Spin Manifolds

Disclaimer: This exposition is profane. It is the result of me trying to be productive during a raging headache, and ending up with comic relief.

I recommend this to be read when you are looking for a laugh, or already feeling silly.

The motivation for the field of differential geometry can be thought of as a large group of punk math students screaming:

“We won’t wait for the world to let us study change! We want to differentiate on whatever we want! Give us more interesting manifolds than just \(\mathbb{R}^n\). Fuck the Euclidean system!”

A rough and ready definition of an \(n\)-dimensional manifold is a topological space which looks locally like what the government wants \(\mathbb{R}^n\). With this in mind, my punk math friends, let’s figure out how to study change.

First, how the hell do we assemble the tangent spaces at various points of a manifold into a coherent whole?

I hope we can agree that the total derivative of a \(C^{\infty}\) function should change in a \(C^{\infty}\) manner from point to point.

We create this differentiability for functions over our manifold M by considering behavior in the overlaps of coordinate neighborhoods.

Consider an arbitrary point \(p \in M\), and two distinct neighborhoods \(U_\alpha\) and \(U_\beta\) which both contain \(p\).

Screenshot from 2014-11-28 18:05:51

The transition map \(\tau_{\alpha, \beta}:= \psi_\alpha^{-1}\psi_\beta\) (fuck you I’ll put composition in any order I want, it’s a free country)

is the type \(\tau_{\alpha, \beta}: \mathbb{R}^n \to \mathbb{R}^n\). This map better be \(C^\infty\) (differentiable arbitrarily often), or you’re going to have a bad time doing calculus on that motherfucker.

There are many sorts of manifolds, with different requirements for continuity of these transition maps.

In our case, we want to do calculus without a hitch, so we’ll insist that all of our transition functions must be \(C^\infty\), and refer to such manifolds as “smooth”.

We’re working in a category, where our

  • objects = smooth manifolds
  • morphisms = \(C^\infty\)-maps between these smooth motherfuckers

We call this category “Diff”, which is the category-cool version of saying “let’s do some differential topology up in this bitch.”

When you hear “Let’s do calculus” your “We’re working in Diff!” alarm bells should go off.

Alright, I’m getting off track. How do break out of the cage imposed by the Euclidean system? Let’s start with some tangent bundles.

That’s right kids, it’s a fucking fiber bundle!

Our manifold \(M\) is the base space, and the fiber (over each point \(p\) in \(M\)) is just the tangent plane over that point, which we’ll denote as \(T_p(M)\) because I’m lazy and don’t want to type “the tangent plane over the point \(p\) in our manifold \(M\)” each goddamn time.

If you’re visual thinker like me, here are some pictures to make you feel good about yourself.

tangent3 tangent4 img8

Oooh, ahhhh. So shiny.

Alright, I’m going to keep being lazy and give some more shorthand names, claiming that it’s for clarity:

Let \(V^n\) := the set of all column vectors of height \(n\)
and \(U\) := an open subset in \(\mathbb{R}^n\)

Recall that we’re working with a fiber bundle, the tangent bundle \(T(M) = U \times V^n\)

The base space is our manifold \(M\)
the fibers are our tangent plane \(T_p(M) = p \times V^n\)

Of course, this tells you jack shit about how to actually compute the tangent plane given any point on a surface, but you can go look in any standard multivariable calc book for that stuff.

Motherfucking Modules

Think the tangent bundle is general as fuck? You obviously haven’t read enough Grothendieck. The tangent bundle is an example of a motherfucking module bundle.

A module bundle exactly what you think it is:
a bundle with fibers that are motherfucking modules.

These are usually called vector bundles, but my ring-theorist-friend convinced me that I have to use modules all the time to get an intuition for them, so I’m trying that. In his words: who needs inverses anyway when we can formally append them. I’m not used to modules yet, so I’ll probably commit the sin of switching between modules and vector spaces as we go.

But I’m getting ahead of myself, talking about module bundles before defining a module. Let’s back the fuck up.

What is a motherfucking module?

a vector space – requiring property of invertibility = a module

If you’re feeling like a tight-ass today, here’s the rigorous-as-fuck (RAF) definition of a module in stuff, structure, properties form (aka Baez-style).

Let \(R\) be a ring and \(1_R\) be its multiplicative identity.

