The Configuration Space of All Lines In the Plane

Time for another classic post from the old site, while I'm at it… As we have seen in a bunch of previous posts, the notion of configuration space has held a prominent place in my mathematical explorations, because, I can't emphasize enough, it's not just the geometry of things that you can directly see that are important; it's the use of geometric methods to model many things that is.

Consider all possible straight lines in the plane (by straight line I mean those that are infinitely long). This collection is a configuration space—a catalogue of these lines. What does that mean? Of course, in short words, it's a manifold, and we've talked about many of those before. But it always helps to examine examples in detail to help develop familiarity. Being a manifold means we can find local parametrizations of our collection by $n$-tuples of real numbers. How do we do this for lines in the plane?

Finding Some Coordinates: The Slope-Intercept Equation

First off, almost anyone reading this, regardless of mathematical background, probably was drilled at one time or another on the following famous equation, called the slope-intercept equation:
\[
y = mx + b.
\] The variable m if you recall is the slope of a line, and b is the y-intercept which means where the line crosses the y-axis. However few people realize that what they're doing, really, is indexing every non-vertical line in the plane with some point in another plane (called, say, the mb-plane rather than the old traditional xy-plane). That is, we have a correspondence between points in one abstract mb-plane to the lines in the xy-plane, where m is the slope, and b the y-intercept. It doesn't get all lines, though—vertical lines have infinite slope, and you can't have a coordinate on any plane with an infinite value. That is, you have charted out the space of all possible nonvertical lines with points in a different plane.

Of course how would we catch the vertical lines? We chart it out using different coordinates. There's the possibility of using inverse slope and x-intercept, which merely means we write the equation using x as a function of y. In other words, all non-horizontal lines are given by:
\[
x=ky + c.
\] It's essentially obtained by reflecting everything about the diagonal line x = y and finding the usual slope and intercept. In other words we can chart out all non-horizontal lines on this new kc-plane. So you could represent the collection of all lines in the plane by having two charts... an atlas or catalogue of all lines, with these two sheets, the mb-plane and the kc-plane.

However, note that most lines in the plane—ones that are neither vertical nor horizontal, can be represented in both forms. That is, they correspond to points on both sheets. So, we would say in our usual terminology that these two sheets, or charts, determine a manifold of all possible lines; we just need to check that the overlap is smooth. But let's stop to think about what that means. We can think of it as trying to glue together both sheets to form just a single catalogue. The method of gluing is that we glue together the points that represent the same line in the plane. There is a nice, exact formula for the gluing, which is very simply determined: $(m,b)$ and $(k,c)$ represent the same line if and only if $y = mx + b$ and $x = ky + c$ are equations of the same line. All we have to do to convert from one to the other is solve for x in terms of y, that is, invert the functions. So we solve
\[
y = mx + b \iff y-b = mx \iff y/m - b/m = x \iff x = (1/m) y - b/m
\] that is, if $k = 1/m$ and $c = -b/m$, then $(m,b)$ and $(k,c)$ represent the same line in the two sheets. So to "glue" the sheets together, we glue every $(m,b)$ in the mb-plane to $(1/m,-m/b)$ in the kc-plane. Obviously the sheets need to be made of some very stretchable material, because it is going to be awfully hard to glue points together. Actually it's pretty hard to physically do this so don't try this at home, but just try to imagine it (don't you just love thought experiments?). For example, points in the mb-plane $(1,1)$, $(2,2)$, $(3,3)$, and $(4,4)$ get glued to the corresponding points $(1,-1)$, $(1/2,-1)$, $(1/3,-1)$, and $(1/4,-1)$. You glue them in a very weird way, but if you suppose for a moment, that you allow all sorts of moving, rotating, shrinking, stretching in this process (topological deformations), but without tearing, creasing, or collapsing, you can preseve the "shape" of this space, and yet make it look like something more familiar. This would be our new catalogue of lines. In addition, the catalogue has an additional property: nearby points in the catalogue translate to very similar-looking lines.

So, What Is It?

One should wonder what kind of overall "shape" our nice spiffy catalogue has, after gluing together the two possible charts we've made for it. As it turns out, its shape is the Möbius strip! That's right, the classic one-sided surface (without, as it turns out, its circle-boundary).

That is to say, if you give me a point on the Möbius strip, it specifies one and only one line in the plane. One would not, initially, be able see why non-orientability enters the picture. But a little interpretation is in order. First, if we take a particular line and rotate it through 180 degrees, we get the same line back. Everything in between gives every possible (finite) slope. It so happens that as far as slopes of lines is concerned, $\infty=-\infty$, and if you go "past" this single projective infinity, as they call it, you go to negative slopes. In other words, if you start on a journey on rotating a line through 180 degrees, from vertical back to vertical, you come back to the same line, except with orientation reversed (because what started out as pointing up now points down).

