Covering Maps and Picard’s Little Theorem.

Math
Here is a pretty wild theorem that generalizes Liouville's theorem in complex analysis: Picard's Little Theorem: If a holomorphic, entire function $f:\mathbb C \to \hat{\mathbb C}$ misses three points, it is constant. This theorem is remarkable, and I had never really heard of it, since its proof relies on a certain covering. The following argument can be made elementary (by describing a certain covering) but instead we will spend some time talking about some implications of the Uniformization Theorem, and finish with an "easy" lifting argument. Note that Picard's Little theorem works just as well for $f: \mathbb C  \to \mathbb C$, since we can remove a point, and rotate to stereographically project, so we replace three points with two points. Uniformization Theorem: every simply connected Riemann surface is the…
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Fourier Transform on Finite Abelian Groups

Math
the main goal of this quick note is to simultaneously misrepresent the work done by Harmonic Analysts and by working under some dubious assumptions, shed some light on how we can define a Fourier transform  on finite abelian groups.   Suppose we are given a function $f:\mathbb R \to \mathbb R$ that is $2 \pi$ periodic, and suppose further that it is a function that deserves to be integrated. Then, we know from the grapevine that we can rewrite $f$ in terms of its "Fourier series:" $$f(x)=a_0+\sum_{n=1}^{\infty}[a_n \cos(nx)+b_n \sin(nx)]$$ and lest anyone is nauseated by trigonometric functions, let's quickly generalize out of this to functions $f: \mathbb R \to \mathbb C$ by rewriting $\cos(x)$ and $\sin(x)$ in terms of $e^{inx}$. In other words, we are now thinking about $$L^2(\mathbb R):=\{f:\mathbb…
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The Ham Sandwich Theorem Via Borsuk Ulam

Math
I've recently been doing some research for my Senior thesis in applications of characteristic classes to plane equipartition problems.These types of problems follow a general procedure known as the Configuration Space-Test Map paradigm. I'll try to demonstrate how these basic arguments go, using what is probably the most famous theorem in equivariant topology: The Borsuk Ulam Theorem: Every $\mathbb Z_2$-equivariant map $f:S^n \to \mathbb R^n$ vanishes. The theorem can be proven in a more general language that I will try to flesh out in future blog posts, but I will prove the two dimensional case using just basic covering space theory here and apply it to the two-dimensional ham sandwich theorem: The Ham-Sandwich Theorem: Any two masses in $\mathbb R^2$ can be simultaneously  equipartitioned by a single line. The idea will be…
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What No Algebra Teacher Ever Told Me.

Math
These are a bunch of Topological proofs for facts in  Algebra. Have you ever had the experience of  insisting to yourself that the fundamental theorem of algebra is really a topological result, only to forget the argument? Have you ever proven Cayley Hamilton by muttering something about cofactors? Showing Bezout's Identity by stating the  Euclidian algorithm (whose proof eludes lazy undergraduates everywhere.) I have. I am therefore writing this blogpost to once again weasel out of actually doing any math. Theorem (Bezout's Identity): For any coprime integers $a,b \in \mathbb Z$, there exist integers $c,d$ so that $ac-bd=1$. Proof: recall from my previous blogpost "a cute application of mapping class groups" that a member of $\pi_1(\mathbb T)$, or $\mathbb Z \oplus \mathbb Z$ can be realized as a simple closed curve, if…
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A Cute Application of Mapping Class Groups

Math
Let's start with a fairly innocuous question: Question: What elements of $\pi_1(\mathbb T^2)$ can be represented by a  (homotopy class of a) simple closed curve? A naive guess might be that with some effort, every element can be obtained by some simply closed curve, but further inspection shows that this cannot be the case. For example, first drawings of the element $(2,0)$ will show that the curve crosses itself. On the other hand, there are elements that surely can be represented: $(1,0), (2,1), (3,2)$ and so on. In fact, given any $(a,b)$ with $\mathrm{gcd}(a,b)=1$, we have the following procedure: which fails for when $(a,b)$ have a common factor, since it produces multiple curves! However, we claim that indeed these are the only elements of $\pi_1(\mathbb T)$ that can be represented by a…
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Nakayama Says It’s Enough To Know The Derivative

