The (2,11) torus knot complement. At first I figured making nice images of torus knots might look “boring,” but I think this is my favorite knot so far! Image
Here as well the picture is drawn by finding a parameterization to the normal tube (of some small radius) about the knot with respect to the spherical metric on S^3, then stereographically projecting from a point on the knot so we are viewing from “inside” the knot complement.
And here’s the (4,11) torus knot as well! Image

• • •

Missing some Tweet in this thread? You can try to force a refresh
 

Keep Current with Steve Trettel

Steve Trettel Profile picture

Stay in touch and get notified when new unrolls are available from this author!

Read all threads

This Thread may be Removed Anytime!

PDF

Twitter may remove this content at anytime! Save it as PDF for later use!

Try unrolling a thread yourself!

how to unroll video
  1. Follow @ThreadReaderApp to mention us!

  2. From a Twitter thread mention us with a keyword "unroll"
@threadreaderapp unroll

Practice here first or read more on our help page!

More from @stevejtrettel

29 Jan 20
I’m back! Having a busy month moving and settling in @Stanford, but this hasn’t totally kept me away from making cool pictures :) Here’s a depiction of an Anosov mapping class on a torus 1/n
Let’s unpack a bit what this means. One way to picture a homeomorphism from the torus to itself is a way of “putting clothes” on the torus: if the domain is some toroidal Christmas sweater and the codomain our torus, a homeomorphism tells us which part of the sweater goes where.
There are an infinite number of ways to put our torus in its Nordic knit - but this infinitude comes in two flavors. First - having already put the sweater on the torus, many new “clothings” can be built by bunching it up a little here, or stretching a little there.
Read 24 tweets
8 Nov 19
Better video (and explanation!) of this week's mathematical conspiracy; a thread. 1/n
The space of quadratic polynomials can be identified with R^3; we think of the function ax^2+bx+c as the point (a,b,c). The polynomials with integer coefficients form the cubical lattice. 2/n
The discriminant of ax^2+bx+c, namely b^2-4ac determines whether the polynomial has two real roots (positive discriminant) or complex conjugate complex roots (negative discriminant). The discriminant=0 surface is a cone. 3/n
Read 10 tweets

Did Thread Reader help you today?

Support us! We are indie developers!


This site is made by just two indie developers on a laptop doing marketing, support and development! Read more about the story.

Become a Premium Member ($3/month or $30/year) and get exclusive features!

Become Premium

Too expensive? Make a small donation by buying us coffee ($5) or help with server cost ($10)

Donate via Paypal

Thank you for your support!

Follow Us on Twitter!

:(