§ Hairy ball theorem from Sperner's Lemma (TODO)

§ 1D proof of Sperner's: Proof by cohomology

§ 2D proof of Sperner's: Proof by search

§ Proof of hairy ball by sperner's lemma [TODO]

§ Why hairy ball is interesting: Projective modules

The reason I care about the hairy ball theorem has to do with vector fields. The idea is to first think of smooth vector fields over a smooth manifold. What algebraic structure do they have? Indeed, they are a vector space over R\mathbb R. However, it is difficult to exhibit a basis. Naively, for each point pMp \in M, we would need a basis TpBTpMT_p B \subset T_p M as a basis. This brings in issues of smoothness, etc. Regardless, it would be uncountable in dimension. On the other hand, let's say we allow ourselves to consider vector fields as modules over the ring of smooth functions on a manifold. That is, we can scale the vector field by a different value at each point. We can hope the ""dimension"" of the module is much smaller. So, for example, if we think of R2\mathbb R^2, given some vector field Vvxx^+vyy^V \equiv v_x \hat x + v_y \hat y, the functions vxv_x and vyv_y allow us to write basis! Create the vector fields Vxx^V_x \equiv \hat x and Vyy^V_y \equiv \hat y. Then any vector field VV can be written as V=vxVx+vyVyV = v_x V_x + v_y V_y for functions vx,vyv_x, v_y in a unique way! However, as we know, not all modules are free. A geometric example of such a phenomenon is the module of vector fields on the sphere. By the hairy ball theorem, any vector field must vanish at at least a single point. So if we try to build a vector field pointing "rightwards" (analogous to x^\hat x) and "upwards" (analogous y^\hat y), these will not be valid smooth vector fields, because they don't vanish! So, we will be forced to take more than two vector fields. But when we do that, we will lose uniqueness of representation. However, all is not lost. The Serre Swan theorem tells us that any such module of vector fields will be a projective module. The sphere gives us a module that is not free. I'm not sure how to show that it's projective.

§ Simple example of projective module that is not free.

  • Let KK be a field. Consider RK×KR \equiv K \times K as a ring, and let MKM \equiv Kbe a module on top of RR.
  • MM is a projective module because MKRM \oplus K \simeq R(that is, we can direct sum something onto it to get the some iR\oplus_i R)
  • On the other hand, MM itself is not free because MiRM \neq \oplus_i R for any ii. Intuitively,MM is "half an RR" as MKM \simeq K while RK×KR \simeq K\times K.
  • The geometric picture is that we have a space with two points {p,q}\{p, q\}. We have a bundleon top of it, with MM sitting on pp and 00 (the trivial module) sitting on top of qq.When we restrict to pp, we have a good bundle MM.
  • But in total over thr space, we can't write the bundle as M×{p,q}M \times \{p, q\} because thefibers have different dimensions! The dimension over pp is dim(M)=1dim(M) = 1 while over qqis dim(0)=0dim(0) = 0.
  • What we can do is to "complete" the bundle by adding a copy of MM over qq, so that we canthen trivialise the bundle to write M×{p,q}M \times \{p, q\}.
  • So, a projective module corresponds to a vector bundle because it locally is like a vector space,but may not be trivialisable due to a difference in dimension, or compatibility, or some such.