Finding Gemstones: on the quest for mathematical beauty and truth
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What happens in characteristic p?

There is a reason that the real numbers and complex numbers are so popular. They have the nice property that $1+1+1+\cdots+1$ is not equal to zero, no matter how many times you add $1$ to itself.

Strange things happen in fields of prime characteristic $p\neq 0$, such as the field $\mathbb{Z}/p\mathbb{Z}$ of integers taken modulo $p$. In these fields, $1+1+\cdots+1=0$ for any number of $1$’s which is a multiple of $p$. We get non-intuitive but strangely beautiful identities like $(x+y)^p=x^p+y^p$, since the coefficient $\binom{p}{i}$ is divisible by $p$ for $1\le i\le p-1$.

And, we get a representation of the symmetric group which is indecomposable but not irreducible. Allow me to explain.

The symmetric group $S_n$ is the group of permutations of $\{1,2,\ldots,n\}$ under composition. We can represent these elements as $n\times n$ permutation matrices, by assigning to the permutation $\pi:\{1,2,\ldots,n\}\to \{1,2,\ldots,n\}$ the matrix whose $(i,\pi(i))$th entry is $1$ for all $i$, and whose other entries are all $0$. For example, for $n=3$:

  • The permutation $(123)$ is assigned the matrix: \[\left(\begin{array}{ccc} 0 & 0 & 1\\ 1 & 0 & 0\\ 0 & 1 & 0 \end{array} \right)\]

  • The identity permutation $(1)(2)(3)$ is assigned the identity matrix: \[\left(\begin{array}{ccc} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{array} \right)\]

  • The permutation $(12)(3)$ is assigned the matrix: \[\left(\begin{array}{ccc} 0 & 1 & 0\\ 1 & 0 & 0\\ 0 & 0 & 1 \end{array} \right)\]

And so on. This gives an action of $S_n$ on the vectors in $\mathbb{R}^n$ by multiplying by the associated matrix. For instance, $(123)$ sends the vector $(2,4,-3)$ to $(-3,2,4)$ by permuting the coordinates.

There are a few things that are “fixed” by this action. For one, any permutation in $S_3$ maps the vector $(1,1,1)$ to itself. In fact, it fixes the one-dimensional subspace spanned by this vector: every vector of the form $(r,r,r)$ is fixed under any permutation. Let’s call this subspace $U$.

Decomposing vector spaces as the direct sum of two subspaces is fun, so let’s look at the space perpendicular to the all $1$’s vector: the space of vectors whose entries sum to $0$. In $\mathbb{R}^3$, this would be the plane $x+y+z=0$. Notice that every permutation fixes this space too: any permutation of a vector whose sum is $0$, such as $(1,1,-2)$, also has sum $0$. Let’s call this subspace $W$.

So, we can write $\mathbb{R}^n$ as $U\oplus W$, where the action of $S_n$ on $U$ and $W$ give two smaller matrix representations of $S_n$.

This worked out quite nicely when dealing with real numbers. But what happens if we are instead considering vectors whose entries are from a field $k$ of characteristic $p$, for some $p$ dividing $n$? For instance, consider $k^3$ where $k$ has characteristic $3$. Then the vector $(1,1,1)$, an element of the subspace $U$ from above, has sum $1+1+1=0$. But that means that $(1,1,1)$ is in $W$ as well! The spaces that used to be perpendicular are now nested; $U$ is contained in $W$ altogether.

In fact, this representation is now indecomposable; it cannot be written as a direct sum of two smaller representations. It is not irreducible, however; the sub-representation $W$ is still well-defined, but it is not a direct summand of the entire space!

Strange things happen in characteristic $p$.