Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let M be a (real) smooth manifold, $p \in M$ and. The space of (linear) derivations $D:C^{\infty}(p) \to \mathbb{R}$ (ie, maps satisfying D(f+g) = D(f)+ D(g) and D(fg)=D(f)g(p) + f(p)D(g)) on the algebra $C^{\infty}(p)$ of differentiable functions defined on some neighbourhood of $p$ is then a n-dimensional vector space (this is one way to define the tangent space $T_p M$ after all).
It is easy to see that if we consider instead derivations $D:C(p) \to \mathbb{R}$ on the space $C(p)$ of continuous functions, then the space of derivations is trivial.

My question is: when M is a complex (or analytic) manifold, what is the dimension of the space of derivations on holomorphic (or analytic) functions defined near p?
I've once heard that this space is infinite dimensional. Is this true? (and if it is there's a simple proof or some reference material?)

share|improve this question
2  
The space of derivations $C^\infty(U)\to\mathbb R$ is not $n$-dimensional. Indeed, derivations $C^\infty(U)\to C^\infty(U)$ are vector fields, essentially, and then you can evaluate at any point of $U$. –  Mariano Suárez-Alvarez Mar 5 '10 at 21:57
1  
@Mariano: Surely, the question must be about linear maps $D$ satisfying $D(fg)=D(f)g(p)+f(p)D(g)$. –  Harald Hanche-Olsen Mar 5 '10 at 22:25
1  
I think there may be a confusion of terminology, perhaps: it sounds like what the original poster may have heard was concerning the space of all derivations from the ring O(M) to itself, rather than from the ring into the ground field. –  Yemon Choi Mar 5 '10 at 22:33
3  
If oner works in a neighborhood of a point, there is no difference between smooth and analytic (or algebraic) case. The statements and proofs are the same. The space of derivations at a point is the tangent space, and has the dimension equal to the dimension of the manifold. On the other hand, derivations of the algebra of functions (without fixing the point) form an infinite dimensional space (Lie algebra) of vector fields. –  Pavel Etingof Mar 6 '10 at 0:50
2  
@Pavel: Yes, but in the $C^\infty$ case, the derivations can be seen as a module over $C^\infty$, and this module has finite rank. It may not have a basis, if the manifold doesn't admit a global frame (is that right terminology?), but the finite rank follows by a partition of unity argument. Try to extend that to an analytic manifold, and you run into trouble for lack or partitions of unity. This could be what's behind the problem statement. –  Harald Hanche-Olsen Mar 6 '10 at 4:43

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.