on S1 and S3 are induced by the product of complex numbers and quaternion numbers, respectively, where we regard complex numbers as 2-dimnesional vectors and quaternion numbers as 4-dimnesional vectors. For instance, when n=1, first we need to identify S1 with complex numbers of length 1. Then we use the product of two complex numbers to define the multiplication on S1. The point here is that the product of two complex numbers of length 1 is still of length 1. The multiplication on S3 follows from the same ideas, but you may have to read some references on quaternion numbers to see how to define product of 4-dimensional vectors.

### Question 2. Is Sn an H-space?

In addition to the cases above, the answer is YES when n=7. The multiplication on S7 is induced by the product of Cayley numbers. Again you may have to read references on Cayley numbers to see how to define product of 8-dimensional vectors.

The questions above are called Hopf invariant one problem. Adams solved this problem in 1950s in his famous paper. Homotopy theory was shown her powers in solving this problem. Adams' answer is as follows.

Theorem 1: Sn is a topological group if and only if n=0,1,3.

Theorem 2: Sn is an H-space if and only if n=0,1,3,7

Theorem 2 tells that if n is not equal to 0,1,3,7, there is no way to make a continuous multiplication on Sn such that Sn has the identity element under this multiplication. As a good exercise, you may try to look at why we could not make a continuous multiplication on the sphere S2. You may regard S2 as adding the infinite point to complex numbers. For sure we can define a multiplication on complex numbers. Say vector addition. However whatever multiplication defined on complex numbers is impossible continuously extended to S2.

Theorem 1 tells that S7 is never a topological group under whatever multiplication. In fact, whatever multiplication defined on S7 with identity is never associative. Homotopy theory also tells that whatever multiplication defined on S3 with identity is never commutative.

Now you may want to know how to prove the theorems above. Adams' ideas are as follows: By assuming that Sn admits a multiplication with identity, one gets certain "operations" on so-called homology of certain space. After a long discussion, Adams obtained a contradiction when n is not equals to 0,1,3,7. Thus he proved Theorem 2.

For Theorem 1, homotopy theory shows that any topological group admits a so-called classifying space. By assuming that S7 adimits a topological group structure, one gets a classifying space for S7. Again homotopy theory shows that this is impossible by considering so-called the cohomology of the resulting classifying space.