# base points of multiplicity $>1$

Let $S$ be a smooth projective surface (I am mostly intrested in the case when $S$ is a product of curves, say $S=\mathbb{P}^1 \times \mathbb{P}^1$ but probably this is not important).

Consider a family of curves $X \subset S \times T$ parametrised by a variety $T$ of dimension 2 (the fibres $X_t$ are distinct). Denote the projection $X \to S\$ as $p_S$ and $X \to T\$ as $p_T$. Define a map $\tau: X \to \mathbb{P}TS$ which to every point $x \in X$ associates the projectivisation of the tangent vector of $X_{p_T(x)}$ at $p_S(x)$.

My question is: how can one show that there exists $s \in S$ such that the map $\tau$ is non-constant on $p^{-1}(s)$?

In other words, how can one show that there is a point $s \in S$ such that not all curves passing through it touch each other, but on the contrary, their tangent spaces sweep the tangent space of $S$ at this point?

update: Consider the map $\sigma: \mathbb{P}(TX/T) \to \mathbb{P}(TS)$ induced by the projection $p_S: X \to S$. If $\sigma^{-1}(v)$ were finite for some $v \in TS$, then the projection of $v$ to $S$ would be the answer to the question. If the image of the map $\sigma$ is of dimension 3 then almost all fibres are of dimension 0, since $\mathbb{P}(TX/T)$ is of dimension 3. Perhaps one can show that $\mathrm{dim}\ \mathrm{Im}\ \sigma=3$?

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On $S = \mathbb P^1\times \mathbb P^1$ you have $TS=\mathcal O(2,0)\oplus \mathcal O(0,2)$, so $\mathbb PTS$ is not trivial. So I am not sure what examples you have in mind. – Lev Borisov Feb 12 '14 at 16:43
I might be missing something obvious, but can't the structure group of the tangent bundle of a Cartesian square of a curve be reduced to $\mathbb{G}_m$ (we get the cocycle with values in $\mathbb{G}_m$ by putting the respective cocycle for the tangent bundle of the curve on the diagonal of the matrix)? The image in $H^1(S,\mathrm{PGL}_2)$ would then be trivial – Dima Sustretov Feb 12 '14 at 17:21
This assumption is not crucial to the statement, so I edited the question accordingly. – Dima Sustretov Feb 12 '14 at 17:31
I am confused by the statement. Let $\pi:\mathbb{P}(TS)\to S$ be the projectivized tangent bundle. Let $i:X\to S$ be an unramified morphism from a smooth curve. Let $\tau:X\to \mathbb{P}(TS)$ be the canonical lift. Since $\pi\circ \tau$ equals $i$, which is non-constant, thus also $\tau$ is non-constant. Are you asking something else? Are you assuming that $\mathbb{P}(TS)$ is isomorphic to $S\times \mathbb{P}^1$? Do you really want to consider $\text{pr}_{\mathbb{P}^1}\circ \tau$? – Jason Starr Feb 12 '14 at 18:14
MO has a glitch in TeX: some of the math expressions tend to overlap with the following text, as you can see at the end of "Denote the projection $X\to S$ as..." The problem can be fixed by adding extra space "\ " just before the ending dollar sign. Strangely, the glitch occurs in the questions, but not in the comments. – Wlodek Kuperberg Feb 12 '14 at 19:49

As it stands, the statement is not true. Suppose $S$ has a fibration $f:S\rightarrow B$ onto a curve; pick up any curve $C$, and put $T=B\times C$, $X=S\times C$. Embed $X$ in $S\times T$ by $(s,c)\mapsto (s, f(s),c)$. Then given $s\in S$ there is only one curve of the family passing through $s$, hence one tangent direction. You need some hypothesis to avoid this trivial situation.

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yes, the fibres $X_t$ are supposed to be distinct, I have clarified this. – Dima Sustretov Feb 13 '14 at 10:42
In your Edit, you do not use that $X$ has dimension $3$. Take $X=S={\mathbb P}^1\times{\mathbb P}^1$ with $T={\mathbb P}^1$, $p_T$ the second projection, and $p_S=1_X$. Then $p'(x)$ is surjective. How does "this mean that $\tau$ is surjective on $p^{-1}(s)$" ? – Sasha Anan'in Feb 13 '14 at 12:06
You are right, I delete this part. – abx Feb 13 '14 at 12:46

I guess I post an answer to close the issue.

The statement is true in characteristic 0, but not in positive characteristic. Conterexamples in the latter case are easy to come by an $S=\mathbb{A}^1 \times \mathbb{A}^1= \mathbb{A}^2$, say, using curves with projections on one of the $\mathbb{A}^1$ factors everywhere ramified.

In characteristic 0 the argument is as follows, working over $\mathbb C$ (one would have to use formal completions of local rings and formal power series, put on the level of ideas the proof is the same).

So suppose the statement is not true. Then one can find a point $y$ and a neighbourgood $z$ such that for any $z \in U$ and any curve $X_t$ incident to $Q'$ the image of $T_x X_t$ in $T_y S$ depends only on $y$. In other words all curves $X_t$ are integral with respect to a distribution on $S$. But by uniqueness of solutions of ordinary differential equations, there can be at most one integral curve incident to each $z \in U$, which contradicts that dimension of $T$ is 2.

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