Hi all,

I just wanted to show off a program that I made for my Honours dissertation.

https://ariymarkowitz.github.io/Bruhat-Tits-Tree-Visualiser/

Github page

I've written some information about what this is on the GitHub page, I'll paste it below.

About the Bruhat-Tits tree

p-adic numbers

There are a couple of ways of defining the p-adic numbers. One way is to consider the rationals ℚ, and use a different metric (notion of 'distance') from the usual one—under the p-adic metric, rationals x and y are closer together when (x-y) is a multiple of a larger power of p for some given prime p (to be specific, d(x, y) = e^(-v(x - y)), where v(z) is the unique integer such that z = a/b p^(v(z)), where a and b are not multiples of p). We may then take the completion, similar to the real numbers, to give a field and metric space ℚp.

The other way of describing the p-adic numbers is to consider them as the field of infinite expansions

a_(-m) p^(-m) + a_(-m+1) p^(-m+1) + ... + a_(0) p^0 + a_1 p^1 + ...

This is similar to the base-p expansion of a real number, however while real numbers can have infinitely many negative terms and finitely many positive terms, p-adic expansions can only have finitely many negative terms and may have infinitely many positive terms. The p-adic numbers in which all of the powers are nonnegative is called a p-adic integer, and the ring ℤp of p-adic integers is analogous to the integers in the field of rational numbers. Note that the rational numbers are the set of p-adic numbers with finitely many terms in the expansion.

The p-adic numbers allow us to use the properties of metric spaces and geometry to explore number-theoretic properties.

The Bruhat-Tits tree

The Bruhat-Tits tree is a way of visualising the actions of GL(2, K) (invertible 2x2 matrices over K), where K is a local field. In this visualiser, we take K to be ℚp, the p-adic numbers for some prime p. (In actuality, we take K = Q, the rational numbers, since we cannot express the irrational numbers in ℚp with finite precision.)

Let V be the 2-dimensional vector space over K. A lattice is a subset of V that is a free submodule of ℤp of rank 2; in other words, a subset of V that is closed under linear combinations with coefficients in ℤp, and is generated by 2 basis vectors. Lattices are similar to vector spaces, but there is an

I tried to prove that it didn't a fixed point so it has to be a translation isometry but got nowhere.

So instead I tried to use process of elimination. It can only be a translation, rotation or identity isometry and I already proved it cant be a rotation isometry.

Can someone please help me prove that f ∘  f is a translation isometry or that f ∘  f isn't an identity isometry?

What are some good roguelikes with an isometric view?

https://www.reddit.com/r/isometric/comments/gr3q2a/my_first_shot_at_isometric_illustration/frzy9i5/?context=10000

The 2-dimensional hyperboloid, a model for hyperbolic space, is the set I={(x,y,z) , x^(2) + y^(2) - z^(2) = -1} equipped with the lorentzian scalar on R^(3). In Martellis Geometry of Surfaces, Section 2.3, I read that the isometry group of I is generated by reflections r^(S) along subspaces S of I. Such subspaces are defined as the intersection of a 2-dimensional vector subspace in R^(3) with I, S = W intersect I. This might seem like a stupid question, but how do we define these reflections? Of course I understand how a 'Eucledian' reflection works, but if I reflect the whole of R^(3) along any 2-dimensional plane that is not vertical, I is not preserved in this reflection. So what do I miss?

Furthermore, how can we express the correlation between O^(+)(2,1), the group of linear isomorphisms of R^(3) that preserve the lorentzian scalar, which also equals Isom(I), and this group generatred by all reflections r^(S) along subspaces S?

Hi. I'm not sure if "isometry" is even the right word to use here, but here goes.

I'm reading some papers on debiasing word embeddings and a recurring topic/problem in these papers is the issue of "bias by neighbors." This basically refers to the phenomenon that debiasing techniques may be able to reduce the bias component of word embeddings (e.g., by making gender-neutral words such as "computer programmer" equidistant from "male" and "female" etc.) but how bias is still reflected in the neighborhood of the word embeddings. For example, the words "nurse" and "receptionist" still being close together.

My intuition is that since debiasing techniques don't seem to change the inherent structure of the embeddings, there must be some inherent structure that isn't being addressed.

Any feedback is appreciated. Thanks.

Papers for reference:

1. Man is to Computer Programmer as Woman is to Homemaker? Debiasing Word Embeddings (Bolukbasi et al., 2016)
2. Lipstick on a Pig: Debiasing Methods Cover up Systematic Gender Biases in Word Embeddings But do not Remove Them (Gonen and Goldberg, 2019)

It's been my white whale for 10 years, here we go:

(100%) It was 3D (100%) Some levels had multiple tiers/floors to them, that were all in view for that level (90%) It had quite a light/white colour scheme for the most part (80%) There were creatures/monsters/aliens something like that to avoid (50%) It had like cone-like spikes as part of some of the levels

Feel free to ask any clarifying questions!

so I understand the proof that any isometry, f, of R^n can be written as the composition of at most n+1 reflections. so in the proof I learnt, you basically just find reflections G_1,...,G_n such that when they are all composed together with f, fixes 0 and the 'n' basis vectors, giving us the identity. what I don't understand is the corollary to that theorem, which states that if the isometry, f , fixes a point then it can be written as a composition of at most n reflections. I would understand it, if it fixed 0, but I don't get why it would work for an arbitrary point, as you would still need to find reflections to fix 0 and the n basis vectors.

let f be an isometry on R^3 defined by F(x,y,z)=(-y,x,z+1). I know that F can be composed of by at most 4 reflections. how do I show it can't be composed of any less.

so I understand the proof that any isometry, f, of R^n can be written as the composition of at most n+1 reflections. so in the proof I learnt, you basically just find reflections G_1,...,G_n such that when they are all composed together with f, fixes 0 and the 'n' basis vectors, giving us the identity. what I don't understand is the corollary to that theorem, which states that if the isometry, f , fixes a point then it can be written as a composition of at most n reflections. I would understand it, if it fixed 0, but I don't get why it would work for an arbitrary point, as you would still need to find reflections to fix 0 and the n basis vectors.