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.


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