## The Cup Product of the Klein Bottle ($\mathbb{Z}/2$)

I didn’t want to use a brute force method, so I thought for a while about how to compute the cup product of the Klein bottle. This is what I came up with. I thought I’d share it.

Take coefficients in $\mathbb{Z}/2$. We compute the cup product by computing the intersection form (which defines the intersection product), then Poincare duality.

The Klein bottle is compact and $\mathbb{Z}/2$-orientable, so Poincare duality applies: $$a^* \smile b^* = [a.b][K]$$

where $[K]$ is the fundamental class of the Klein bottle, $[a.b]$ is the number of intersections of the cycles $a$ and $b$ (thought of as sub-manifolds embedded in the Klein bottle). Note that $a^*$ and $b^*$ denote the cocycles corresponding to the cycles $a$ and $b$ respectively.

We view the Klein bottle as the fiber bundle $$S^1 \to K \to S^1$$ Fix a fiber circle, $F$, and denote the base circle as $B$.

By pictoral argument: $F$ has zero transverse intersection with itself, and intersects the base circle $B$ once. By wiggling the base circle by an infinitesimal amount $\epsilon$, we see that $B$ intersects transversely with itself once.

Another perspective:

We may more succinctly state the intersection product as the following matrix:

$$\begin{pmatrix} \beta \smile \beta & \beta \smile \gamma \\ \gamma \smile \beta & \gamma \smile \gamma \end{pmatrix} \simeq \begin{pmatrix} [B.B] & [B.F] \\ [F.B] & [F.F] \end{pmatrix} = \begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix}$$

From the above matrix, we see that $\beta^2 = \beta\gamma = 1[K]$, and $\gamma^2 = 0[K]$. And, of course, $\beta^3 = 0$, because $H^3(K) = 0$. This gives us the ring structure of $H^*(K; \mathbb{Z}/2)$: $$(\mathbb{Z}/2\mathbb{Z})[\beta, \gamma]/(\beta^2 + \beta\gamma, \gamma^2, \beta^3)$$

P.S. If we wanted to be totally rigorous, we’d have to show that the cocycles $\beta$ and $\gamma$ form the basis of the $H^1(K; \mathbb{Z}/2) \simeq \mathbb{Z}/2 \oplus \mathbb{Z}/2)$. Until then, we don’t quite have that the intersection matrix gives us the cup product of $H^*(K; \mathbb{Z}/2)$.

## What is the difference between homotopy and coherent homotopy?

This question had been bugging me for a while, and I have been unable to find a source that is suited to the beginning topologist. Eric Peterson kindly answered this for me, and I found his explanation so astounding that I wish to share on the off chance that you, dear reader, will similarly benefit from this visually rich narrative. All errors are mine and not his.

Coherence is essentially about the existence of diagram categories. For instance, suppose you have some homotopy class $A \wedge A \to A$, which I’d like you to think of as a multiplication map on $A$. You can think of this as a homotopy class

$$S^0 \to F(A \wedge A, A)$$

in the appropriate function spectrum. Then, given any map of this signature, you can Continue reading What is the difference between homotopy and coherent homotopy?

## Spf $E^*[[x]]$: Your walk through a flower garden

Inspired by the extraordinary expository style of Dr. Kazuya Kato, I’ve started reading parts of a (translated) Japanese children’s book when I’m stuck on a tough paper or concept — revisiting the concept with such a dreamlike world in mind usually unfolds an illustrative perspective. A misty world which begs to be put into firm ground via prolonged formal and concrete afterthought.

He embraces that teaching can be poetic and tantalizing, providing not a definition but a deep and creative hint that causes an exploratory shift in perspective, allowing you to walk down the path to the conclusion yourself. I wanted to try to exposit with this philosophy: confusion is expected and encouraged as impetus for reaching understanding. With that in mind, step into your flower garden.

Planted in a line of earth ($\text{Spec }R$)
there are flowers, $C$, whose heads are smooth projective genus 1 curves
with stems that can retract into the ground,
s.t. the flower meets the earth at one point (a marked point).

## Oriented Cobordism Cohomology

Edit: When I say cobordism, I mean oriented cobordism unless stated otherwise. Also note that I accidentally flip-flopped $\Omega^*$ and $\Omega_*$ — $\Omega^n$ should be cobordism classes of maps from manifolds of codimension $n$ to $X$, and  $\Omega_n$ is cobordism classes of maps from manifolds of dimension $n$ to $X$.

Let’s say $M_1, M_2,$ and $X$ are differentiable manifolds. We have a map $f_1$, and a map $f_2$.

Let’s think of a movie: Continue reading Oriented Cobordism Cohomology