## The diagonal fallacy and James’ Submodule Theorem

The diagonal fallacy is my name for the mistaken belief that the direct sum decomposition of $\mathbb{R}^2$ as the direct sum $\langle (1,0) \rangle \oplus \langle (0,1) \rangle$ is in some way unique. It is frequently committed by undergraduates, many of whom have the even more mistaken belief that $\oplus$ somehow means $\cup$. (Probably they are being mislead by their deep insight that $\oplus$ and $\cup$ are both categorical coproducts.)

James’ Submodule Theorem is a fundamental theorem in the representation theory of the symmetric group. To give an illustrative example, fix a prime $p$, and let $M$ be the permutation module for $\mathbb{F}_pS_n$ acting on the set $\Omega$ of all ordered pairs $(i,j)$ with $1 \le i ,j \le n$ and $i\not= j$. Thus the elements of $M$ are formal linear combination of the elements of $\Omega$ with coefficients in the finite field $\mathbb{F}_p$. Let $\langle \ , \ \rangle$ be the bilinear form on $M$ defined on basis elements so that

$\langle (i,j), (i',j') \rangle = \begin{cases} 1 & (i,j) = (i',j') \\ 0 & (i,j)\not= (i',j'). \end{cases}$

Write $U^\perp$ for the orthogonal complement of a subspace $U$ of $M$ with respect to this form. Let

$b = \sum_{\sigma \in S_3} \sigma\; \mathrm{sgn}(\sigma)$.

The Specht module $S^{(n-2,1,1)}$ can be defined as the $\mathbb{F}_pS_n$-submodule of $M$ generated by the element

$e = (2,3)b = (2,3) - (3,2) + (3,1) - (1,3) + (1,2) - (2,1).$

Writing $U$ for $S^{(n-2,1,1)}$, a special case of James’ Submodule Theorem states that if $V$ is a $\mathbb{F}_pS_n$-submodule of $M$ then either $Vb = 0$, in which case $V \subseteq U^\perp$, or $Vb = \langle e \rangle$, in which case it follows from the definition of $U$ that $V \supseteq U$.

A corollary is that in any direct sum decomposition of $M$ there is a unique indecomposable summand containing $U$, namely the unique summand $Y$ such that $Yb \not= 0$. This summand $Y$ is unique up to isomorphism, by the Krull–Schmidt theorem.

If $p$ is odd then $Y$ is in fact unique as a submodule of $M$, since then $Y$ is a summand of the submodule $\langle (i,j)-(j,i) : 1 \le i < j \le n\rangle$ of $M$, and this module is indecomposable if $p$ divides $n$ and otherwise semisimple with two non-isomorphic summands (so no `diagonal' submodules), namely the Specht modules $S^{(n-1,1)}$ and $S^{(n-2,1,1)}$. When $p=2$ the situation is more complicated, and I think it would be interesting to know whether $Y$ is unique as a submodule of $M$.

The following example shows one way uniqueness can fail.

Example. Let $F$ be a field and let $A = F[x]/(x^2)$. Then $A$ has a two-dimensional module $\langle v, s \rangle$ on which $x$ acts by $vx = s$, $sx = 0$, and a one-dimensional module $\langle w \rangle$ on which $x$ acts by $wx = 0$. There are distinct direct sum decompositions of $M = \langle v, s, w \rangle$, both with a unique summand containing $\langle s \rangle$, namely

$M = \langle v, s \rangle \oplus \langle w \rangle = \langle v + w, s \rangle \oplus \langle w \rangle.$

Note that $Mx = \langle s \rangle$, but evidently there is no ‘unique submodule theorem’ applicable to the image of the map $x : M \rightarrow M$.