# Mathematics Prelims

## March 15, 2009

### Path Homotopy

Filed under: Topology — cjohnson @ 12:10 pm

Over the next few posts we’re going to be discussing homotopy groups in topology.  Homotopy groups will be our window into the world of algebra, giving us the opportunity to use machinery from algebra to do interesting things with topological spaces (and vice versa).  Before trying to put an algebraic structure on a topological space though, we need to discuss paths.

A path is a mathematical way of connecting two points in space.  In the case of the Euclidean plane, for instance, there is a line segment between any two points.  There are also all sorts of curves that start at one point and end at another.  These are paths as well, provided they can be represented as continuous functions.  That is, a path in a topological space $X$ is a continuous function $\rho : [0, 1] \to X$.  We call the points $\rho(0)$ and $\rho(1)$ the initial and terminal points of the path.  The choice of the interval [0, 1] is semi-arbitrary, since we could just as easily have used ${}[-\pi, e]$ or a finite portion of a ray in some other space.  However, [0, 1] is convenient and it’s not very difficult to reparameterize a path that uses some other interval to “fit” in [0, 1].  For example, if we had a path $p : [-\pi : e] \to X$, we could just take $\theta(x) = (x + \pi)/(\pi + e)$, then if we define $\rho = p \theta$, we’d have the same path represented as $\rho : [0, 1] \to X$.

Now, between two points, say $x_0, x_1 \in X$, there may be several paths.  We would like to know if two paths connecting these points are essentially the same thing or not.  That is, if one path can be deformed into the other.  We saw last time that homotopy gave us a way to deform functions, but we have to take a little bit of care when it comes to paths.  You see, we would like to use homotopy to bend one path into another, but with the restriction that all of the intermediate functions are paths as well.  Our homotopy should be anchored down at the path’s endpoints.

If $\rho_1$ and $\rho_2$ are paths with initial point $x_0$ and terminal point $x_1$, we say that $\rho_1$ and $\rho_2$ are path homotopic if there exists a homotopy $\{ h_t : [0, 1] \to X \}$ (recall we may think of a homotopy as a family of functions) such that for every $t$, we have that $h_t$ is also a path from $x_0$ to $x_1$.  Recalling the video from last time, we had a homotopy between two curves in the plane.  This was not an example of a path homotopy because, for one thing, the endpoints of the paths changed as we bent one curve into the other.

Modifying that example slightly, however, we can construct a path homotopy.  In this case our space is $\mathbb{R}^2$ and our two paths are given by $(-\cos(\pi t), \sin(\pi t))$ and $(2t-1, \cos(3 \pi t - \pi/2))$.  Using the straight-line homotopy we then have the following.

Notice how here the points (-1, 0) and (1, 0) are the endpoints for all of the paths in the homotopy.

Of course, not all paths are homotopic to one another.  In the special case of $\mathbb{R}^n$ all paths are homotopic, just as before, but if we simply remove the origin from $\mathbb{R}^2$, it’s easy to construct two non homotopic paths.

Here the blue path and the red path are simply the portions of the unit circle connecting (-1, 0) and (1, 0).  However these paths are not path homotopic as we would have to either break a path or move through the origin to deform one path into the other, just as before.  Even though these two curves are not path homotopic, they are in fact homotopic in $\mathbb{R} \setminus 0$ as the following video demonstrates.

The purple path bends the blue path into the red path while avoiding the origin.  Again, this is not a path homotopy since the endpoints aren’t anchored down.

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## 1 Comment »

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