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1) Consider the algebra $\mathcal T (M) = \bigoplus \limits _{p, q \ge 0} \mathcal T ^{p, q} (M)$, where $\mathcal T ^{p, q} (M)$ is the space of tensor fields of type $(p,q)$. I should endow it with some topology, but I do not know what to choose: the topology of pointwise convergence is one option, the compact-open topology another one, but I am open to other suggestions as well.

Is it possible to recover both the topological and the differential structure of $M$ from $\mathcal T (M)$? (This would be some sort of Gelfand-Naimark in a smooth setting; in fact, the component $\mathcal T ^{0, 0} (M)$ endowed with a suitable topology would lead to precisely some version of Gelfand-Naimark.)

A related question is: do we really need to consider all the types of tensors above? Given that covariant and contravariant are dual to each other (and we are working with manifolds of finite-dimension), shouldn't these two classes contain the same amount of information, either one being redundant for the issue under discussion? It seems that one could restrict to either $\bigoplus \limits _{q \ge 0} \mathcal T ^{0, q} (M)$, or to $\bigoplus \limits _{p \ge 0} \mathcal T ^{p, 0} (M)$.

2) I am curious about the same question but for a different algebra: if $\mathcal X (M)$ is the Lie algebra of the smooth vector fields on $M$, let $\mathcal U (M)$ be its universal enveloping algebra. Does $\mathcal U (M)$ encode all the information about $M$? Can $M$ be recovered from $\mathcal U (M)$? (This latter algebra, though, does not contain the smooth functions, so I have some doubts about it.)

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    $\begingroup$ What you want is Connes' reconstruction theorem. See Branimir Cacic's answers to mathoverflow.net/questions/16833/… and mathoverflow.net/questions/191720/commutative-spectral-triple $\endgroup$ – user1504 Jan 19 '16 at 20:48
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    $\begingroup$ @user1504 To be pedantic: This gives you a condition for when an algebra is of the form $C^\infty(M)$ for some smooth manifold $M$ (which is difficult). If I understood the question right, it is about the next step, where we already know that the algebra in question is the algebra of functions of some manifold and try to reconstruct it algebraically. $\endgroup$ – Bertram Arnold Jan 19 '16 at 21:59
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    $\begingroup$ That may be true, but one might as well know about Connes' theorem when asking this sort of question. Not totally clear from the 2nd paragraph if the OP did. $\endgroup$ – user1504 Jan 19 '16 at 22:09
  • $\begingroup$ @user1504: No, I wasn't aware of Connes's theorem, your pointers to the other two questions are very helpful, thank you. They tell me that my question is intelligently posed, provided that I add some supplementary structure on $M$ (in Connes's case, a Riemannian structure). $\endgroup$ – Alex M. Jan 20 '16 at 12:21
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Look at $M=S^7$. This has 28 smooth structures, and their tangent bundles are all trivial - compare Parallelizability of the Milnor's exotic spheres in dimension 7 . Thus their tensor algebras do not carry more information about the smooth structure than $C^\infty(M)$.

On the other hand, for a compact smooth manifold, the algebra of smooth functions is sufficient to recover the smooth structure: Every maximal ideal of this algebra is of the form $\{f\in C^\infty(M): f(x) = 0\}$ for some $x\in M$, and the topology on $M$ is recovered as the weakest one on the space of maximal ideals such that $\mathfrak{m}\mapsto [f]\in C^\infty(M)/\mathfrak{m}\cong \mathbb{R}$ is continuous for all $f\in C^\infty(M)$. So this space is a compact topological manifold with a subsheaf $C^\infty$ of the sheaf of continuous functions, and we take a chart $U\to\mathbb{R}^n$ to be smooth if all of its coordinate functions are in $C^\infty(U)$.

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  • $\begingroup$ 1) Is it obvious (or easy to prove) that the spectrum of $C^\infty(M)$ is a topological manifold and not just a topological space? 2) Since the coordinate functions are not necessarily defined on $M$, did you mean them to belong to $C^\infty(U)$ where $U$ is the domain of the said chart? $\endgroup$ – Alex M. Jan 19 '16 at 19:36
  • $\begingroup$ ad 1: It is homeomorphic to $M$, so it certainly is a topological manifold. ad 2: This hinges on the construction of the subsheaf $C^\infty$: Say that a continuous function $f: U\to \mathbb{R}$ is smooth if for all smooth functions $\chi$ such that supp $\chi\subset U$, the function $f\chi$ is in $C^\infty(M)$. $\endgroup$ – Bertram Arnold Jan 19 '16 at 19:42

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