What are the derivations of the algebra of continuous functions on a topological manifold?

A supermanifold is a locally ringed space (X,O) whose underlying space is a smooth manifold X, and whose sheaf of functions locally looks like a graded commutative algebra which is an exterior algebra over the sheaf of smooth functions. The wikipedia article has a detailed description. The n-lab post is a little thinner, but by pointing is out I've probably set in motion events that will alter it to be the superior article. Ahh, Heisenberg Uncertainty on the internet.

What neither of these article tell you is why must the manifold be smooth?

Once I thought I understood the reason. Now I'm not so sure. The reason I was brought up with came from looking at some examples of maps of supermanifolds. For simplicity let X be just an ordinary manifold regarded as a (trivial) supermanifold. Let $\mathbb{R}^{0|2}$ be the supermanifold whose underlying manifold is just a point and whose ring of functions is the exterior algebra $A=\mathbb{R}[\theta_1, \theta_2]$, where $\theta_i$ is an odd generator. The algebra A is four dimensional as a real algebra.

A map of supermanifolds $\mathbb{R}^{0|2} \to X$ is the same as a map of algebras $O(X) \to A$ where $O(X)$ is functions on X (of a to-be-determined type). Since X is a trivial supermanifold this is the same as pair of linear maps $a,b$:

$$ f \mapsto a(f) + b(f) \theta_1 \theta_2$$

It is easy to see that $a$ all by itself is an algebra map $O(X) \to \mathbb{R}$. This is the same thing as evalutation at a point $x_0 \in X$. Once I knew a reference for this, but I can't seem to find it. I think it is true as stated, that algebra homomorphisms $O(X) \to \mathbb{R}$ are in bijection with the points of the manifold $X$, where $O(X)$ is either smooth or continuous functions, without taking into account any topology. If this is false, then we should take into account the topology as well. In any event, $a(f) = f(x_0)$ is evaluation at $x_0$. What is $b$?

By using the algebra property of the above assignment we see that $b$ must satisfy: $$b(fg) = f(x_0) b(g) + b(f) g(x_0).$$ In other words, $b$ is a derivation of $O(X)$ at the point $x_0$.

So now the argument goes something like, varying over all the map from $\mathbb{R}^{0|2}$ to $X$ we see that the functions must have all first derivatives at all points and so must be $C^1$-functions. Using more odd manifolds $\mathbb{R}^{0|4}$, $\mathbb{R}^{0|6}$, etc. we see that they must have arbitrarily high derivatives and so must in fact be smooth.

Or do they?

In order to really make this sort of argument hold water I think you need to know something like "there aren't any derivations of the algebra of continuous functions". You need the naive continuous version of the theory to be trivial or obviously useless. Otherwise I don't see a good motivation for why one must work smoothly, why it is the only interesting situation. So my question is:

What exactly are the derivations (taking into account topology or not) of the algebra of continous functions on a topological manifold X?

This is surely equivalent to knowing the local question, i.e. when $X = \mathbb{R}^n$. Why is the naive theory of continuous supermanifolds badly behaved?

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    $\begingroup$ Possible typo: "A map of supermanifolds R^{0|2} --> X is the same as a map of algebras A --> O(X)." Should that be O(X) --> A? $\endgroup$ – David E Speyer Feb 16 '10 at 14:50
  • $\begingroup$ yes. that's a typo. I'll edit it. $\endgroup$ – Chris Schommer-Pries Feb 16 '10 at 19:03
  • $\begingroup$ Hmmm. Yes. I see that this was way too easy a question I should have put more thought into it first. $\endgroup$ – Chris Schommer-Pries Feb 16 '10 at 20:13
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    $\begingroup$ Hi Chris! For some reason the nLab entry on supermanifolds hasn't improved yet (hopefully it will, some day). But maybe that's because your question here isn't really about supermanifolds? Because what has improved is the entry on derivations! :-) That now freatures an example section with the answer to your question: ncatlab.org/nlab/show/derivation#DerOfContFuncts But this might also indicate an answer concerning supermanifolds: I suppose one can set their theory up not just for smooth functions, but for any "Fermat theory" ncatlab.org/nlab/show/Fermat+theory $\endgroup$ – Urs Schreiber Feb 18 '10 at 8:53

