Let $f: X \to Y$ be a morphism of schemes over a field. I'd like to know why the sections of the sheaf $f ^* \Omega _Y$ are given by $d \phi$ for $\phi \in \mathscr O_Y$.

Does this mean that e.g. $f ^* \Omega _Y(X)=\Omega _Y(Y)$, without any additional constraints? Why can the sections of $f ^* \Omega _Y$ be seen this way?

$f^*$ is defined by $f ^* \mathscr F = f^{-1} \mathscr F \otimes_{f^{-1}\mathscr O_Y} \mathscr O_X$, and it is known that $(f^* \mathscr O_Y^{\oplus n})(U)= (\mathscr O_X^{\oplus n})(f^{-1}(U))$, so since $\Omega_Y$ is locally-free it follows that $\Omega_X$ is locally-free too. This seems "close" to what we want to show. How does the correspondence between sections of $f ^* \Omega _Y$ and elements $d\phi$ work?


1 Answer 1


It is true that there is a canonical morphism of sheaves of $\mathcal O_X$-Modules $\phi: f^*\Omega_Y \to \Omega_X$.
By adjunction this morphism corresponds to an easier to understand morphism of $\mathcal O_Y$-Modules $\psi:\Omega_Y\to f_*\Omega_X$ :
Given an open subset $U\subset X$ the morphism $\psi_U:\Omega_Y(U)\to f_*\Omega_X(U)=\Omega_X (f^{-1}U)$ associates to a differential form $\omega \in \Omega_Y(U)$ the lifted differential form $f^*(\omega)\in \Omega _X(f^{-1}(U)$, just as in differential topology.

Beware however that $f^*\Omega_Y$ and $\Omega_X$ are not isomorphic in general and even their spaces of global sections $f^*\Omega_Y(X)$ and $\Omega_X(X)$ need not be isomorphic:
For example if $X$ is a smooth complete curve of genus $g\neq 1$ over the field $k$ and if $f:X\to Y=\operatorname {Spec}(k)$ is the morphism to a point, then $f^*\Omega_Y=f^*\mathcal O_Y=\mathcal O_X$ so that $$f^*\Omega_Y(X)=\mathcal O_X(X)=k\neq \Omega_X(X)=k^g $$
Finally let me mention that your question "Does this mean that e.g. $f ^* \Omega _Y(X)=\Omega _Y(Y)$ ?" also has a negative answer :
If $p:P= \operatorname {Spec}(k) \to Z$ is the inclusion of a point into a smooth projective curve of degree $g\gt1$ we have $$p ^* \Omega _Z(P)=\mathcal O_P(P)=k\neq \Omega _Z(Z)=k^g$$

  • $\begingroup$ Thank you for your answer. Gathmann claims that the canonical morphism $f^* \Omega_Y \to \Omega_X$ is given by $d \psi \mapsto d(\psi \circ f)$. I wonder what does this formula mean, in particular why the elements of $f^* \Omega_Y$ where denoted by $d (\text{function in }O_Y)$. Is there an intuitive explanation for what are the sections of the pullback when e.g. $f$ is an inclusion $X \to \Bbb P^n$? $\endgroup$
    – Emolga
    Oct 28, 2016 at 6:03
  • 1
    $\begingroup$ The intuitive explanation is that a differential form on $X$ is the linear form on $TX$ obtained by restricting a linear form on $T\mathbb P^n$ to $TX$. However the definition in algebraic geometry is very formal and uses an algebraic gadget called "Kähler differentials" which has as essential virtue that $d(uv)=udv+vdu$ for elements of the base ring. Regular local functions $\psi$ thus have a differential $d\psi$ and Gathmann's formula for the morphism $f^*\Omega_Y \to \Omega_X$ is indeed correct. The reason for that formalism is the possibility of study of varieties with singularities. $\endgroup$ Oct 28, 2016 at 6:59

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