Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $X$ be an algebraic variety, $S \subset X$ its singular locus, and $x \in S$. Say that $x$ is good if for any neighborhood $U$ of $x$, any top differential form $\omega$ on $U \setminus S$ and some resolution of singularities $U' \rightarrow U$, the pullback of $\omega$ to $U'$ is regular.

An obvious necessary condition is that $S$ is of co-dimension at least 2. An example of a good point is $(0,0,0)$ in the variety defined by $x^2+y^2+z^2=0$. An example of a singular point which is not good is $(0,0,0)$ in the variety defined by $x^4+y^4+z^4=0$.

Here is the question:

Is the notion of good point related to other properties of singular points?

We are mainly interested in the case when $X$ is complete intersection.

share|improve this question
add comment

1 Answer

up vote 5 down vote accepted

The condition you want is very close to having rational singularities.

Definition: A variety $X$ of characteristic zero has rational singularities if it is:

  1. Cohen-Macaulay and,
  2. for some resolution of singularities $\pi : Y \to X$ we have $\pi_* \omega_Y = \omega_X$.

Now, let me explain why condition 2. is essentially the one you described. If $S$ is codimension 2, then $\omega_X$ is actually defined to be the pushforward of $\Omega^{\dim X}_{X\setminus S}$. In other words, $\omega_X$ is exactly those top forms coming from the smooth locus. Since $Y$ is a resolution and hence smooth, $\Omega^{\dim Y}_Y = \omega_Y$.

The statement that $\pi_* \omega_Y = \omega_X$ simply says that any form from the smooth locus of $X$, also comes from a form on $Y$. Which is exactly what you want.

It is worth noting that this definition is independent of the choice of resolution. In particular, if it holds for one resolution, it holds for all.

For further reading I would suggest Singularities of Pairs by János Kollár.

share|improve this answer
Thank you very much for your answer. There is one thing that I did not understand. Do you mean that if $X$ is Cohen-Macaulay and regular outside subvariety of co-dimension 2 then it is good? This will mean that the variety defined by $x^4+y^4+z^4=0$ is good, and I think that it is not true. Or do you mean that if $X$ is Cohen-Macaulay and have rational singularities than it is good. Thank you again, Rami Is it true that the variety defined by $x^4+y^4+z^4=0$ does not have rational singularities? –  Rami Feb 3 '12 at 22:37
I've tried to clarify things. But also, if $X$ has rational singularities it is good. Rational singularities implies Cohen-Macaulay and being regular outside a set of codimension –  Karl Schwede Feb 3 '12 at 23:50
Sorry for bothering you again, but I still have some qustions: 1. When you discuss direct image do you mean direct image in the derived category or in the abilian category of sheaves? 2. You say " $ω_X$ is actually defined to be the pushforward of $Ω^{dim X}_{X∖S}$." Why it is true? Do you mean that it is true whenever S is of co-dimension 2 or only under certain conditions that are satisfied in this case? what condition? I only see why it is true in the case that X is Gorenstein and only on the level of the abilian category. This should be enough for me but I still want to understand. –  Rami Feb 4 '12 at 4:00
I think $\omega_X$ is the unique line bundle on $X$ extending $\Omega^{d}_{X^{\textrm{sm}}}$, where $d$ is the dimension of $X$ and $X^{\textrm{sm}}$ is the nonsingular locus of $X$. The existence and uniqueness of $\omega_X$ are actually not so easy; see Chapter 6.4 in Liu's "Algebraic geometry and arithmetic curves" for a general discussion. In general, if you have a good enough morphism $X\to Y$ of schemes, there exists a unique line bundle $\omega_{X/Y}$ (called the relative dualizing sheaf of $X\to Y$ I think) extending the line bundle $\Omega^r_{X^{sm}/Y}$, where $r$ is the rel. dim. –  Ari Feb 4 '12 at 7:34
1. I don't mean the derived pushforward. I mean the ordinary pushforward. In characteristic zero, one also has Grauert-Riemenschneider vanishing which makes $R \pi_* \omega_Y = \pi_* \omega_Y$. 2. This is not trivial, but it's true as I stated it, the Gorenstein condition is not needed. Whenever S has codim 2, $\omega_X$ is the pushforward of the top forms from the smooth locus, the point is that $\omega_X$ is always S2 and it agrees with the top forms on the smooth locus, thus it agrees with the pushforward from the smooth locus. –  Karl Schwede Feb 4 '12 at 12:54
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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