If you fix an origin and declare that it correspond to a certain special line in the plane, and then select a "core circle" for the Möbius strip, then as you travel around this circle, the distance traveled represents rotation angle for this special line. Traveling from the origin along the core circle and making one full loop should correspond to rotating the special line by 180 degrees. If you instead move up or down on the core circle, you instead end up sliding the line along a perpendicular direction, without changing its angle. So moving up and down the strip corresponds to parallel sliding of lines, and moving around the strip along a circle corresponds to rotating a line.

The Derivation

The specific formula we use is \[
F(m,b) = \left(\cot^{-1} m, \frac{b}{\sqrt{m^2 + 1}}\right),
\] which sends the line to the angle it makes with the $y$-axis, and its signed perpendicular distance to the origin (the sign is determined by $b$). For the other chart, \[ F(k,c) = \begin{cases}\left( \cot^{-1} \left( \frac{1}{k}\right), \dfrac{c}{\operatorname{sgn}(-k)\sqrt{k^2+1}}\right) \quad & \text{ if } k \neq 0 \\
(0,-c) & \quad \text{ if } k = 0
\end{cases}
\] We'll show how we got this in an update to this post, or perhaps a "Part 2." This can be readily checked by simply substituting the transition charts for $(m,b)$ and $(k,c)$. However, the extra case for $k=0$ here is simply gotten by taking the limit as $k$ goes to zero from above in the other case. What proves that it is a Möbius strip is that, if we take the limit as $k$ goes to zero from below, it will approach $(\pi,c)$ instead of $(0,-c)$. This would make it discontinuous, unless we decide to identify $(\pi,c)$ with $(0,-c)$: the $c$ going to $-c$ means we take the strip at $\pi$ and flip it around to glue it to the strip at $0$ (see this post for another example of defining a Möbius strip this way). Technically, we need an infinitely wide Möbius strip for this, but we can always scrunch it down into a finite-width strip without its boundary circle (using something like arctangent). It's just that the closer you get to the edge, the quicker things go off to infinity.

The animation is an example "path" through "line space," The blue dot travels around the white circle, and the line in the plane that corresponds to it is the blue line. The red line is a reference line perpendicular to the blue one, and always passes through the origin. The distance the blue dot from the core circle (in turquoise) indicates how far from the origin that the blue and red lines intersect. Because of the "scrunching down," though, the closer to the edge of the strip we get, the more dramatic the change in distance the blue line is from the origin. Here it is again in the plane with the Möbius strip as a picture-in-picture reference:

The more general object here we are describing is closely related to the notion of Grassmannian manifold, which are all $k$-dimensional subspaces in an $n$-dimensional vector space (the only difference is that Grassmannians only consider spaces through the origin).

Singular Value Decomposition Delights, Part 2

First off, congratulations to math homie and guest blogger Adam on his advancement to candidacy! Now back to our explorations from last week. So the goal of the previous vector bundle problem is to consider a vector bundle over the circle, represented as planes over an interval such that the fibers at one end of the interval are identified with the other end using some linear transformation T. In order to visualize this, we shall adopt the following approach: we put our base space circle in the xy-plane. arrange the fiber planes perpendicularly to circle (planes θ = constant), like so:

and then we choose a reference circle in one of these planes (say, of radius 1). Then as we move around the main circle, we consider the surface swept out as we carry that circle with us and apply the interpolated transformation

A(t) = exp(tξ) Σ^texp(–tη)

to that circle as we go. When we come around to the beginning, this circle should be deformed by the original transformation T. We do the experiment with the matrix

0.498502 0.214052
0.0433899 0.60181

which is a rotation of -2.62394 radians (-150.3°), then stretching with singular values 0.69543 and 0.41803, and finally applying a rotation of 2.24233 radians (128.5°). We start with the identity matrix, and as we move around, rotate the appropriate fraction t, times of the first angle, stretch by the second factor to the t power, and finally rotate by the same fraction of the second angle. So at time t = 1, a full revolution about the main circle, we get full operator T applied to it. This is the surface swept out:

And if we kept going, it would spiral inward more and more, which when doing a cutaway view (essentially, starting only with, say, 3/4 or 1/2 of the initial circle), we get the original picture shown in Part 1. But actually, it is more interesting to see what happens if we just unfurl it in a straight line (in the spirit of the original definition as an identification space)... and let the mapping continue past t = 1. The result is the title picture, which shows what happens when we proceed to t = 3 (which corresponds to the transformation T³). Forget donuts, let's do pasta (more comprehensively, here)!

Of course, we know that this shows our bundle is trivial—the only way to get a nontrivial bundle this way is to have a transformation with a negative singular value, in which case, since we can't raise negative numbers to real powers and stay in the reals, makes the above construction fail, as it should (since a nontrivial bundle fails to have a continuous global frame). So what good is this, really, besides making things pretty? As it turns out, such oddly behaving, yet topologically trivial bundles, do have some interesting, different geometry. It is encoded in the notion of connection—a connection will reveal nontrivialities in the geometry that topology cannot (in fact, Riemannian geometry starts out by considering a connection specifically on the tangent bundle, the Levi-Civita [LAY-vee CHEE-vee-tah] connection). This is an interesting topic for yet another part (I'm not promising it soon, however!).

