The notion of Kähler differential is a very general way to encode a notion of differential form: something that is dual to a derivation or vector field.
Conceptually, in dual language of algebras, a symmetry of a commutative algebra $A$ is an automorphism $g\colon A\to A$, i.e., $g(a b)=g(a)g(b)$. The ‘infinitesimal’ symmetries are the derivations $X\colon A\to A$, with $X(a b)=X(a)b+X(b)a$. The module of Kähler differentials $\Omega^1_K(A)$ parametrizes derivations, in the sense that every derivation $X$ corresponds uniquely to a morphism of $A$-modules $\mu_X: \Omega_K^1 (A)\to A$.
Kähler differentials are traditionally conceived in terms of an algebraic construction of a certain module $\Omega_K^1(Spec R)$ on a given ordinary ring $R$. On spaces modeled (in the sense described at space) on the site CRing$^{op}$, such as varieties, schemes, algebraic spaces, Deligne-Mumford stacks, this produces the correct notion of differential form in this context. This is the case discussed in the section
below.
The definition, concrete as it is, applies of course also to function rings on spaces not modeled on $CRing^{op}$, such as rings $C^\infty(X)$ of smooth functions on a smooth manifold. One might expect that the module of Kähler differentials $\Omega_K(Spec C^\infty(X))$ of $C^\infty(X)$, regarded as an ordinary ring, does reproduce the familiar notion of smooth differential forms on a manifold. But it does not. This is discussed in the section
This shows that the concrete algebraic construction of Kähler differential forms over plain rings, traditionally thought of as their very definition, does in fact not correctly capture their nature. There is another definition – obtained from the nPOV – which does capture the situation correctly:
In fact, already the definition of module has to be freed from it concrete realization in the context of ordinary rings, to exhibit its true nature. What this is has been established long ago in
and in Jon Beck’s thesis, and is discussed in more detail in the entries module and Beck module : Beck and Quillen noticed that the category $R Mod$ of modules over an ordinary commutative ring $R$ is canonically equivalent to the category $Ab(CRing/R)$ of abelian group objects in the overcategory $CRing/R$ of all rings, over the given ring $R$:
Under this equivalence an $R$-module $N$ is sent to the square-0-extension ring $R \oplus N$ that is canonically equipped with a ring homomorphism $R \oplus N \to R$ and with a unital and associative and commutative product operation
that makes it first an object in the overcategory $CRing/R$ and in there an abelian group object, hence an object in $Ab(CRing/R)$.
Conversely, every module arises this way, up to isomorphism. So this gives an equivalent way of defining modules over rings.
And this is the right definition. Notably, this definition does not assume anything about the ring $R$. It does not even assume that $R$ is a ring at all! It could be anything.
Concretely: for $C$ any category of test objects – so that we may think of objects in the opposite category $C^{op}$ as function rings on the test objects $C$ – we may define the category of module s over an object $R \in C^{op}$ by the above equation:
Notice that this is now a definition. And that $R$ could be anything, and the definition still makes sense.
The category of all modules over all possible objects $R$ is then nothing but the codomain fibration
where $I$ is the interval category and fiberwise (over $C^{op}$) we form abelian group objects.
This turns out to be the correct category theoretic definition of module (as discussed there). In fact, this is is the special case of the higher categorical definition that works for $C$ any (∞,1)-category. In that case the construction $Ab(C^{op}/R)$ of abelian group objects in the overcategory is generalized (straightforwardly! and in fact even more elegantly) to the notion of tangent (∞,1)-category.
With the above correct notion of module in hand, all the other concepts of deformation theory, notably those of derivations and of Kähler differentials follow straightforwardly
given an $R$-module $N$ regarded as an object $p_N : R \oplus N \to R$; the derivations on $R$ with coefficients in $N$ are precisely the sections $\delta : R \to R \oplus N$ of $p_N$.
The assignment $Spec R \mapsto \Omega_K(Spec R)$ of modules of Kähler differentials is the assignment universal with respect to derivations, which means that
is the left adjoint to the above projection $p_C : Mod_C \to C^{op}$:
this means that every derivation $\delta : R \to \mathcal{N}$ (being a section in $C$ of the module which is the overcategory element $\mathcal{N} \to R$) is identified conversely with a morphism $\Omega_K(R) \to \mathcal{N}$ in the category of abelian group objects in the overcategory $C^{op}/R$:
Notice that in all of the above now, $C$ is still a completely arbitrary category.
