(Apologies in advance for any imprecision in the following; I am a computer scientist and regret never having taken an actual course on topology.)


One way to define the pure braid group $P\_n$ is as follows: consider a pure braid to be a set of $n$ non-intersecting arcs in $x,y,t$-space which are monotone in the $t$ direction, such that the $i$th arc connects $(i,0,0)$ to $(i,0,1)$. Sets of arcs which can be deformed continuously into each other without any arcs intersecting are considered equivalent pure braids.


Consider a generalization where the endpoints are not integer positions on two lines, but arbitrary fixed positions in the $t = 0$ and $t = 1$ planes. It seems clear to me that this does not change the topology of the space [1], so I will call such a collection of arcs an *embedded braid*. I'm looking at characterizing "optimal" embedded braids that minimize a certain functional $F$. For concreteness, let $F$ be the total length of the arcs (the actual functional I need to use is slightly different, but this should be close enough to carry over the results).


If the requirement that arcs do not intersect is removed, the space of embedded braids can be given an affine structure over which $F$ is convex, and it is immediately apparent that there is a single local minimum which is the global minimum: connect each pair of endpoints with a straight line. What can we say about the local minima of $F$ if we retain the non-intersection property? My intuition says that each topologically distinct braiding (corresponding to a particular element of $P\_n$) forms a connected component with a unique local minimum, but I cannot tell how to begin proving this. It would be true if one could show that the connected components are convex subsets of the space; this does not hold under the "obvious" affine structure where we treat each arc and each $t$ independently, but that leaves the possibility of some other choice of affine structure which works. Are there some nice proof techniques that would help proving uniqueness of local minima in this context? Would any ideas or analogies from the theory of minimal surfaces help?


As Andrew Stacey mentions, the connected components are *open* subspaces; I believe existence of local minima can be guaranteed by considering the *closure* of the component seen as a subset of the space of embeddings that allow intersection. (Following Kevin Walker's comment, I realize this is called compactification.) This would include embeddings "on the boundary", whose arcs can intersect but which are only an infinitesimal displacement away from non-intersecting embeddings with the right topology. As a concrete example, the following braid,



```

\   /
 \ /
  \
 / \
 \ /
  \
 / \
/   \

```

on being pulled tight, approaches a configuration where the arcs are infinitesimally close together in the middle; the minimal length is attained by the intersecting embedding where the two arcs share a common point.


[1] Since the endpoints are fixed, we can associate each such generalized braid uniquely with a pure braid by shrinking the $t$ dimension slightly and composing each arc $i$ with a fixed arc from its endpoints to $(i,0,0)$ and $(i,0,1)$ respectively.



Minimal-length embeddings of braids into R^3 with fixed endpoints
UPDATE.


I revisited the question and realized that verifying the local CAT(0) property is not that easy. When I wrote the original answer, I was under impression that removing any collection of codimension 2 subspaces (more precisely, their tubular neighborhoods) from $\mathbb R^n$ leaves one with a locally CAT(0) space. This is true in $\mathbb R^3$ but there are counterexamples in $\mathbb R^4$. *This* particular subset might satisfy the condition but this does not seem to follow from any "generic" argument.


Further, there are some discouraging examples. First, approximating by arcs by "ropes" with fixed-size square sections does not actually work: in this case there are non-unique minimal configurations. So one really needs to deal with zero-width ones. Second, if you consider 3 arcs and allow two of them intersect while deforming the braid (but still disallow intersections with the 3rd one), then again, you can have two distinct minimal configurations in the same equivalence class.


Below is the original answer (and I suggest that it is unaccepted).




---


I can prove uniqueness of a local minimum for another length-like functional
(similar but not equal to the sum of lengths). I believe that it should work the same way
for the sum of lengths, but unfortunately the underlying geometric theory does not
seem to exist (yet?).


First let me reformulate the problem. Let $X$ denote the set of possible horisontal cross-sections of braids.
This is the set of $n$-tuples of distinct points in $\mathbb R^2$. Geometrically this is $(\mathbb R^2)^n$
minus a collection of codimension 2 subspaces corresponding to positions where some two points coincide.


Actually I prefer another formalization: the arcs forming the braid are ropes of nonzero width
and with square cross-sections.
More precisely, the horizontal section of every rope is an $\varepsilon\times\varepsilon$ square (with sides parallel
to the coordinate axes), and these sections should not overlap. So $X$ is $\mathbb R^{2n}$ minus a union of polyhedral
heighborhoods of codimension 2 subspaces. This formalization makes the local structure simpler,
and the original one is the limit as $\varepsilon\to 0$.


An embedded braid is a path $f:[0,1]\to X$. We want to minimize the functional
$$
 L = \int\_0^1 \left(\sqrt{(df\_1/dt)^2+1}+\dots+\sqrt{(df\_n/dt)^2+1}\right) dt,
$$
over all paths between two given positions $a,b\in X$, in a given connected component
of the space of such paths. Here $f\_i=f\_i(t)$ are 2-dimensional coordinates.
Another functional,
$$
 L' = \int\_0^1 \sqrt{(df\_1/dt)^2+\dots+(df\_n/dt)^2+1}\ \ dt,
$$
is easier to deal with, because this is the Euclidean length of the corresponding
path $(f(t),t)$ in $X\times\mathbb R\subset \mathbb R^{2n+1}$. Let us work with $L'$.


The connected component of the set of paths is the homotopy class. Fixing the homotopy class of a path is the same as
fixing endpoints of its lift to the [universal cover](http://en.wikipedia.org/wiki/Covering_space) of the space.
(In the case of zero-width ropes, you first take the universal cover and then the completion in order to add "boundary cases".)
And local length minimizers are called geodesics. So we want to show that, in the universal cover of $X\times\mathbb R$,
every pair of points is connected by a unique geodesic.


Our space $X$ (and hence $X\times\mathbb R$) is a locally [CAT(0) space](http://en.wikipedia.org/wiki/CAT%28k%29_space).
This can be shown using standard curvature tests for polyhedral spaces.
A globalization theorem says that a complete, simply connected, locally CAT(0) space is globally CAT(0),
in other words, it is a Hadamard space. So the universal cover is a Hadamard space.
Hadamard spaces feature uniqueness of geodesics, hence the result.


Now what about the original functional $L$? It is also a length functional, but for a non-Euclidean norm on $\mathbb R^{2n+1}$.
So, to carry over the proof, one needs to develop an analog of CAT(0) spaces modelled after Finsler spaces rather than Riemannian. I think it should be possible - after all, all this CAT(0) business is about convexity of distance, and this convexity is there in all normed vector spaces.
But I have not heard of such generalizations (maybe this is just my ignorance).


