Kawamata-Viehweg vanishing for non-Gorenstein singularities

Question

Let $Y$ be a normal projective algebraic variety with rational singularities. Assume that there is an effective divisor $\Delta$ on $Y$ such that $\omega_Y+\Delta$ is a Cartier divisor. Does the analog of Kawamata-Viehweg theorem hold for $\omega_Y+\Delta$? In other words, if $L$ is a very ample line bundle on $Y$, is it true that $L\otimes (\omega_Y+\Delta)$ has no higher cohomology? Note that if $Y$ is Gorenstein, then one can take $\Delta=0$ (and in this case the statement indeed follows immediately from Kawamata-Viehweg).

Answer 1

You don't need the $\Delta$. In fact, it is true without the $\Delta$ and may or may not be true with it. It depends on the singularities of the pair $(Y,\Delta)$.

From the question I am guessing you want to work in characteristic zero.

KV Vanishing with rational singularities Let $Y$ be a projective variety over an algebraically closed field of characteristic zero with rational singularities and $\mathscr L$ a nef and big line bundle on $Y$. Then $$ H^i(Y, \omega_Y\otimes \mathscr L)=0 $$ for $i>0$.

Proof Let $f:X\to Y$ be a resolution of singularities and consider $\mathscr M:=f^*\mathscr L$. It is easy to see that $\mathscr M$ is nef and big on $X$ and that $\mathscr M^{-1}=f^*\mathscr L^{-1}$, so $Rf_*\mathscr M^{-1}\simeq Rf_*\mathscr O_X\otimes \mathscr L^{-1}\simeq \mathscr L^{-1}$.

Next apply KV vanishing on $X$ for $j<\dim Y=\dim X$: $$ H^j(Y, \mathscr L^{-1})\simeq \mathbb H^j(Y, Rf_*\mathscr M^{-1}) \simeq H^j(X,\mathscr M^{-1}) =0 $$ Now the desired statement follows by Serre duality. (For which you only need that $Y$ is Cohen-Macaulay, which follows from being rational, and not that it is Gorenstein, which does not as you know). $\ \ \square$

Note that it is also true that $Rf_*\omega_X\simeq f_*\omega_X\simeq \omega_Y$. In case you need to deal with $\omega$ directly. You don't need $\omega_Y$ to be invertible anywhere for this.

---

Regarding the case when you actually have a $\Delta$ that you need to add. There are a few cases when KV vanishing still holds:

1. If $(Y,\Delta)$ is semi log canonical. This includes the case klt that Dave mentions in the comments, but it has the advantage that it allows actual divisors with coefficient $1$ while klt requires that the coefficients of $\Delta$ are strictly less than one and I have a suspicion that you have an integral divisor, not one with fractional coefficients. Unfortunately, this is still pretty restrictive, but has the advantage that it does not require any positivity assumptions about $\Delta$. There are various version of Kodaira type vanishing theorems in the literature. The most general I know is Cor 1.3 in this paper.
2. If $\Delta$ has some positivity properties, you are good. For instance if $\Delta$ is a nef Cartier divisor, then you can just replace $L$ with $L\otimes \mathscr O_Y(\Delta)$ and be on your way. Then again, from what you write it looks like $\Delta$ is only Cartier if $Y$ is Gorenstein, so indeed you need something else.
3. You can try to combine the previous two cases: If you can write $\Delta=\Delta_1+\Delta_2$ such that $(Y,\Delta_1)$ is log canonical and $\Delta_2$ is a nef Cartier divisor, then you're OK by replacing $L$ with $L\otimes \mathscr O_Y(\Delta_2)$ and apply the log canonical KV vanishing to $(Y,\Delta_1)$. Not knowing more about your specific situation this seems to be your best bet.

---

One more thing: Please do not write things like $L\otimes (\omega_Y + \Delta)$ ! This mixes things that live in different categories (line bundles vs. divisors). Of course, here everyone knows what you mean, but then why not write what you mean. That you wrote makes no sense, really. You cannot add a sheaf to a divisor and then tensor it with a line bundle. It hurts my eye to look at it, but more importantly if you carry on this level of disregard for proper notation you will end up writing something that people will misinterpret and nobody wants that. (For example if you have a map $f:Y\to Z$, then $f_*\Delta$ and even $\mathscr O_Z(f_*\Delta)$ could be completely different from $f_*\mathscr O_Y(\Delta)$, or similarly, $f_*K_Y$ is not the same as $f_*\omega_Y$.