Homeomorphisms of the torus, part II (matrices and linear algebra)

12 Oct

So we know what a torus is, and what a homeomorphism is, from our first post in this series.  Hopefully you remember this picture, which demonstrates how we can tile the plane using the unfolded torus.  (If not just check out the first post.)


We also had a picture showing curves on the torus, represented on the grid.

Left: follow the numbers to see the knot.  Right: look at the bottom-most green line.

Left: follow the numbers to see the knot. Right: look at the bottom-most green line.

Remember we said that every curve on the torus could be represented by drawing it on the torus and cutting (like in the left) and then unfolding it into a line (as in the right).  The left picture is what happened when I cut up the torus, while the right picture is six copies of the left pictures, as they appear in the plane.  When I glue all six squares together, I can see the slope of the line, as in the bottom most green line.

Notice that we can characterize this curve by the slope of the line- here’s it’s 2/3.  This is the same as 4/6 or 6/9, and you can see this on an actual torus by pulling the strings so that you get back to 2 times one directions and 3 times the other.  Or just believe me if you aren’t into making models (I build things out of paper and tape and floss very often) [Clean floss].

So we can think of curves on the torus as rational points \frac{p}{q} in the plane.  Great!

Here’s a fact: we can define a homeomorphism of the torus by describing where the (1,0) and (0,1) base curves are sent (in our case the red and blue curves).  Why is this fact true?  Because if I know where the red and blue curves go, I can figure out where any other point of the torus goes by using their coordinates (like if a point is at (1/3, 1/9) in the plane, I can send that point to 1/3(f(red)) + 1/3(f(blue)) and this’ll be continuous).  Curves have to be sent to curves, so I need to assign a rational point \frac{p}{q} to each base curve- remember, we associate a rational number to each curve of the torus.  I can do this using a matrix \left( \begin{array}{cc} p_{red} & p_{blue} \\ q_{red} & q_{blue} \end{array} \right).

In this manner we can describe a homeomorphism of the torus by a matrix that describes where each base curve is sent via its columns, and hence we can figure out where any point is sent by writing its coordinates in terms of the base curves.

Example: the identity homeomorphism sends the red curve back to itself, and the blue curve back to itself.  So we want to say that (1,0) maps to (1,0), and (0,1) maps to (0,1).  This means we have \left( \begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}\right), which is, coincidentally, the identity matrix.  You can check that this does what we want it to by multiplying the matrices \left( \begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array} \right) \left(\begin{array}{c} 1 \\ 0 \end{array} \right)=\left( \begin{array}{c} 1 \\ 0 \end{array} \right).  Sweet!

Remember that our original idea was that somehow we could classify all homeomorphisms of the torus.  So far we’ve made an equivalence between homeomorphisms of the torus and these 2 by 2 matrices.  Next I’m going to lay down some facts from linear algebra, which are given without proof because honestly I forgot how to prove this stuff/all my linear algebra.

  • Trace is invariant under conjugation.  Conjugation is multiplying a matrix by another as well as its inverse: so when we conjugate A by B, we get B^{-1}AB.  Trace is the sum of the diagonal elements of your matrix, so the trace of the identity matrix is 2.  So this fact says that tr(A) = tr(B^{-1}AB).
  • The determinant is the product of the eigenvalues.  Determinant for a matrix A=\left( \begin{array}{cc} a & b \\ c & d \end{array}\right) is given by the formula $det A = ad-bc$.  The eigenvalues of a matrix are the numbers \lambda that solve the equation (a-\lambda)d-b(c-\lambda) =0.  We are just laying down the facts, yo.
  • Eigenvalues scale eigenvectors.  This means for any \lambda that we found earlier, there’s a vector \textbf{v}=\left( \begin{array}{c} x_1 \\ x_2 \end{array} \right) so that A\textbf{v} = \lambda \textbf{v}.  This is sort of the definition of eigenvalue and eigenvector.

Woof.  Lots of words, lots of terms and definitions.  The next post in this series will tie together the picture of the torus and all this linear algebra and actually classify all the homeomorphisms (I’ll give you a hint; it’s by trace) as well as give some properties of them.


5 Responses to “Homeomorphisms of the torus, part II (matrices and linear algebra)”

  1. emma October 13, 2013 at 6:36 am #

    I understood very little, but learned for the first time what a homeomorphism is 🙂

    • yenergy October 14, 2013 at 10:30 am #

      Well I’m glad you got something out of it! I’m trying to make this stuff more understandable but I know that I don’t do a great job of it. Thanks for stopping by and commenting!

      • emma October 14, 2013 at 10:35 am #

        I have no background in maths (Biology major here) so it’s not your fault! However, I recently switched to engineering so I’m trying to hone my math chops. Although, I don’t think I will ever go down the mathematician route. 🙂


  1. Homeomorphisms of the torus, part IV (topology of the identity component) | Baking and Math - December 30, 2013

    […] I for a definition of homeomorphism and torus and Part II for a bit more linear algebra. […]

  2. An open problem in group theory | Baking and Math - February 25, 2014

    […] may remember that I once did a series of posts 1, 2, 4, on the homeomorphisms of the torus.  You don’t need to read all the posts to get this […]

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