# Extra Algebraic $(1,1)$ cycles on a complex surface

Suppose $$x,y,w,z$$ are homogeneous coordinates of $$\mathbb{CP}^3$$, and $$\begin{eqnarray} X_t := \left(F_t = x f_2 +y g_2 +t F_3 = 0 \right) \end{eqnarray}$$

be a family of degree 3 hypersurfaces in $$\mathbb{CP}^3$$, and $$f_2, g_2$$ and $$F_3$$ are generic homogeneous polynomials of degree 2,2 and 3.

What I often read in physics literature is that on $$t=0$$, the line $$x=y=0$$ lies entirely inside $$X_0$$, and we have a new" algebraic $$(1,1)$$ cycle, and if we deform $$X_0$$ to some $$X_t$$ (with small $$|t|$$), that cycle transform into a mixed cycle.

I want to understand that, i.e. I want to know

1-Why $$x=y=0$$ is a new $$(1,1)$$ cycle. Usually, I read this is because this cycle isn't a complete intersection of a hyperplane in $$\mathbb{CP}^3$$ with $$X_0$$. But I'm not sure whether this is enough to say that this algebraic cycle is independent of the other cycles.

2- Why when I (infinitesimally) deform $$X_0$$ to $$X_t$$, the cycle transforms into a mixed one. Again in physics, people would say, well $$x=y=0$$ is just three points inside $$X_t$$. But it would be nicer to see this more in a more intrinsic" way.

I tried to find the answers by using the Picard-Fuchs equation, but no success yet.

For example, we have the correspondence (the following theorem is due to Griffiths if I'm right),

$$\begin{eqnarray} Prim^{1,1}(X_t) = \left(\frac{\mathbb{C}[x,y,z,w]}{<\frac{\partial F_t}{\partial x},\frac{\partial F_t}{\partial y},\frac{\partial F_t}{\partial z},\frac{\partial F_t}{\partial w}>} \right)_{2}, \end{eqnarray}$$ i.e. the prime elements of $$H^{1,1}(X_t)$$ correspond to homogeneous polynomials of degree 2 of the right hand side in the above equation.

In this case, we find $$dim_{\mathbb{C}} Prim^{1,1}(X_t) = 6$$, which I think it's true since $$h^{1,1}(X_t) = 7$$, so primimitivity must put one contraint and we get 6 independent prime lements.

Now, I expect that $$dim_{\mathbb{C}} Prim^{1,1}(X_0)$$ to jump (because I must have extra cycles), but it stays the same!