An unexpectedly delightful application of the Singular Value Decomposition

We begin this entry with a bit of commentary on numerical analysis. If there's anything that I think that should be taught in math classes more, it is the Singular Value Decomposition of a matrix. Every real n × m matrix A can be factored as A = UΣV* where U and V are orthogonal, and Σ is "diagonal" (but can be rectangular). It is in fact more useful than the much more often discussed eigenvalues and eigenvectors (which often are computed for symmetric matrices where the two concepts coincide, anyway). It would probably do a great amount of good for people's intuition in linear algebra classes. For some reason, it is completely ignored in a lot of abstract linear algebra books, somehow having acquired a reputation of being "too applied," and thus would sully the purity of their most precious subject! I never heard it talked about in any upper division or graduate math class, until I'd somewhat belatedly discovered that numerics was really the best place to go to explore my interests (better late than never). But the SVD is absolutely ubiquitous in science. The first place I'd heard of it was a linguistics class, where the prof talked about some paper on latent semantic analysis. As I understood it, the SVD was essentially used to compute axes of word meanings(!!) and the information used to categorize the words. That was really a long time ago, so forgive me, linguists, if this seems a gross mischaracterization. In any case, it also is intimately connected with least squares (briefly mentioned in the last entry).

At this point, the sufficiently attention-spanned reader (all too rare these days!) is most certainly scratching their head (yes, this blog will use singular they) wondering what that has to do with the very out-sized picture. First of all, the picture generated actually is a trivial case in which the SVD isn't actually needed. But if it wasn't for the SVD, I would have never been inspired to make a picture of my findings, so it deserves the tribute. It is inspired by trying to explore the concept of orientation, and to generalize constructions like the famous Möbius Strip:

We can generalize the Möbius strip by the use of vector bundles and identification spaces. We get the topological twist in a Möbius strip by gluing the ends of the usual strip with a flip. If we allow ourselves some imagination, one way to generalize is to first imagine the Möbius strip as now being constructed from a strip that is infinitely tall ([0,1] × ℝ) to get the Möbius vector bundle. Now when we "glue" the ends together, we can still flip it (identify (0,x) with (1,-x)). Or, we can stretch it and identify, say, (0, x) to (0, 2x), the "strained cylinder" (as Roger Penrose calls it in The Road to Reality). If we have to restrict ourselves to a finite picture, what happens here is that it goes around in a loop and reconnects to itself twice as high, so it looks discontinuous. But it's not. It turns out to be, in fact, a cylinder and it is in fact orientable.

Taking a cue from a problem in Spivak's excellent 5-volume series, A Comprehensive Introduction to Differential Geometry, we can imagine that instead of starting out with just a line for the vertical dimension, we could have a whole space ℝⁿ, and identify the initial and final spaces by an invertible linear transformation T : ℝⁿ → ℝⁿ, namely, identifying (0,v) to (1,Tv). If T has positive determinant, then we have not made anything topologically new: it's S¹ × ℝⁿ. But to actually (constructively) prove this is a more difficult task. But one way it can be done is by choosing a global frame, n continuous vector fields that are a basis at every point. And one way to do that is to somehow connect the transformation T to the identity. This can be done any number of ways, since, in fancy-schmancy speak, the orientation-preserving elements of the general linear group form a path-connected component containing the identity. But one quickly implementable realization is to use the SVD! Namely, factor the matrix T as UΣV* with U and V orthogonal and Σ the singular values, which must all be positive if the matrix has positive determinant. Since it is a diagonal matrix of positive numbers, Σ can be raised to any real power t. And U, V are respectively exponentials of some antisymmetric matrices ξ and η (which are not unique). So given some basis vectors v_i at 0, we consider the section

t ⟼ (t, exp(tξ)Σ^texp(–tη)v_i).

At time t = 0 it is just (t, v_i). At time t = 1, it is (1, UΣV*v_i) = (1, Tv_i), which is identified with (0,v_i). (Warning: Although we could make things more notationally uniform by writing Σ^t = exp(tσ) where σ is the diagonal matrix consisting of the logarithms of the original singular values, we must resist the temptation to compute the result as exp(t(ξ+σ–η)), despite it being called an "exponential" map, as that only works when all the matrices commute!!) The path is continuous and so we have proved that there is indeed a continuous, globally defined frame, and thus our bundle is trivial. Beautiful? Well, where are the pictures? The picture above is given by considering the case n = 2, that of a plane, and the transformation by scaling by 2. If we take circular set in the initial plane, we can see how this transformation acts (here the U and V are the identity and Σ is just twice the identiy) and make it go 'round and 'round (without identifying), we get something that is reminiscent of nested tori (but not really, as it is just one torus spiraling into itself!). We're not done with the awesomeness of this example quite yet though—I've got some other math to get back to! But there are more interesting visualizations from this example, so stay tuned!