(stuff): A left \(R\)-module \(V\) consists of an abelian group \((V, +)\)

(structure): and an operation \(R \times V \to V\) ; aka we’re closed under this bitch.

(properties): such that \(\forall r,s \in R\) and \(x, y \in V\)

distributive as fuck:

  1. \(r(x+y) = rx + ry\)
  2. \((r + s)x = rx + sx\)

associative as fuck:

  1. \((rs)x = r(sx)\)

and of course, the multiplicative identity holds, because rings are not pathological little fucks (okay, fine maybe the ring of integers of \(\mathbb{Q}(\sqrt{-5})\) are little fucks but they still obey this property)

  1. \(1_Rx = x\)

BUNDLES of Motherfucking Modules

We can look at these module bundles in at least 2 ways. More if you get creative, but you heard what happened in Don’t Hug Me I’m Scared.

1. We have a module space assigned to each point in a manifold in such a way that as we move smoothly over the manifold, the module twists smoothly.

2. Locally a module bundle is trivial, so what we care about are the transition functions. These must lie in the automorphism group of our module.

What properties do we need to slap on these motherfuckers to make them play nice with spin representations?

Our manifold \(M\) must satisfy:

  1. orientable
  2. Riemmanian
  3. 2nd Stiefel-Whitney class vanishes

If all of these conditions hold, we say \(M\) is spin, because who the fuck wants to write “manifold that is orientable, Riemannian, and whose 2nd Stiefel-Whitney class vanishes” every time they talk about slapping a spin structure on a manifold.

Go forward unabashed, even if you don’t know what these words mean. I’m about to tell you!


We want our bundle to be orientable, which we can define as a continuation of our previous module bundle definitions:

1. If we can choose a frame which varies smoothly over the manifold (and is always a frame), our bundle is orientable.

2. If our transition functions all lie in the special orthogonal group, SO(n), our bundle is orientable.


We need our manifold’s structure group, SO(n), to lift to spin. If it lifts to spin, we can describe the Dirac operator globally!


But you shouldn’t fucking believe everything I say — I hope you’re asking yourself “what the fuck does that even mean?”

What the fuck is a spin GROUP?

Praise the Flyng Spaghetti Monster, the concept of a Spin group is hella geometric.

Below are illustrations of the \(n\)-dimensional Spin group, \(\text{Spin}(n)\), as a subobject of their corresponding \(C\ell\) algebra:

hackpad.com_aNjuUb3nCm4_p (1)

What the fuck is a spin STRUCTURE?

The structure group of any principal \(G\)-bundle \(E \to B\) is just \(G\).

It should come as no surprise that a spin structure on our manifold \(M\) is composed of

  • a principal \(Spin(M)\)-bundle, \(\text{Spin}(M) \to M\)
  • a bundle morphism \(\psi\) from \(\text{Spin}(M) \to M\) to \(SO(M) \to M\)

Screenshot from 2014-10-04 22:35:31

which restricts fiberwise to the covering homomorphism \(\text{Spin}(M) \to \text{SO}(M)\).


We study the notion of curvature and its relation to topology under the name ‘Riemannian geometry’, which is where most of the ‘geometry’ in ‘differential geometry’ comes out to play.

Don’t forget that we’re in \(\text{Diff}\)! Our manifolds are smooth motherfuckers.

Slap a Riemannian metric on that motherfucker and we get smoothly varying choices of inner product on tangent spaces.

Fuck yes! Let’s introduce a Riemannian metric on the manifold. In other words, let us pass from \(GL_n\) to the maximal compact subgroup \(O_n\).

orientation = reducing the structure group from \(O_n\) to \(SO_n\)
spin structure = then lifting the structure group to the universal covering group \(Spin_n \to SO_n\).

When does the structure group lift to Spin?

Recall that transition functions must satisfy the cocycle condition.

Thus, lifts of the transition functions must also satisfy this condition. When this is possible, we can attach a spin bundle by specifying that its transition functions = the lifts of the transition functions of the cotangent bundle.

Henceforth, I’ll assume that you motherfuckers are cool with homology and homotopy. I’m also hella tired right now, so the rest of this is a bit wibbly.

All I’m tryna say is: the structure group lifts Spin to if there is no obstruction.

There is no obstruction if the 1st and 2nd Stiefel-Whitney classes vanish (i.e. \(w_1 = w_2 = 0\)).