By allowing $C$ – the collection of test spaces – to be a general (∞,1)-category, the above story gives the following completely general nPOV on the nature of Kähler differentials:
For $C$ any (∞,1)-category of test spaces, write $p_{C^{op}} : Mod := T_{C^{op}} \to C^{op}$ for the tangent (∞,1)-category of its opposite category.
the fiber of $Mod \to C^{op}$ over $R \in C^{op}$ is the (∞,1)-category $R Mod$ of modules over $R$;
for $(p_{\mathcal{N}} : \mathcal{N} \to R) \in R Mod$, a derivation on $R$ with coefficients in $\mathcal{N}$ is a section $\delta : R \to \mathcal{N}$ of $p_{\mathcal{N}}$.
The assignment of modules of Kähler differentials or cotangent complexes is the left adjoint
of the tangent (∞,1)-category projection $p_{C^{op}}$.
Its value $\Omega_R(R)$ on an object $R \in C^{op}$ is the module of Kähler differentials on $Spec R \in C$.
We spell out very concretely definitions of Kähler differentials for special concrete choices of base category $C$ as special cases of the above general story. We start with the familiar cases and then work our way up to more general or richer cases.
In terms of the above discussion, we now take $C = CRing^{op}$ to be the opposite category of the category of ordinary (commutative unital) ring. In fact without changing anything of the discusson we may assume that the ring $R$ in question is equipped with a ring homomorphism $k \to R$ from a ring or field $k$. This makes $R$ a $k$-algebra, and we shall often speak of algebras in the following, where we could just as well speak of rings.
Suppose $A$ is a commutative algebra over a field $k$. We may define Kähler differentials either by an explicit construction or by a universal property. In fact there are two explicit constructions.
The simplest construction, maybe, is as follows. The module of Kähler differentials $\Omega^1_K(A)$ over $A$ is generated by symbols $d a$ for all $a\in A$, subject to these relations:
$d c = 0$ when $c$ is a ‘constant’, that is, an element of $k$ regarded as an element of $A$.
$d(a b)=(d a)b+a(d b)$.
$d(a+b)=d a+d b$.
$(d a)b = b(d a)$.
In particular there are only finite sums in the module of Kähler differentials.
Another more sophisticated construction of $\Omega^1_K(A)$ is given below. But turning to the universal property, note that we can define derivations from $A$ to any $A$-module $M$: they are $k$-linear maps $X : A \to M$ satisfying the product rule:
Then $\Omega^1_K(A)$ may be defined as the universal $A$-module equipped with a derivation. In other words, there is a derivation
and if $X:A\to M$ is any derivation from $A$ to some $A$-module $M$, then there is a unique $A$-module morphism
such that the following diagram commutes:
We say that $X$ factors through $d$.
We can replace the commutative algebra $A$ more generally by a morphism of commutative unital rings $f:R\to S$. Then the module of Kähler differentials is the $S$-module $\Omega^1_K(S/R)$ corepresenting the functor
that assigns to every $S$-module $M$ the set of derivations on $S$ with values in the (bi)module $f_* M$, where $f_*:S Mod\to R Mod$ is the restriction of scalars.
In other words, $Der_R(S,f_*M)\cong Hom_S(\Omega^1_K(S/R),M)$. In a diagram: for every $R$-derivation $X\colon S \to M$ there is a unique morphism (of $S$-modules) $\mu\colon \Omega^1_K(S/R) \to M$ making the following diagram commute:
This framework also gives another construction of the module of Kähler differentials, instead of the generators and relations definition given above.
Let $I$ be the augmentation ideal, i.e. the kernel of the multiplication map
Then $\Omega^1_K_{S/R}= I/I^2$ and there is a canonical induced map $d: S\to \Omega^1_{S/R}$ given by $d s = [1\otimes s - s\otimes 1]$.
Furthermore, if $R$ is in characteristic zero, one may introduce Kähler $p$-forms , which are elements of the $p$-th exterior power $\Omega^p_K_{S/R}:=\Lambda_R^p \Omega^1_K_{S/R}$. The module of Kähler differentials readily generalizes as a sheaf of Kähler differentials for a separated morphism $f:X\to Y$ of (commutative) schemes, namely it is the pullback along the embedding of the ideal sheaf of the diagonal subscheme $X\hookrightarrow X\times_Y X$.
Compare the role of universal differential envelope and Amitsur complex for analogous constructions in the noncommutative case. The appropriate extension of the module of relative Kähler differentials to the derived setting is the cotangent complex of Grothendieck–Illusie.
The module of Kähler differentials on $R$ is isomorphism to the first Hochschild homology of $R$
Under mild conditions the anaous statement is true for higher Kähler differentials and higher Hochschild homology: this is the Hochschild-Kostant-Rosenberg theorem.
We have seen that we define Kähler differentials $\Omega^1_K(A)$ for any commutative algebra $A$.
