
> 
> What are the derivations of the algebra of *continuous* functions on a topological manifold?
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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](http://en.wikipedia.org/wiki/Supermanifold) has a detailed description. The n[-lab post](http://ncatlab.org/nlab/show/supermanifold) 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? 
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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?



Derivations of C(X)? or Why Must Supermanifolds be Smooth?
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.


