I don't know algebraic geometry but hopefully the following makes sense: let $X$ be the set of all pairs of plane projective (say) conics $(F,G)$ s.t. $F,G$ have no common components; I imagine this is a dense open subset of the space of pairs of conics. By Bezout's Theorem each pair of conics intersect in exactly $4$ points counting multiplicities. The intersection may be viewed as a multiset, or a sum of dirac measures on $\mathbb{P}^2$. Equip $\mathbb{P}^2$ with any reasonable metric.
Question: Is the map $(F,G)\mapsto F\cap G$ continuous w.r.t. the Wasserstein (earth mover's) distance on $\mathbb{P}^2$? If not, is it continuous at least in some sense?
Question2: Is the topology induced by earth mover's distance the same as $(\mathbb{P}^2)^4/S_4$, where $S_4$ acts on $(\mathbb{P}^2)^4$ by permuting coordinates?
I’m answering Question 2 first, since it’s the simpler one. As hinted in my comment, the answer is “yes”.
Let $X=(\mathbb{P}^2)^4/S_4$ and let $q$ (resp. $W$) be the natural quotient topology (resp. the topology coming from Wasserstein distance). Since the projection $(\mathbb{P}^2)^4 \rightarrow (X,W)$ is clearly continuous, so is the identity map $f: (X,q) \rightarrow (X,W)$.
Now, $(X,q)$ is compact metric, and $(X,W)$ is metric hence Hausdorff, and $f$ is a continuous bijection. It’s then easy to see that $f$ is closed, hence open, hence a homeomorphism, and thus $q=W$.
Now, onto the original Question. The base field is always $\mathbb{C}$, and all varieties are endowed with the analytic topology. Please ask if some parts are unclear, I think it’s all correct but I’m not necessarily precise enough.
I’m also not considering continuity at very degenerate cases, where one of the conics is a double line and the other one is the reunion of two distinct lines.
First, I use $\mathbb{P}^5$ as my space of plane projective conics: $[a:b:c:d:e:f]$ represents the conic $ax^2+by^2+cz^2+dxy+eyz+fzx=0$.
There is a map $\gamma: (\mathbb{P}^2)^3 \rightarrow (\mathbb{P}^5)^2$, mapping triplets $([a:b:c],[d:e:f],[g:h:i])$ to the couple $((ax+by+cz)(dx+ey+fz)=0,(ax+by+cz)(gx+hy+iz)=0)$.
There is a map $\delta: (\mathbb{P}^2)^3 \rightarrow (\mathbb{P}^5)^2$, mapping triplets $([a:b:c],[d:e:f],[g:h:i])$ to the couple $((ax+by+cz)^2=0,(dx+ey+fz)(gx+hy+iz)=0)$.
The set $\mathcal{C}_2$ of interest is the complement of the reunion of the image of $\gamma$, the diagonal embedding of $\mathbb{P}^5$ into its square, and the image of $\delta$ and its symmetric. In particular, this reunion has an empty interior (being of Zariski dimension at most $6$), so $\mathcal{C}_2$ is indeed open and dense in $(\mathbb{P}^5)^2$.
To show that $I: (F,G) \in \mathcal{C}_2 \rightarrow F \cap G \in X$ is continuous, we use the following criterion.
Lemma: Let $f: U \rightarrow V$ be a function, where $U,V$ are metric spaces and $V$ is compact. Then $f$ is continuous if and only if its graph is closed in $U \times V$.
Proof: it follows from the easy fact that a sequence $y_n \in V$ diverges iff it has at least two distinct limit points.
So we’re reduced to the following situation: given pairs $(f_n,g_n) \in \mathcal{C}_2$ (to be precise, the conics are $f_n(x,y,z)=0$ and $g_n(x,y,z)=0$) and $a_n,b_n,c_n,d_n \in \mathbb{P}^2$ such that
then $\alpha,\beta,\gamma,\delta$ are the intersection points of $f$ and $g$ counted with multiplicity.
Clearly $f,g$ vanish at $\alpha,\beta,\gamma,\delta$.
We discuss the simple intersection points first.
Lemma: let $f,g: \mathbb{C}^2 \rightarrow \mathbb{C}$ be two conics vanishing at $(0,0)$. Assume that the intersection multiplicity of $f,g$ at $0$ is one. Then,
Proof: $(f,g):\mathbb{C}^2 \rightarrow \mathbb{C}^2$ is a diffeomorphism near $0$. The conclusion then follows from an “effective openness theorem” along the lines of: Let $f : B(0,R) \rightarrow \mathbb{R}^n$ (the ball being closed and in $\mathbb{R}^n$ too) be such that $f(0)=0$ and $d_0f$ is invertible. Let $\mu >0$ be the minimum of its norm on the unit sphere. Let $C$ be an upper bound on $d^2_xf(u,v)$ for $u,v$ in the unit sphere and $x \in B(0,R)$. Let $\varepsilon > 0$ be such that $\varepsilon < \mu/C$, then every $y \in B(0,\varepsilon(\mu-C\varepsilon)/3)$ is the image of a unique $x \in B(0,\varepsilon)$.
This theorem implies that every simple intersection point of $f,g$ appears among the $\alpha,\beta,\gamma,\delta$.
Suppose now (up to renaming) that $\alpha=\beta$ is a simple intersection point of $f,g$. Then the injectivity statement above shows (after choosing suitable affine coordinates etc) that $a_n=b_n$ for all large enough $n$. Then $\nabla f_n(a_n)$ and $\nabla g_n(a_n)$ are collinear for all large enough $n$, so $\nabla f(\alpha)$ and $\nabla g(\alpha)$ are collinear and $\alpha$ is not simple, a contradiction.
Hence every simple intersection point appears exactly once.
This solves the problem in all cases (by extensive checks), unless the intersection of $f,g$ contains two double points $p$ and $q$. This is what we now assume. The only way to have an issue is if $p\neq q$ and $\alpha=\beta=\gamma=\delta=p$.
Suppose $f,g$ are both singular, this implies that they’re both products of two linear polynomials – studying cases, this means that either $f$ is a square, or $g$ is. In this case, after changing coordinates, we can assume $f=x^2$, $g=yz$, then we can replace $g$ with $f+g$.
In other words, we can assume that $g$ is smooth. Up to changing a little bit the coordinates and the sequence (but nothing consequential, in particular it changes nothing in the limit), we can assume that $a_n=p=[0:0:1]$ for all $n$.
We let $0 \in Z$ be a holomorphic manifold such that $g_z$ parametrizes smooth conics conics near $g$ that go through $p$, with $g_0=g$, and $g_n$ corresponds to $g_{z_n}$.
Choose a small open subset $D \subset \mathbb{P}^1$ such that for all directions $\Delta \in D$, the line with direction $\Delta$ going through $p$ cuts the conic $g_z=0$ at a unique other point $Q(\Delta,z) \in \mathbb{P}^2$ (this $Q$ is a holomorphic function), with $q=Q(\Delta_0,0)$.
If $D,Z$ are chosen small enough, then $Q$ remains in a neighborhood of $q$, so never takes the values $a_n,b_n,c_n,d_n$. In particular, the function $F_n=f_n(Q(\cdot,z_n))$ doesn’t vanish on $D$. But they locally uniformly converge to $f(Q(\cdot,0))$, which is not constant to $0$, but has a zero. This contradicts Hurwitz’s theorem, thus solving out the last “nondegenerate enough case”.