I recently read somewhere that for a cubic surface $S$ in $\mathbb{P}^3$, the classes $[\Delta] - [\Delta']$ generate the vanishing cohomology $H^2(S,\mathbb{Q})_{van}$, taken over all pairs of lines $\Delta$ and $\Delta'$ on $S$. Recall that $H^2(S_0,\mathbb{Q})_{van}= \ker (j_*:H^{n-1}(S_0,\mathbb{Q}) \rightarrow H^{n+1}(S,\mathbb{Q})$, where $S_0 \xrightarrow{j} S$ is a smooth hyperplane section. Note that the vanishing cohomology is generated by the vanishing cycles. Could someone please explain this to me? I'm very new to intersection theory and would like to understand this.
I should add some context. This is part of an exercise in Chapter 3 of Voisin's Hodge Theory and Complex Algebraic Geometry Volume II. The hint recommends the use of Theorem 3.27. This chapter also provides all of the relevant notation. Aside from doing it this way, I would be interested in seeing any other way we can compute the vanishing cycles.
Not sure what better explanation you'd like, than the one in the book.
If $ j : S \rightarrow \mathbb{P}^3 $ denotes the inclusion of the smooth cubic surface, then it is clear that for any two lines $ \Delta, \Delta' $ on $ S $, the class $ [\Delta] - [\Delta'] $ belongs to $ H^2(S, \mathbb{Q})_{van} $. The fact that this class is not zero follows because:
$ \bullet $ The self intersection $ (\Delta, \Delta) = -1 $.
Proof: Use the genus formula to get $ 2g(\Delta) - 2 = (\Delta, \Delta + K_S ) $ where $ K_S $ is the canonical divisor of $ S $. By adjunction, $ K_S = -H.S $, here $ H $ is a hyperplane in $ \mathbb{P}^3 $. Hence by Bezout's theorem, $ (\Delta, K_S) = -1 $. Knowing $ g(\Delta) = 0 $, we get the desired self-intersection.
$ \bullet $ $ [\Delta] - [\Delta'] \neq 0 $
Proof: Because $ (\Delta- \Delta' , \Delta- \Delta') = -2 -2(\Delta, \Delta') $ and the last is equal to $ -4 $ or $ -2 $ depending respectively whether the lines intersect or do not.
Now the monodromy action is irreducible therefore it suffices to show that the nonzero subspace generated by the classes $ [\Delta] - [\Delta'] $ is invariant under monodromy. But this follows because the monodromy acts by self-homeomorphisms of $ S $ and preserves the intersection form on $ S $, hence sends a line to necessarily a line again.