Why is basic team not enough to identify manifolds?

I have actually listened to that some concerns in group theory protect against identifying all manifolds upto homotopy making use of the basic team stable. Does any person recognize what are those concerns?

Many thanks,. K.

0
2019-05-07 15:06:55
Source Share
Answers: 5

The following (renowned) inquiry is probably in the spirit of the OP is inquiry :

If $M$ and also $N$ are shut aspherical $n$ - manifolds with isomorphic basic teams, are $M$ and also $N$ homeomorphic? (This is called the Borel Conjecture.)

Scott Carter is instance highlights the demand for the presumption that the manifolds are shut. Yet the inquiry over recommends that probably the basic team suffices to identify the topological sort of a shut aspherical manifold.

0
2019-12-03 01:35:41
Source

This would certainly have belonged as a comment to MattE solution, yet I do not have the online reputation to leave remarks. The factor it is difficult to algorithmically identify topological manifolds (approximately homotopy) is, as MattE really plainly clarified, that words trouble for teams can be installed in the inquiry of identifying manifolds. Without a doubt, intend that you intend to determine if a finitely offered team $G$ is unimportant. First, there are numerous building and constructions that generate a portable 4 - manifold whose basic team is isomorphic to $G$ ; pick your favored building and construction and also allow $M$ be the resulting manifold. Observe that the junction pairing in homology is determinable, so we might equally as well think that we understand what this is for our manifold $M$. It ends up that the homotopy sort of a merely attached manifold is entirely established by the junction kind on the 2nd indispensable homology team (which any kind of junction kind can be become aware). So if on the side we construct the merely attached manifold $N$ with junction kind accompanying the among $M$, after that determining whether the homotopy sorts of $M$ and also $N$ synchronize will certainly determine if the basic team of $M$ is unimportant or otherwise, i.e. if $G$ is the unimportant team or otherwise.

There are some concerns on just how you calculate the junction pairing and also just how you locate the manifold $N$ offered the junction kind, and also there are methods to streamline all these inquiries, yet I will certainly not enter into this, in the hope that what I created will certainly be adequate! Certainly, if you desire even more information, do not hesitate to ask, and also I will certainly attempt to address!

0
2019-05-11 20:10:43
Source

For a far better set of instances take the as soon as pierced torus and also a set of trousers. Both have basic teams devoid of ranking 2. They are both homotopy matching to a number 8, yet as manifolds they are not homeomorphic given that one has 3 border parts and also the various other has one border part.

There are merely attached shut manifolds of measurement 4 and also greater that are not homeomorphic. In also measurements they can be identified by their center dimensional homology junction kinds.

0
2019-05-09 10:57:11
Source

Well, there are no factors to anticipate that manifolds with equivalent π 1 are homotopy equal, and also as a whole, they without a doubt aren't. Also for shut manifolds of dealt with measurement : claim, both CP 2 and also S 4 have unimportant basic team.

0
2019-05-09 10:42:52
Source

I assume that what you are considering is the insolvability of the so - called word trouble in group theory. What this suggests is (to name a few points) that it is not feasible to write an algorithm (i.e. a computer system program) which can inform whether 2 teams, each offered by finitely several generators and also relationships, are or aren't isomorphic.

If you are offered a manifold (offered claim by gluing with each other numerous open rounds), you can get. such a discussion of its basic team, yet (at the very least when the measurement is 4 or greater - - - and also I assume that 4 is the appropriate bound below) actually you can get an approximate discussion this way, therefore the insolvability of words trouble suggests that, in technique, you do not have any kind of means to identify what the basic team of your manifold is (also whether it is unimportant!)

Therefore, you can not locate an algorithm to establish whether 2 manifolds (of measurement 4 or greater) are homotopic (allow along homeomorphic or diffeomorphic), given that homotopic manifolds will certainly have isomorphic. $\pi_1$s, therefore we can make use of such an algorithm to address words trouble, by inscribing 2 finitely offered teams as the $\pi_1$ of some manifolds, and afterwards using our theoretical manifold algorithm. Adhering to Pete Clark's comment listed below, I'm no more certain concerning the accuracy of the coming before declaration. What holds true is that there is no basic algorithm to establish the basic team of a manifold (in measurement $\geq 4$), or perhaps to establish if a manifold is merely attached. (Whether this in fact bans a category is unclear to me ; with good luck, a person with even more experience will certainly consider in.)

Consequently, a whole lot (although not all) examinations of greater dimensional manifolds limit focus to merely linked manifolds. After that there is no blockage to category originating from concerns of the uncomputatibility of $\pi_1$. Certainly,. as others have actually kept in mind in their solutions, there can be several non - diffeomorphic, non - homeomorphic manifolds, and also non - homotopic manifolds of an offered measurement, so the trouble with $\pi_1$ is much from the only trouble when it involves identifying manifolds. Yet it is absolutely one trouble, and also you are proper that (at the very least from the perspective I'm clarifying below) it is a trouble of group theory.

Included : See Damiano's response to this inquiry for a specific description of the partnership in between words trouble for teams and also the category trouble for manifolds.

0
2019-05-09 10:42:39
Source