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I know that the amalgamated free product of two groups $G\star_K H$ has a certain topological meaning. What about a semi-direct product $H \rtimes G$ ?

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Perhaps you could describe the topological meaning of the amalgamated free product you refer to first. This may help in garnering an analogous description for the semi-direct product. –  user31714 Jan 12 '13 at 20:12
@user31714 I guess that Curiosity is referring to the fact that for every two spaces with $\pi_1$ respectively $G$ and $H$ you can build up the pointed product of the two spaces whose $\pi_1$ is the free product of $G$ and $H$. –  Giorgio Mossa Jan 12 '13 at 20:49
If X and Y have homeomorphic subspace A then the fundamental group of the space obtained by gluing X and Y along A can computed using the fundamentals groups of X,Y and A using amalgamated free products. I am referring to Seifert-Van Kampen theorem and I may be not rigorous in my argument above but I hope I described the idea. –  Curiosity Jan 12 '13 at 20:55
So are you asking what topological space has fundamental group $H\rtimes G$ if $H$ and $G$ are fundamental groups of some other topological spaces? –  Alexander Gruber Jan 12 '13 at 21:16
@AlexanderGruber, Yes that is what I am asking. –  Curiosity Jan 12 '13 at 22:43

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up vote 10 down vote accepted

The wording of your question is rather vague but hopefully the following will hint at least a little at one example of how the semi-direct product of the fundamental groups of topological spaces can arise naturally in common constructions. I apologise in advance if any of the definitions below aren't known to you. They should all be easily searchable on wikipedia.

If you don't know anything about fiber bundles (or the more general family of maps fibrations) then to put it very loosely, a map $p\colon E\rightarrow B$ is a fiber bundle if for all $b\in B$, the preimage of a neighbourhood $U\ni b$ is homeomorphic to the product of some space $F$ and $U$. That is, $p^{-1}(U)\cong F\times U$. We often write that the sequence of maps $F\rightarrow E\rightarrow B$ is a fibration sequence with fiber $F$ (the first map in this sequence is just inclusion of $F$ in $E$).

The proper definition puts some restrictions on the map $p$ but the above definition is good enough to give an intuitive feel for what fiber bundles 'look like'. The important part you should take away from the definition is that the space $E$ can be seen as a kind of 'twisted product' of the spaces $F$ and $B$. You should hopefully already be able to see a loose parallel between fiber bundles and semi-direct products of groups.

A very useful property of fiber bundles of 'nice spaces' (roughly speaking, if all the spaces involved are CW-complexes then we're fine) is that the homotopy groups of the fibration sequence fit in to a long exact sequence. Now, suppose that the space $F$ is connected and $\pi_2(B)=0$. Then from this long exact sequence in homotopy, we can extract the short exact sequence $$\pi_2(B)=0\rightarrow\pi_1(F)\rightarrow\pi_1(E) \rightarrow\pi_1(B)\rightarrow 0=\pi_0(F)$$ and now it should be clear that if this short exact sequence splits, then the splitting lemma for non-ableian groups tells us that $\pi_1(E)\cong \pi_1(F)\rtimes \pi_1(B)$.

The conditions on the homotopy groups of $B$ and $F$, and the requirement that the short exact sequence splits are rather specialised. So, this is far from a universally useful construction, but hopefully it's the sort of link between fundamental groups and semi direct product of groups that you were looking for.

Following the comments of KotelKanim below, the conditions on the homotopy groups of $B$ and $F$ are not necessary as long as the fiber bundle involved has a section. This means that there exists a map $s\colon B\to E$ such that $p\circ s=\mbox{Id}_B$.

Suppose the bundle $p$ has a section $s$ and let $\delta\colon\pi_n(B)\to\pi_{n-1}(F)$ be the connecting homomorphism in the associated long exact sequence of homotopy groups. Suppose $\ker\delta\neq\pi_n(B)$. Then by exactness at $\pi_n(B)$ we have $\mbox{Im}\, p_*\neq \pi_n(B)$ and so let $h$ be an element of $\pi_n(B)\setminus\mbox{Im}\, p_*$. We have $h=\mbox{Id}_{\pi_n(B)}(h)=p_*(s_*(h))$ but then this implies that $h\in\mbox{Im}\, p_*$ which is a contradiction. It follows that $\ker\delta=\pi_n(B)$ and so $\mbox{Im}\,\delta=0$.

Hence we get the augmented long exact sequence in homotopy $$\cdots \rightarrow \pi_{n+1}(B)\rightarrow 0\rightarrow\pi_n(F)\rightarrow\pi_n(E) \rightarrow\pi_n(B)\rightarrow 0 \rightarrow \pi_{n-1}(F) \rightarrow \cdots$$ because $\delta$ factors through $0$ and so we have a split short exact sequence for each $n$ which, by the splitting lemma, gives us the isomorphism $\pi_n(E) \cong \pi_n(F) \oplus \pi_n(B)$ for $n\geq 2$ (as higher homotopy groups are abelian) and for $n=1$ gives us the previous semi-direct product.

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It seems to me, that if the fiber bundle has a section, then the long exact sequence will also have "sections" which make it into a bunch of short split exact sequences. This means that all homotopy groups of the total space will be semi-direct product of the respective homotopy groups of the base and the fiber. –  KotelKanim Apr 23 at 13:23
You would also need $\pi_{n+1}(B)$ and $\pi_{n-1}(F)$ to be trivial in order to have the above short exact sequence at $n$. –  Daniel Rust Apr 23 at 13:31
really? I think that the existence of a section at each stage also forces the connecting homomorphisms to be trivial. –  KotelKanim Apr 23 at 13:39
Ah yes you're right, If the bundle has a section then the LES breaks into split SESs for each $n$. Good to know. I'll add it to the answer :). –  Daniel Rust Apr 23 at 14:02

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