When do blowups ''commute''? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T07:49:17Z http://mathoverflow.net/feeds/question/106590 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/106590/when-do-blowups-commute When do blowups ''commute''? Spinorbundle 2012-09-07T10:29:13Z 2012-09-07T19:51:34Z <p>Let $M$ be a manifold (variety, scheme, your favorite object) and let $N_1,N_2$ be two submanifolds (subvarieties, closed subschemes, ideal sheafes, etc.) such that $N_1 \cap N_2 \neq \emptyset$. Denote by $\text{BL}_{N_i}M$ the blowup of $M$ along $N_i, i=1,2$ and let $\text{BL}_{N_1}\text{BL}_{N_2}M$ be the blowup of $\text{BL}_{N_1}M$ along the proper transform $N_2^\prime$ of $N_2$. (Define $\text{BL}_{N_2}\text{BL}_{N_1}M$ vice versa). </p> <blockquote> <p>Question: Under which conditions does the following hold: $\text{BL}_{N_1}\text{BL}_{N_2}M \cong \text{BL}_{N_2}\text{BL}_{N_2}M$ ?</p> </blockquote> <p>This should be true if we are in the situation $N_1 \subseteq N_2 \subseteq M$, since blowups commute with restriction'' but is it in genreal true? If so, is there some reference available? If not, is there a criterium when blow ups commute (e.g. something like transversal intersection or) or/and can you come up with an easy counterexample?</p> http://mathoverflow.net/questions/106590/when-do-blowups-commute/106593#106593 Answer by Sándor Kovács for When do blowups ''commute''? Sándor Kovács 2012-09-07T11:08:26Z 2012-09-07T11:08:26Z <p>This generally fails even under nice circumstances.</p> <p>Let $M$ be a smooth projective variety over $\mathbb C$ of (say) dimension $3$ such that it does not contain a rational curve. For instance an abelian threefold. Let $N_1$ and $N_2$ be two smooth projective curves in $M$ intersecting in (at least) a point and for simplicity assume that the intersection is transversal in exactly one point. Finally, let $B=BL_{N_1}BL_{N_2}M$ and $B'=BL_{N_2}BL_{N_1}M$. </p> <p>The preimage of $N_1$ in $B$, $E_1$ is a $\mathbb P^1$-bundle (ruled surface) over $N_1$ and the preimage of $N_2$, $E_2$ is a $\mathbb P^1$-bundle over $N_2$ blown-up at a point. Similarly the preimage of $N_2$ in $B'$, $E_2'$ is a $\mathbb P^1$-bundle over $N_2$ and the preimage of $N_1$, $E_1'$ is a $\mathbb P^1$-bundle over $N_1$ blown-up at a point. </p> <p>If $B\simeq B'$, then $E_1\simeq E_2'$ and $E_2\simeq E_1'$, but this cannot happen if $N_1\not\simeq N_2$.</p> <hr> <p>Actually, in the case you claim the two operations commute, you've got a problem: If $N_1\subseteq N_2$, then what is the proper transform of $N_1$ on ${BL}_{N_2}$?</p> http://mathoverflow.net/questions/106590/when-do-blowups-commute/106601#106601 Answer by Karl Schwede for When do blowups ''commute''? Karl Schwede 2012-09-07T12:43:27Z 2012-09-07T19:51:34Z <p>Sándor is exactly right, this generally isn't true. However, there is a related statement that is always true, instead of blowing up strict transforms of $N_1$ and $N_2$, you should blow up total transforms.</p> <hr> <h2>Blowing up total transforms</h2> <p>Suppose that $I_1$ and $I_2$ are the ideals defining $N_1$ and $N_2$ in $M$. Let $$Y_1 = Bl_{I_1} M = Bl_{N_1} M.$$ Now define the <em>total transform of</em> $I_2$, denoted $\overline{I_{2}}$, to be the ideal formed by extending $I_2$ to $Y_1$, in other words $\overline{I_2} = I_2 \cdot O_{Y_1}$ (note that by the universal property of blowing up $I_1 \cdot O_{Y_1}$ is an invertible sheaf).</p> <p>$\overline{I_2}$ need not define a manifold, or even a reduced scheme, but we can still blow it up. Set $$Y_{1,2} = Bl_{\overline{I_2}} Y_1.$$ I claim: </p> <p><strong>Theorem:</strong> <em>We have</em> $Y_{1,2} = Y_{2,1}$ <em>where the second object is obtained in the same way but blowing up</em> $I_2$ <em>first.</em></p> <p>There are at least two ways to see this. </p> <ol> <li>You can do this from the universal property of blowing up.</li> <li>Since you already assumed that $M$ is a manifold, it is integral. Thus the charts of a blowup can be computed as follows. Suppose $M = \text{Spec } A$ is affine for simplicity. If $\langle x_1, \dots, x_n\rangle$ generate an ideal $I$, then as $i = 1,...,n$ varies, $Y_{I,i} = \text{Spec } A[x_1/x_i, \dots, x_i/x_i, \dots, x_n/x_i]$ are affine charts covering the blowup. $Y_I = Bl_I Y$.</li> </ol> <p><strong>EDIT:</strong> As Dustin Cartwright points out in a comment below, blowing up $I_1$ followed by blowing up $\overline{I_2}$ is the same as blowing up $I_1 \cdot I_2$. </p> <hr> <h2>A normal crossings example</h2> <p>I'd like to give you an example in $\mathbb{A}^4 = \text{Spec } k[x,y,u,v]$ showing the difference between blowing up strict transforms and total transforms, even when the $N_i$ intersect transversally (are in normal crossings). Let $N_1 = V(x,y)$ be a plane and let $N_2 = V(y,u,v)$ be a line. </p> <p>Suppose we blowup the line $N_2$ first, then the strict transform and the total transform of $N_1$ coincide. In particular, $Y_{2,1}$ is the object you were considering. </p> <p>On the other hand, let's blow up $N_1$ first, in this case the strict transform $\widetilde{N_2}$ of $N_2$ is a line, but the total transform $\overline{N_2}$ of $N_2$ is two lines. $\overline{N_2}$ contains the strict transform but also one of the new $\mathbb{P}^1$'s lying over the origin of $N_1 \subseteq \mathbb{A}^4$. The blowups of the strict and total transform of $N_2$ on $Y_1$ are thus manifestly different. </p> <p>One can check that $Y_{1,2} = Y_{2,1}$ if you did the total transform blowup, and so blowing up the strict transforms does not commute.</p> <p><strong>Note:</strong> If I recall correctly, you can also do the example when two planes are <em>kissing.</em> $N_1 = V(x,y)$, $N_2 = V(u,v)$. The same sort of thing happens, but it's even messier. I've seen this example in various papers on resolution of singularities.</p> <hr> <h2>In the literature</h2> <p>The difference between strict and total transforms plays a key role in modern resolution of singularities algorithms. Even though conceptually strict transforms are much easier to think about, they are much <em>harder</em> to compute or control. Thus modern algorithms tend to use total transforms instead in <em>SOME</em> places, and then peel off any copies of the exceptional you can, while possibly leaving embedded components. </p> <p>See for example: </p> <ol> <li>Section 7 of <a href="http://arxiv.org/pdf/math/0206244v2.pdf" rel="nofollow">This paper by Bravo-Encinas-Villamayor</a></li> <li>Example 2.3 of <a href="http://arxiv.org/pdf/1006.1915.pdf" rel="nofollow">This paper by Howard Thompson</a></li> </ol>