What's the relationship between Kahler differentials and ordinary differential forms?

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    I would have thought this got mentioned in some book on either commutative algebra or differential geometry... I'm not an expert, but the universal property of Kahler differentials should be suggestive. – Yemon Choi Nov 19 '09 at 9:00
  • I agree, but I just can't find where. – Abtan Massini Nov 19 '09 at 9:43
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    There is some discussion of this in section 8.1 of "Elements of noncommutative geometry" by Gracia-Bondia, Varilly, and Figueroa. See in particular the example starting on page 324. – Zach Conn Jun 20 '11 at 23:26
  • @ZachConn: I believe there's an erratum in that book. Their proposition 8.1 is asserting what has been proven below to be false. If you look at the proof, the problem is that they are identifying the double dual of $\Omega_k^1(C^\infty(M))$ with itself, which is a mistake. – Bruno Stonek Apr 27 '16 at 14:57
up vote 46 down vote accepted

Let $M$ be a differentiable manifold, $A=C^\infty (M)$ its ring of global differentiable functions and $\Omega^1 (M)$ the A-module of global differential forms of class $C^\infty$. The A-module of Kähler differentials $\Omega_k(A)$ is the free A-module over the symbols $db$ ($b \in A$) divided out by the relations

$d(a+b)=da+db,\quad d(ab)= adb+bda,\quad d\lambda=0 \quad(a,b\in A, \quad \lambda \in k)$

There is an obvious surjective map $\quad \Omega_k(A) \to \Omega^1 (M)$ because the relations displayed above are valid in the classical interpretation of the calculus (Leibniz rule). However, I do not believe at all that it is injective.

For example, if $\: M=\mathbb R$, I see absolutely no reason why $\mathrm{d}\sin(x)=\cos(x)\mathrm{d}x$ should be true in $\Omega_k(A) $ (beware the sirens of calculus). Things would be worse if we considered $C^\infty$ functions which, unlike the sine, are not analytic. The same sort of reasoning applies to holomorphic manifolds and also to local rings of differentiable or holomorphic functions on manifolds.

To sum up: the differentials used in differentiable or holomorphic manifold theory are a quotient of the corresponding Kähler differentials but are not isomorphic to them. (And I think David's claim that they are isomorphic is mistaken.)

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    Dear John, this is entirely possible. The definition I use is the same as in Weibel's book on Homological Algebra (8.8.1.),Qing Liu's (6.1.1.)and Hartshorne's (II,8) on Algebraic Geometry, Matsumura's on Commutative Ring Theory (9.25),Bourbaki's Algebra ( III,$10,11) and actually in all the books on the subject that I know of. What other definition of Kähler differentials do you have in mind? Greetings, – Georges Elencwajg Nov 19 '09 at 20:36
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    Georges is quite correct in that it is not trye that $d\mathrm{sin}=\mathrm{cos}dx$ in the module of Kähler differentials for $A$. – Mariano Suárez-Álvarez Nov 19 '09 at 20:40
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    If I understood correctly the thread at $n$-cafe basically concludes with the ordinary de rham differentials being the double dual of kahler differentials. Is that right? – Saal Hardali Jan 2 '16 at 8:44
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    @Georges I understand that you may have strong feelings about the n-cafe, and you are entitled to them. However, I'm not sure that your comment addresses the issue. The mathematical content of Saal's question is whether the map $\Omega_k(A) \to \Omega^1(M)$ can be identified with the map $\Omega_k(A) \to \Omega_k(A)^{**}$, where $N^* = Hom_A(N,A)$ is the $A$-linear dual. This is not addressed by David Speyer's answer because perhaps every $A$-linear map $\Omega_k(A) \to A$ annihilates $d\sin(x) - \cos(x)dx$, by contrast with his construction $\Omega_k(A) \to K$ using an ultrafilter. – Tyler Lawson Apr 22 '16 at 17:53
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    I'll mention that it is true that $\Omega^1(M)$ is the double dual of $\Omega_k(A)$. It follows from proposition 4.7 (vector fields are derivations on $C^\infty(M)$) and exercise 11-7 (tensor fields are $C^\infty(M)$-linear functions on $T(M)$) in Lee's smooth manifolds. – Denis Nardin Apr 22 '16 at 18:00

There is a discussion of this issue at the $n$-category cafe. I'd encourage people who were interested in this question to head over there and see if they can lend some insight.

