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](https://doi.org/10.1007%2FBF01388734), 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](http://doi.org/10.1307/mmj/1029004602), 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?



Morse theory on outer space via the lengths of finitely many conjugacy classes
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.


