5
$\begingroup$

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...

$\endgroup$

2 Answers 2

13
$\begingroup$

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 $C'$.

$\endgroup$
5
  • $\begingroup$ 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... $\endgroup$
    – janmarqz
    Dec 9, 2009 at 16:28
  • 1
    $\begingroup$ juan, are you referring to the Phantom Tollbooth? $\endgroup$
    – HJRW
    Dec 10, 2009 at 6:29
  • $\begingroup$ i just check on Phantom Toollbooth (google) but i still don't see how my comments make you -Prof- think what i am refering... $\endgroup$
    – janmarqz
    Dec 10, 2009 at 17:53
  • 1
    $\begingroup$ There's a character in the book called "The Mathemagician". $\endgroup$
    – HJRW
    Dec 10, 2009 at 18:06
  • $\begingroup$ we all are mathemagiZians :) $\endgroup$
    – janmarqz
    Dec 11, 2009 at 1:44
1
$\begingroup$

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

$\endgroup$
1
  • 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 $\mathbb{R}^3$. $\endgroup$
    – HJRW
    Dec 11, 2009 at 5:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.