Math
This is a quick note on an idea that I think is worth pursuing, but I've mentioned in a previous post (although I was sort of blasé about it.) There is nothing really to prove here, but I feel that Nakayama's Lemma is hard to conceptualize, and the Jacobson Radial generalization makes an intuitive result  obscure. All modules in this post are finitely generated. Nakayama's Lemma: If $M$ is an $R$ module with $(R,\mathfrak{m})$ local, then $M=N+\mathfrac{m}M$ implies that $N=M$. The idea here is that finitely generated modules are closely related to vector spaces in a sense that can be made precise: Corollary:  $a_1, \dots,a_n$ generate $M$ as a module if and only if their images in the quotient generate $M/\mathfrac{m}M$ as an $A/\mathrfac{m}$-vector space. The forward direction is easy…
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“Projective Geometry Is All Geometry”

Math
The title  is a quote attributed to  the mathematician William Cayley.   The goal of this post is to merely organize and make sense of the fact that all three of the principal continuous geometries are contained in projective geometry. Apriori, this is really remarkable since the definition of projective space (and its symmetries) as  linear subspaces of $\mathbb R^n$ has very little to do with the other geometries we consider.  For me, the most interesting fact here is not that the underlying spaces (euclidian, spherical etc.) can be embedded in projective space, but rather that there exist embeddings that preserve the symmetries of each respective geometry, a statement that will be made precise immediately: Definition 1: A pair $(Y:H)$ is said to be a geometry (in the sense of…
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Open Mapping Theorem for Sufficiently Nice Lie Groups

Math
Today, I was in class, and was given a question of the form "show that $h:G \to H$ cannot be surjective, and I asked my professor, "can we assume some basic functional analysis? The open mapping theorem implies that this is is open, and hence a homeomorphism, which it is not." A fellow student remarked that manifolds aren't Linear, which I accidentally misheard as "please quit mathematics immediately." Luckily, I am very bull-headed and I've spent the day reformulating the proof for the open mapping theorem in Banach spaces to Lie groups. I think it works--  I was unable to find any such result in my books on Lie theory, and I didn't want to google it initially (we live with our contradictions), but after proving the lemma, I looked it up…
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Adjoints in an Inner Product Space.

Math
I've heard many times "a vector space is naturally isomorphic to its double dual," with reference to the "canonical isomorphism" (evaluation.) However, in the same breath, there is no "easy" way to construct an isomorphism from a vector space to its usual dual (in a sense that can  be made formal.) This is really pretty sad, since in finite dimensions they are always isomorphic. But, on the other hand, if one works in the category of inner product spaces, a real vector space is canonically isomorphic to its dual space. In fact, the isomorphism is exactly what we would expect: Proposition 1: Given a finite dimensional real vector space $V$ equipped with inner product $(\dot,\dot)$, there is an isomorphism $\phi: V \to V^*$ given by $\phi(x):= y \mapsto (x,y)$. First note that this makes…
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A Geometric Reason to Care About Discrete Valuation Rings

Math
Perhaps those that have taken an introductory commutative algebra class have come across the profoundly opaque definition of a discrete valuation ring (DVR), thinking to themselves "why have the math gods condemned me?" From a purely algebraic perspective, it is a remarkable that a ring $A$ is a valuation ring if and only if it is a domain so that for any element in the fraction field for $A$, $x$ or $x^{-1}$ is contained in  $A$. Even further, if a valuation ring is Noetherian (and not a field) it is a DVR. Again, algebraically, there are a lot of reasons to care about these rings, but today isn't about algebra. While there are a huge number of equivalences we can speak of, but instead we will make use of the following definition:   Definition:…
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