0
$\begingroup$

I am working on extremely large, symmetric matrices of counts, and attempting to identify patterns/shapes within them. Wavelets are a popular tool in image processing, and have some nice statistical properties when applies to discrete random variable.

The problem I face is this: 2D wavelet transforms on the diagonal contain redundant values, making the statistical interpretation of their coefficients difficult.

HOWEVER, i noticed that the lower triangular portion of my dataset (i.e. the dataset without any redundant values) has a natural analogue as a Möbius strip!

The steps taken to remove the redundant entries in my matrix are illustrated in the image below:

  1. Remove the upper right half of the matrix.
  2. Slice the image diagonally through the middle.
  3. Twist and flip the lower right portion and "glue" it to the upper left portion so that the dashed black lines are aligned and pointing in the same direction.
  4. Give the square a half twist and "glue" the yellow sides so that the arrows align in the same direction.

If you're incredulous that these steps are justified, see this video about redundant ordered pairs.

This is all well an good, but is there a natural extension of wavelets (or ANY orthogonal transformation) onto a non-orientable surface? If so, how does one interpret a multi-resolution analysis?

My intuition is that horizontal detail wavelets will carry over, but the non-orientable property will complicate vertical and diagonal detail wavelets.

I am interested in sources that discuss the topic. Web searches haven't turned up anything. I have emailed the topology faculty at my school, and have yet to hear back.

$\endgroup$
  • $\begingroup$ Wavelets require a meaningful topology. Matrices without any additional structure are do not have a meaningful topology because basis permutations do not change the information in the matrix. The fact that you impose a topological structure does not make it any more meaningful. So either you are not giving the entire picture, or using a wavelet transform (or any other transform with a concept of localisation) is mostly meaningless. $\endgroup$ – Jazzmaniac Jan 19 '18 at 22:09
1
$\begingroup$

Bear with me, as this is really only tangential¹ to my area of expertise, but:

Maybe the question is not how to extend wavelets to work on non-orientable surfaces, but to interpret said surface as something where wavelets natively work well.

Enter differential geometry (differential topology? I don't know.):

As you've notice: if you only consider a part of your moebius strip, let's say, everything but the original "yellow line", then it looks perfectly much like structure that you can work with: a (discrete) two-dimensional vector space. I presume you have some explicitly vector-space notation of what your matrix represents!

In fact, the idea here would be to treat your surface as a vector bundle: You take a circle (which is, if you think about it, a one-dimensional vector space itself, with some kind of "modulo" characteristics). To every point on that circle, you assign yet another vector space: a one dimensional line, tangential to that circle.

Now, take a part of your circle (an arc), and find the product space of that with your vector bundle: That's something that should look topologically like your original vector space.

This way of looking at your moebius strip should allow you to use your transforms like you used to use it on the original matrix.


¹ hrhr, a pun.

$\endgroup$
  • $\begingroup$ lol -lol ! (those are orthogonal LOLs). Thanks for the response! I have to step out, but I will look into this and credit you with the answer if this bears out. Much appreciated! $\endgroup$ – Lewkrr Jan 19 '18 at 23:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.