Don’t take my word for it! You should be asking: what the hell are these classes and what do they have to do with structure groups?

The conditions \(w_1\) and \(w_2\) can be interpreted geometrically as follows.

Let \(E\) be a vector bundle over a manifold \(M\). Then \(E\) is orientable iff the restriction of \(E\) to any circle embedded in \(M\) is trivial.

If \(M\) is simply-connected and \(\text{dim}(M) < 4\), then \(E\) is spin iff the restriction of \(E\) to any 2-sphere embedded in \(M\) is trivial.

Why? \(H_2(M, Z/2)\) is generated by embedded 2-spheres (when \(\pi_1 = 0\) and \(\text{dim}(M) > 4\)).

Let \(O_n\) be the orthogonal group. Before I go, I want to tell you something enticing.

  1. \(\pi_1(O_n) = \mathbb{Z}/2\) — orientations and \(w_1\) (1-connected cover)
  2. \(\pi_2(O_n) = \mathbb{Z}/2\) — spin structure and \(w_2\) (3-connected cover)
  3. \(\pi_3(O_n) = 0\)
  4. \(\pi_4(O_n) = \mathbb{Z}\) — string structure and \(p_1/2\)

Let’s say that \(X\) and \(Y\) are some smooth motherfuckin’ manifolds. When is a map \(X \xrightarrow{f} Y\) null-homotopic?

When that shit LIFTS. \(f\) better induce a zero map between the homotopy groups.

\(H^n(X; G) \simeq [X; K(G,n)]\)

If this isomorphism isn’t up your alley, then I highly suggest you revaluate your life decisions because it is sick as fuck! If you’ve never seen it before, have some John Baez!

\(Y_1 \to B^2(\pi_2(Y_1)) = B^2\pi_2(Y)\)

\(c \in H^2(Y_1, \pi_2(Y))\)

\(Y_n\) lifts to \(Y_{n+1}\) if a cohomology class in \(H_{n+1}(X, \pi_{n+1}(Y))\).

“I’m not good at math”

A 10 year old girl and her father sat in the back of my car as I drove them home after Thanksgiving.

During this snippet of the conversation, I felt like I was punched in the chest.


Do you like literature?

Yeah … all the time when I space out.

Do you like literature?

I love stories!

Do you like spelling?

No, I’m bad at spelling.

What makes you say that?

I don’t understand why letters go next
to each other in the order they do.

Imagine that you were taught spelling, and not shown any stories.
Do think you would like literature?

If I didn’t see the stories, how could I like literature?

What you’re learning right now via memorization,
that is to math as spelling is to literature.

What do you mean?

Do you imagine pictures when you read?

Yeah, I love stories!

Do you like patterns?

I like patterns!


They’re pretty!

I love patterns too. I spend all of my time
imagining pictures and moving shapes just like you do.

I thought you did math.

Moving, stretching, and constructing shapes is a form of math.
It’s called geometry.

Whoa! Really? I’m learning geometry in class,
but we just memorize the formulas for volume.

Memorizing formulas is like spelling practice instead of literature.

Can you be my tutor for real geometry? For 4 hours everyday!

*Her dad: No, sweetie, she’s probably very busy.*

I can give your dad the information of a few people who would be better qualified than me
to teach you the literature of shapes.

*She falls asleep*

I forgot to mention to her (wrt her worry of being bad at multiplication) that two of the greatest mathematicians of all time said they were unable to add without mistakes.

As for myself, I must confess, I am absolutely incapable even of adding without mistakes.
— Jules Henri Poincaré

I’ve always been weak in arithmetic.
— Alexander Grothendieck

Notes on Covering Spaces as Extensions

This post assumes knowledge of fiber bundles, the group action functor, groupoids, and basic vector calculus. I am in the process of learning the topics discussed below, and I deeply appreciate constructive feedback.

How does a big space cover a little one?

Given a covering space \(E \to B\) we can uniquely lift any path in the base space (once you choose a starting point) to a path in \(E\). Conversely, we can create a covering space of \(B\) by letting the fiber over \(b\) be \(F(b)\).