Here is a sketch of a proof that $d (e^x) \neq e^x dx$ in the Kahler differentials of $C^{\infty}(\mathbb{R})$. More generally, we should be able to show that, if $f$ and $g$ are $C^{\infty}$ functions with no polynomial relation between them, then $df$ and $dg$ are algebraically independent, but I haven't thought through every detail.

Choose any sequence of points $x_1$, $x_2$, in $\mathbb{R}$, tending to $\infty$. Inside the ring $\prod_{i=0}^{\infty} \mathbb{R}$, let $X$ and $e^X$ be the sequences $(x_i)$ and $(e^{x_i})$. Choose a nonprincipal ultrafilter on the $x_i$ and let $K$ be the corresponding quotient of $\prod_{i=0}^{\infty} \mathbb{R}$.

$K$ is a field. Within $K$, the elements $X$ and $e^X$ do not obey any polynomial relation with real coefficients. (Because, for any nonzero polynomial $f$, $f(x,e^x)$ only has finitely many zeroes.) Choose a transcendence basis, $\{ z_a \}$, for $K$ over $\mathbb{R}$ and let $L$ be the field $\mathbb{R}(z_a)$.

Any function $\{ z_a \} \to L$ extends to a unique derivation $L \to L$, trivial on $\mathbb{R}$. In particular, we can find $D:L \to L$ so that $D(X)=0$ and $D(e^X) =1$. Since $K/L$ is algebraic and characteristic zero, $D$ extends to a unique derivation $K \to K$. Taking the composition $C^{\infty} \to K \to K$, we have a derivation $C^{\infty}(\mathbb{R}) \to K$ with $D(X)=0$ and $D(e^X)=1$. By the universal property of the Kahler differentials, this derivation factors through the Kahler differentials. So there is a quotient of the Kahler differentials where $dx$ becomes $0$, and $d(e^x)$ does not, so $dx$ does not divide $d(e^x)$.

I'm traveling and can't provide references for most of the facts I am using aobut derivations of fields, but I think this is all in the appendix to Eisenbud's Commutative Algebra.

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    It is a standard fact that given a field extension $L/K$ with $K$ of characteristc zero, a subset $S\subseteq L$ is algebraically independent over $K$ iff $\{ds:s\in S\}$ is linearly independent in $\Omega_{L/K}$ over $L$. If $\mathcal M$ is the field of meromorphic functions on $\mathbb C$, one can use this to show that $e^x$ and $x$ are linearly independent in the $\mathcal M$-vector space $\Omega_{\mathcal M/\mathbb C}$. – Mariano Suárez-Álvarez Dec 25 '09 at 16:24
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    Don;t you hate when people say "it is a standard fact that..." without giving references? The above claim is stated and proved, for example, in Matsumura's Commutative Algebra as theorem 87. – Mariano Suárez-Álvarez Dec 25 '09 at 16:30

UPDATE: My answer essentially just gives the definition of Kahler differentials and differential forms and misses the point of the question. Georges' answer addresses the relationship between the two. As David before me, I also encourage you to vote Georges' answer up and mine down.

Let $M$ be a smooth manifold and $p$ a point in $M$. The usual definition of the tangent space to $M$ at $p$ is as the vector space of linear maps $D: C^{\infty}(M) \to \mathbb{R}$ satisfying the Leibniz rule $$D(fg) = D(f)g(p) + f(p)D(g)$$ Equivalently, let $I$ be the ideal of $C^{\infty}(M)$ consisting of all functions vanishing at $p$. Then $T_p M$ is the dual of the vector space $I/I^2$ (which you hence call the cotangent space to $M$ at $p$). Indeed, $D(f) = 0$ for every $f \in I^2$, and conversely any linear map $r: I/I^2 \to \mathbb{R}$ gives rise to a derivation $D(f) := r(f-f(p))$.

