Click the image to disable camera rotation. An older version of this doodle can be found here and an *even older* version can be found here.

__What the Heck is this Thing and Why are There an Ungodly Number of Sliders?__

A Brieskorn variety is the locus of points in $z\in \mathbb{C}^n$ satisfying
$$\sum_{i=1}^n z^{p_i}=0$$
Where $\{p_i\}$ is a set of positive pairwise coprime integers. Why do these things matter? It turns out that taking the intersection of a Brieskorn variety with a small (2n-1)-sphere centered at the origin will yield a manifold with the same homology groups as the standard sphere. In high dimensions, this intersection sometimes turns out to be a manifold which is homeomorphic but not diffeomorphic to a standard sphere. If $n=3$ and our exponents are 2, 3, 5 we get the Poincare dodecahedral space. For other triples of exponents, we get examples of 3-manifolds which have the geometry of the universal cover of $SL(2,\mathbb{C})$.

At this point I have enough geometrical knowledge to know what these statements mean, but not *why* they are true. That's why I decided to make this visualization (though, admittedly, it turned out to me more of an aesthetic endeavor than an intellectual one).

The scheme I use to visualize this structure is fairly straightforward. First of all, $n=3$ and our exponents are 7,8,9. We aren't visualizing the entire thing, only a 3D cross section (most of the sliders are for selecting this cross section). A general way of rendering level-sets of scalar function $f$ is to treat the function $\frac{f}{||\nabla f||}$ as a distance field and raymarch it. Here, we take
$$f(z)=\Bigg|\sum_i z^{p_i}\Bigg|$$
This function will be zero exactly when $z$ is in the Brieskorn variety, so I conjecture that raymarching $\frac{f}{||\nabla f||}$ will result in a shape at least somewhat-similar to the Brieskorn variety. Given the 3-plane which we want to intersect the variety with, this gives us a recipe to visualize that intersection. But how do we go about selecting that 3-plane?

Initially, in the two shaders I have linked at the top, I had the orientation of the 3-plane fixed and I just translated it around a bit with the interactivity. What I noticed which I thought was very odd was that the structure appearing on the screen always had the symmetry group which was the dihedral group on $p_1$ vertices. Eventually, I discovered that if you change the orientation of the plane which is intersecting the variety, you get new symmetry based on the rest of the exponents!

This made me realize that I wanted to be able to explore all possible 3-planes that could intersect the structure. This requires a huge number of parameters, seeing as the dimension of the Grassmannian $Gr(k,m)$ is $k(m-k)$. Recall from earlier that we had the complex dimension $n=3$, implying the real dimension of the ambient space is $m=6$, and the dimension of our k-planes is $k=3$ so we have 9 total parameters to describe a point in the Grassmannian $Gr(3,6)$. The way I obtain the exact 3-plane from the slider values is by orthonormalizing a triangular matrix filled with the slider values (the values must first be exponentiated to make sure we get a wide range of planes after normalization, but this is the basic idea). The 6 sliders below that simply correspond to a 6-vector which translates the 3-plane around in $\mathbb{C}^3$. The bottom slider simply zooms the camera in and out.