Let $V_1$ and $V_2$ be two distinct smooth subvarieties of the smooth variety $X$ which are regularly embedded. I would like to find a reasonable criteria which guaranties the smoothness of the Blow-up $\widetilde{X}$ of $X$ along the union of $V_1$ and $V_2$.

For example, is $\widetilde{X}$ smooth if $V_1$ and $V_2$ meet transversally?

  • $\begingroup$ Take two lines in $\mathbb{P}^3$. Then the blow up is smooth if and only if the lines do not intersect. $\endgroup$ – J.C. Ottem Nov 19 '10 at 13:50
  • $\begingroup$ In the example of two lines in a threefold these subvarities don't intersect transversally and the blow-up is not expected to be smooth. I would like to see a non-trivial example or a useful criteria for the smoothness of the blow-up. $\endgroup$ – Passenger Nov 19 '10 at 14:14
  • 1
    $\begingroup$ It seems to me that with more or less the same proof you can show that if a subvariety $V$ of a smooth variety $X$ is complete intersection, then the blow-up of $X$ along the (reduced) ideal of $V$ is smooth if and only if $V$ is smooth $\endgroup$ – Francesco Polizzi Nov 19 '10 at 14:36
  • $\begingroup$ Francesco, do you mean a subvariety of codimension more than one? $\endgroup$ – roy smith Nov 19 '10 at 14:51
  • $\begingroup$ @roy Yes, of course I was thinking in codimension at least 2. Thank you! $\endgroup$ – Francesco Polizzi Nov 19 '10 at 16:50

Appearently, the blow-up IS smooth if $V_1$ and $V_2$ intersect transversally. In this case we have that $$ Bl_{V_1 \cup V_2} X = Bl_{\bar{V_1}}Bl_{V_2}X =Bl_{\bar{V_2}}Bl_{V_1}X $$where $\bar{V_i}$ denotes the proper transform of $V_i$. This is essentially Proposition 2.9 in Kiem and Moon's article http://arxiv.org/abs/1002.2461.

| cite | improve this answer | |

In general the answer is no.

Take $X=\mathbb{A}^3$ with coordinates $x,y,z$ and let $V_1$ and $V_2$ be two lines meeting in one point, for instance

$V_1:=\{x=y=0\}, \quad V_2:= \{x=z=0\}$.

Then the ideal of $V_1 \cup V_2$ is $I=(x, yz)$ and the equation of the blow-up $\widetilde{X}$ of $X$ along $I$ are given in $X \times \mathbb{P}^1$ by

$\lambda x - \mu yz=0$,

where $[\lambda : \mu]$ are homogeneous coordinates in $\mathbb{P}^1$. In the chart $\mu=1$ the blow-up is then given by

$\textrm{Spec }k[x,y,z, \lambda]/(\lambda x - yz)$,

hence it has an isolated singularity at the origin.

The other chart $\lambda=1$ is instead smooth, so this is actually the unique singular point of $\widetilde{X}$.

| cite | improve this answer | |
  • $\begingroup$ If I'm not completely mistaken the blow-up here is still Cohen-Macaulay. Is there a counterexample where the blow-up is not CM? $\endgroup$ – J.C. Ottem Nov 19 '10 at 15:28
  • $\begingroup$ In this example the blow-up has a unique isolated hypersurface singularity, in particular it is Gorenstein (hence Cohen-Macauley). Any birational morphism $f \colon Y \to X$ can be obtained as the blow up of some sheaf of ideals $I$ in $X$, so a priori everything can happen. However, the ideal can be horrible. Actually, I do not know if there exists any example with non-CM blow-up in the situation proposed in the question... $\endgroup$ – Francesco Polizzi Nov 19 '10 at 17:12
  • 4
    $\begingroup$ This example shows that maybe we do not want to call V_1 and V_2 transversal. $\endgroup$ – O.R. Nov 29 '10 at 13:41

Let $V$ be a finite-dimensional complex vector space, let $A$ be a subspace, and let $\alpha:V\to V/A$ be the projection. The blow-up of $PV$ along $PA$ can then be identified with

$\{(L,M)\in PV\times P(V/A): L\leq \alpha^{-1}(M)\}$

Now suppose we have another subspace $B$. I think that the blow-up along $PA\cup PB$ is just the fibre product of the blow-ups along $PA$ and $PB$, namely

$X=\{(L,M,N)\in PV\times P(V/A)\times P(V/B) : L\leq \alpha^{-1}(M)\cap\beta^{-1}(N)\}$

We have $PA\cap PB=P(A\cap B)$, and this intersection is transverse iff $A+B=V$. If so, then the map $(\alpha,\beta):V\to V/A\times V/B$ is surjective, so the spaces $\alpha^{-1}(M)\times\beta^{-1}(N)=(\alpha,\beta)^{-1}(M\times N)$ form a vector bundle over $P(V/A)\times P(V/B)$, whose associated projective bundle is $X$; this shows that $X$ is smooth. Thus, the question has an affirmative answer at least for linear subspaces of projective space.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.