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Jordan-quiver isomorphism classes are indexed by partitions

Statement

Every object of $J_{k}$ is isomorphic to a unique direct sum $I_{λ} = I_{λ_{1}} \oplus \dots \oplus I_{λ_{s}}$ , where $λ$ is a partition. The unique simple object is $S = I_{1}$ .

Facts & Assumptions

Given: A field $k$ and a finite-dimensional vector space $V$ with a nilpotent endomorphism $x$ , defining an object of $J_{k}$ .

[F1]

Every finite-dimensional nilpotent endomorphism has a basis that is a concatenation of Jordan strings, including the empty basis on the zero space (Every finite-dimensional nilpotent endomorphism has a basis of Jordan strings).

[F2]

For a nilpotent endomorphism, the numbers of Jordan blocks of each size are determined by the ranks of its powers (Ranks of shifted powers determine Jordan form up to block order).

[F3]

In a Jordan string of length $m$ , $x v_{1} = 0$ and $x v_{j} = v_{j - 1}$ for $j > 1$ ; its matrix is $J_{m} (0)$ (Jordan blocks, Jordan strings, and their endpoints).

[F4]

A partition is a finite weakly decreasing sequence of positive integers; its parts record a multiset uniquely (Partitions, English diagrams, and conjugation).

[F5]

$J_{k}$ is a category whose objects and morphisms are finite-dimensional nilpotent representations and their intertwiners (Nilpotent representations of the Jordan quiver).

[F6]

Here a simple object means a nonzero object whose only subobjects are the zero subobject and the identity subobject.

[F7]

Subobjects in $J_{k}$ correspond to invariant subspaces, via their images (Jordan-quiver categories are abelian; finite-field models are finitary).

Proof

technique · direct

1.1F1F3F4F5

Apply [F1] to $x$ . The spans of its Jordan strings are invariant, their concatenation is a basis of $V$ , and by [F3] each span is isomorphic as a representation to the block $I_{m}$ whose length is that string's length. Sorting these finitely many positive lengths weakly decreasing gives a partition $λ$ by [F4] and an isomorphism $(V, x) ≅ I_{λ}$ . The zero space gives the empty partition.

2.1F2step 1.1

If $I_{λ} ≅ I_{μ}$ , the intertwining isomorphism conjugates their nilpotent matrices, so the ranks of every power agree. By [F2], the multiplicities of blocks of each length agree; hence the multisets of parts and the partitions are equal.

3.1F3F4F5F6F7step 1.1∎

If a string has length $m > 1$ , the span of its initial vector is a nonzero proper invariant subspace, hence a proper nonzero subobject by [F7]; therefore $I_{m}$ is not simple by [F6]. The object $I_{1} = S$ is nonzero and has only the zero and whole subspaces, so it is simple by [F6] and [F7]. Any decomposition in step 1.1 with at least two summands has a proper nonzero invariant summand and hence is not simple. A single summand is simple only when it has length one. Thus $S$ is the unique simple object.

Depends on

Used by

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Sources

Olivier Schiffmann, Lectures on Hall Algebras, arXiv:math/0611617v2, §2.1 (standard reference, not scraped)
S. Treil, Linear Algebra Done Wrong, Chapter 9, Theorems 4.2–4.3 (standard reference, not scraped)