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Let $G$ be a finitely generated subgroup of a product of two finite rank free groups $F_m \times F_n$. If there is a Lipschitz retraction $F_m \times F_n \to G$ with respect to word metrics, then $G$ is undistorted in $F_m \times F_n$, and the Dehn function of $G$ has a quadratic upper bound.

Suppose now that we require only that $G$ is undistorted in $F_m \times F_n$. Is it still true that the Dehn function of $G$ has a quadratic upper bound?

flag	asked Sep 14 at 16:00 Lee Mosher 5,916●1●9●26

	Lee: All subgroups G are either free or direct products of free groups or are not finitely-presented. Among latter, I do not know any examples of undistorted ones, do you? In fact, the only undistorted subgroups of semi simple Lie groups that I know are retracts, non-uniform lattices and Anosov (in the sense of Guichard and Wienhard) hyperbolic groups (the latter could in the end turn out to be retracts, this is unknown). – Misha Sep 14 at 17:32
	...I forgot one more class of undistorted subgroups: Some polycyclic groups which are peripheral subgroups of non-uniform lattices. However, one should regard them as lattices in solvable Lie groups. – Misha Sep 14 at 17:41
	why you closed my question, please reopen it – Hassan Jolany Dec 21 at 17:47

Robert Young · Accepted Answer · 2012-09-14 17:45:21Z UTC

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The Bieri-Stallings subgroup of $F_n\times F_n$ is undistorted and finitely generated, but not finitely-presented, so in some sense it has an infinite Dehn function. It's the kernel of the map $F_n\times F_n\to \mathbb{Z}$ which sends each generator to 1, and it's generated by elements of the form $g_ih_j^{-1}$ where $g_i$ and $h_j$ are generators of the two different factors.

answered Sep 14 at 17:45

Robert Young
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Welcome to MO Robert! – Andy Putman Sep 14 at 18:23

How does one know, or find a reference for, the statement that this subgroup is undistorted? – Lee Mosher Sep 14 at 18:32

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There's a proof for the theorem that Yves mentions in Olshanskii, Sapir, "Length and area functions on groups and quasi-isometric Higman embeddings" (Theorem 2). For this subgroup, you can construct words explicitly -- if $$w=w_1(g_1,\dots, g_n)w_2(h_1,\dots,h_n)$$ and $w$ lies in the kernel, we can rewrite $w$ as $$w=w_1(g_1 h_1^{-1},\dots, g_n h_1^{-1})(h_1 g_1^{-1})^k w_2(h_1g_1^{-1},\dots,h_n g_1^{-1})$$ where $k$ is the sum of the exponents in $w_1$. This is a product of generators of the subgroup and its length increases by at most a constant. – Robert Young Sep 14 at 19:37

That's pretty. – Lee Mosher Sep 14 at 21:18

Yves Cornulier · Answer 2 · 2012-09-14 17:44:19Z UTC

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The answer is no: actually by Baumslag-Roseblade (JLMS 1984) either $G$ is commensurable to a product of free groups (hence has linear or quadratic Dehn function), or is not finitely presented (so the Dehn function is infinite, or not defined, as you wish). The latter case occurs if $H$ is an infinite word hyperbolic group and $f:F_m\to H$ is a non-bijective surjection and $H$ is the fibre product $[(g,h)\in F_m\times F_m:f(g)=f(h)]$.

answered Sep 14 at 17:44

Yves Cornulier
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	Sorry, I can't type braces so I put brackets. – Yves Cornulier Sep 14 at 17:44
	@Yves, I have the same question here as for Robert Young's answer: how does one obtain nondistortion in these examples? – Lee Mosher Sep 14 at 18:36
	In general the Dehn function of $H$ is equivalent to the distortion of $G$. This is not completely formal (there's a little Van Kampen diagram cuisine) but in the case of $H=\mathbf{Z}$ it's simple to verify by hand. I saw the general result written somewhere but I can't remember right now. – Yves Cornulier Sep 14 at 18:42

Dehn function for undistorted subgroups of a product of free groups

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