advanced35 min studyLesson 518 in Mathematics

Induced representations and Frobenius reciprocity

Extend a subgroup action with explicit left-coset conventions and compute multiplicities by restriction.

On this page

What you’ll understand

  • Construct induction as a tensor product over the subgroup algebra.
  • Verify Frobenius reciprocity by explicit inverse maps.
  • Calculate an induced character and cross-check its decomposition by restriction.

Restriction keeps a representation space and forgets some symmetry operators. Induction moves in the other direction, building a larger group action from a subgroup action. It is not arbitrary extension of the old matrices: it creates one copy of the original space per left coset, with transitions between copies carrying subgroup operators. Frobenius reciprocity makes this construction practical by translating maps out of the induced space into maps respecting only the subgroup. We keep left actions and coset conventions explicit throughout.

One copy per left coset

Let H be a subgroup of a finite group G and W a finite-dimensional complex left representation of H. Define induction using the group algebra of G tensored over the group algebra of H with W. The balancing relation identifies a subgroup factor on the right of a group element with that subgroup element acting on the vector. Left multiplication by G on the first factor is well defined because it respects this relation. It supplies the induced representation without any arbitrary choice of extended matrices.

Choose representatives for left cosets of the form tH. Every group element is uniquely a representative followed by an element of H, so the balanced tensor product has one copy of W for each representative. To act by g on the copy indexed by t, write gt as another representative times h and apply h to the vector in that destination copy. This shows the dimension is the index times the original dimension. A different representative list changes coordinates but not the equivalence class of the representation.

Ind⁡HGW=C[G]⊗C[H]W,gh⊗w=g⊗ρW(h)w,dim⁡Ind⁡HGW=[G:H]dim⁡W\operatorname{Ind}_H^GW=\mathbb C[G]\otimes_{\mathbb C[H]}W,\qquad gh\otimes w=g\otimes\rho_W(h)w,\qquad \dim\operatorname{Ind}_H^GW=[G:H]\dim W
All modules are left modules. The group algebra is a left G, right H bimodule, and the representatives label left cosets tH.↔ Scroll wide equations horizontally.

Reciprocity is an explicit correspondence

An equivariant map from the induced space to a G-representation V is determined by what it does on tensors whose first factor is the identity. Restricting there gives an H-equivariant map from W to the restriction of V. Conversely, given such an H-map, send a general pure tensor to the group element acting on its image under that map. The balancing relation makes this well defined precisely because the original map respects H. The two procedures undo one another and are linear.

This proves Frobenius reciprocity, an isomorphism of Hom spaces rather than merely an equality of dimensions. For irreducible V over the complex numbers, their common dimension is its multiplicity in the induced representation. Taking characters expresses the same identity through normalized inner products on the two different groups. Each group uses its own order in the denominator. The construction and adjunction also exist over arbitrary fields, but our conversion into ordinary irreducible multiplicities uses the semisimple complex setting.

Hom⁡G(Ind⁡HGW,V)≅Hom⁡H(W,Res⁡HGV),T⟼[w↦T(1⊗w)]\operatorname{Hom}_G(\operatorname{Ind}_H^GW,V)\cong\operatorname{Hom}_H(W,\operatorname{Res}_H^GV),\qquad T\longmapsto[w\mapsto T(1\otimes w)]
The inverse sends an H-map A to the map g tensor w ↦ ρV(g)A(w). This formula verifies the side and inverse conventions directly.↔ Scroll wide equations horizontally.

Trace counts cosets that return to themselves

To compute the induced character at a group element, only coset copies sent to themselves contribute diagonal-block traces. A representative t contributes the subgroup character at the conjugate t inverse times g times t, provided that conjugate lies in H. Summing over the fixed cosets gives the character. Equivalently sum over all group elements satisfying that membership condition and divide by the subgroup order, since each fixed coset has exactly that many representatives. Conjugating within H does not change its character value.

