Proving Compactness of a Topological Group Using Subgroups and Quotient Spaces

[Lie]

Hello!

Could anyone help me to resolve the impasse below?

Th: Let G be a topological group and H subgroup of G. If H and G/H (quotient space of G by H) are compact, then G itself is compact.

Proof: Since H is compact, the the natural mapping g of G onto G/H is a closed mapping. Therefore if a family S of closed subsets of G has the finite intersection property, then so does {g(F): F in S}. So that G/H is compact, then \bigcap_{F \in S} g(F) \neq \varnothing . But as I conclude that \bigcap_{F \in S} F \neq \varnothing \; ?
If necessary we also know that G/H is Haudorff space.

Thankful! :)

 

[mathwonk]

what if the quotient splits the group? I.e. what if G is siomorphic to H x G/H?

Then in general, just off the top of my head, maybe this is true locally, i.e. maybe the map G-->G/H is a locally trivial fibration

 

[Lie]

what if the quotient splits the group? I.e. what if G is siomorphic to H x G/H?

No!

Then in general, just off the top of my head, maybe this is true locally, i.e. maybe the map G-->G/H is a locally trivial fibration

I don't understand!

 

[mathwonk]

assume it splits. then could you do it?

 

[Lie]

assume it splits. then could you do it?

Can I do that G/H is not a group? Remember that H is only one subgroup. It is not normal!

 

[micromass]

Hello Lie!

Since H is compact, the the natural mapping g of G onto G/H is a closed mapping.

I wonder how you know this, don't you need some kind of of Hausdorff property for this?

EDIT: never mind about this, you don't need Hausdorff.

Anyway, the following could help you:

Definition: We call f a proper map if it is continuous, closed, surjective and if every f^{-1}(y) is compact (=we say that the fibers are compact).

Theorem: If f:X\rightarrow Y is proper and if Y is compact, then X is compact.

Proof: Note that closedness of f is equivalent with:

For all open U containing f^{-1}(y), there exists an open neighbourhood W of y, such that f^{-1}(W)\subseteq U.
​

This equivalence is easy to see by taking W=Y\setminus f(X\setminus U).

So, now take an open cover of X. For each y in Y, we know that f^{-1}(y) is non-empty and compact and thus covered by a finite number of our covers. Let U_{y,1},...,U_{y_n} be the elements of our cover. Then by the above, there exist W_{y,i} such that

f^{-1}(W_{y,i})\subseteq U_{y,i}

This forms an open cover of Y and thus we can take an finite subcover. This finite subcover is the one we're looking for...

 

[Lie]

Hello Lie!

I wonder how you know this, don't you need some kind of of Hausdorff property for this?

EDIT: never mind about this, you don't need Hausdorff.

OK! I don't really need Hausdorff.

Anyway, the following could help you:

Definition: We call f a proper map if it is continuous, closed, surjective and if every f^{-1}(y) is compact (=we say that the fibers are compact).

The natural mapping g of G onto G/H is a closed mapping, continuous, surjective and every g^{-1}(xH) is closed. Why it's proper?

Theorem: If f\colon X \to Y is proper and if Y is compact, then X is compact.

This result is interesting, but I would try to use my argument above.

Thankful! :)

 

[micromass]

g^{-1}(xH)[/itex] is closed.

We have g^{-1}([x])=xH. Now H is compact, and thus every translation xH of H is compact as well.

This result is interesting, but I would try to use my argument above.

Very well, but I doubt it's going to work.

 

[Lie]

We have g^{-1}([x])=xH. Now H is compact, and thus every translation xH of H is compact as well.

Indeed! :)

Very well, but I doubt it's going to work.

Tip of Hewitt & Ross! ;)

 

[micromass]

Tip of Hewitt & Ross! ;)

What about this: Take (F)_{F\in S} with the fip. I now form all finite unions of this collection and I get a collection (F^\prime)_{F^\prime\in S^\prime}. Clearly this has the fip and the intersection of this collection is nonempty if and only if the original collection is nonempty.

By compactness we know that

\bigcap_{F^\prime\in S^\prime}{g(F^\prime)}

is nonempty, thus take an [x] in it. By compactness of xH it follows that of \bigcap_{F^\prime\in S^\prime}{F^\prime}=\emptyset, then there exist a finite number whose intersection is empty. But because we have chosen the F^\prime to be closed under unions, it follows that Hx\cap F^\prime=\emptyset for some F^\prime, but then g(Hx)=[x] is not an element of

\bigcap_{F^\prime\in S^\prime}{g(F^\prime)}

contradiction...

Hope I didn't make any silly mistakes here...

 

[Lie]

What about this: Take (F)_{F\in S} with the fip. I now form all finite unions of this collection and I get a collection (F^\prime)_{F^\prime\in S^\prime}. Clearly this has the fip and the intersection of this collection is nonempty if and only if the original collection is nonempty.

By compactness we know that

\bigcap_{F^\prime\in S^\prime}{g(F^\prime)}

is nonempty, thus take an [x] in it. By compactness of xH it follows that of \bigcap_{F^\prime\in S^\prime}{F^\prime}=\emptyset, then there exist a finite number whose intersection is empty. But because we have chosen the F^\prime to be closed under unions, it follows that Hx\cap F^\prime=\emptyset for some F^\prime, but then g(Hx)=[x] is not an element of

\bigcap_{F^\prime\in S^\prime}{g(F^\prime)}

contradiction...

Hope I didn't make any silly mistakes here...

Excuse me! I was very busy lately and I could not move from here. Not found any error! Thank very much. :)