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Formally: If \(B\) satisfies a few topological constraints*, the functor \(\text{Fiber}\) is an equivalence of categories

\(\text{Fiber}: \text{Cov}/B \to \text{Set}^{\Pi_1(B)}\)

between the category of covering spaces \(\text{Cov}/B\) and the functor category \(\text{Set}^{\Pi_1(B)}\).

*\(B\) is locally path-connected and semi-locally simply-connected

Covering spaces \(E \to B\) are classified by functors, \(F: \Pi_1(B) \to \text{Set}\). The fundamental groupoid is the group of automorphisms of the appropriate fiber functor. A fiber functor is a forgetful functor from the category of finite sets with \(G\)-action to the category of finite sets.

Sidenote: \(\infty\)-groupoids = spaces up to homotopy. Thus, we can think of \(\Pi_1\) as capturing the information of a space up to 1-homotopy, and truncating the information about higher homotopies.

What is the analogue of a covering space for groupoids?

It’s called a discrete fibration. This is a functor \(p: E \to B\) that satisfies the unique path lifting lemma (for any morphism \(f: x \to y\) in \(B\) and \(\tilde{x} \in E\) lifting \(x\), there’s a unique morphism \(\tilde{f}: \tilde{x} \to \tilde{y}\)).

Sheaves and fibrations are generalizations of the notion of fiber bundles. Discrete fibrations \(E \to B\) are also classified by functors \(B \to \text{Set}\).

As curious persons, we like to take things apart to see how they work.

We can convert a decomposition of a space in terms of simple pieces \(\xrightarrow{into}\) a collection of vector spaces (or modules) and linear transformations (or homomorphisms).

Let’s take something simple apart to get an intuition we can abstract off of: our favourite trivial fiber bundle, the cylinder.

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The sequence \(0 \to F \to E \to B \to 0\) is an example of a chain complex.

The Extension Problem: \(0 \to F \to ? \to B \to 0\)

The extension problem is essentially the question: Given the end terms \(F\) and \(B\) of a short exact sequence, what possibilities exist for the middle term \(E\)?

Screenshot from 2014-10-29 22:19:42

Extensions of the group \(B\) by the group \(F\), that is, short exact sequences \(1 \to F \to E \to B \to 1\), are classified by 2-functor from \(B \to \text{2Aut}(F)\).

Examples of this are the more familiar classifications of central group extensions using \(H^2\) or \(\text{Ext}\).

What does this category \(\text{2Aut}(F)\) look like?

  • objects: \(F\)
  • morphisms: automorphisms of \(F\)
    Screenshot from 2014-10-29 23:00:00
  • 2-morphisms: morphisms between the automorphisms of \(F\)
    Screenshot from 2014-10-29 22:59:18

For more on this, I recommend Dr. Baez’s Lectures on n-Categories and Cohomology, which will explains why “…generalizing the fundamental principle of Galois theory to fibrations where everything is a group gives a beautiful classification of group extensions in terms of nonabelian cohomology.”

Excited yet? Let’s look at chain complexes a bit more carefully.

If you have vector calculus running through your veins and at your fingertips, then you’re already familiar with a differential complex:

\(\mathbb{H_1} \xrightarrow{grad} \mathbb{H}_{curl} \xrightarrow{curl} \mathbb{H}_{div} \xrightarrow{div} \mathbb{L}_2\)

(where \(\mathbb{H}_{curl}, \mathbb{H}_{div}\) are the domains for the curl and div operators respectively.)

You’ll note that the composition of any two consecutive maps is zero. This is a key point of the definition of a chain complex (a sequence of modules connected by homomorphisms such that the composition of any two consecutive maps it zero).

In the category of groups, this is equivalent to the question: What groups \(B\) have \(A\) as a normal subgroup and \(C\) as the corresponding factor group?

Screenshot from 2014-10-26 21:08:36
\(ker(f)=0\), \(im(f) = ker(g)\), \(im(g) = C\), Source

In general, for a short exact sequence: \(0 \to A \xrightarrow{f} B \xrightarrow{g} C \to 0\)

\(A\) is a subobject of \(B\), and the corresponding quotient is isomorphic to \(C\):

\(C \cong B/f(A)\) , (where \(f(A) = im(f)\)).

What are these 0’s for?

0 represents the zero object, such as the trivial group or a zero-dimensional vector space. The placement of the 0’s forces \(f\) to be a monomorphism and \(g\) to be an epimorphism.

