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Let $F_n$ be the free group on letters $\{x_1,\ldots,x_n\}$ and let $X_n$ be the (reduced) outer space of rank $n$. Points of $X_n$ thus correspond to pairs $(G,\mu)$, where $G$ is a finite connected metric graph of total edge-length $1$ with no valence $1$ or $2$ vertices and no separating edges and $\mu\colon F_n \rightarrow \pi_1(X_n)$ is the conjugacy class of an isomorphism.

The space $X_n$ was introduced in the paper

M. Culler and K. Vogtmann, Moduli of graphs and automorphisms of free groups, Invent. Math. 84 (1986), no. 1, 91–119.

The main theorem of this paper is that $X_n$ is contractible. They prove this in the following way. Let $W_0$ be the following set of elements of $F_n$:

  • $x_i$ with $1 \leq i \leq n$, and
  • $x_i x_j$ and $x_i x_j^{-1}$ with $1 \leq i < j \leq n$.

For each $w \in W_0$, and each point $(G,\mu)$ of $X_n$, let $|(G,\mu)|_w$ be the length of the shortest unbased loop in $G$ representing the conjugacy class $w$. Define

$$|(G,\mu)| = \sum_{w \in W_0} |(G,\mu)|_w.$$

They use this as a sort of "Morse function" on $X_n$, and study its level sets. One of their key results is as follows. A rose in $X_n$ is a point $(R,\mu)$, where $R$ is a metric graph with $1$ vertex and $n$ edges, each of length $1/n$. Let $(R_0,\mu_0)$ be the rose where $\mu_0$ identifies the loops in $R_0$ with the basis elements $x_1,\ldots,x_n$.

Proposition (Prop 6.2.5 in the above paper): The rose $(R_0,\mu_0)$ is the unique minimum of the above Morse function among roses, i.e., if $(R,\mu)$ is another rose then $|(R_0,\mu_0)| \leq |(R,\mu)|$ with equality if and only if $(R,\mu) = (R_0,\mu_0)$.

I remark that Prop 6.2.5 has the opposite inequality to the one I wrote, but I think this is a typo and the proof seems to give the above. The proof of this proposition is pretty simple and convincing.

However, it seems to be contradicted in the following paper:

J. Smillie and K. Vogtmann, Length functions and outer space, Michigan Math J. 39 (1992) 485–493.

In fact, Theorem 1 of this paper seems to say that not only is the above proposition false, but it remains false if $W_0$ is replaced by any other finite set of elements of $F_n$. The authors of this paper do not say anything about this contradicting anything in the previous paper (even though they share an author, so presumably if I am right they were aware of this).

Question: Am I misinterpreting any of these papers? If not, what exactly is going on here?

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1 Answer 1

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You don't misunderstand, it's a subtle point that I'm sure I'll get wrong here too. You might find the proof of a slightly more general statement in Krstić and Vogtmann's "Equivariant Outer Space and automorphisms of free-by-finite groups" illuminating: basically the idea is that for any rose (I believe in fact for any marked metric graph, but let's not get ahead of ourselves), there exists a finite set of conjugacy classes whose lengths pin down that particular rose completely.

The problem, or the result of the Smilie–Vogtmann paper, is that the lengths of those same conjugacy classes will not work for every marked graph at once. The reason there's no contradiction is that the minimizing marked graph is unique, while the others need not be.

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    $\begingroup$ Ah, that's really enlightening! Thanks!!! ps: I'm not registered so I can't upvote, but I really appreciate this answer! $\endgroup$
    – Sarah
    Commented Jul 9, 2023 at 1:51

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