Intuition for Blow-up.

Question

If I blow up a complex manifold along a submanifold, can you give me a picture to have in mind for the blown-up manifold? Can you also tell me why this is the right picture?

Answer 1

The following is more or less the description you can find in Griffiths and Harris's Principles of Algebraic Geometry on page 182.

For the case of a point in a complex manifold, the idea is to take a local neighborhood homeomorphic to a disc $\Delta$ in $\mathbb{C}^n$ centered at 0, and take the projection $\pi: \tilde{\Delta} \longrightarrow \Delta$ where $\tilde{\Delta} = \{(z,l) | z_il_j = z_j l_i \,\forall i,j \}\subset \mathbb{C}^n\times \mathbb{P}^{n-1}$ where $z \in \Delta$ and $l\in \mathbb{P}^{n-1}$. (If you have trouble seeing this as a manifold, perhaps recall that there is an embedding of $\mathbb{P}^n\times \mathbb{P}^m$ into $\mathbb{P}^{(n+1)(m+1)-1}$ and work out the defining equations in that space).

Away from $z = 0$ the projection $(z,l) \mapsto z$ is going to be one-to-one. In fact it is a homeomorphism.

However at $z=0$ we see that $\pi^{-1}(0) = \{ (0,l)\} \ \cong \mathbb{P}^{n-1}$ since of course $0=0$. Now the trick is to understand how lines through $z=0$ in $\Delta$ lift to $\tilde{\Delta}$ at $z=0$. To do this, take the limit of the preimage of a point travelling along a line in $\Delta$ towards $0$. You will see that it goes to $(0,l)$ where $l$ is the equivalence class of the line.

Explicitly, the line has equation $t(a_1,...,a_n)$ for $a_i\in \mathbb{C}$ not all zero and $t\in \mathbb{C}$. If $t\neq 0$ then $\pi^{-1}(t(a_1,...,a_n)) = (t(a_1,...,a_n),[a_1:...:a_n])$. The limit as $t\rightarrow 0$ is clearly (0,[a_1:...:a_n]) in $\tilde\Delta$ and 0 in $\Delta$.

If we have a curve $C$ through $0$ in the manifold, we define the total transform of $C$ to be the homeomorphic preimage of $\pi^{-1}(C-\{0\})$ plus the points in the fibre over $0$ that correspond to the different angles at which $C$ approaches $0$. In the zariski topology this is the closure of $\pi^{-1}(C-\{0\})$ (since these points are the limits of points in the preimage, as i described above).

To make the blow up of the manifold, one attaches $\tilde\Delta$ to $\Delta$ away from $z=0$ by the homeomorphism. Away from $0$, the other charts remain the same.

Here is some pictures from an undergrad paper, I think this helps get an intuition for how blowing up separates the slopes at 0. Here we have a node $y^2 -x^2(1+x) = 0$ and a cusp $x^2 − y^3 = 0$ in $\mathbb{C^2}$ (be careful since this is only the real picture). In the first case, the blow up separates the curve going through $0$ by taking the preimage of $0$ to two points corresponding to the slopes of the curve through $0$. In the second case, the curve approaches 0 from one direction.

Note that the resulting total transforms are not singular.

Here is another picture of the same thing with a local picture of the blow up of the disc, which you can find in this great paper.

For the case of a higher dimensional submanifold, the intuition remains the same. As you can see from the wikipedia article, it is defined locally by equations that are the same as the blow up of a dimension 0 submanifold. You are taking a projection $\tilde M \longrightarrow M$ that is a homeomorphism everywhere except at the submanifold, and when you lift a curve that intersects the submanifold, we define the points in the preimage of the submanifold to be the ones corresponding to the slope at which the curve intersects it.

1 http://math.berkeley.edu/~aboocher/emma.pdf (picture)

Answer 2

Not a picture and not a complex manifold, but I hope this helps. Take the curve $y^2 = x^2 (1 + x)$ in $\mathbb{R}^2$. It has a singularity (self-intersection) at the origin. The blow up embeds the curve in $\mathbb{R}^3$ and detaches the curve at the singularity: think about it as picking up one branch and lifting it above the other so that they don't intersect anymore. We are "blowing up" the curve at the singularity.