Kahler differentials and Ordinary Differentials

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.)


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.


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 $x$.