Blogs

I was going through a problem in Lee’s Topological Manifolds book1 (problem 7-12). It states that the infinite broom has a strong deformation retract to $(0, 0)$ but not for $(1, 0)$ (only a deformation retract). Below is some of my playful observations inspired by this problem. (Not restricted to the infinite broom space.) Inifinite Broom Note that the deformation retract to $(0, 0)$ itselfs imply the deformation retract to $(1, 0)$, since one can define the deformation retract to $(1, 0)$ by “first retract to $(0, 0)$, then push it to the point $(1, 0)$ along the line joining these two points”....

I am reviewing some fundamental algebra, and I just learnt something beautifully suggesting the connection of ring theory and geometry, which made me eager to learn some algebraic geometry (which I was not very interested when I was in undergrad). It is an exercise in Aluffi’s book1. Lets’ jump into it. The problem Let $K$ be a compact topological space, and let $R$ be the ring of continuous real-valued functions on $K$, with addition and multiplication defined pointwise....

Long time ago I read a quote from Habu1(as shown in the screenshot): The true talent is the ability to carry on with passion, even when it probably won’t pay off. (not translated word by word) I was a teenager and it did not make any sense to me. How could passion be the true talent? Look at those geniuses like Gauss, Mozart, John von Neumann! What they have done is not something that could have been done by normal people....

Warning You need to know some basic group theory terminology to appreciate(I hope you do) the following content. Burnside’s lemma Here is the statement of the Burnside’s lemma: Let $G$ be a group that actions on a set $X$. Denote $X^{g}$ the set of fixed points of $g$ i.e. $\{x \in X | g \cdot x = x\}$, then the number of orbits of the action is equal to $\dfrac{1}{|G|}\sum\limits_{g \in G} |X^g|$....

Problem Let $I_{m} = \int^{2\pi}_{0} \cos(x)\cos(2x)\dots\cos(mx) dx$. For $m$ in $1, 2, \dots, 10$, for which $m$ is $I_m \neq 0$ ? Solution $I_m$ is non-zero if and only if $m \equiv 0, 3 \pmod{4}$. Main idea The official(seemingly? It is in the putnam problem book and a few solutions I found online do the same.) solution is to substitute $\cos x = \frac{e^{ix} + e^{-ix}}{2}$ followed by grouping the terms into $\cos x \cos (2x) \dots \cos (mx) = e^{\text{something}}$, and analyze the something....