When W is trivial, the induced representation is the permutation action on left cosets, recovering fixed-coset counts. A nontrivial W retains information about the subgroup action within a returning coset and can produce negative or complex traces. Induction does not usually preserve irreducibility: distinct G-types can restrict to overlapping H-types. Reciprocity measures exactly that overlap and often computes a decomposition much faster than building all induced matrices. A dimension check remains essential because confusing index with subgroup order changes the constructed space.

Induce a nontrivial order-two character to S3

Take H={e,(12)}⊂S3H=\{e,(12)\}\subset S_3 and its sign character η=(1,−1)\eta=(1,-1). Compute the induced character on the three S3S_3 classes and decompose it.

  1. At the identity, induction has dimension [S3:H]=3[S_3:H]=3. A three-cycle has no conjugate in H, so its induced character is zero.

  2. For a transposition g there are two x with x−1gx=(12)x^{-1}gx=(12), since its centralizer has order two. Each contributes minus one; dividing their sum by ∣H∣=2|H|=2 gives minus one. The row is therefore (3,−1,0)(3,-1,0).

  3. Weighted inner products with trivial, sign, and standard are (3−3)/6=0(3-3)/6=0, (3+3)/6=1(3+3)/6=1, and 6/6=16/6=1. Hence Ind⁡HS3η≅ε⊕U\operatorname{Ind}_H^{S_3}\eta\cong\varepsilon\oplus U. Restricting U to H gives eigenvalues 1,−11,-1, so it contains eta once, in agreement with reciprocity.

The result

The induced action is three dimensional and reducible. It contains one sign and one standard summand, with no trivial summand.

Explore the model

Interactive lab

Interactive lab · modular

Count cosets before adding their internal spaces

Pause and Reset, set both modulus m=6m=6 and step a=2a=2, and predict the subgroup orbit {0,2,4}\{0,2,4\} of the cyclic group. Reset and set both m=6,a=1m=6,a=1; predict the full six-element orbit. The first subgroup has two cosets, so inducing a two-dimensional representation from it would create dimension four. The instrument displays one additive orbit and no internal representation vectors; enumerate the other coset {1,3,5}\{1,3,5\} separately. Control changes retain time, while Reset clears both parameters and time.

Count cosets before adding their internal spacesAddition 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. Current readings: Residue r0: 0; Orbit period · m/gcd(m,a): 12; Distinct residues visited: 1 of 12. All parameters are adjustable with the labeled controls below.Repeated addition modulo 12: the current residue is 0, and the orbit period is 12.01234567891011+ 5mod 12An additive orbitn = 0rₙ = 0Period = 12Cycle starts at 0:0 → 5 → 10 → 3 →8 → 1 → 6 → 11 →4 → 9 → 2 → 7 → 0Exact residues · two additions per second
Additive orbitCurrent residue / next edge
Residue r0
0
Orbit period · m/gcd(m,a)
12
Distinct residues visited
1 of 12
t = 0 s

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 forInduction multiplies the original dimension by the number of cosets, not by the subgroup size.

Check your understanding

With the stated left-module convention, how does an H-map A: W → Res V extend to a G-map from Ind W to V?

Try it yourself

Let H=A3=⟨r⟩⊂S3H=A_3=\langle r\rangle\subset S_3 and let θ(r)=ω=e2πi/3\theta(r)=\omega=e^{2\pi i/3}. Find the character and decomposition of Ind⁡A3S3θ\operatorname{Ind}_{A_3}^{S_3}\theta.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

The index is two. A transposition has no conjugate in H, while conjugation by an odd permutation exchanges r and its inverse.

Compare with the solution

The identity value is two and the transposition value is zero. At r, the three even conjugators contribute ω\omega and the three odd conjugators contribute ω2\omega^2; dividing by three gives ω+ω2=−1\omega+\omega^2=-1. The row is (2,0,−1)(2,0,-1), exactly the standard irreducible character. Restricting that standard action to H gives θ⊕θ2\theta\oplus\theta^2, so theta occurs once and reciprocity confirms the multiplicity.

Take it with you

  • The balanced tensor relation fixes the induction convention.
  • Frobenius reciprocity identifies concrete Hom spaces.
  • Induced traces come only from coset copies returning to themselves.

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