I would like to know if there is a formula for the relative canonical bundle when performing a blow-up along an "authorized" subvariety. More precisely:

Let $X$ be a projective, irreducible, normal variety and let $Y \subset X$ be a smooth subvariety such that $X$ is normally flat along $Y$. Let $p : Z = Bl_{Y}(X) \rightarrow X$ be the blow-up of $X$ along $Y$. Is there a formula which relates $p^{*} K_X$, $K_Z$ and the exceptional divisor of the blow-up (and maybe something else)? I think that when $Y$ is a point, there is a well-known formula.



2 Answers 2


Franz, to complement Karl's answer let me remark that actually you cannot expect a formula without further conditions. I will use your notation.

If $X$ is singular, $K_X$ may not be Cartier or even $\mathbb Q$-Cartier (i.e., that some non-zero integer multiple would be Cartier). Without this it is hard to make sense of the pull-back $p^*K_X$. There are various ways to deal with this, but at the beginning you might not want to deal with those difficulties. One way to get around this problem is to add a divisor $\Delta$ to $K_X$ so that $K_X+\Delta$ is $\mathbb Q$-Cartier and compare that to $K_Z+\Delta_Z$ where $\Delta_Z=p^{-1}_*\Delta$, the strict transform of $\Delta$ on $Z$. However, then you get slightly different values.

So, anyway, perhaps you would want to assume that $K_X$ is $\mathbb Q$-Cartier before you do anything else. Next, let $E=\cup E_i$ denote the exceptional divisor of $p$ (it is not necessarily irreducible!). Then obviously there exists a formula: $$ K_Z\sim p^*K_X + \sum a_i E_i.\tag{$\star$} $$ Notice that if $K_X$ is itself not Cartier, then the $a_i$ may not be integers. See this for some examples.

Anyway, the point I want to make is that the $a_i$ that may appear are strongly related to the singularities of $X$ along $Y$. You can think of them as a measure of how bad those singularities are: the smaller (usually more negative) the $a_i$ the worse the singularity. The ones that are somewhat nicer and manageable are the ones for which $a_i\geq -1$. For more on this you should look up a survey or a book on the minimal model program and the singularities that it can deal with.

Finally, as Karl already pointed out, pretty much anything can happen as far as what values the $a_i$ can take. I guess there is an upper bound, but that is the easy direction. Otherwise, take your favorite smooth projective variety $T$ with your favorite embedding into a projective space and let $X$ be the cone over $T$, and $Y$ the vertex. If you blow up $X$ at $Y$, then you get a smooth variety and the exceptional divisor $E$ will be isomorphic to $T$. Now take your formula from $(\star)$ and apply adjunction. You'll get: $$ (a+1)E|_{E}\sim K_T. $$ From the construction you see that $E|_E$ is isomorphic to the inverse of the hyperplane class coming from the original embedding of $T$, so by choosing your $T$ and the embedding wisely you can get all kinds of values for $a$. For instance if $T\subset \mathbb P^n$ is a degree $d$ hypersurface, then $K_T\sim (d-n-1)H$ and $E|_E\sim -H$, so you get that $$ a= n-d. $$ This already shows that any integer is possible. Taking quotients as in here gives you fractions.

  • $\begingroup$ Dear Sandros, I really thank you for your detailed answer. In fact, as your example shows, when X is a hypersurface and X is equimultiple along Y (which is equivalent to normal flatness for hypersurfaces) there seems to be a formula K_Z = p^*K_X + (n-d)E (d is the multiplicity of X along Y and n the codimension of Y in X). In fact I know this formula is true if Y is a point and X is a hypersurface. I am pretty sure it is true if X is a hypersurface equimultiple along Y and Y is smooth. When X is no longer a hypersurface, a good replacement for equimultiplicity is normal flatness. $\endgroup$
    – Franz
    Mar 22, 2011 at 22:37
  • $\begingroup$ Thus I wondered if there were also a formula in this case. $\endgroup$
    – Franz
    Mar 22, 2011 at 22:37
  • $\begingroup$ Hi Franz, this sounds plausible, but I am a little doubtful that you would get such a simple formula in general. However, it seems to work for hypersurfaces. $\endgroup$ Mar 23, 2011 at 5:49

If $X$ is smooth and you are blowing up a smooth subvariety $Y$ of codimension $n$, then $K_Z = p^*K_X + (n-1)E$. This is an exercise in Chapter II, Section 8 of Hartshorne.

If $X$ is not smooth at least at the generic point of $Y$, then I don't think much can be said. If $X$ is smooth at the generic point of $Y$ though, then at least one of the components of $K_Z - p^*K_X$ will be $(n-1)E$ (that will be the component dominating $Y$).

EDIT: Just to give an example of what things can occur, consider the singularity $k[x,y,z]/(x^n + y^n + z^n)$ for $n \geq 2$. Then blowing up the origin resolves the singularities but the relative canonical is divisor $K_Z - p^*K_X = (2-n)E$. In particular, every coefficient $\leq 0$ occurs. If you are willing to look at higher dimensions, then every integer appears as a coefficient.

By the way, if $X$ is not Gorenstein (or quasi-Gorenstein/1-Gorenstein), then $p^* K_X$ can be substantially harder to understand. For some different definitions see the paper of de Fernex and Hacon, Singularities on normal varieties.

Can I ask a dumb question, what does it mean to be normally flat?

EDIT: Nevermind, I answered my own question. Normal flatness is a condition in Hironaka's proof of resolution of singularities. I knew I had heard it somewhere before.

  • 2
    $\begingroup$ Hi thanks for your answer. Normal flatness is the notion developped by Hironaka to decide what kind of sequences of blow up you have to do to get a resolution od singularities. It means that the normal cone of X along Y is flat over Y. $\endgroup$
    – Franz
    Mar 22, 2011 at 20:59
  • 1
    $\begingroup$ Franz, thanks. By the way, I don't think the modern resolution of singularities algorithms (which are of course based on Hironka's) use the notion of normal flatness. See for example any number of things written by Villamayor et. al., or Bierstone-Milman, or Wlodarczyk, or Kollar's book. $\endgroup$ Mar 22, 2011 at 21:05
  • $\begingroup$ @ Franz and Schwede. The question and the answers seems to consider $Z$ to be normal. May I ask please: is it easy to see from the assumtpions? $\endgroup$
    – pi_1
    Apr 5, 2021 at 17:42
  • $\begingroup$ If you blowup a regular subscheme of a regular scheme, you always obtain a regular and hence normal scheme. Some relevant computations are in Hartshorne Chapter II, Section 8. $\endgroup$ Apr 13, 2021 at 21:39

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