Representation Qit
Exercise PLAN-114
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
A student claims: for any representation (phi, V) of a finite group G and any X in L(V), the symmetrized operator Xbar = (1/|G|) sum_g phi(g) X phi(g)^{-1} is always block diagonal with respect to the isotypical decomposition AND each diagonal block is a scalar multiple of the identity. Decide whether the second half of this claim is true in general. Give an explicit counterexample using the action of S_2 = {e, (12)} on (C^2)^{(x)2} by permuting the two tensor factors (so phi((12)) = F, the swap), and explain precisely what feature of the decomposition makes the diagonal block fail to be scalar.
Hint: mathematical route
When an isotypical component has multiplicity greater than one, the symmetrized operator's diagonal block is identity-on-irrep tensor an arbitrary multiplicity-space operator, so it need not be a scalar - a concrete S_2 swap counterexample.
- Schur forces scalar diagonal blocks only when each isotypical component is a single irrep copy (multiplicity one).
- For S_2 on (C^2)^{(x)2}, Vtriv is the 3-dim symmetric subspace (trivial irrep, multiplicity 3) and Vsgn the 1-dim sign irrep.
- With X = |00><11|, F X F = X so Xbar = |00><11|, a rank-one non-scalar operator supported on Vtriv.
- Multiplicity 3 makes the Vtriv block 1_{W_triv} (x) X_triv with X_triv arbitrary on C^3, hence not scalar.
Reference solution
The block-diagonal part of the claim is correct, but the assertion that every diagonal block is a scalar multiple of the identity is FALSE in general; it holds only when each isotypical component is a single copy of an irrep (multiplicity one). Counterexample: let S_2 act on (C^2)^{(x)2} by phi(e) = 1 and phi((12)) = F (swap). The isotypical decomposition is (C^2)^{(x)2} = Vtriv (+) Vsgn, where Vtriv is the 3-dimensional +1 eigenspace of F (symmetric subspace, spanned by |00>, |11>, (|01>+|10>)/sqrt 2) carrying three copies of the trivial irrep, and Vsgn is the 1-dimensional -1 eigenspace (spanned by (|01>-|10>)/sqrt 2) carrying the sign irrep. Now take X = |00><11| (a rank-one operator). Symmetrizing: Xbar = (1/2)(X + F X F^{-1}) = (1/2)(|00><11| + F|00><11|F) = (1/2)(|00><11| + |00><11|) = |00><11|, because F|00> = |00> and F|11> = |11> (both symmetric), so F X F = X. The operator Xbar = |00><11| maps entirely within the symmetric (trivial-isotypical) subspace Vtriv, so it is block diagonal (its Vsgn block and the off-diagonal blocks are zero, consistent with Schur part (i)). However, restricted to the 3-dimensional Vtriv it is the rank-one operator |00><11|, which is NOT a scalar multiple of the 3x3 identity (it is not even diagonal in the symmetric basis and has rank 1, trace 0). The reason the diagonal block is not scalar is multiplicity: Vtriv contains the trivial irrep three times, so Hom_G(triv, Vtriv) is 3-dimensional and the symmetrized operator restricted to Vtriv is of the form 1_{W_triv} (x) X_triv with X_triv an arbitrary 3x3 operator on the multiplicity space C^3 - only this multiplicity-space factor is scalar when the multiplicity is 1. Hence Schur's Lemma forces scalar diagonal blocks only in the multiplicity-free case; with multiplicity greater than 1 the diagonal block is merely of the form (identity on the irrep) (x) (arbitrary operator on the multiplicity space).
Assessment rubric
- Correctly judges the claim: block-diagonal part holds but diagonal-block-scalar part is false in general (1 weight)
- Gives the explicit S_2 action on (C^2)^{(x)2}, the isotypical decomposition Vtriv (3 copies of trivial) (+) Vsgn, and a concrete X whose symmetrization stays non-scalar on Vtriv (1 weight)
- Verifies the symmetrization computation (e.g. Xbar = |00><11| via F|00>=|00>, F|11>=|11>) showing a rank-one, non-scalar diagonal block (1 weight)
- Explains that multiplicity > 1 is the cause: the diagonal block is 1_W (x) X_k with X_k arbitrary on the multiplicity space, scalar only when multiplicity is 1 (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: 1dbd4d008c55610bb6ff9b91740c7f08a9dbcc5b016fc059ca82c24fc064abea
Item digest: be8cfff02ab1e6793f2ec533ebd7da483867295b6376480ae2ff6811b33f7663
Graph snapshot: qcb-snapshot-ffa644f0da3031a1ce754b9a
Authored support steps: 4. This is not measured graph distance or model difficulty.