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.
No comments:
Post a Comment