Note: For the record, exterior algebras and derivations are irrelevant to my question. However, I have a hard time assessing what I want to ask and I find it is the easiest to do so using a direct example. Hence exterior algebras.

When studying derivations of the exterior algebra over a smooth manifold $M$, $\Omega(M)$, one often encounters statements such as

  • A derivation is local, eg. for $\omega\in\Omega(M)$, $(D\omega)|_U=D(\omega|_U)$, where $D$ is a derivation of the exterior algebra, and $U$ is an open set of $M$.

  • A derivation is algebraic if it vanishes on the grade 0 subspace. An algebraic derivation is ultralocal, in the sense $D\omega|_p$ depends only on $\omega_p$ if $D$ is an algebraic derivation.

Many of these statements reference or use the fact that $\Omega(M)$ is an associative, anticommutative $\mathbb{R}$-, and $C^\infty(M)$-algebra, that is also $\mathbb{N}$-graded, however for example the two I singled out rely on the fact that an element of $\Omega(M)$ is a map $M\rightarrow\wedge T^*M$ (that is also a section).

If I wanted to abstract studying derivations of the exterior algebra into studying derivations of an $R$-algebra ($R$ is a commutative ring) that is associative, anticommutative and $\mathbb{N}$-graded, I could do it, but then statements such as "$D$ is local" or "$D$ is ultralocal" would make no sense, since if $\omega$ is an element of this algebra, things such as $\omega|_U$ or $\omega_p$ don't make sense.

I realize that the exterior algebra has two additional operations in this case, a "restriction", which maps a pair $(U,\omega)$ ($U$ is open and $\omega$ is a differential form on $M$, but it can also be a differential form defined on any region that contains $U$) to $\omega|_U\in\Omega(U)$ and an "evaluation", which maps the pair $(p,\omega)$ (where $p\in M$ and $\omega$ is once again either a form on $M$ or a form defined on a region that contains $U$) to $\omega_p\in\wedge T_p^*M$, but it is even hard for me to come up with a rigorous description of these maps. I'll attempt:

Evaluation is the following map: $$\text{Eval}:M\times\Omega(M)\rightarrow\wedge T^ *M$$ such that $\text{Pr}_1(M\times\Omega(M))=\pi_{\wedge T^*M}\circ\text{Eval}$, where $\text{Pr}_1(M\times\Omega(M))$ is the projection to the first factor.

Let the "space of local differential forms" be $\Omega_{\text{loc}}(M)=\bigsqcup_{U\in\tau_M}\Omega(U)$, so an element of $\Omega_{\text{loc}}(M)$ is a pair $(U,\omega)$ where $U\in\tau_M$ is open and $\omega$ is defined on $U$ only. The restriction is then a map $$ \text{Res}:\tau_M\times\Omega(M)\rightarrow\omega_{\text{loc}}(M) $$ such that $\text{Pr}_1(\tau_M\times\Omega(M))=\text{Dom}\circ\text{Res}$ and $\text{Eval}(p,\cdot)\circ\text{Res}=\text{Eval}(p,.)\circ\text{Pr}_2(\tau_M\times\Omega(M))$ where $\text{Eval}$ has been extended to a partial function on $\Omega_{\text{loc}}(M)$ and $p$ is an arbitrary point of $M$ for which "Eval" as a partial function is defined.

I have no idea whether my "Res" and "Eval" functions capture the "locality" of the exterior algebra well, and although no references are made to elements of the exterior algebra being functions, I have referenced $M$, even though for a general $R$-algebra, I'd have no access to $M$.

So basically, my question is, how can I tell if an $R$-algebra/ring I am studying in the abstract can be realized as some kind of algebra of maps? Is there a name for such algebras/rings?


You have an algebra $A$ over a field $k$ and you want to know whether it is isomorphic to an algebra of functions on some set $M$, right? But functions into what? For instance, $A$ is trivially isomorphic to the set of functions from a one-element set into $A$.

If you want it to be isomorphic to an algebra of functions from $M$ into $k$, then I can give a more meaningful answer (though it still may not be what you want). Any algebra $A$ of functions from some set $M$ into $k$ comes with evaluation homomorphisms $\hat{p}: A \to k$ given by $\hat{p}(f) = f(p)$, for all $p \in M$. So a necessary condition is that there is a separating family of homomorphisms from $A$ into $k$. "Separating" means that the intersection of their kernels is $\{0\}$.

In fact this condition is sufficient, too. Given such a family, suppose it is indexed by a set $M$, i.e., the homomorphisms are $\{\phi_p: p \in M\}$. Then define a map $\Gamma: A \to k^M$ (where $k^M$ is the algebra of functions from $M$ into $k$) by $(\Gamma a)(p) = \phi_p(a)$. If the family is separating this will be a monomorphism.

I'm not an algebraist, so this answer may be totally off base, sorry.

  • $\begingroup$ I did not totally think this through, but I'd say, I want to know when is an $R$-algebra ($R$ is a commutative ring) isomorphic to a $C^ k(M)$-algebra, where $C^k(M)$ is the ring of $C^k$-functions over a $C^k$-manifold $M$, where $k$ might be $0,1,...,\infty,\omega$. But I guess we can also have $M$ be a topological space without it having to be a topological manifold. (I guess isomorphism of $R$ and $C^k(M)$ is also needed for that.) $\endgroup$ – Bence Racskó Jan 2 '17 at 22:01
  • $\begingroup$ Hmm. But if you want that much structure, I wonder what extra value is brought by having an abstract characterization. Why not just assume the algebra you're working with is of this type? $\endgroup$ – Nik Weaver Jan 2 '17 at 23:41
  • $\begingroup$ I don't know if there is any value, that's it. I don't "need" this for anything, I am just interested. I feel the thoughts behind this inquiry would eventually lead to sheaf theory, which is something I am very very roughly aware of. I figured there must be some palpable characteristics that distinguish a "localizable" algebra from a general one. I might be wrong, but that's why I asked. And that's why my question is so vague, it is very hard for me to express via words what I'm actually curious about. $\endgroup$ – Bence Racskó Jan 3 '17 at 0:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.