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I am reading a text in which, at some point, the author define the blow up of $\mathbb{C}^2$ at $(0,0)$ as "a complex manifold $\hat{\mathbb{C}}^2$ obtained by identifying two copies of $\mathbb{C}^2$ in the following way: $$ (x,t)\sim(s,y)\iff s=\dfrac{1}{t};y=tx $$ for $t\neq 0,s\neq 0$, where $(x,t)$ and $(s,y)$ are the coordinates of the mentioned copies.

I cannot understand this definition. Can someone explain to me what it means?

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What part don't you understand? –  Qiaochu Yuan Dec 12 '12 at 23:26
Well, in what sense does this relation defines a manifold? How does this relates to the origin $(0,0)$? –  Marra Dec 12 '12 at 23:36
The author is taking the quotient by the equivalence relation. –  Qiaochu Yuan Dec 12 '12 at 23:38
Oh. I'm not very proficient with manifolds; does an equivalence relation define a manifold? –  Marra Dec 12 '12 at 23:41
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2 Answers

up vote 4 down vote accepted

An alternative definition of the given blow-up is the following. Take $\mathbb{C}^2\times\mathbb{CP}^1$ with coordinates $((z,w),[s:t])$ and take the submanifold $X=\{zt=ws\}$. We can restrict the projection $\pi:\mathbb{C}^2\times\mathbb{CP}^1\to\mathbb{C}^2$ to $X$, obtaining $\pi_X:X\to\mathbb{C}^2$.

We note that $\pi_1((z,w),[s:t])=(z,w)$, but if $z\neq0$ (or $w\neq 0$), then $zt=ws$ implies that $t=ws/z$ (or $s=zt/w$), hence $[s:t]$ is determined by $(z,w)$ if $(z,w)\neq(0,0)$. That is to say, $\pi_X:X\setminus \pi^{-1}((0,0))\to\mathbb{C}^2\setminus\{(0,0)\}$ is a bijection (and actually a biholomorphism).

What is the set $\pi^{-1}((0,0))$? Well, if $z=w=0$, then the equation $zt=ws$ is true for every choice of $[s:t]\in\mathbb{CP}^1$, hence $\pi^{-1}((0,0))=\mathbb{CP}^1$.

Now, let us show that this is the same as your definition (the affine part, at least): consider the mapping $F:X\to\mathbb{CP}^2\times\mathbb{CP}^2$ given by $$F((z,w),[s:t])=\{[zs:t:s],[s:tw:t])\}$$ then, as $zt=ws$, we have that $w=zt/s$. Hence, the image of $F$ in the affine set $s,t\neq 0$ is given by the equations $s=1/t$ and $y=tx$.

The last space, $F(X)$, can be easily seen to be biholomorphic to the gluing you defined in the question: take the maps $f_1, f_2:\mathbb{C}^2\to\mathbb{CP}^2\times\mathbb{CP}^2$ given by $f_1(a,b)=([a:b:1],[1:ab^2:b])$ and $f_2(a,b)=([a^2b:1:a], [a:b:1])$; they are charts for $F(X)$, indeed they are injective and holomorphic and $f_2^{-1}f_1(a,b)=(1/b, ab)$, if $b\neq0$. Every manifold is obtained by gluing the charts with the relations induced by the transition functions, therefore, the described quotient is exactly the blow-up we described in the beginning.

As for the geometric meaning, the blow-up of $\mathbb{C}^2$ at a point can be viewed as the following operation: we erase the point and in some way we separate the directions stemming from the point, by inserting, in the place of the point, the manifold (a $1$-dimensional manifold in this case) corresponding to all the possible directions from the point (which is, actually, the normal bundle of the point as a submanifold of $\mathbb{C}^2$); such a set of directions corresponds to $\mathbb{CP}^1$ (indeed, as the line at infinity of $\mathbb{CP}^2$, the complex projective line is exactly the possible directions of a line in the complex projective plane).

You could imagine it as the plane spiraling up, like an helicoidal shape, but in the end the fiber over the origin (the center of the helicoid has to be compact... so it is actually a bit hard to figure in your mind.

Please, if this helped you, feel free to ask if you want me to be more specific on some aspect.

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Well now, I think I'm starting to understand it. So the blowup is a manifold defined on the set $\mathbb{C}^2\times \mathbb{CP}^1$. Correct me if I'm wrong, but, as a line tends to $(0,0)$ in $\mathbb{C}^2$, we are "forcing" its value, at zero, to be its "complex inclination" as the curve tends to infinity? –  Marra Dec 13 '12 at 0:35
Oooh no, ignore what I said above, I think I got this. –  Marra Dec 13 '12 at 1:09
The blow-up is a sub manifold of $\mathbb{C}^2\times\mathbb{CP}^1$ together with its (of the submanifold) projection onto $\mathbb{C}^2$.An example: $C=\{z^3+z^2-w^2=0\}$ in $\mathbb{C}^2$ is singular in the origin ($3z^2+2z=0$ and $2w=0$ yield $z=w=0$), now consider $\pi:X\to\mathbb{C}^2$ and take $\pi^{-1}(C\setminus\{(0,0)\})=C'$. The closure $\overline{C}'$ is a smooth complex submanifold of $X$ (it is given by the equations $s^3w+ts^2-t^3=0$ and $zt=ws$). The idea is that the two branches at the origin have been separated in the blow-up, having different tangent directions. –  wisefool Dec 13 '12 at 15:16
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Maybe this will help.

You can think about the blowup in the following way. It is a space $X$ with a surjective map $\pi:X\to \mathbb C^2$ which away from the origin (and in general away from the blowup locus) is an isomorphism. In other words, above every point that is not the origin in $\mathbb C^2$ has precisely one preimage under $\pi$.

At the origin is where the action happens. Above $(0,0)$ you should see an entire copy 0f $\mathbb P^1$. The way I like to think about this is that you're parametrizing lines through the origin by their slope.

The description you cite should line up with this.

You can also think about the blowup $X$ as a subset of $\mathbb C^2\times \mathbb P^1$. You can be really concrete with this in coordinates if you want, since we have coordinates on both $\mathbb C^2$ and $\mathbb P^1$. So $X = \{((x_0,x_1),(y_0:y_1)): x_0y_1=x_1y_0\}$. This also lines up with the description you gave.

Yes this is a complex manifold of dimension, in fact its also an algebraic variety. So yeah, the upshot is that when you blowup you're changing the geometry of your space near the center of the blowup (in this case the origin) but away from this so-called exceptional locus, you have an isomorphism.

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