The following special case deserves special attention:
The algebra $A = C^\infty(X)$ of smooth functions on some smooth space $X$ (a smooth manifold or a generalized smooth space) is in particular a commutative algebra. So one might think that its Kähler differentials form the ordinary differential forms on $X$ – in analogy to the case when $A$ consists of the algebraic functions on an affine algebraic variety in which case Kähler differentials are often taken as a definition of 1-forms.
However, when $A = C^\infty(X)$ consists of smooth functions on a manifold, the ring theoretic Kähler differentials do not agree with the ordinary smooth 1-forms on this manifold! (Unless $X$ is, for instance, the point, of course). However, there is a canonical map from the Kähler differentials to the ordinary 1-forms.
But there is a solution to this, and an explanation for why something goes wrong:
Smooth spaces such as manifolds are not modeled on the category $C =$ CRing${}^{op}$, as varieties are. Instead, they are modeled on the category $C = \mathbb{L}$ of smooth loci, which is $= C^\infty Ring^{op}$ the opposite of the category of C-infinity rings.
In particular, the algebra $A = C^\infty(X)$ of smooth functions on a manifold carries naturally the structure of such a $C^\infty$-ring. This does have “underlying” it the ordinary commuttative ring of functions that forget the $C^\infty$-ring structure, but forgetting this structure is precisely what makes the definition of Kähler differentials fail to reproduce that of ordinary smooth 1-forms.
If we do regard $C^\infty(X)$ as a C-infinity ring, the its Kähler differentials do agree with ordinary 1-forms on $X$.
We discuss how Kähler differential forms relate to the ordinary notion of differential forms.
Since there are only finite sums in the module of Kähler differentials, the usual $d f=f' d t$ works only if $f$ is a finite polynomial in $t$, say, if $A$ is $C^\infty(\mathbb{R})$ (smooth maps) or $\mathbb{R}[\![t]\!]$ (power series). For example, let $f = t^n$ then
However, we have
as Kähler differentials. Intuitively, the reason is that $d$ cannot pass through the infinite sum
However, the only proof we know that $d e^t \ne e^t d t$ is quite tricky: in fact it uses the Axiom of Choice!
It would be desirable to either find a proof that avoids the Axiom of Choice, or show that axioms beyond ZF are necessary for this result.
To avoid this annoying property of Kähler differentials we can proceed as follows. Given a commutative algebra $A$, let $Der(A)$ be the $A$-bimodule of derivations.
Define $\Omega^1(A)$ to be the dual of $Der(A)$:
in other words, the set of $A$-module maps $\omega : Der(A) \to A$, made into an $A$-module in the usual way. There is a derivation
given by
for $X \in Der(A)$.
Now, suppose $A=C^\infty(M)$ where $M$ is any smooth manifold. Then elements of $\Omega^1(A)$ can be identified with ordinary smooth 1-forms on $M$:
from the Hadamard lemma it follows that $Der(C^\infty(M)) = \Gamma(T M)$ is precisely the $C^\infty(X)$-module of vector fields on $X$, and 1-forms are the $C^\infty(X)$-linear duals of vector fields, by definition.
And in this case, one can show that any derivation $X: A \to M$ factors through $\Omega^1(A)$ when $M$ is free, and in particular if $M=A$.
We can expand on this remark as follows. Quite generally, for any commutative algebra $A$ over a field $k$, we have
by the universal property of Kähler differentials, which identifies derivations $A \to A$ with $A$-module morphisms $\Omega^1_K(A) \to A$.
Using the definition
(of ordinary smooth 1-forms in the case that $A = C^\infty(M)$) we have that these are the linear bidual of the Kähler differentials:
There is always a homomorphism from a module to its double dual, so we have a morphism
In the case when $A = C^\infty(M)$ this map is onto but typically not one-to-one, as witnessed by the fact that $d e^t = e^t d t$ in the ordinary 1-forms $\Omega^1(A)$ but not in the Kähler differentials $\Omega^1_K(A)$. However, one can show that when $M$ is a free $A$-module, any derivation $X: A \to M$ not only factors through $\Omega^1_K(A)$ (as guaranteed by the universal property of Kähler differentials), but also $\Omega^1(A)$. More generally this is true for torsionless $M$, i.e. $A$-modules that inject into their double duals, since $\Omega^1(A)$ is torsionless. So $\Omega^1_K(A)$ is the universal differential module for all modules, while $\Omega^1(A)$ is the universal differential module for torsionless $A$-modules.
Martin Gisser? Couldn’t resist the above uncredentialed drive-by edit… Is it torsionlessness that ultimately discerns differential geometry from algebraic geometry? (Need to fill in more dots between torsionless modules and “geometric modules” in the sense of Jet Nestruev.)
Eric: Does this universal property mean that there is some diagram in some category for which the Kähler differentials can be thought of as a (co)limit?
