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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 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 at 17:21
    
This assumption is not crucial to the statement, so I edited the question accordingly. –  Dima Sustretov Feb 12 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 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 at 19:49

1 Answer 1

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 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 at 12:06
    
You are right, I delete this part. –  abx Feb 13 at 12:46

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