For those of you who have been waiting for me to get from Radon’s theorem to the Colorful Carathéodory theorem and then to Tverberg’s theorem, your patience will be rewarded with today’s post. The proof I will be presenting is due to Karanbir Sarkaria, but the presentation mostly comes form Gil Kilai’s blog post on the result, which in turn is based on the article “Carathéodory’s Theorem, Colorful and Applicable” by Imre Bárány and Shmuel Onn.

To prove Tverberg’s theorem, we will need some nontrivial algebra in the form of tensor products. Recall that if and are respectively - and -dimensional vector spaces over some field , then the tensor product is defined (up to isomorphism) to be the vector space together with a bilinear map , having the following universal property: for any bilinear map , there exists a unique map such that

Let denote the -vector space of all functions . We use the notation to denote the function

We define a bilinear function by

,

where and . I claim that is a basis for . Indeed,

Let be another -vector space and be a bilinear function. Define to be the linear map satisfying

For any , we have that

where the last equality follows from the bilinearity of .

In fact, a pure tensor can be identified as a rank-$latex $ matrices over . Indeed, define the map from into this space by

Theorem.(Tverberg) Let . For any set containing at least points, there exist pairwise-disjoint subsets such that

*Proof. *Let be distinct points in . Without loss of generality, we may assume that . Set , define to be the -dimensional affine space

,

and embed in by . Define a dimensional vector space by taking to be the standard basis elements and setting . Consider the tensor product . We define subsets of by

I claim that . Indeed, since , we have that

for every . The Colorful Carathéodory Theorem tells us that there exists a point , for , such that , so we can write , where and each is of the form . We can use the indices to obtain a partition of defined by

where we allow the be empty. We can rewrite the convex combination for as

Since is a basis for , we have that, for any fixed,

Since , we see that

,

which implies that , for all , since is linearly independent in . Since this equality holds for , we conclude that the vectors in defined by

are all equal. Since the last coordinate of , for , is , we see that . Since , we see that , for every . Scaling by , we obtain that the vector . Defining by deleting the last coordinate of , we see that

,

since all the are equal.