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 $X$?



Can you ``perturb'' a submanifold to intersect transversally with any other smooth submanifold of projective space?
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.