Here's a proof from definition. (I don't think it has anything to do with compactness.) Let us show that derivation $\delta:C(X)\to C(X)$ vanishes for any topological manifold $X$. Indeed, $\delta(1)=0$ as usual, so it suffices to show that whenever $f\in C(X)$ vanishes at $x\in X$, so does $\delta(f)$. For this, it is enough to write $f$ as a product $f=g_1g_2$ with $g_1(x)=g_2(x)=0$. Take $g_1=\sqrt{|f|}$, $g_2=f/g_1$, and define $g_2$ to be zero where it is undefined.

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    $\begingroup$ Just as a remark: a very general result of Banach algebra theory says that if $A$ is a commutative Banach algebra then every derivation $D:A\to A$ has range contained in the Jacobson radical of $A$ -- this is Marc Thomas' resolution of the Singer-Wermer conjecture. ams.org/mathscinet-getitem?mr=90d:46075 (If one knows beforehand that $D$ is continuous, then the proof is much easier and can be given/found in 1st courses on Banach algebras.) $\endgroup$ – Yemon Choi Feb 16 '10 at 18:49

Let $X$ be a topological space and let $d:C(X)\to C(X)$ be a not-necessarily continuous derivation. Let $f\in C(X)$ and let $x\in X$. I want to show that $d(f)(x)=0$, and to do so it is enough to consider the case in which $f$ is real-valued and non-negative, for every element of $C(X)$ is a $\mathbb C$-linear combination of such functions. But in that case there exists a function $g\in C(X)$ such that $f=g^2$, and then $d(f)(x)=2g(x)d(g)(x)=0$ because $g(x)=0$.

(Note that by [Sakai, Shôichirô. On a conjecture of Kaplansky. Tôhoku Math. J. (2) 12 1960 31--33. MR0112055 (22 #2913)] a derivation of a $C^*$-algebra is automatically continuous, so imposing continuity in this context does not change much)

  • $\begingroup$ Hmm. I edited the details in while t3suji was typing, I guess. $\endgroup$ – Mariano Suárez-Álvarez Feb 16 '10 at 14:52
  • $\begingroup$ I'm sorry --- it was not clear to me from your post that you were going to give a proof, and I was worried that you'd just leave a reference. $\endgroup$ – t3suji Feb 16 '10 at 15:09

Assuming that the manifold has a smooth structure (not guaranteed, I know) then can't you argue as follows? Any derivation of the continuous functions must restrict to a derivation of the subalgebra of smooth functions and we know what all of those are (i.e. differentiation), and that they don't extend to the algebra of continuous functions. Hmm, to complete the argument as simply as possible then I want to use the fact that smooth functions are dense in continuous ones so I guess I also want to work with continuous derivations.

But, of course, I can remove my first assumption since any manifold locally has a smooth structure and derivations are local phenomena! And I also note that Mariano quotes a result that says any derivation must be continuous. Voila!

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    $\begingroup$ You would need to know all the derivations from smooth functions to continuous ones, not just the derivations from smooth functions to smooth functions, for this argument to work. Fortunately, I think you do! $\endgroup$ – David E Speyer Feb 16 '10 at 15:35
  • $\begingroup$ To localize as suggested in your second paragraph, one needs---I think---knowing something like that $d(f)(x)=0$ whenever $f$ is locally constant at $x$ (which is usually stated in this context by saying that $d$ does not increase supports). Is this true here? $\endgroup$ – Mariano Suárez-Álvarez Feb 16 '10 at 15:46
  • $\begingroup$ @Mariano: that should follow from the existence of bump functions just as in the smooth case. $\endgroup$ – Andrew Stacey Feb 16 '10 at 21:00
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    $\begingroup$ @David: hang on, isn't the question about pointwise derivations? That is, linear maps A -> R that satisfy the derivation law? So I don't understand what you mean by "all the derivations from smooth functions to continuous ones". Surely if I have a ring hom B -> A and a derivation A -> R then I get a derivation B -> R. Maybe I've misunderstood something, though. $\endgroup$ – Andrew Stacey Feb 17 '10 at 9:31

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