Sheaf of relative Kähler differentials intuitively

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.

• If you haven't before, check out 21.2.20 in Vakil's FOAG – Wojowu Nov 16 '20 at 18:06
• 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. – Isak the XI Nov 16 '20 at 20:33

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.

• 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? – Isak the XI Dec 22 '20 at 23:12
• another remark: what do you mean by " the $k$'th order sheaf principal parts" $\mathcal{P}^k_{X/Y}$? – Isak the XI Dec 22 '20 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.

I write another post to reply to questions. I hope this is ok for the members of this forum.

Question 1. "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? – Isak the XI"

Answer 1. If you read Chapter III.10 in Hartshorne on smooth morphisms, you will find that for any morphism $$f:X\rightarrow S$$ where $$X,S$$ are irreducible schemes of finite type over a field $$k$$ and where $$X$$ is integral, it follows $$f$$ is "smooth of relative dimension $$n$$ iff $$f$$ is flat and $$\Omega^1_{X/S}$$ is locally free of rank $$n$$.

(This is Definition 1 and Example III.10.0.2).

The "generic smoothness theorem" (Corollary III.10.7) says that if $$k$$ is algebraically closed of characteristic $$0$$ and if $$X$$ is "non singular", there is an open non-empty subscheme $$U\subseteq S$$ where the induced morphism $$f_U:f^{-1}(U)\rightarrow U$$ is smooth. Since $$X$$ is integral it follows $$V:=f^{-1}(U)$$ is integral hence $$f_U$$ is smooth of relative dimension $$n$$ iff $$\Omega^1_{V/U}$$ is locally trivial of rank $$n$$. By Theorem III.10.2 this implies that for any point $$u\in U$$ it follows the fiber $$f^{-1}(u)\otimes \overline{\kappa(u)}$$ is equidimensional of dimension $$n$$ and regular. Hence the locus $$V$$ and $$U$$ where $$\Omega^1_{X/S}$$ is locally trivial says something about the fibers of the morphism $$f$$. Outside of $$U$$ it follows the fiber is no longer "regular".

Question 2. "another remark: what do you mean by " the k'th order sheaf principal parts"? – Isak the XI"

Why must nilpotent elements be allowed in modern algebraic geometry?

you will find information on nilpotent elements in algebraic geometry and jet bundles/principal parts. In the thread I mention a construction of a locally ringed space $$(X^m, \mathcal{O}_{X^m})$$ that is closer to Hartshornes definition of an algebraic variety in Chapter I in his book. This construction can be done for a scheme $$X$$ that has an open affine cover $$Spec(A_i)$$ where $$A_i$$ is a Hilbert-Jacobson ring. The locally ringed space $$X^m$$ has non-trivial nilpotent elements in its structure sheaf. I explain in the thread why the points in the space $$X^m$$ can be thought of as solutions to systems of polynomial equations in fields. Hence the construction can be used when teaching algebraic geometry (the post was an answer to a question on teaching algebraic geometry to students with little knowledge on "ringed spaces") - it is more "intuitive".

Question 3."As far as I understood the issue correctly for usual schemes or (as relative version) for morphisms f:X→S between usual schemes, to know "everything" about "ramifications" of f it suffice to study only Ω1X/Y"

Answer: You should not trust wikipedia on this. I believe the EGAIV reference I gave is the best reference on this issue.

Note: If $$H\subseteq G$$ is a closed subgroup of an affine algebraic group $$G$$ of finite type over an algebraically closed field $$k$$, it follows the quotient $$G/H$$ is a smooth quasi projective scheme of finite type over $$k$$. In this case the tangent bundle $$T_{G/H}$$ and cotangent bundle $$\Omega^1_{G/H}$$ determine each other. There is a canonical isomorphism

$$\Omega^1_{G/H}\cong T_{G/H}^*$$

of locally free sheaves/geometric vector bundles. Since the theory of algebraic groups and Lie algebras is a well developed theory and since $$T_{G}(e)\cong Lie(G)$$ ($$e\in G$$ is the identity), people usually use the tangent bundle $$T_G$$ and $$T_{G/H}$$ when working with algebraic groups and their quotients. The study of rings of differential operators and D-modules on a flag variety $$G/P$$ is a much studied subject, and here one uses the tangent bundle $$T_{G/P}$$ and the "universal enveloping algebra" $$U(\mathcal{O}_{G/P}, T_{G/P})$$.

• I can't see the deleted question, but it is actually not rare that moderators delete posts outside their field of expertise because some of their standards are not achieved. Sometimes it is justified, sometimes not and there is not much you can do about it. I was unhappy about it several time too, but ultimately this is how MSE works. I believe here on MO this is less frequent, because the quality of questions and answers is usually better, so moderators are a bit less rigid. – Nicolas Hemelsoet Dec 23 '20 at 13:56
• The person was not a moderator. – hm2020 Dec 23 '20 at 14:01