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 $L$? )

  • 3
    $\begingroup$ 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 $q_0$. $\endgroup$ Nov 20 '18 at 14:41
  • $\begingroup$ Thanks! My main issue is with $V_q$. $\endgroup$
    – Giulio
    Nov 20 '18 at 14:43
  • $\begingroup$ 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). $\endgroup$
    – Giulio
    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.

  • $\begingroup$ 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!! $\endgroup$
    – Giulio
    Nov 20 '18 at 20:33

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