These chain complexes generalize past where our geometric intuition can follow, but it’s nice to be grounded in examples.

“Two objects \(X\) and \(Y\) are connected by a map \(f\) between them. Homological algebra studies the relation, induced by the map \(f\), between chain complexes associated with \(X\) and \(Y\) and their homology. This is generalized to the case of several objects and maps connecting them.” – Source

Curious as to why we need the the composition of consequent maps to be 0 and how this fits into homology? I recommend Dr. Ghrist’s notes on Homology from Elementary Applied Topology.

How can a big commutative algebra define a little one?

Classifying how a big space will cover a little one amount to classifying how a little commutative algebra can “sit inside” a big one.

We can study the ways a little thing \(k\) can site inside a bigger thing \(K\) (\(k \hookrightarrow K\)) by keeping track of the symmetries of \(K\) that fix \(k\). These “fixing symmetries” form a subgroup of the symmetries of K. \(Gal(K/k) \subseteq Aut(K)\).

Let \(k\) and \(K\) fields such that \(k \hookrightarrow K\).

In this case, \(K\) is called an i>extension field of \(k\) (denoted \(K/k\)).

I’m going to interject here and give a bit of context:

Field extensions are fiber bundles with zero-dimensional fibers (covering spaces).

A fiber bundle with zero-dimensional fibers has a total space that “gives multiplicity” to the points of the base space.

(This means that a multiple-valued function on the base space has a good chance of being interpretable as a single-valued function on the total space.)

Screenshot from 2014-10-26 21:28:26

But now I’m getting ahead of myself, without grounding you in classical commutative algebra!

Now that you have a glimpse of the importance of studying extensions, let’s get back to those:

Let \(Aut(K/k)\) be the set of all \(k\)-automorphisms of \(k\)

\(Aut(K/k) = {\sigma in Aut(K) : \sigma\vert_k = Id_k}\)

But why is the restriction of \(\sigma\) to \(k\) an automorphism of \(k\)?

If \(k\) is a splitting field of a family \(\mathcal{P}\) of polynomials, then any action \(\sigma\) preserves the set of roots of \(\mathcal{P}\), and since \(k\) is generated by these roots, the action \(\sigma\) preserves \(k\) (i.e. \(\sigma k = k\)).

Then \(Aut(K/k)\) is a group, the automorphism group of \(K/k\), or the Galois group of \(K/k\), denoted as \(Gal(K/k)\).

the category \(\text{Fields}\):

  • objects: Fields, i.e. \(F_1, F_2\)
  • morphisms: Field extensions, \(F_1 \xrightarrow{\phi} F_2 := F_2/F_1\) (note that morphims are all injective)

The Fundamental Theorem of Galois: Given a polynomial, the intermediate fields of the splitting field are in one-to-one correspondence with the subgroups of the Galois group.

Suppose \(G\) acts transitively on \(X\). Pick any figure \(x\) of type \(X\) and let H be its “stabilizer”: the subgroup consisting of all guys in \(G\) that map \(x\) to itself. Then we get a one-to-one and onto map

\(f: X \to G/H\)

sending each figure \(gx\) in \(X\) to the equivalence class \([g]\) in \(G/H\).

We can use this to set up a correspondence between sets on which \(G\) acts transitively and subgroups of \(G\). This is one of the principles lurking behind Galois theory

— John Baez, The Tale of Groupoidification

If we’re looking for amazing applications of Galois theory within, say, arithmetic, then we might as well read Kronecker or Hecke.

If we interpret Galois theory in a very expansive way, then the Erlangen Program, and Cartanian geometry, are legitimate consequences.

Thanks to David Yang and Semon Rezchikov for helping me correct my visualization of a category of representations, and thanks to Aaron Slipper for teaching me basic Galois Theory!


Classifying Spaces Made Easy
Lectures on n-Categories and Cohom.
Motivic Spectral Sequence
Topology for Physicists
Covering Spaces
Homology: Elementary Applied Topology
Exact Sequence

Postscript: We have the beginnings of a map of things that can be extended to define other things:

Fields \(\leftrightarrow\) Groups
Base Spaces \(\leftrightarrow\) Covering Spaces
Groupoids \(\leftrightarrow\) Discrete Fibrations
Principle Fiber Bundles \(\leftrightarrow\) Structure Groups