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Let $\mathcal{D} \approx \mathbb{P}^{\delta_d}$ be the space of homogeneous degree $d$ polynomials in three vriables, where $\delta_d = \frac{d(d+3)}{2}$. Let $$ X \subset \mathcal{D} \times \mathbb{P}^2$$

be a smooth embedded complex submanifold, not necessarily closed. Given a point $p\in \mathbb{P}^2$, we get a hyperplane $$\tilde{H}_p \in \mathcal{D} \times \mathbb{P}^2.$$ Note that a point $p$ first of all gives a hyperplane $H_p$ in $\mathcal{D}$ (which is the space of degree $d$ polynomials passing through the point $p$). This gives us a hyperplane $$ \tilde{H}_p := H_p\times \mathbb{P}^2 \in \mathcal{D} \times \mathbb{P}^2.$$ Let us further define $H_p^* \subset H_p$ to be the space of degree $d$ curves such that $p$ is a smooth point of the curve. Similarly define $$ \tilde{H}_p^* := H_p^*\times \mathbb{P}^2 \in \mathcal{D} \times \mathbb{P}^2.$$

Is it true that for almost all choices of $p \in \mathbb{P}^2$, $\tilde{H}^*_p$ is transverse to $X$?

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2 Answers

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My first answer was wrong. The answer is no.

First observe that the embedding $X\to \mathcal D\times\mathbb P^2$ is a bit distracting. The map $X\to \mathbb P^2$ plays no role in the question of transversality to a given $\tilde H_q$. So I think about the problem like this:

You have smooth maps

$\mathbb P^2\leftarrow H\to \mathcal D\leftarrow X$,

where $H\subset \mathcal D\times \mathbb P^2$ is as in my wrong answer (i.e. the space of all pairs $(f,q)$ where $q$ is on the curve defined by $f$). The projection $H\to \mathbb P^2$ is a submersion, so the fiber $H_q$ over any $q$ is a manifold, and the question is whether for a dense set of $q$ the map $H_q\to \mathcal D$ is transverse to the map $X\to \mathcal D$.

(We say that two maps $A\to B\leftarrow C$ are transverse if whenever points $a$ and $c$ both go to the point $b$ then the tangent space $T_bB$ is spanned by the images of $T_aA$ and $T_cC$.)

If the map $H\to \mathcal D$ were a submersion, then the answer would be yes (for any smooth $X$ and any map $X\to \mathcal D$), by the following argument:

The submersion $H\to \mathcal D$ is transverse to any $X\to \mathcal D$. Thus the fiber product $Y=H\times_{\mathcal D}X$ is a manifold, and a little bit of playing with tangent spaces yields that those $q$ for which $H_q\to \mathcal D$ is transverse to $X\to \mathcal D$ are precisely the regular values of the projection $Y\to \mathbb P^2$. In particular the transversality holds for a dense set of choices of $q$.

But $H\to \mathcal D$ is not a submersion; this fails at precisely those points $(f,q)$ such that $q$ is a non-smooth point of the curve defined by $f$.

And there are counterexamples with $d=2$. Let's work in an affine plane in $\mathbb P^2$ with coordinates $(x,y)$. For each $q=(x_0,y_0)$ the quadratic equation $(x-x_0)(y-y_0)=0$ defines a curve through $q$. Let $X\subset \mathcal D$ be this two-dimensional family. For any $(x_0,y_0)$ the manifold $H_q$ intersects $X$ non-transversely.

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Thank you for the answer. But I have a question about the counter example. Here your X also depends on the point q. So as you change q, both the X and the H move. In my question, X does not depend on q. As you change q only H should change. X should remain fixed. So do you still think this is a counter example to my question? –  Ritwik Oct 21 '11 at 19:08
    
I did not express myself very well. I meant that $X$ is parametrized by pairs $q=(x_0,y_0)$, and that for any given point $q$ the manifold $H_q$ fails to be transverse to $X$ (where it meets $X$, namely at the point in $X$ corresponding to $q$). Does that make sense? –  Tom Goodwillie Oct 22 '11 at 0:04
    
I have edited the question slightly. According to your argument, the answer to my question should now be yes, since $\pi_{D}$ is a submersion. Am I correct? –  Ritwik Oct 25 '11 at 2:48
    
Yes, I believe so. –  Tom Goodwillie Oct 25 '11 at 11:10
    
One further thing. Requiring $\pi_{\mathcal{D}}$ to be a submersion is a sufficient criteria, but not a necessary one. I simply need $\pi_{\mathcal{D}}$ to be transverse to $X$ (which is guaranteed if it is a submersion). –  Ritwik Oct 25 '11 at 13:22
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Recall that for a smooth map $f:Y\to Z$ we say that $z\in Z$ is a critical value for $f$ if there is a point $y\in Y$ such that $f(y)=z$ and such that the induced map of tangent spaces $T_yY\to T_zZ$ is not surjective. A point in $Z$ is called a regular value of $f$ if it is not a critical value of $f$ (in particular if it is not in the image of $f$ at all). The usual tool in differential topology for showing that something can be perturbed to make something transverse to something is Sard's Theorem, which says that the set of critical values of a $C^\infty$ map $f$ always has measure zero and in particular that the set of regular values is dense. Of course, in your case you have complex analytic rather than just $C^\infty$.

Let $H\subset\mathcal D\times \mathbb P^2$ be the union over $p\in \mathbb P^2$ of $H_p\times p$. This is a submanifold, and furthermore the projection $H\to \mathbb P^2$ is a submersion. (That is, every point of $\mathbb P^2$ is a regular value for this map.) It follows that the fiber product $X\times_{\mathbb P^2}H$ is a manifold. Consider the projection $$ X\times_{\mathbb P^2}H\to \mathbb P^2. $$ I claim that the regular values of this map are precisely those $q$ such that $X$ is transverse to $\tilde H_q$. That gives what you want.

EDIT: This answer is wrong. I will have to think about it some more.

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Thank you. This is a very neat proof! I assume that in this case I can also conclude that the regular values form an OPEN dense subset of $P^2$? –  Ritwik Oct 19 '11 at 2:58
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No. Its complement is the image of a closed set (the set of critical points), but that does not guarantee that it itself is closed unless $X$ is compact. –  Tom Goodwillie Oct 19 '11 at 10:22
    
I am sorry, but your last statement.... that the regular values of the projection map are those q such that \tilde{H}_q is transverse to X........is it obvious? –  Ritwik Oct 20 '11 at 18:19
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