Is $H^0(C,\Omega_X)\cong H^0(C,\Omega_X\otimes\mathcal{O}(-p))$?

If $$C$$ is a complex curve, then any point is a hypersurface. To a point $$p$$ in $$C$$, suppose we have $$w(p)=0$$ for all $$w\in\Omega_X$$, then do we have $$H^0(C,\Omega_X)\cong H^0(C,\Omega_X\otimes\mathcal{O}(-p))$$?

In general, what do $$\mathcal{O}(-p)$$ and $$\mathcal{O}(p)$$ look like?

• @reuns how to define the isomorphism – Danny Mar 27 at 4:34
• Tell me if it makes sense to you : $X$ a complex Riemann surface, $M$ its field of meromorphic functions, for each $U\subset X$ open and $D$ a divisor on $X$, $O(D)(U) = \{ f \in M, Div(f|_U) + (D\cap U) \ge 0\}$, $O(U)$ is a ring, the others are $O(U)$ modules, $\Omega(D)(U)$ the meromorphic one-forms on $U$ with poles prescribed by $D$, $O(-p)$ is the "sheaf" mapping each $U$ to $O(-p)(U)$, $\Omega \otimes O(-p) : U \mapsto \Omega(U) \otimes_{O(U)} O(-p)(U)$, – reuns Mar 27 at 5:40
• $\omega \otimes f\mapsto f \omega$ is an isomorphism $\Omega(U) \otimes_{O(U)} O(-p)(U)\to \Omega(-p)(U)$, $H^0(X,\Omega\otimes O(-p)) = \{ \nu, \forall U \subset X, \nu \in( \Omega\otimes O(-p))(U)\}$, and so $H^0(X,\Omega\otimes O(-p))= H^0(X,\Omega(-p))$. And since you assumed $\Omega(-p)(X) = \Omega(X)$.. About a one-form $\omega= fdg,f,g \in M$ having a zero at $p$ : it means in local chart $\omega = g(z)df(z)$ and $g(z)f'(z)$ has a zero at $p$. – reuns Mar 27 at 5:46
• @reuns In general, for a complex manifold $X$, can we generalise that $\Omega_X\otimes O(D)\cong \Omega(D)$? – Danny Mar 28 at 15:21
• In dimension $\ge 2$ there is no meromorphic function having a pole at a single point, the basic element of divisors of meromorphic functions are the hypersurfaces ( submanifolds of codimension $1$), so you need to be careful what kind of $D$ you meant and see if you find an isomorphism. – reuns Mar 28 at 15:25