Character inner products and orthogonality
Derive multiplicities from an averaged Hom-space action and compute a complete decomposition from three trace values.
On this page
The question for today
What you’ll understand
- Derive the character inner product from invariant homomorphisms.
- Apply complex character orthogonality with class-size weights.
- Check irreducibility using the sum of squared multiplicities.
A character records one number for each group element: the trace of its representing matrix. Traces forget coordinates but retain enough information to recover every complex finite-group representation up to equivalence. The reason is not that arbitrary matrices are determined by their traces. It is that complete reducibility restricts the possibilities and Schur’s lemma makes irreducible characters orthonormal. The resulting inner product turns decomposition into a finite weighted sum, with complex conjugation and conjugacy-class sizes doing essential work.
Trace becomes a class function
A character is constant on conjugacy classes because conjugating a group element conjugates its representing matrix, and trace is unchanged by similarity. Its value at the identity is the dimension. Direct sums add characters. For a finite-order complex matrix, every eigenvalue is a root of unity, so replacing the element by its inverse conjugates the trace. This is consistent with choosing an invariant Hermitian form and making all representing matrices unitary, although traces remain the same in the original basis.
Define the inner product on complex class functions to be linear in its first argument and conjugate linear in its second, with the group average as normalization. If a sum is organized by conjugacy classes, each representative must carry the number of elements in that class. Omitting these weights changes the inner product for nonabelian groups. Omitting conjugation breaks positivity for complex-valued characters, a failure already visible for a nontrivial character of the cyclic group of order three.
Average the action on a space of maps
The group acts on linear maps from W to V by sending a map to the action on V, then that map, then the inverse action on W. The fixed maps are exactly the intertwiners. Averaging these operators gives a projection onto that fixed subspace, and the trace of a projection is its rank. The trace of the operator on the space of maps is the character of V times the conjugate character of W. Thus the inner product of the two characters equals the dimension of the intertwiner space.
For inequivalent irreducibles this dimension is zero, and for equivalent irreducibles it is one by the complex form of Schur’s lemma. Orthogonality follows without choosing decomposition vectors. If a general representation contains an irreducible with some multiplicity, taking its character inner product with that irreducible character returns the multiplicity. The result must be a nonnegative integer, providing a strong diagnostic when a proposed character or class-size list has been entered incorrectly. Rounding a substantially nonintegral answer would conceal an error.
A norm tests multiplicities, not dimension
The permutation representation of a group acting on a finite set has character equal to the number of fixed points, since each fixed basis vector contributes one diagonal entry. For the natural action of the symmetric group on three letters, the three class values are three, one, and zero: the identity fixes all points, a transposition fixes one, and a three-cycle fixes none. The fixed constant vector gives a trivial summand, and subtracting its character leaves the standard two-dimensional action.
Its remaining class values are two, zero, and minus one. With class sizes one, three, and two, its squared norm is one, confirming irreducibility. The original permutation action has squared norm two despite dimension three, because it contains two distinct irreducible summands once each. A representation containing two copies of a single irreducible instead has squared character norm four. These distinctions make character norms useful for finding repeated modes and symmetry-allowed linear maps before constructing any preferred basis.
Work it through
Decompose the three-point permutation action
For , use class sizes and characters , , , and to find the multiplicities in P.
The trivial multiplicity is . The sign multiplicity is . The standard multiplicity is .
The candidate standard character has norm . It is the character of the invariant zero-sum plane, so the norm criterion proves that this actual representation is irreducible.
The recovered dimension is , and adding the two character rows yields on every class. Also .
The permutation representation is the direct sum of the trivial representation and the standard two-dimensional irreducible, with no sign summand.
The living notebook
Explore the model
Interactive lab · modular
Fixed basis vectors in a cyclic permutation
Pause and Reset, then set both modulus and additive step . Predict that one Step, advancing to time , highlights residue and that the translation permutation fixes no residue. Reset and set both ; predict every residue is fixed by that translation and the selected orbit remains at zero. The lab shows one additive orbit; count fixed points of the full permutation on paper to obtain characters and . Control changes retain time unless Reset is used.
- Residue r0
- 0
- Orbit period · m/gcd(m,a)
- 12
- Distinct residues visited
- 1 of 12
Addition modulo m starts at zero: rₙ = na mod m. Playback advances two additions per second. The orbit has period m/gcd(m,a), including period one for a = 0. This is not a cryptographic security test or a primality algorithm. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.
Look forA permutation character counts all fixed basis vectors, while an orbit display follows only one chosen basis label.
Pause & reason
Check your understanding
Your turn
Try it yourself
A genuine representation has character on the same three classes. Decompose it using the displayed irreducible rows and verify the answer's dimension and squared norm.
Take a moment to make a prediction or work it out on paper. Open a hint when you need one.
Give me a hint
Take three separately weighted inner products, including the class-size factor three on the transpositions.
Compare with the solution
The multiplicities are for the trivial row, for sign, and for standard. Thus , of dimension four. Its norm is , matching . The negative trace at a transposition is permitted; a character is not a probability distribution.
Keep the idea
Take it with you
- Characters are class functions and identity values give dimensions.
- The normalized conjugated inner product measures intertwiner dimensions.
- Squared character norm counts squared multiplicities.
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