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In working of the unbounded derivation of C*-algebras. I observed the following: For topological manifold $M$, the number of closed, linear independent, unbounded derivation it admitted on $C(M)$ is exactly the dimension of $M$.

Of course this is true for smooth manifold. But I found that it may holds for arbitrary manifold. I try to google it but seems like no positive results. I would like to know if my result is known and well-studied. Thank you in advance.

Edit By linear independent, I mean the following: derivations $\delta_{1},...\delta_{n}$ with common domain $U$ is linear independent if $\sum_{i=1}^{n}c_{i}\delta_{i}=0$ for $c_{1},...c_{n}$ belong to the C*-algebra, then $c_{1}=...=c_{n}=0$.

The precise statement of what I observe is: For each dense subalgebra $U$ of $C(M)$, let $N_{U}$ be the number of closed, linear independent derivation that $U$ admitted. Then the maximum of $N_{U}$ among all $U$ in $C(M)$(if exists) is equal to the dimension of the $M$.

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    $\begingroup$ In general, given a "nice" commutative unital algebra $A$ over the complex numbers, the object which tells you "how many" derivations there are from $A$ to other symmetric $A$-bimodules is $\Omega_A$, the Kähler module of $A$. There are functional-analytic versions of $\Omega_A$ when $A$ is a Banach or Frechet algebra and one is dealing with continuous everywhere-defined derivations. I am not sure if it is possible to set up a theory for closed densely-defined derivations from a given $A$, or whether one should restrict to some chosen dense subalgebra of $A$ like I did in my previous comment. $\endgroup$
    – Yemon Choi
    Commented Jul 11, 2021 at 12:58
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    $\begingroup$ Sorry that I should make definition clear. By "linear independent" I mean the derivation is linear independent over the ring $C(M)$. So $D_{h}$ is linear dependent to $D_{h'}$ for any $h, h'\in C(T)$. $\endgroup$
    – Ken.Wong
    Commented Jul 11, 2021 at 16:28
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    $\begingroup$ It would help if you used terminology that is more standard among a (pure) mathematical audience. Note that "linear dependence" and "linear independence" are somewhat subtle when you are dealing with modules over rings: a basic example seen in homological algebra, whose details unfortunately escape me right now, is that you can have a module M over a ring R which is generated by three elements of M, such that no two of those elements will generate M on its own, yet the three elements are not "linearly independent" over R. The relevant words here are "syzygy" and "homological dimension" $\endgroup$
    – Yemon Choi
    Commented Jul 11, 2021 at 17:37
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    $\begingroup$ What is true is that under some natural but non-trivial restrictions on the algebra $A$, its Kaehler module is projective as an $A$-module, which in turn implies that locally it has a notion of "rank" (which is what you are thinking of as dimension). Then the fact that the rank of the Kaehler module of $A$ equals the "dimension" of the algebraic variety associated to $A$ is a well-known idea, although there are technical details that may need to be addressed in particular settings. $\endgroup$
    – Yemon Choi
    Commented Jul 11, 2021 at 17:41
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    $\begingroup$ I agree with Yemon that it's still unclear what "linear independent" means. For example, there are many derivations with incompatible domains. Is it a priori assumed that the derivations in consideration have a common core? $\endgroup$
    – Narutaka OZAWA
    Commented Jul 12, 2021 at 8:22

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