Kahler differentials and Ordinary Differentials

Question

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

Answer 1

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.)

Answer 2

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.

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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.

Answer 3

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 $x$.

Answer 4

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.

Answer 5

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

Answer 6

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?)