Work over \({\mathbb Q}\) and let
The Chinese remainder theorem gives
Multiplication by \(z\) is an operator \(H\). In the ordered local bases \(1,\varepsilon\) and \(1,\eta,\eta^2\), its matrix is
Reading the drawing
The packet at z equals one has basis one and epsilon, with multiplication by H minus one moving one to epsilon to zero. The packet at z equals minus one has basis one, eta and eta squared, with multiplication by H plus one giving a three-term nilpotent chain.
Drawing labels and data 16 labels
Read the original labels as selectable text. Tables preserve grid relationships; other short labels are grouped compactly.
- packet at \(z=1\)
- packet at \(z=-1\)
- \(1\)
- \(\varepsilon\)
- \(0\)
- \(H-1\)
- \(H-1\)
- \(1\)
- \(\eta\)
- \(\eta^2\)
- \(0\)
- \(H+1\)
- \(H+1\)
- \(H+1\)
- two independent basis vectors
- three independent basis vectors
Evidence note from the source edition
These arrows are the actual nilpotent actions \(H-1\) and \(H+1\). Each nonzero displayed basis vector belongs to the indicated packet. Only the last nonzero vector in each chain is an ordinary eigenvector.
The projectors can be written explicitly
Modulo \((z-1)^2(z+1)^3\), put
Then \(e_+\equiv1\pmod{(z-1)^2}\) and \(e_+\equiv0\pmod{(z+1)^3}\). Consequently
The packet dimensions are two and three. The ordinary eigenspaces each have dimension one; the generalized eigenspaces have dimensions two and three.
Read this as
Canonical idempotents isolate the full primary pieces, including their nilpotent structure. Semisimplicity is not required, and replacing a packet by an ordinary eigenspace can discard most of it.