Why does a blowup of $\mathbb{A}^2$ about the origin result in the Möbius band?

Question

I am new to learning about blowups, so I was searching for an easy example to wrap my mind around the concept. I came across this document https://homepage.univie.ac.at/herwig.hauser/Publications/ggt05-hauser.pdf and on the top of page 5, it gives a brief example of a blowup of $\mathbb{A}^2$. It says, “for $Z=0$ the origin in the real plane $\mathbb{R}^2$, we obtain (cum grano salis) for $\tilde{\mathbb{A}^2}$ the Möbius band in $\mathbb{R}^3$.” Here, the blowup is being done with $Z=0$ as its center.

I‘m not seeing why this is what the blowup should be, but it’s entirely possible that can be attributed to my lack of current understanding on the topic. If someone could explain it, it would be much appreciated! I would also love to hear other “simple/beginner” examples of blowups.

Answer 1

The subset $\{(r\cos\theta, r\sin\theta, \theta) \mid r\geq 0, \theta \in [0, \pi) \} \subset \mathbb{R}^2\times \bigl([0, \pi]/ (0=\pi) \bigr)$ projecting down to the first two coordinates exhibits this subset as a blowup of $\mathbb{R}^2$ at the origin. This is the `double spiral staircase' formed by a line rotating around the origin from $0$ to $\pi$, and the top and the bottom identified, thus the Mobius band.

Answer 2

Intuitively, you are replacing the point at which you blow up with "the straight lines passing through that point" with the direction in which you follow the line not part of the data. Since you are on $\mathbb{A}^2$ this corresponds kind of to enlarge the point to a circle and then quotienting the circle over the antipodal map. Thus, with one go around the circle you would wrap twice around your blowup, which you can see as the fact that you need to "go around the boundary of the Moebius strip twice to come back to the starting point".

Notice however that the blow up is not compact. Hopefully, someone with a deeper understanding in algebraic geometry than me can explain better the technicalities involved.