§ Hairy ball theorem from Sperner's Lemma (TODO)
- Let $\Delta$ be an n-dimensional simplex with vertices $v_0, v_1, \dots, v_n$.
- Let $\Delta_i$ be the face opposite to vertex $v_i$. That is, $\Delta_i$ is the face with all vertices except $v_i$.
- The boundary $\partial \Delta$ is the union of all the $n+1$ faces of $\Delta_i$ (i is from $0$ to $n$).
- Let $\Delta$ be subdivided into smaller simplicies forming a simplciial complex $S$.
- Sperner's lemma : Let the vertices of $S$ be labelled by $\phi: S \rightarrow \Delta$ (that is, it maps all vertices of the simplicial complex $S$ to one of the vertices of the simplex $\Delta$), such that $v \in \Delta_i \implies \phi(v) \neq i$. Then there is at least one $n$-dimensional simplices of $S$ whose image is $\Delta$ (That is, there is at least one n-dimensional-sub-simplex $T \subseteq S$such that vertices of $T$ are mapped to $\{0, 1, \dots, n\}$). More strongly, the number of such sub-simplices is odd .
- We can see that the map $\phi$ looks like some sort of retract that maps the complex $S$ to its boundary $\Delta$. Then Sperner's lemma tells us that there is one "region" $T \subseteq S$ that gets mapped onto $\Delta$.
§ 1D proof of Sperner's: Proof by cohomology
- For 1D, assume we have a line with vertex set $V$ and edges $E$. Let the vertex at the beginning be $v_0$ and the vertx at the end be $v_1$. That is, $\Delta \equiv \{v_0, v_1\}$and $S \equiv (V, E)$ is a subcomplex of $\Delta$ --- that is, it subdivides the line $\Delta$into smaller portions. Let $\phi: S \rightarrow \Delta$ be the labelling function.
- create a function $f: \Delta \rightarrow \mathbb F_2$ that assigns $0$ to $v_0$ and $+1$ to $v_1$: $f(v_0) \equiv 0; f(v_1) \equiv 1$. Use this to generate a function on the full complex $F: S \rightarrow F_2; F(v) \equiv F(\phi(v))$.
- From $F$, generate a function on the edges $dF: E \rightarrow F_2; dF(\overline{vw}) = F(w) + F(v)$. See that this scores such that $dF(AB) = +1$, $dF(BA) = +1$, $dF(AA) = dF(BB) = 0$. (Recall that the arithmetic is over $F_2$) So, $dF$ adds a one every time we switch from $A$ to $B$ or from $B$ to $A$.
- However, we also see that $dF$ is generated from a "potential function "f". Hence we have the identity $\sum_{e \in E} dF(e) = f(v_1) - f(v_0) = 1 - 0 = 1$. Hence, we must have switched signs an odd number of times.
- Since we start from $A$, that means we must have switched from $A$ to $B$ an odd number of times.
§ 2D proof of Sperner's: Proof by search
- Start from an edge in the bottom $ef$ labeled $BC$. We are looking for a simplex labeled $ABC$.
- To start: Pick some vertex above $ef$, say $g$. If this is labeled $A$, we are done. If not, say this is labeled $B$. So we get triangle $efg=ABB$. Launch our search procedure from this triangle $efb$.
- Find the triangle adjacent to $efg$ along the edge $eg=AB$ (the other $AB$ edge, not the one we started with). If this adjacent triangle $egh$ has $h=A$ we are done. If not, move to the triangle $egh$.
- See that we will either find a triangle labeled $ABC$, or we will keep running into triangles labeled $ABB$.
- We cannot ever repeat a triangle in our path; to repeat a triangle is to start with some edge $xy$and then to pick a vertex $z$ such that $xyz=efg$ where $efg$ was already picked. This must mean that the edge $ef$ was already picked. [TODO ]
§ 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 $\mathbb R$.
However, it is difficult to exhibit a basis. Naively, for each point $p \in M$, we would
need a basis $T_p B \subset T_p M$ 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 $\mathbb R^2$, given some vector field $V \equiv v_x \hat x + v_y \hat y$,
the functions $v_x$ and $v_y$ allow us to write basis! Create the vector fields $V_x \equiv \hat x$
and $V_y \equiv \hat y$. Then any vector field $V$ can be written as $V = v_x V_x + v_y V_y$
for functions $v_x, v_y$ 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 $\hat x$) and "upwards" (analogous $\hat y$),
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 $K$ be a field. Consider $R \equiv K \times K$ as a ring, and let $M \equiv K$be a module on top of $R$.
- $M$ is a projective module because $M \oplus K \simeq R$(that is, we can direct sum something onto it to get the some $\oplus_i R$)
- On the other hand, $M$ itself is not free because $M \neq \oplus_i R$ for any $i$. Intuitively, $M$ is "half an $R$" as $M \simeq K$ while $R \simeq K\times K$.
- The geometric picture is that we have a space with two points $\{p, q\}$. We have a bundle on top of it, with $M$ sitting on $p$ and $0$ (the trivial module) sitting on top of $q$. When we restrict to $p$, we have a good bundle $M$.
- But in total over thr space, we can't write the bundle as $M \times \{p, q\}$ because the fibers have different dimensions ! The dimension over $p$ is $dim(M) = 1$ while over $q$is $dim(0) = 0$.
- What we can do is to "complete" the bundle by adding a copy of $M$ over $q$, so that we can then trivialise the bundle to write $M \times \{p, q\}$.
- 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.