Is a quadratic differential literally a square of a differential?

Question

I have seen definitions of quadratic differentials as functions $\omega : TS \to \mathbb{C}$ that satisfy $\omega(\lambda v) = \lambda^2 \omega(v)$, and that we can write them as $\omega = f_i dz_i^2$ where $f_i = f_j \left(\frac{dz_i}{dz_j}\right)^2$. Does that mean that the $dz^2$ is literally the square of $dz$? I.e. that for a point $p \in S$ and $v \in T_pS$ that $(dz^2)|_p(v) = (dz|_p(v))^2$? As such, those quadratic differentials would satisfy the square of the scalar multiple thing, along with $$dz^2(v + w) = dz^2(v) + 2 dz(v) dz(w) + dz^2(w).$$

Answer 1

A quadratic differential may be viewed as a symmetric two-form, $\phi_z(u,v)=a(z) dz(u)dz(v)$ (a bit like a Riemannian metric). The point is that it takes values in ${\Bbb C}$ and that it is possible to multiply complex numbers (no complex conjugation). So your two last formulae are correct. You could e.g. equally well write the last as: $$ \phi(v+w,v+w)=\phi(v,v)+2\phi(v,w)+\phi(w,w)$$ in terms of the two form.

Later edit: The formula mixing $f_i$ and $f_j$ is also fine when interpreting $z_i(z_j)$ as a change of coordinate charts.