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The blow-up $\tilde{X}$ of $\mathbb{A}^3$ at the line $\ell: x_1=x_2=0$ is by definition the closure in $\mathbb{A}^3 \times \mathbb{P}^1$ of the graph of the function $f: \mathbb{A}^3 - \ell \rightarrow \mathbb{P}^1$, that takes a point $(x_1,x_2,x_3)$ to $(x_1:x_2)$. As it turns out, $\tilde{X}$ is the closed set of $\mathbb{A}^3 \times \mathbb{P}^1$ given by the equations $x_1 y_2 = x_2 y_1$. The exceptional set of $\tilde{X}$ is isomorphic to $\mathbb{A}^1 \times \mathbb{P}^1$.

Question: The exceptional set of the blow up of $\mathbb{A}^3$ at a point is isomorphic to $\mathbb{P}^2$, which is interpreted as the directions of all lines in $\mathbb{A}^3$ through that point. What is the analogous interpretation of the blow-up described above? In particular, how can one understand the geometric meaning of its exceptional set?

Arctic Char
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Manos
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  • Here's a related question about visualizing the blow-up of a plane at a point. Maybe it helps. –  Jan 19 '16 at 00:37
  • @Rahul: I am ok with visualizing blowing up at points. What i am trying to understand is blowing up at subvarieties. – Manos Jan 19 '16 at 00:38
  • Is it not the same as the product of the blow-ups of the normal spaces at each point $p\in\ell$? –  Jan 19 '16 at 01:04
  • @Rahul: How do you see that? How do the normal spaces come into play? – Manos Jan 19 '16 at 01:07
  • It seems to me that the exceptional set is the collection of all lines of fixed $x_3$ value through the points $(x_1, x_2, x_3)$ on $l$. (Which may be what Rahul is saying, since this is the collection of all lines in the "orthogonal" space to $l$ at that point.) This makes sense if you consider the "limit" in the blowup as you approach the exceptional divisor from the direction of some line and at some fixed height, analogously to the case of blowing up a point in the plane. – Elle Najt Jan 19 '16 at 01:27
  • @AreaMan: That's a nice explanation, thanks. – Manos Jan 19 '16 at 01:32

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$\newcommand{\Cpx}{\mathbf{C}}\newcommand{\Proj}{\mathbf{P}}$Over the complex numbers, blowing up a smooth variety $Y \subset X$ may be viewed geometrically as the following procedure:

Let $\pi:\nu \to Y$ denote the normal bundle of $Y$ in $X$, and let $E = \Proj(\nu) \to Y$ be its projectivization (i.e., remove the zero section and quotient by the $\Cpx^{\times}$-action given by scalar multiplication in the fibres). There is a tautological line sub-bundle $\tau \subset \pi^{*}\nu \to E$ whose fibre over a point $[p]$ of $\Proj(\nu)$ is the line represented by $[p]$. (That is, $[p]$ represents a normal line to $Y$ at some point of $Y$, and the fibre of $\tau$ at $[p]$ is that normal line to $Y$.)

It's straightforward to check that:

  • The complement of the zero section of $\nu$ is biholomorphic to the complement of the zero section of $\tau$.

  • The total space of $\tau$ is obtained from the total space of $\nu$ by removing the zero section and gluing in the exceptional divisor $E = \Proj(\nu)$.

  • The preimage in total space of $\tau$ of some point $y$ in $Y$ is obtained by blowing up the normal space $\nu_{y}$ at the origin.

Working locally in the horizontal directions, if $Y = \Cpx^{k} \subset X = \Cpx^{n}$, blowing up $Y$ amounts to blowing up the origin in $\Cpx^{n-k}$ and crossing with $\Cpx^{k}$.

Andrew D. Hwang
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  • What is $\pi^* \nu$? (I'm trying to interpret this as a pullback of the bundle $\nu$ via $\pi$, but I'm confused because $pi : \nu \to Y$...) – Elle Najt Jan 19 '16 at 01:40
  • I'm also using $\pi$ to denote projection from $E = \Proj(\nu) \to Y$, so $\pi^{*}\nu$ refers to pulling back $\nu$ over its own projectivization. :) – Andrew D. Hwang Jan 19 '16 at 13:44