Let $f: X \to Y$ be a separated morphism between $k$-varieties or more general schemes of finite type. The most common way in standard literature on algebraic geometry to define the sheaf of relative Kähler differentials $\Omega_{X/Y}$ is to observe that the diagonal map $\Delta: X \to X \times_Y X$ is a closed embedding (we assume $f$ separated) and let $I \subset O_{X \times_Y X}$ ideal sheaf define image $\Omega(X)$. The sheaf of relative Kähler differentials is defined as

$$ \Omega_{X/Y}:= \Delta^* (I/I^2) $$

and I'm interested in the geometric motivation behind this definition. As is so often the case, the origins are in differential geometry.

Let $f: X \to Y$ a equidimensional surjective map between connected $k$-manifolds $Y, X$ for $k= \mathbb{R}, \mathbb{C}$ and we moreover assume that every fiber $F:= f^{-1}(y) \subset X$ for $y \in Y$ is also a connected submanifold of same dimension. Most natural examples: a vector bundle $X$ over a manifold $Y$ or a fibration with 'nice' behavior on fibers. 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, for every $x \in f^{-1}(y)$, $(T_{X/Y})_x$ is the tangent space of the fiber at $x$. The relative space of Kähler differentials $\Omega_{X/Y}$ is defined as the dual of $T_{X/Y}$ and sits in the sequence which we obtain if we dualize the previous sequence above of tangent spaces:

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

Now let us translate the first definition from modern algebraic geometry also to framework of differential geometry:

Let $f: X \to Y$ as above surjective map between connected manifolds with equidim submanifold fibers and now let us embed $X$ via diagonal map $\Delta: X \to X \times_Y X$ into the fiber product (a problem: does the fiber product exist as manifold of $X,Y$ nice enough?).

Since $\Delta(X) \subset X \times_Y X$ is a smooth submanifold, we can pick local coordinates $( x_1 , ... , x_n )$ around a $x \in \Delta(X)$ such that $ \Delta(X) $ is locally defined by $x_{k + 1} = ... = x_n = 0$; then with this choice of coordinates

$$T_x X \times_Y X = k{\frac{\partial}{\partial x_1} \vert _x,... \frac{\partial}{\partial x_n} \vert _x }$$

$$T_x \Delta(X) = k{\frac{\partial}{\partial x_1} \vert _x,... \frac{\partial}{\partial x_k} \vert _x }$$

$$(N_{X \times_Y X/\Delta(X)})_x = k{\frac{\partial}{\partial x_{k+1}} \vert _x,... \frac{\partial}{\partial x_n} \vert _x }$$

and the ideal sheaf is locally generated by $x_{k+1}, ..., x_n $. The bundle $N_{X \times_Y X/\Delta(X)}$ is the normal bundle of embedding $\Delta$ and fits in following canonical sequence:

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

Also, it is well known that the pairing

$$(I/I^2)_x \times (N_{X \times_Y X/\Delta(X)})_x \to k$$

is perfect and therefore $I/I^2 \cong (N_{X \times_Y X/\Delta(X)})^{\vee}$. As pointed out above in algebraic geometry we define sheaf (or bundle in more old fashioned language) of relative Kähler differentials $\Omega_{X/Y}$ as $ \Omega_{X/Y}:= \Delta^* (I/I^2) $.

Therefore it is reasonable to conjecture although I nowhere found a proof that if there is any justice n this world then these two definitions of relative differential bundles should coinside in the setting of differential geometry. That is for $f: X \to Y$ surjective map between connected manifolds with equidim submanifold fibers, we should have

$$T_{X/Y} \cong \Delta^* N_{X \times_Y X/\Delta(X)}$$

Can we write down an explicit isomorphism and understand what is geometrically going on there? I would like also remark that I asked almost identical question in MathStackExchange

and suppose that possibly MO might be a better place to ask about it.

  • $\begingroup$ If you haven't before, check out 21.2.20 in Vakil's FOAG $\endgroup$
    – Wojowu
    Nov 16, 2020 at 18:06
  • $\begingroup$ In 21.2.20 Vakil only refer to the second definition I described in my question. In 21.2.21 there is a reason why these two definitions should morally coinside in case $f: X \to Y$ with $Y = \{*\}$. Below I gave a sketch why these two definitions might coinside in setting of differential geometry. $\endgroup$
    – user267839
    Nov 16, 2020 at 20:33

2 Answers 2


This question may have interest for many people. 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. There is a long exact sequence continuation of the sequence S3.

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 of dimension $n$. 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).

