Motivation behind the sheaf of relative Kähler differentials

Question

I'm interested in the geometric motivation behind the bundle (or in more general framework the sheaf) of relative differentials $\Omega_{X/Y}$ of a morphism $f: X \to Y$ smooth $k$-varieties. Differential geometry provides following picture of relative differentials:

Let $f: X \to Y$ a equidimensional surjective map between connected manifolds $Y, X$ and moreover assume that every fiber $F:= f^{-1}(y) \subset X$ for $y \in Y$ is also a connected submanifold of same dimension. We obtain an exact sequence of tangent spaces

$$ 0 \to T_{X/Y} \to T_X \to f^*T_Y \to 0 $$ where $T_{X/Y}$ is the kernel of induced map of tangent bundles. intuitively what is really going on there is that for every $x \in f^{-1}(y)$, the $(T_{X/Y})_x$ is the tangent space of the fiber at $x$. The relative space of Kähler differentials $\Omega_{X/Y}$ defined as the dual of $T_{X/Y}$ and sits in the sequence which we will obtain if we dualize the sequence above of tangent spaces:

$$ 0 \to f^*\Omega_Y \to \Omega_X \to \Omega_{X/Y} \to 0 $$

Another definition of relative Kähler differentials which is more common to use in modern algebraic geometry works as follows:

Let us embedd $X$ as the image of the diagonal $\Delta: X \to X \times_Y X$ and assume that the ideal sheaf $I \subset O_{X \times_Y X}$ defines the closed image $\Delta(X) \subset X \times_Y X$.

This gives us another sequence of tagent spaces

$$ 0 \to T_{\Delta(X)} \to T_{X \times_Y X} \to N_{X \times_Y X/X} \to 0 $$

with normal bundle $N_{X \times_Y X/X}$. It's a basic fact that the dual of $N_{X \times_Y X/X}$ is $I/I^2$ and most books on algebraic geometry define the sheaf relative Kähler differentials by

$$\Omega_{X/Y} := I/I^2$$

Since this definition not uses that $f$ is a map of smooth maps this is a far generalization of the old fashion setting from differential geometry. Now, if there is any justice n this world then these two definitions of relative differentials should coinside if we deal with $X, Y$ and $fY$ nice enough.

Therefore, if $f$ a surjective map between conneted manifolds such that every fiber is a connected submanifold, why the tangent bundle $T_{X/Y}$ and the pullback of normal bundle $\Delta^* N_{X \times_Y X/X}$ of $\Delta(X) \subset X \times_Y X$ are canonically isomorphic?

Can we write down an explicit isomorphism and understand what is geometrically going on there?

Answer 1

Given a linear map $g \colon V \rightarrow W$ between finite dimensional vector spaces, we can form the exact sequence

$$ 0 \rightarrow V \xrightarrow{\Delta} V \times_{W} V \rightarrow (V \times_{W} V )/ \operatorname{im}(\Delta) \rightarrow 0 $$

where $V \times_{W} V = \{ (v,v') \, | \, g(v) = g(v') \}$ and $\Delta(v) = (v,v)$ is the diagonal embedding. Then we can see that $(V \times_{W} V) / \operatorname{im}(\Delta)$ is canonically isomorphic to $\ker g \subseteq V$ via the subtraction isomorphism $[(v,v')] \mapsto v - v'$ (whose inverse is given by $v \mapsto [(v,0)]$).

This applies fiberwise to the situation you describe for manifolds. Let's assume that $f$ is a surjective submersion (this is enough to guarantee that the fibered product actually exists in the category of smooth manifolds). Fix $x \in X$ and pull back the short exact sequence via $\Delta$ to obtain $$ 0 \rightarrow \Delta^{*} \left( T(\operatorname{im} \Delta) \right)|_{x} \rightarrow \Delta^{*} T(X \times_{Y} Z)|_{x} \rightarrow \Delta^{*} (N_{\operatorname{im}{\Delta} \hookrightarrow X \times_{Y} X})|_{x} \rightarrow 0 $$

Using the identifications $$ T(X \times_{Y} X)|_{(x,x')} = \{ (v,v') \in T_x X \times T_{x'} X, \, | \, df|_{x}(v) = df|_{x'}(v') \}, \\ \Delta^{*} T(X \times_{Y} Z)|_{x} = T_x X \times_{T_{f(x)} Y} T_x X, \\ \Delta^{*} \left( T(\operatorname{im} \Delta) \right)|_{x} = \{ (v,v) \in T_x X \times T_x X \} \cong T_x X, \\ T_{X/Y}|_{x} = \ker \left( df|_{x} \right) $$ we see that we are in the algebraic situation described in the beginning of the answer (with $V = T_x X, W = T_{f(x)} Y$ and $g = df|_{x}$).

