Suppose that $f: X \to Y$ is a degree $n$ map of Riemann surfaces. If we are given a holomorphic differential $1$-form $\omega$ on $X$, how do we define the pushforward $f_*\omega$? I can't seem to find the definition written down anywhere so any references / insight would be helpful. Thanks!
What is the pushforward of differential forms?
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Pushforward of differential forms is a standard operation, described in many expository texts.
A fairly detailed presentation is given in Chapter VII of
- Werner Greub, Stephen Halperin, Ray Vanstone. Connections, Curvature, and Cohomology. Volume I. De Rham Cohomology of Manifolds and Vector Bundles. Pure and Applied Mathematics 47A (1972), Academic Press.
Of course, for a covering map (like in your case) the pushforward map can be defined much more easily: given an arbitrary $y∈Y$, pick a sufficiently small neighborhood $U$ of $y$ so that the preimage of $U$ under $f$ is a projection from the product of $U$ with a discrete set $S$. The projection map identifies each of the leaves indexed by $S$ with $U$ via a diffeomorphism (in our case, a biholomorphic map), so we can transport $ω$ along these diffeomorphisms and sum over all $s∈S$. Since $f$ has degree $n$, the cardinality of $S$ equals $n$ and the sum is finite. The result is clearly independent of the choice of $U$. For a local diffeomorphism, we can instead cover $X$ by a locally finite cover $V$ such that every $V_j$ maps injectively into $Y$, then take similar sums as described above (no longer with a fixed choice of $U$).
This construction works for any finite degree holomorphic map, and there is no need to require the covering to be regular. If the map is ramified, first remove the ramification points, which are isolated in $Y$.
Here is a construction of push-forward (of sorts), assuming that $f: X\to Y$ is an $n$-fold regular ramified covering, i.e. there exists a group $G$ (of order $n$) of biholomorphic automorphisms of $X$ such that $f(p)=f(q)\iff Gp=Gq$.
Replace the holomorphic 1-form $\omega$ by its $G$-average: $$ \alpha= \frac{1}{n}\sum_{g\in G} g^*\omega. $$ Now, $\alpha$ (unlike $\omega$) is $G$-invariant.
Define a 1-form $\beta$ on $Y$ as follows: If $y\in Y$ is not a ramification point of $f$, $v\in T_yY$, then pick $x\in f^{-1}(y)$. There exists $u\in T_xX$ such that $df(u)=v$. Set $$ \beta(u):=\alpha(v). $$ If $y$ is a ramification (aka critical) point of $f$, then $\beta(y)=0\in T^*_yY$.
It is a nice exercise to check that this construction does yield a 1-form $\beta$ on $Y$ (one needs to verify smoothness and independence of the choice of $x$; the latter is ensured by the $G$-invariance of $\alpha$). Also, if $\alpha\in \Lambda^{1,0}(X)$, then so is $\beta$, if $\alpha$ is holomorphic, so is $\beta$, etc.
Then one can regard $\beta$ as a push-forward of $\alpha$.