Example: Ramification. For a holomorphic map $f:X \rightarrow Y$ of complex projective manifolds the same constructions applies because of a classical result saying $f$,$X$ and $Y$ are algebraic. Moreover any coherent analytic sheaf $E$ on $X$ is algebraic. Hence for $f$ we get a short exact of $\mathcal{O}_X$-modules

S6 $f^*\Omega^1_{Y/k} \rightarrow \Omega^1_{X/k} \rightarrow \Omega^1_{X/Y} \rightarrow 0.$

Here $\mathcal{O}_X$ is the sheaf of regular functions on the complex algebraic manifold $X$. If the map $f$ is flat it follows $\Omega^1_{X/Y}$ is locally free of rank $n$ iff the fibers of $f$ are complex projective manifolds of dimension $n$. In general the coherent sheaf $\Omega^1_{X/Y}$ is generically locally free, hence it is related to the "ramification" of the map $f$. Hence the cotangent sheaf $\Omega^1_{X/Y}$ and more generally the $k$'th order sheaf of principal parts $\mathcal{P}^k_{X/Y}$ is used in the study of the ramification and discriminant of the morphism $f$

Example: Etale morphisms. In Hartshorne Ex. III.10.3 they prove that a morphism $f:X\rightarrow Y$ of finite type over a field $k$ is etale iff $f$ is flat and $\Omega^1_{X/Y}=0$.

Exampe: If $X$ is a complex projective manifold it follows the sheaves $\Omega^1_{X/\mathbb{C}}$ and $T_{X/\mathbb{C}}$ are locally free of finite rank and dual of each other: $\Omega^1_{X/\mathbb{C}} \cong T_{X/\mathbb{C}}^*$. In the non smooth situation there is a canonial morphism $i:\Omega^1_{X/\mathbb{C}} \rightarrow T_{X/\mathbb{C}}^*$ but $i$ is not an isomorphism in general. The relative cotangent bundle $\Omega^1_{X/Y}$ is a coherent $\mathcal{O}_X$-module in general.

  • $\begingroup$ Could you elaborate a bit more your explanations that since $\Omega^1_{X/Y}$ is generically locally free, hence it is related to the "ramification" of $f$. Which picture one should have in mind? Like in covering theory that a ramified cover is almost everywhere unramified (that is a usual cover) exept on a "small" set? In algebro geometric setting "almost everywhere" probably means on a open subscheme (=a generic property)? That's the analogy, right? $\endgroup$
    – user267839
    Dec 22, 2020 at 23:12
  • $\begingroup$ another remark: what do you mean by " the $k$'th order sheaf principal parts" $\mathcal{P}^k_{X/Y}$? $\endgroup$
    – user267839
    Dec 22, 2020 at 23:13

supplement/ an "almost" answer: I noticed that OP's of several related questions (1, 2) asked about similar issue. The best explanation I found there was that for a smooth manifold $X$ the tangent space $TX$ of $X$ is canonically isomorphic to normal bundle of diagonal embedding $\Delta: X \to X \times X$. This is the special case of my approach with $f: X \to Y$ where $Y:= \{* \}$ is a point. This argument is explicitely given in 21.2.21 from Ravi Vakil's FOAG notes linked by Wojowu.

Explicitely for manifold $X$ the isomorphism $TX \cong \Delta^* N_{X \times_Y X/\Delta(X)}$ is given by $(x, v) \mapsto (x,x v, -v)$. This was the case $f: X \to Y$ for $Y:= \{* \}$.

Seemingly this can be generalized if $f: X \to Y$ is sufficiently nice. Say $f$ is a vector bundle but I'm still interested to know if the argument comming now can be generalized to the case where $f$ is a fibration where each fiber is a submanifold. We can reduce it to local case where $X= Y \times \mathbb{R}^n$. Then $X \times_Y X$ exist as manifold and can be explicitely described by $$X \times_Y X= \{(x,z) \ \vert \ f(x)= f(z)\} = \{(y, v), (y, w) \ \vert \ v, w \in \mathbb{R}^n \} = \coprod_{y \in Y} (f^{-1}(y) \times f^{-1}(y)) $$ Then $\Delta(X)= \{(y, v), (y, v) \ \vert \ v \in \mathbb{R}^n \}$ and the isomorphism $T_{X/Y} \cong \Delta^* N_{X \times_Y X/\Delta(X)}$ can be defined fiberwise by establishing an isomorphism between $(T_{X/Y})_{y,v}$ and $(\Delta^* N_{X \times_Y X/\Delta(X)})_{y,v}$ separately for every $y \in Y$ under application of the case obove for $Y:=y$ and $X:= f^{-1}(y)$.

The verification that this morphism is continuous with respect the variation of $y \in Y$ looks laborious but straight forward.


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