# More math

Last night, I learned, once more, the definition of absolute continuity. Formally, a function $f : X \to Y$‘s being absolutely continuous is its for any $\epsilon > 0$, having a $\delta > 0$ such that for any finite number of pairs of points $(x_k, y_k)$ with $\sum |x_k - y_k| < \delta$ implies $\sum |f(x_k) - f(y_k)| < \epsilon$. It is stronger than uniform continuity, a special case of it. I saw that it implied almost everywhere differentiability and is intimately related to the Radon-Nikodym derivative. A canonical example of a function not absolute continuous but uniformly continuous, to my learning last night afterwards, is the Cantor function, this wacky function still to be understood by myself.

I have no textbook on this or on anything measure theoretic, and though I could learn it from reading online, I thought I might as well buy a hard copy of Rudin that I can scribble over to assist my learning of this core material, as I do with the math textbooks I own. Then, it occurred to me to consult my math PhD student friend Oleg Olegovich on this, which I did through Skype this morning.

He explained very articulately absolute continuity as a statement on bounded variation. It’s like you take any set of measure less than $\delta$ and the total variation of that function on that set is no more than $\epsilon$. It is a guarantee of a stronger degree of tightness of the function than uniform continuity, which is violated by functions such as $x^2$ on reals, the continuity requirements of which increases indefinitely as one goes to infinity and is thereby not uniformly continuous.

Our conversation then drifted to some lighter topics, lasting in aggregate almost 2 hours. We talked jokingly about IQ and cultures and politics and national and ethnic stereotypes. In the end, he told me that введите общение meant “input message”, in the imperative, and gave me a helping hand with the plural genitive conjugation, specifically for “советские коммунистические песни”. Earlier this week, he asked me how to go about learning Chinese, for which I gave no good answer. I did, on this occasion, tell him that with all the assistance he’s provided me with my Russian learning, I could do reciprocally for Chinese, and then the two of us would become like Москва-Пекин, the lullaby of which I sang to him for laughs.

Back to math, he gave me the problem of proving that for any group $G$, a subgroup $H$ of index $p$, the smallest prime divisor of $|G|$, is normal. The proof is quite tricky. Note that the action of $G$ on $G / H$ induces a map $\rho : G \to S_p$, the kernel of which we call $N$. The image’s order, as a subgroup of $S_p$ must divide $p!$, and as an isomorphism of a quotient group of $G$ must divide $n$. Here is where the smallest prime divisor hypothesis is used. The greatest common divisor of $n$ and $p!$ cannot not $p$ or not $1$. It can’t be $1$ because not everything in $G$ is a self map on $H$. $N \leq H$ as everything in $N$ must take $H$ to itself, which only holds for elements of $H$. By that, $[G:N] \geq [G:H] = p$ which means $N = H$. The desired result thus follows from $NgH = gH$ for all $g \in G$.

Later on, I looked at some random linear algebra problems, such as proving that an invertible matrix $A$ is normal iff $A^*A^{-1}$ is unitary, and that the spectrum of $A^*$ is the complex conjugate of the spectrum of $A$, which can be shown via examination of $A^* - \lambda I$. Following that, I stumbled across some text involving minors of matrices, which reminded me of the definition of determinant, the most formal one of which is $\sum_{\sigma \in S_n}\mathrm{sgn}(\sigma)\prod_{i=1}^{n}a_{i,\sigma_{i}}$. In school though we learn its computation via minors with alternating signs as one goes along. Well, why not relate the two formulas.

In this computation, we are partitioning based on the element that $1$ or any specific element of $[n] = \{1, 2, \ldots, n\}$, with a corresponding row in the matrix, maps to. How is the sign determined for each? Why does it alternate. Well, with the mapping for $1$ already determined in each case, it remains to determine the mapping for the remainder, $2$ through $n$. There are $(n-1)!$ of them, from $\{2, 3, \ldots, n\}$ to $[n] \setminus \sigma_1$. If we were to treat $1$ through $i-1$ as shifted up by one so as to make it a self map on $\{2, 3, \ldots, n\}$ then each entry in the sum of the determinant of the minor would have its sign as the sign of the number of two cycles between consecutive elements (which generate the symmetric group). Following that, we’d need to shift back down $\{2, 3, \ldots, i\}$, the presentation of which, in generator decomposition, would be $(i\ i+1)(i-1\ i) \ldots (1\ 2)$, which has sign equal to the sign of $i$, which is one minus the column we’re at, thereby explaining why we alternate, starting with positive.