Now let $X$ be a scheme over a field $k$ (you can generalize this to a morphism of schemes) and $x$ a closed point. Consider the local ring $\mathcal{O}_{x, X}$ of functions regular at $X$. Then the stalk at $x$ of the sheaf of Kahler differentials $\Omega^1_X$ corepresents the functor taking an $\mathcal{O}_{x,X}$-module $\mathcal{F}_x$ to $\mathrm{Der}(\mathcal{O}_{x,X}, \mathcal{F}_x)$. In particular, $$\mathrm{Der}(\mathcal{O}_{x,X}, k) \cong \mathrm{Hom}(\Omega^1_{X,x}, k)$$

It is in this sense that you think of $\Omega^1_{X,x}$ as the cotangent space to $X$ at $x$. Indeed, in this case $\Omega^1_{X,x} \cong m/m^2$ where $m \subset \mathcal{O}_{x, X}$ is the ideal of functions vanishing at $x$.

  • I think "represents" is not to be taken in it's literal sense in the sentence "the stalk at x of the sheaf of Kähler differentials represents derivations". But it can easily be rewritten to eventually get a representation of $Der(\mathcal{O}_{X,x}, k)$. – Konrad Voelkel Nov 19 '09 at 16:15
  • You're right: that choice of words was unfortunate. I have clarified it now. – Alberto García-Raboso Nov 19 '09 at 16:51
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    Dear Alberto, thank you for your update but of course I strongly discourage anybody to vote you down! I am incredibly impressed by the intellectual honesty and fair-play that I am witnessing from David and you: this is MUCH more admirable than any result on Kähler differentials. – Georges Elencwajg Nov 20 '09 at 19:42

UPDATE The previous answer that was here was pretty much completely wrong. Thanks to Georges for several corrections. I encourage everybody to vote my answer down and his up.

If $M$ is a $C^{\infty}$ manifold and $A$ is the ring $C^{\infty}(M)$ then there is a natural map from the Kahler differentials $\Omega_{A/\mathbb{R}}$ to the $C^{\infty}$ one-forms. This is surjective but far from injective. For example, $d \sin x \neq \cos x dx$ in the Kahler differentials. The basic problem is that Kahler differentials are only linear for finite sums, so they can't "see" nonpolynomial relations.

If you replace $C^{\infty}$ by complex analytic and $M$ is an open simply connected set in $\mathbb{C}^n$, then the Kahler differentials map to the holomorphic $(1,0)$-forms. This is still true for any complex manifold if interpreted as a statement about sheaves; let $\mathcal{H}$ be the sheaf of holomorphic functions and define the sheaf of Kahler differentials by sheafifying the presheaf $U \mapsto \Omega_{\mathcal{H}(U)/\mathbb{C}}$; then we again have a map from Kahler differentials to holomorphic $(1,0)$ forms.

In algebraic geometry, if $M$ is a smooth complex algebraic variety, one usually considers the sheaf gotten by sheafifiying the presheaf $U \mapsto \Omega_{\mathcal{O}(U)/\mathbb{C}}$. (We are now using the Zariski topology.) These are, by definition, the algebraic $(1,0)$ forms. They are not isomorphic to the holomorphic $(1,0)$ forms but, if $M$ is projective, they have the same cohomology by GAGA.

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    Dear David, unless I misunderstand you I think the statements in the differentiable and holomorphic case are not quite correct. Would you care to have a look at my answer below? Friendly, – Georges Elencwajg Nov 19 '09 at 17:58
  • I think I'm right. I was a little confused by your answer because you spent a lot of time doing constructions with local rings, which I don't think you need to do. I'd be glad to hear about any error. Is it not true that any 1-form on a C^{\infty} manifold can be written as sum f_i d g_i, with f_i and g_i C^{infty} functions? I think this follows from the local fact and partitions of unity. Or am I wrong that two such expressions are equal if and only if their equality follows from the Liebnitz rule and linearity? Or is there some other error? – David E Speyer Nov 19 '09 at 20:41
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    I think you are confusing me with Alberto (just below), since I didn't spend any time at all on constructions with local rings! MY post is the last one. It is perfectly true that 1-forms on M are, as you say, sums of global forms of the type fdg ( Kähler differentials surject on classical 1-forms as I mention in my post). The problem is, as you suggest yourself, that the equality of honest 1-forms cannot be proved just by linearity and Leibniz. To repeat my example, how would you prove in the case of M=R(the reals) that d(sin x)= (cos x)dx (true as one-forms) is true in the Kähler module ? – Georges Elencwajg Nov 19 '09 at 22:16
  • Dear David, your gentlemanly reaction to my comment is incredibly gracious and generous. Let me use this opportunity to thank you and your colleagues for this (addictive!) wonderful site, whose lightning-fast success (1600 users in a little over a month) is more than deserved. – Georges Elencwajg Nov 20 '09 at 7:42
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    Dear David, I see some people (yourself maybe?!) have voted you down. This is ridiculous: I am voting you up ! – Georges Elencwajg Nov 20 '09 at 19:50

