Exploring a Little Theorem about Metric Groups

[jostpuur]

Hello, I came up, and proved, a little theorem. My question is, that do you know this theorem from any other context, or if it has other similar forms (or if it is incorrect).

Suppose we have two metric groups A and B, and we know that A is simply connected. If we have a group homomorfism $$H:A\to B$$, then with certain assumptions (I won't list all the assumtions because they are not imporant for the idea), the homotopy group of the B is precisly the kernel of H.

 

[matt grime]

Since you believe the assumptions are unimportant, I can only conclude you think that this is true generically. And that is obviously not true: let H be the trivial homomorphism.

 

[jostpuur]

I thought they would not be important for someone, who knows this theorem, to recognize it. But I'll throw all assumptions I used. So the metric groups are $$(A,+_A,d_A)$$ and $$(B,+_B,d_B)$$. We assume that the sums are continuous in the given metrics, and also that for all $$x,y,z\in A$$, $$d_A(x,y)=d_A(x+z,y+z)$$. Both sets are path-connected. The homomorphism is continuous, open and surjective. All points of the kernel are isolated in the kernel.

The simplest application of the theorem is a homomorphism $$H:\mathbb{R}\to S^1$$, that curls the real line to a circle. The kernel is isomorphic to $$\mathbb{Z}$$, and so is the homotopy group of the circle.

 

[DeadWolfe]

I think that it is true that if G is a Lie group, and H is it's covering space, then the kernel of the covering map is the fundamental group.

I think that there's no way that what you are suggesting could possibly be true though. It would imply the fundamental group of a Lie Group is always totally disconnected, and if you removed the condition that points in the kernel be isolated, then it would imply that the kernel of any covering map is the fundamental group.

 

[matt grime]

I thought they would not be important for someone, who knows this theorem...

Hypotheses are part of the theorem. We have no idea what *you* think is obvious, implicit, or natural. So don't presume to tell us what is not important. We can't guess if you are correct, or just getting some assumptions right by fluke. If we all adopted you attitude then everything would be tautologically true.

 

[matt grime]

The simplest application of the theorem is a homomorphism $$H:\mathbb{R}\to S^1$$, that curls the real line to a circle. The kernel is isomorphic to $$\mathbb{Z}$$, and so is the homotopy group of the circle.

This is the application of deck transformations. That Z and R are (metric) groups is immaterial.

 

[Office_Shredder]

Hypotheses are part of the theorem. We have no idea what *you* think is obvious, implicit, or natural. So don't presume to tell us what is not important. We can't guess if you are correct, or just getting some assumptions right by fluke. If we all adopted you attitude then everything would be tautologically true.

If someone said "That theorem where if a funciton is negative at one point, and positive at another one, it's zero somewhere in between", you'd probably realize they were talking about Bolzano's theorem (even though you might not remember the name, I had to google it myself personally :tongue2: ). But it would be reasonable to expect someone who knows the theorem already to know that he's talking about a continuous function, defined on an interval [a,b] where f(a)<0, f(b)>0 or vice versa

 

[matt grime]

Intermediate value theorem.

And I do not accept any premise of your assertion. The OP claimed to have proved a theorem without asserting any of the hypotheses. You cannot do that. Or do you know the name of the theorem he's proved? Assuming it has a name...

trivially every conjecture is true modulo any exceptions. Such assertions are the bane of internet forums like this where any crank, such as the Russian name (victor sokraine?) in the number theory forum here, can post something without proof and with obvious counter example and try to obfuscate with the 'oh but obviously I meant to exclude that case' line.

The point was that the OP asserted that the assumptions were *unimportant*, which is clearly not true.

 

[mathwonk]

i agree that is was ridiculous to expect us to evaluate or recognize this theiorem from the opriginal non description,a nd also that it appears to be the standard theorem from covering spaces relating the fiber of the universal cover to the fundamental group of the base.

 

[jostpuur]

I corrected my OP by giving the assumtions in the next post. It wasn't really the point to start arguing about their imporance.

This is the application of deck transformations. That Z and R are (metric) groups is immaterial.

I don't know yet about deck transformations (I checked the MathWorld page on it, but naturally don't understand it yet), but even when fundamental group of a circle can be obtained in other ways, that doesn't change the fact that it also would follow from this theorem I mentioned.

It would imply the fundamental group of a Lie Group is always totally disconnected

My understanding on Lie groups is still quite poor, so I don't understand what this is all about fully (hopefully I'm getting onto a course about Lie groups on the fall). But I cannot help making some guesses. Is there some theorem that tells, that for a given Lie group B, there exists another one A that is simply connected, and you can have a group homomorphism from A to B? That just a guess, I'm trying to make sense out of the DeadWolfe's comment.

If the fundamental group of B is not totally disconnected, wouldn't this theorem simply imply, that then such A and H as stated in the assumptions, would not exist?

 

[matt grime]

I corrected my OP by giving the assumtions in the next post. It wasn't really the point to start arguing about their imporance.

but it was the assumptions you missed out that made the theorem true, and the ones you put in (metric group and homomorphism) don't tell you anything at all.