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. :)