To see what is going on geometrically, it might be useful to start with the case where $Y = \textrm{pt}$ so that the fibered product is just a regular product $X \times X$ and the relative tangent bundle is the regular tangent bundle of $X$. Take $X = \mathbb{R}$ and identify each equivalence class in a fiber of the normal bundle of $X$ inside $X \times X$ with a slanted line. The isomorphism then identities each such slanted line with its intersection with the (say) $x$ axis (which is the "unslanted, regular" tangent space).

Answer 2

In general if $k \rightarrow A \rightarrow^f B$ is an arbitrary sequence of maps of commutative unital rings, there is a canonical right exact sequence of $B$-modules

S1. $B\otimes \Omega^1_{A/k} \rightarrow \Omega^1_{B/k} \rightarrow \Omega^1_{B/A} \rightarrow 0$.

In Matsumura, Commmutative ring theory Thm 25.1 you may find this construction. If $B$ is $0$-smooth over $A$ it follows when you add a $0$ on the left it follows S1 is split exact. When you dualize S1 you get the sequence

S2. $0 \rightarrow Der_A(B) \rightarrow Der_k(B) \rightarrow^{Tf} B\otimes_A Der_k(A)$

where $Tf$ is the "tangent morphism" of the map $f$. If $X:=Spec(B)$ and $Y:=Spec(A)$ it follows the relative tangent bundle $T_{X/Y}$ is the sheafification of the module $Der_A(B)$. When you sheafifiy S2 you get the sequence

S3. $0 \rightarrow T_{X/Y} \rightarrow T_{X/k} \rightarrow^{Tf} f^*T_{Y/k}$

and S3 is an exact sequence of $\mathcal{O}_X$-modules in general.

If $k$ is the field of complex numbers and $X,Y$ are finitely generated and regular algebras over $k$ it follows $T_{X/k}$ and $f^*T_{Y/k}$ are finite rank locally free sheaves on $X$. The "relative tangent sheaf" $T_{X/Y}$ is related to properties of the morphism $f$.

Definition. Assume $X,Y$ are smooth irreducible schemes of finite type over $k$ the field of complex numbers (this means the local rings of $X,Y$ are regular at all points). The morphism $f$ is "smooth of relative dimension $n$" iff

1. $f$ is flat
2. $dim(X)=dim(Y)+n$
3. The cotangent sheaf $\Omega^1_{X/Y}$ is locally free of rank $n$.

By Hartshorne, Prop III.10.2 this implies that the fibers $f^{-1}(y):=X_y$ of $X$ at any point $y\in Y$ are regular. Hence the cotangent sheaf $\Omega^1_{X/Y}$ "measures" when the fibers of $f$ are regular/smooth of the same dimension $n$.

Example: Let $\pi:X:=\mathbb{V}(\mathcal{E}^*)\rightarrow Y$ be a finite rank geometric vector bundle on $Y$. Locally it follows the sheaf $\mathcal{E}^*$ is a free $A$-module $E:=A\{x_1,..,x_n\}$ on the elements $x_i$ hence locally the map $\pi$ is the following map:

S4. $A \rightarrow B:=A[x_1,..,x_n]$

where the latter is a polynomial ring on $n$-variables. In this case the relative module of Kahler differentials $\Omega^1_{B/A}$ becomes

S5. $\Omega^1_{B/A}\cong B\{dx_1,..,dx_n\}$

which is the free $B$-module of rank $n$ on the elements $dx_i$. Hence the relative cotangent sheaf $\Omega^1_{X/Y}$ is in this case locally trivial of rank $n$ and the morphism $\pi$ is smooth of relative dimension $n$. The fibers are affine spaces and they are smooth. Hence the relative cotangent sheaf is related to properties of the morphism. The relative tangent sheaf is the dual of the relative cotangent sheaf, but when you dualize you loose information. Hence the cotangent sheaf is more fundamental.

Note: When you write down the relative tangent sequence, this is not a sequence of "tangent spaces", it is an exact sequence of coherent sheaves on $X$. The relative "tangent bundle" is not a vector bundle in general. In your situation it is a "coherent sheaf". If the relative cotangent sheaf is locally trivial it follows the fibers of your morphism are smooth manifolds (if you consider holomorphic maps of complex projective manifolds).