# S^3 (the most basic prime manifold) is prime

So a while ago, I was reading Hatcher’s notes on 3-manifolds. In there, he defines what it means for a manifold to be prime and states, casually, that the 3-sphere $S^3$ is prime. He later says that it follows immediately from Alexander’s Theorem as, and I quote: Every 2-sphere in $S^3$ bounds a 3-ball. And that’s it. Done.

Wait, what?!

Elsewhere, Hatcher expands his above statement: …every 2-sphere in $S^3$ bounds a ball on each side…[and h]ence $S^3$ is prime. Again, though, it isn’t accompanied by anything, and while this is clearly a trivial result, I just couldn’t see it for the longest time…I knew that it followed from a number of things, e.g. the fact that $S^3$ is the identity of the connected sum operation, that $S^3$ is irreducible (and that every irreducible manifold is prime), that one gets the trivial sum $M\# S^3=M$ by splitting along a 2-sphere $S$ in $M^3$ which bounds a 3-ball in $M$, etc. Even so, I didn’t want to leverage some enormous machinery to deduce the smallest of results and what I really wanted was for someone to tell me what I was missing. So I never stopped thinking about this, even after moving forward, until finally – it just clicked!

I figure other people who are as visualization-impaired as I may benefit from seeing this explained in greater depth, so in lieu of typing a blog post containing something new and attention-worthy, I figure I’d share this instead. Details after the break.

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# Algebraic Geometry Observation II: Sheaf Theory

Observation II. Sheaf theory is hard.

Per my earlier entry: Given a smooth complex algebraic variety $X$, I finally manage to track down a semi-manageable definition for the structure sheaf $\mathcal{O}_X$. But here’s the thing with super-abstract definitions of things:

There’s a big difference between “getting them” and understanding them. In this case, I read the definition a few dozen times, annotated the .pdf file by Elliott with what I found, and felt as if I truly “got it”. Then, I read an example regarding the topological space $X=\mathbb{C}^n$ with the claim that $\mathcal{O}_X=\mathbb{C}[x_1,x_2,\ldots,x_n]$

…say what now?!

Finally, I reread a bunch of stuff and found a cool analogy between arbitrary varieties $X$ and the case where $X=M$ is a smooth $k$-times continuously differentiable manifold of dimension $\dim X=n$. This cleared things up for me.

It really is going to be a long summer. Heh.