Digging deeper: The isotypic components

In last week’s post, we made use of the coinvariant ring \[\mathbb{C}[x_1,\ldots,x_n]/I\] where $I=(p_1,\ldots,p_n)$ is the ideal generated by the positive-degree homogeneous $S_n$-invariants (symmetric polynomials). We saw that this was an $S_n$-module with Hilbert series $(n)_q!$, and claimed that it was the regular representation.

Let’s see why that is, and see if we can understand where the irreducible components occur.

More precisely, our goal is to understand the series \[\sum_{d} H_{\chi^\mu}(d)q^d\] where $H_{\chi^\mu}(d)$ is the number of copies of the $\mu$th irreducible representation of $S_n$ occurring in the $d$th degree component of $\mathbb{C}[x_1,\ldots,x_n]/I$. In Stanley’s paper on invariants of finite groups, he states without proof the answer as the following ``unpublished result of Lusztig’’:

Writing polynomials in terms of invariants and coinvariants

There is a fun little fact regarding polynomials in two variables $x$ and $y$:

Any two-variable polynomial $f(x,y)$ can be uniquely written as a sum of a symmetric polynomial and an antisymmetric polynomial.

(To be more precise, this is true for polynomials over any field of characteristic not equal to $2$. For simplicity, in what follows we will assume that our polynomials have coefficients in $\mathbb{C}$.)

Recall that a polynomial $g$ is symmetric if it does not change upon permuting its variables. In this case, with two variables, $g(x,y)=g(y,x)$. It is antisymmetric if swapping any two of the variables negates it, in this case $g(x,y)=-g(y,x)$.

It is not hard to prove the fact above. To show existence of the decomposition, set $g(x,y)=\frac{f(x,y)+f(y,x)}{2}$ and $h(x,y)=\frac{f(x,y)-f(y,x)}{2}$. Then \[f(x,y)=g(x,y)+h(x,y),\] and $g$ is symmetric while $h$ is antisymmetric. For instance, if $f(x,y)=x^2$, then we can write \[x^2=\frac{x^2+y^2}{2}+\frac{x^2-y^2}{2}.\]

Expii: Learning, connected

It’s been several months since I posted a gemstone, and the main reason is that much of my free-time mathematics energy recently became channeled into a new project: Expii.

Expii (currently beta) is a new online crowdsourced learning site that aims to fill the gaps in users’ understanding of topics, with the goal of making math, science, and other topics easy for everyone in the universe. Its motto? Learning, connected.

With an addictive, game-like format (hence the XP pun) in which users are awarded “fame points” for writing good explanations and “experience points” for successfully making it through tutorials, Expii is more interactive and community oriented than other online learning resources like Wikipedia. It is also more structured than question-and-answer sites like Quora or Stack Exchange, in that the primary “graph structure” for the topics is organized by our team, and users fill in the content in the nodes.

PEMDAS is broken (and how we can fix it)

In the first week of teaching my Calculus 1 discussion section this term, I decided to give the students a Precalc Review Worksheet. Its purpose was to refresh their memories of the basics of arithmetic, algebra, and trigonometry, and see what they had remembered from high school.

Surprisingly, it was the arithmetic part that they had the most trouble with. Not things like multiplication and long division of large numbers - those things are taught well in our grade schools - but when they encountered a complicated multi-step arithmetic problem such as the first problem on the worksheet, they were stumped:

Simplify: $1+2-3\cdot 4/5+4/3\cdot 2-1$

Gradually, some of the groups began to solve the problem. But some claimed it was $-16/15$, others guessed that it was $34/15$, and yet others insisted that it was $-46/15$. Who was correct? And why were they all getting different answers despite carefully checking over their work?

A proof of the existence of finite fields of every possible order

This is our first contributed gemstone! Submitted by user Anon1.

In the following, $p$ denotes a prime. We wish to prove that, for all positive integers $n$, there is a finite field of order $p^{n}$. Step 1. Restating the problem.

Claim: It suffices to show that, for some power of $p$ (call it $q$), there exists a finite field of order $q^{n}$.

Proof. Suppose there is a field $F$ such that $|F| = q^{n}$. The claim is that the solution set to $x^{p^{n}} = x$ in $F$ is a subfield of order $p^{n}$.

Since $q$ is a power of $p$, we have \[p^{n}-1 | q^{n}-1.\] Since $q^{n}-1$ is the order of the cyclic group $F^{ \times }$, we know that $F^{ \times }$ has a unique cyclic subgroup of order $p^{n}-1$. This contributes $p^{n}-1$ solutions to $x^{p^{n}} = x$ in $F$. But $0$ is another solution, since it is not an element of $F^{ \times }$. This gives $p^{n}$ solutions, so this gives all solutions.