Given a smooth morphism $f:(X,\mathcal{O}_X)\to (S,\mathcal{O}_S)$ between two smooth manifolds .I came across some different definition of relative differential in different context
- define it as cokernel $\Omega^1_{X/S} = \text{coker}(f^*:\Omega ^1_S\to \Omega_X^1)$ which is a rank n vector bundle if $f$ is relative dimension $n$ .
- first define the ideal sheaf $\mathcal{I} = \ker(\mathcal{O}_X\bigotimes_{f^{-1}\mathcal{O}_S}\mathcal{O}_X\to \mathcal{O}_X)$ where the map is given by multiplication $x\otimes y \mapsto xy$ then take the quotient $\Omega^1_{X/S} = \mathcal{I}/\mathcal{I}^2$
I try to prove that in some sense these two definitions are almost saying the same thing, the first one is more geometrically and the second one is more algebraically, sorry I don't have any good idea.
I found that this question is related to the exact sequence used to construct the Kodaira-Spencer map .
That is given two manifold $X,Y$ over the base $B$, such that the following diagram commute : $$ \require{AMScd} \begin{CD} X @>{f}>> Y \\ @V{h\circ f}VV @VV{h}V \\ B @>>{\text{id}}> B \end{CD} $$
then we have the following exact sequence :
$$f^*\Omega ^1_{Y/B}\to \Omega^1_{X/B}\to \Omega^1_{X/Y}\to 0 \tag{*}$$
If we take the base $B = \{pt\}$, then $\Omega^1_{X/B} = \Omega^1_{\mathcal{O}_X/\Bbb{C}} = \Omega^1_X$ then this sequence says that $\Omega^1_{X/Y}$ is the cokernel of the pullback of the differential. (that is starting from basic definition (2) we prove that is the same thing in (1).)
To prove (*) we just need to apply the fundamental result(which can be found in all algebraic geometry textbook) that : if $\varphi:B\to C$ is a $A$-algebra morphism then we have the exact sequence:
$$\Omega^1_{B/A}\otimes_{B}C \to \Omega^1_{C/A} \to \Omega^1_{C/B} \to 0$$