Representation Qit
Exercise PLAN-116
primary source nl only. Current content-bound mathematical second pass accepted. No new Lean verification is claimed for this NL view.
Assumptions and conventions
Question
Proposition 3.22 says the decomposition of a representation into isotypical components is unique. A student over-reads this as: 'the decomposition of a representation into irreducible subrepresentations is unique.' Show by counterexample that the finer decomposition into individual irreducible subspaces is NOT unique, even though the isotypical components are. Use the trivial-isotypical component of the S_2 action on (C^2)^{(x)2} (equivalently, any space carrying the trivial representation with multiplicity >= 2) and exhibit two genuinely different splittings into irreducible lines.
Hint: mathematical route
Isotypical components are unique but splitting a multiplicity-greater-than-one component into individual irreducible subspaces is not, as shown by two distinct trivial-line splittings of the symmetric subspace of (C^2)^{(x)2}.
- Proposition 3.22 fixes the isotypical components, not the internal irreducible splitting of a multiplicity > 1 component.
- Vtriv = Sym^2(C^2) (the +1 eigenspace of F) is the unique 3-dim trivial-isotypical component with trivial S_2 action.
- Since S_2 acts as identity on Vtriv, every line is a trivial subrep; two different line-splittings of Vtriv both valid.
- Non-uniqueness comes from the freedom to change basis of the multiplicity space; multiplicity-free decompositions are unique.
Reference solution
The student's reading is wrong: it is the isotypical components that are unique, not the splitting of a multiplicity-greater-than-one component into individual irreducible subspaces. Take the S_2 action on (C^2)^{(x)2} with phi((12)) = F. The symmetric subspace Vtriv = Sym^2(C^2) is the +1 eigenspace of F, a 3-dimensional space on which S_2 acts trivially; it is the trivial-isotypical component and is uniquely determined (it equals the +1 eigenspace of F, basis-independent). But because S_2 acts as the identity on all of Vtriv, EVERY one-dimensional subspace of Vtriv is an irreducible (trivial) subrepresentation, and Vtriv can be written as a direct sum of three trivial lines in infinitely many ways. Concretely, one splitting is Vtriv = span(|00>) (+) span(|11>) (+) span((|01>+|10>)/sqrt 2); a genuinely different splitting is Vtriv = span((|00>+|11>)/sqrt 2) (+) span((|00>-|11>)/sqrt 2) (+) span((|01>+|10>)/sqrt 2). Both are decompositions of Vtriv into three irreducible (trivial) subrepresentations, but the first line span(|00>) of the first splitting is not among the lines of the second splitting, so the two decompositions into irreducibles are different. This does not contradict Proposition 3.22: the isotypical component Vtriv itself is identical in both (it is the full +1 eigenspace of F), and only its internal basis / line-by-line splitting changed. The source of the non-uniqueness is multiplicity: when an irrep occurs with multiplicity n >= 2, the choice of the n individual copies corresponds to a choice of basis of the n-dimensional multiplicity space, and any invertible change of that basis gives an equally valid finer decomposition. Uniqueness is recovered precisely in the multiplicity-free case.
Assessment rubric
- Diagnoses the over-reading: isotypical components are unique, but the splitting of a multiplicity >= 2 component into irreducible subspaces is not (1 weight)
- Identifies Vtriv = Sym^2(C^2) as the unique trivial-isotypical component (the +1 eigenspace of F), 3-dimensional with trivial action (1 weight)
- Exhibits two genuinely different direct-sum splittings of Vtriv into trivial lines and notes a line in one is absent from the other (1 weight)
- Attributes the non-uniqueness to multiplicity (choice of basis of the multiplicity space) and notes uniqueness holds in the multiplicity-free case, consistent with Proposition 3.22 (1 weight)
Graph-grounded prerequisites
These are support/source anchors, not an automatically extracted theorem-dependency proof.
- Operator and semigroup theory (frozen curriculum anchor; see graph registry)
- Matrix analysis (frozen curriculum anchor; see graph registry)
- Tensor networks (frozen curriculum anchor; see graph registry)
Lean view and evidence boundary
No aligned Lean view has been admitted for this native core in this textbook checkpoint. This is not a claim that the underlying mathematics lacks a formal proof. Contributions must preserve this exact target contract and split.
Sources and provenance
- website/research/sources.json
leditzky-2025-repth-qit
Source SHA-256:9f85f45f20ed5a46931ad0dec3b94ee09748bc56e39fae29b0aefe910440b0ae· MIT metadata only - Primary source
Contract: 51057561f0fc75f037dd19eed229350d6fd6055dd9466110f4154df9df57252f
Item digest: ab9df40fd93fe05c0101d8c2d0d0b36cb55443c64c0780a092a57db7377c5e67
Graph snapshot: qcb-snapshot-ffa644f0da3031a1ce754b9a
Authored support steps: 4. This is not measured graph distance or model difficulty.