John Baez: Yeah, take the category all $A$ modules $M$ equipped with a derivation $X : A \to M$, and take the diagram which consists of every object in this category and every morphism, and take the colimit of that, and you’ll get $\Omega^1_K(A)$.
But this is just a cutesy way to say that $\Omega^1_K(A)$ is the initial object of this category.
And this, in turn, is just a cutesy way to say that there is a derivation
such that if $X:A\to M$ is also a derivation, then there exists a unique $A$-module morphism
such that the following diagram commutes:
All this is general abstract nonsense, nothing special to this example! Any universal property involving maps out of an object says that object is initial in some category — and that, in turn, is equivalent to saying that object is the colimit of the enormous diagram consisting of all objects of the same kind! There’s a lot less here than meets the eye.
Eric: Thank you! That actually makes a little sense to me. As trivial as it may seem, the fact that I was even able to ask this question represents tremendous progress :)
Herman Stel: Dear Eric and Prof. Baez. There is a mistake in the explanation by John Baez here. The two latter properties (both being that the derivation $d : A \to \Omega^1_K(A)$ is initial) are correct. The first one is not, though. If $\Omega^1_K(A)$ were the colimit of the huge diagram then for every derivation there would be a morphism from that derivation to the universal derivation, which is not true. Instead, note that an initial object is the vertex of a colimit of the empty diagram in any category (use that $\forall x\in X P(x)$ is true if $X$ is empty).
A $C^\infty$-ring (see generalized smooth algebra) is a ring that remembers that it carries extra smooth structure akin to the smooth structure carried by a ring of smooth functions on a smooth manifold.
If we take a smooth function ring $C^\infty(X)$, regard it as a $C^\infty$-ring and then determine its module of Kähler differentials with respect to the category $C = \mathbb{L} = C^\infty Ring^{op}$, we do recover the ordinary notion of smooth differential forms.
For the moment, this case is described in detail in the entry on Fermat theory.
An ordinary (commutative) ring is precisely a comutative monoid in the category $Ab$ of abelian groups. The case of Kähler differentials over ordinary rings discussed above may therefore be thought of as the case where the category of test objects is taken to be
This has an evident generalization: we may replace here $Ab$ with any category $K$ and consider
In practice $K$ is usually required to be an abelian category, but our definitions so far are general enough not to be concerned about this:
for any such $K$ fixed we follow the general definition, consider the functor
and define the assignment of Kähler differentials
to be the left adjoint of this functor.
For this to make good sense, everything here should be regarded as taking place in (∞,1)-categories, typically modeled by the model structure on simplicial rings.
Then one finds that for $C = sCRing^{op}$ the corresponding notion of module (∞,1)-category reproduces the derived category of ordinary ring modules. This is Example 8.6. in
If in the above setup we choose $K = sAb = [\Delta^{op}, Ab]$ the category of abelian simplicial groups, then $Mon(K)$ is the category of simplicial rings. The category $Mon(K)^{op}$, regarded as a higher category, is the site used in higher geometry in place of $CRing^{}$
For a proof that every derivation of $A = C^\infty(\mathbb{R})$ comes from a smooth vector field on the real line, and an extensive discussion of Kähler differentials versus ordinary 1-forms, see:
See also the discussion at the $n$-Café:
Blog discussion of Week 287. (Summary)
The module of Kähler differentials, MIT OpenCourseWare: 18.726 Algebraic Geometry, Spring 2009.
A detailed discussion of Kähler differentials and their generalization from algebra to higher algebra is in
introducing a setting in which Kahler differentials live quite naturally (but not yet in as much generality as possibly one might hope), see
This paper establishes a relation between the recently introduced notion of differential category and the more classic theory of Kähler differentials in commutative algebra. A codifferential category is an additive symmetric monoidal category with a monad, which is furthermore an algebra modality. An algebra modality for a monad T is a natural assignment of an associative algebra structure to each object of the form T(M). In a (co)differential category, one should imagine the morphisms in the base category as being linear maps and the morphisms in the (co)Kleisli category as being infinitely differentiable. Finally, a differential category comes equipped with a differential combinator satisfying typical differentiation axioms, expressed coalgebraically.
The traditional notion of Kähler differentials defines the notion of a module of A-differential forms with respect to A, where A is a commutative k-algebra. This module is equipped with a universal A-derivation. With this in mind, a Kähler category is an additive monoidal category with an algebra modality and an object of differential forms associated to every object. This object of differential forms satisfies a universal property with respect to derivations. Surprisingly, we are able to show that, under some natural conditions, codifferential categories are Kähler.
Last revised on January 20, 2018 at 15:52:54. See the history of this page for a list of all contributions to it.