# Effective Bertini

Let $$X$$ be a smooth complex projective manifold, and $$L$$ an ample line bundle. By Bertini's Theorem, for every integer $$q$$ big enough there exists an open dense subset $$U_q\subset |qL|$$ such that every divisor $$D$$ in $$U_q$$ is smooth.

Warm up question: is the complement of $$U_q$$ always a divisor?

We can define a bigger open subset $$V_q\subset |qL|$$ as $$V_q:=\{D\in |qL| \; \textrm{s. t.} \; (X,\frac{1}{q}D) \; \textrm{ is klt} \}$$

My question is: what is the dimension of the complement of $$V_q$$ ? (or at least can we bound its asymptotic in $$q$$ ? e.g. is it upper-bounded by $$aq^{\dim X}$$ with $$a$$ a constant which is strictly smaller than the volume of $$L$$? )

• For all $q\geq q_0$, the complement of $U_q$ is a divisor whose degree is known, cf. mathoverflow.net/questions/165672/… Replace $\mathcal{O}(1)$ in that answer by an appropriately positive tensor power of $L$, and use the asymptotic formula to find $q_0$. Nov 20 '18 at 14:41
• Thanks! My main issue is with $V_q$. Nov 20 '18 at 14:43
• Just to make things more explicit, I am interested in the case $q>>0$. Moreover, let me explain a little computation to justify my last guess about the complement of $V_q$. Take $X=\mathbb{P}^2$, and $L=\mathcal{O}(3)$. Fix a line $H$, multiplying by $(q+1)H$ we can embed $| \mathcal{O}(2q-1)|$ into the complement of $V_q$. My guess is that this is kind of the biggest thing that can be outside $V_q$ (but I have no proof for the moment). Nov 20 '18 at 20:37

Regarding the warm up question: No (although for $$q$$ sufficiently large the answer is yes, as Jason Starr comments.) Let $$X = \mathbb{P}^1 \times \mathbb{P}^2$$ and let $$L= \mathcal{O}(1,1)$$. Write homogeneous coordinates on the first factor as $$(u:v)$$ and on the second factor as $$(x:y:z)$$. A divisor $$D$$ in $$H^0(L)$$ is of the form $$u (ax+by+cz) + v (dx+ey+fz)=0.$$ $$D$$ is singular if and only if the matrix $$\begin{bmatrix} a & b& c \\ d & e & f \\ \end{bmatrix}$$ has rank $$1$$. (If this matrix has rank $$2$$, $$D$$ is isomorphic to $$\mathbb{P}^2$$ blown up at a point, when the matrix drops rank, $$D$$ turns into the union of a $$\mathbb{P}^2$$ and a $$\mathbb{P}^1 \times \mathbb{P}^1$$.) The condition that this matrix drops rank is codimension $$2$$.
The complement of $$U_q$$ is called the dual variety to $$X$$. It is usually (no precise meaning attached) true that the dual of a smooth variety is a divisor. For a while, it was an open problem to characterize smooth projective toric varieties whose duals were not divisors; this was solved by Dickenstein, Feichtner and Sturmfels.
• Thanks!! In general, I am more interested in the asymptotic values of $q$ (actually, for $q=1$, $L$ might even have no sections at all!!). And of course, I think $V_q$ is the hard part. Thanks!! Nov 20 '18 at 20:33