Hello everybody, I'm not a mathematician :) so sorry not to write the details here. Basically, I studied engineering and control theory. In nonlinear control theory we often use differentials, either Kahler or ordinary, to study the systems. There has been a long discussion whether those two notions are isomorphic or not. I believe, we have the answer finally. Together with my colleagues we have shown that differential one-froms are isomorphic to a quotient space (module) of Kahler differentials. These two modules coincide when they are modules over a ring of linear differential operators over the field of algebraic functions. It was published as G. Fu, M. Halás, Ü. Kotta, Z. Li: Some remarks on Kähler differentials and ordinary differentials in nonlinear control theory. In: Systems and control letters, article in press, 2011. Available online at http://www.sciencedirect.com/science/article/pii/S0167691111001198

The key ingredient in proofs that derivations are differential forms is the ability to evaluate coefficients: e.g. to go from $\mathrm{d}f(x) = f'(x) \mathrm{d}x$ to $\mathrm{d}f(x)|_a = f'(a) \mathrm{d}x|_a$.

Evaluation maps $\theta_P : \mathcal{C}^\infty(M) \to \mathbb{R} : f \to f(P)$ are examples of ring homomorphisms; abstractly, for any ring homomorphism $\varphi : R \to S$, we can consider the corresponding "evaluation" $$ \varphi_* : \Omega_R \cong R \otimes_R \Omega_R \to S \otimes_R \Omega_R $$ The $S$-module $S \otimes_R \Omega_R$ is isomorphic to the quotient of $\Omega_R$ by the relations $f \mathrm{d}g = \varphi(f) \mathrm{d}g$, and $\varphi_*(f \mathrm{d}g) = \varphi(f) \mathrm{d}g$.

In the case that we use one of the $\theta_P$, $\theta_{P*}$ is evaluating the coefficients of a Kähler differential at $P$, and so the corresponding $\mathbb{R} \otimes_{\mathcal{C}^\infty(M)} \Omega_{\mathcal{C}^\infty(M) / \mathbb{R}}$ looks like the cotangent space to $M$ at $P$.

Write $\mathrm{d}f|_P$ for $\theta_{P*}(\mathrm{d}f)$.

Now, we can apply the usual argument. For $M = \mathbb{R}^n$, to any function $f$, we can take the differential of a Taylor polynomial and get $$ \mathrm{d}f(x)|_P = f'(P) \mathrm{d}x|_P$$ so it's clear that we really do get the cotangent space to $M$ at $P$.

If we take all Kähler differentials and identify differentials that have the same "values" at every point of the manifold, then we get the ordinary differential forms.

Conjecture If $M$ is a compact manifold, then $\Omega_{\mathcal{C}^\infty(M) / \mathbb{R}} \cong \Omega^1(M)$.

(this conjecture has been posted to Are Kähler differentials the same as one-forms for compact manifolds?)

  • Isn't the $sin(x)dx$ example from the accepted answer, put on a circle, a counter-example to the conjecture? – bianchira Mar 6 at 2:50
  • @bianchira: A demonstration would be great! However, my state of knowledge is that compactness is important here, so switching from $\mathbb{R}$ to $S^1$ changes an essential feature of the problem. I had posted my conjecture as a question, but have gotten no answers. – Hurkyl Mar 6 at 5:53

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