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Earlier this year it was asked on MO, "http://mathoverflow.net/questions/198098/are-there-only-countably-many-compact-topological-manifolds" Thanks to Cheeger and Kister, the answer is yes. On the other hand, Manolescu recently debunked the triangulation conjecture. A natural follow-up question asks if there is some other way to enumerate topological n-manifolds, in the sense of creating a Turing machine that will eventually output an example from every homeomorphism class of topological manifolds, given enough time.

Of course, for $n \leq 3$, TOP = PL, so I'm really interested in the cases $n\geq 4$. It's entirely possible that the answer still depends on $n$, so you can interpret the question with either $n$ fixed or variable.

If the answer is no, is it known how hard the problem of enumerating manifolds is? Is it harder than the halting problem?

Edit in response to comments below: I do not mean to jump the gun. To even have a hope that the answer to the question is yes, one would have to have some finitely computable description of topological manifolds. As BjørnKjos-Hanssen indicates in comments, this might take the form of some sequence of approximations. If a direct answer to my question seems out of reach, I would be happy with an answer explaining what is and isn't known. (I also removed the madness about reference to Turing degrees above.)

Earlier this year it was asked on MO, "https://mathoverflow.net/questions/198098/are-there-only-countably-many-compact-topological-manifolds" Thanks to Cheeger and Kister, the answer is yes. On the other hand, Manolescu recently debunked the triangulation conjecture. A natural follow-up question asks if there is some other way to enumerate topological n-manifolds, in the sense of creating a Turing machine that will eventually output an example from every homeomorphism class of topological manifolds, given enough time.

Of course, for $n \leq 3$, TOP = PL, so I'm really interested in the cases $n\geq 4$. It's entirely possible that the answer still depends on $n$, so you can interpret the question with either $n$ fixed or variable.

If the answer is no, is it known how hard the problem of enumerating manifolds is? Is it harder than the halting problem?

Edit in response to comments below: I do not mean to jump the gun. To even have a hope that the answer to the question is yes, one would have to have some finitely computable description of topological manifolds. As BjørnKjos-Hanssen indicates in comments, this might take the form of some sequence of approximations. If a direct answer to my question seems out of reach, I would be happy with an answer explaining what is and isn't known. (I also removed the madness about reference to Turing degrees above.)

Earlier this year it was asked on MO, "http://mathoverflow.net/questions/198098/are-there-only-countably-many-compact-topological-manifolds" Thanks to Cheeger and Kister, the answer is yes. On the other hand, Manolescu recently debunked the triangulation conjecture. A natural follow-up question asks if there is some other way to enumerate topological n-manifolds, in the sense of creating a Turing machine that will eventually output an example from every homeomorphism class of topological manifolds, given enough time.

Of course, for $n \leq 3$, TOP = PL, so I'm really interested in the cases $n\geq 4$. It's entirely possible that the answer still depends on $n$, so you can interpret the question with either $n$ fixed or variable.

If the answer is no, is it known how hard the problem of enumerating manifolds is? Is it harder than the halting problem?

Edit in response to comments below: I do not mean to jump the gun. To even have a hope that the answer to the question is yes, one would have to have some finitely computable description of topological manifolds. As BjørnKjos-Hanssen indicates in comments, this might take the form of some sequence of approximations. If a direct answer to my question seems out of reach, I would be happy with an answer explaining what is and isn't known. (I also removed the madness about Turing degrees above.)

Earlier this year it was asked on MO, "http://mathoverflow.net/questions/198098/are-there-only-countably-many-compact-topological-manifolds" Thanks to Cheeger and Kister, the answer is yes. On the other hand, Manolescu recently debunked the triangulation conjecture. A natural follow-up question asks if there is some other way to enumerate topological n-manifolds, in the sense of creating a Turing machine that will eventually output an example from every homeomorphism class of topological manifolds, given enough time.

Of course, for $n \leq 3$, TOP = PL, so I'm really interested in the cases $n\geq 4$. It's entirely possible that the answer still depends on $n$, so you can interpret the question with either $n$ fixed or variable.

If the answer is no, is it known how hard the problem of enumerating manifolds is? Is it harder than the halting problem?

Edit in response to comments below: I do not mean to jump the gun. To even have a hope that the answer to the question is yes, one would have to have some finitely computable description of topological manifolds. As BjørnKjos-Hanssen indicates in comments, this might take the form of some sequence of approximations. If a direct answer to my question seems out of reach, I would be happy with an answer explaining what is and isn't known. (I also removed the madness about reference to Turing degrees above.)

Earlier this year it was asked on MO, "http://mathoverflow.net/questions/198098/are-there-only-countably-many-compact-topological-manifolds" Thanks to Cheeger and Kister, the answer is yes. On the other hand, Manolescu recently debunked the triangulation conjecture. A natural follow-up question asks if there is some other way to enumerate topological n-manifolds, in the sense of creating a Turing machine that will eventually output an example from every homeomorphism class of topological manifolds, given enough time.

Of course, for $n \leq 3$, TOP = PL, so I'm really interested in the cases $n\geq 4$. It's entirely possible that the answer still depends on $n$, so you can interpret the question with either $n$ fixed or variable.

If the answer is no, is it known how hard the problem of enumerating manifolds is? Is it harder than the halting problem?

Edit in response to comments below: I do not mean to jump the gun. To even have a hope that the answer to the question is yes, one would have to have some finite description of topological manifolds. As BjørnKjos-Hanssen indicates in comments, this might take the form of some sequence of approximations. If a direct answer to my question seems out of reach, I would be happy with an answer explaining what is and isn't known. (I also removed the madness about Turing degrees above.)

Earlier this year it was asked on MO, "http://mathoverflow.net/questions/198098/are-there-only-countably-many-compact-topological-manifolds" Thanks to Cheeger and Kister, the answer is yes. On the other hand, Manolescu recently debunked the triangulation conjecture. A natural follow-up question asks if there is some other way to enumerate topological n-manifolds, in the sense of creating a Turing machine that will eventually output an example from every homeomorphism class of topological manifolds, given enough time.

Of course, for $n \leq 3$, TOP = PL, so I'm really interested in the cases $n\geq 4$. It's entirely possible that the answer still depends on $n$, so you can interpret the question with either $n$ fixed or variable.

If the answer is no, is it known how hard the problem of enumerating manifolds is? Is it harder than the halting problem?

Edit in response to comments below: I do not mean to jump the gun. To even have a hope that the answer to the question is yes, one would have to have some finitely computable description of topological manifolds. As BjørnKjos-Hanssen indicates in comments, this might take the form of some sequence of approximations. If a direct answer to my question seems out of reach, I would be happy with an answer explaining what is and isn't known. (I also removed the madness about Turing degrees above.)