which HNN-extensions are free products? this question is related with another still unsolved about Nielsen-Thruston-reducibility and connected-sum-irreducibility of 3d-torus- bundles...
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This might help.
Lemma If $A$ does not split freely and $C$ is a non-trivial subgroup of $A$ then the HNN extension $G=A*_C$ does not split freely.
The proof uses Bass--Serre theory---see Serre's book Trees from 1980.
Proof. Let $T$ be the Bass--Serre tree of a free splitting of $G$. Because $A$ does not split freely, $A$ stabilizes some unique vertex $v$. But $C$ is non-trivial, so $C$ also stabilizes a unique vertex, which must be $v$. Therefore, $G$ stabilizes $v$, which means the free splitting was trivial. QED
A similar argument shows the following.
Lemma If $ A*_C $ splits non-trivially as an amalgamated free product $ A' *_{C'} B'$ then either $A$ splits over $C'$ or $C$ is conjugate into $C'$.
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$\begingroup$ i must admit that i didn't seek until find, but lot of thanks H.W. i feel that MO put the advance of all "mathemagizians" in a better pace... $\endgroup$– janmarqzCommented Dec 9, 2009 at 16:28
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1$\begingroup$ juan, are you referring to the Phantom Tollbooth? $\endgroup$– HJRWCommented Dec 10, 2009 at 6:29
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$\begingroup$ i just check on Phantom Toollbooth (google) but i still don't see how my comments make you -Prof- think what i am refering... $\endgroup$– janmarqzCommented Dec 10, 2009 at 17:53
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1$\begingroup$ There's a character in the book called "The Mathemagician". $\endgroup$– HJRWCommented Dec 10, 2009 at 18:06
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Let me add an explicit partial solution to the question above: for a torus bundle $E$ over the circle, the fundamental group of $E$ can't be a free product of groups, because if it were, the fundamental subgroup of the fiber would be a free product (by the Kurosch's theorem) which is impossible for the torus, then $E$ isn't a connected sum, hence irreducible.
At least the HNN extensions which are free products can't be torus bundles, and in fact, no other surface bundles unless the surface be the 2-sphere
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1$\begingroup$ Another way of seeing this is to look at $\pi_2$. A non-trivial free product has an essential 2-sphere, whereas it's easy to see that the universal cover of a torus bundle is homeomorphic to $\mathbb{R}^3$. $\endgroup$– HJRWCommented Dec 11, 2009 at 5:02