# Definition of meromorphic differentials

In Diamond, Shurman: A First Course in Modular Forms on p. 77, the meromorphic differentials on an open set $$V\subset \mathbb C$$ of degree $$n$$ are defined as $$\Omega^{\otimes n}(V)=\lbrace f(q)(dq)^n\colon f\text{ meromorphic on }V\rbrace,$$ where $$q$$ is a variable on $$V$$. What is the rigorous definition of the RHS above? Can it be defined as a tensor product of something?

• It's a section of the $n$-th symmetric power of the bundle of meromorphic differentials. – Angina Seng Jun 6 '20 at 20:30

The notation $$\Omega$$ is used for the cotangent bundle of a complex manifold. Its sections are known as holomorphic differential forms. In the case of $$V \subseteq \mathbb{C}$$, the tangent bundle is the trivial line bundle. Therefore, $$\Omega = \Omega^1$$, the differential $$1$$-forms, and given a local coordinate $$q$$, the differential $$1$$-form $$dq$$, which is the dual of the vector field (section for the tangent bundle) $$\frac{\partial}{\partial q}$$, is a basis for this $$1$$-dimensional space.
As with any bundle, one may consider its meromorphic sections, i.e. elements $$\omega \in \Omega(U)$$ where $$U$$ is an open subset such that $$V \setminus U$$ consists of isolated points, and $$\omega$$ has poles in $$V \setminus U$$. Note that in this case this is simply $$\Omega^{(1)}(V) = \{ f(q) dq \}$$ where $$f(q)$$ is a meromorphic function.
Finally, we can tensor bundles. One can think of $$\Omega^{\otimes n}$$ as the $$n$$-power tensor bundle of $$\Omega^{(1)}$$, or as multilinear maps on $$n$$-tuples of vector fields. However, in this degenerate case, this is again a trivial line bundle, spanned by the element $$(dq)^{\otimes n}$$, which is denoted in the book as $$(dq)^n$$. Its meromorphic sections are once more simply of the form $$f(q) (dq)^{\otimes n}$$, where $$f$$ is a meromorphic function.