Griffiths transversality

Remarks

Equivalently, if $s(t)$ is a holomorphic family of classes with $s(t) \in F^p H^k(X_t)$, then the derivative of $s$ in any direction, taken in the fixed space $H$, lies in $F^{p-1} H^k(X_t)$. The idea of the proof: represent $s(t)$ by closed forms of type $(r, k-r)$ with $r \ge p$ on $X_t$. To first order, a change of complex structure replaces $dz_j$ by $dz_j + \sum_k \varphi_{jk} d\bar z_k$, so it lowers the holomorphic degree of a form by at most one. Hence the derivative has components only in types $(r, k-r)$ with $r \ge p - 1$.

Example: for a family of curves and $k = 1$, $F^0 = H$, so condition 2 holds automatically. For Calabi–Yau threefolds and $k = 3$, condition 2 says that the derivative of the line $F^3 = H^{3,0}$ lies in $F^2$, which is a genuine restriction.

Going deeper

For a student

Transversality is the key constraint of variations of Hodge structure; it is built into their definition. Period domains are usually not Hermitian symmetric, and the transversality condition defines a horizontal distribution on them. Curvature estimates along horizontal directions give the Griffiths–Schmid results on period maps, for instance that horizontal holomorphic maps from a disc are distance decreasing for a suitable metric. The derivative of $\mathcal{P}^p$ is computed by the Kodaira–Spencer map followed by cup product and contraction.