Give two definitions of “induced representation”. Why are they equivalent?
Solution
Let \(H\le G\) be a subgroup, let \(k\) be a field, and let \(V\) be a left \(k[H]\)-module.
1. The tensor-product definition is\[ \operatorname{Ind}_H^G V = k[G]\otimes_{k[H]}V, \]with \(G\) acting by left multiplication on the \(k[G]\) factor.
For \(g\in G\) define \[ g\cdot(x\otimes v)=(gx)\otimes v. \] This is well defined because left multiplication commutes with the balancing relation \[ xh\otimes v=x\otimes hv. \]
2. The function-model definition is\[ \mathcal F = \{f:G\to V:f(gh)=h^{-1}f(g)\text{ for all }g\in G,h\in H\}, \]with\[ (g_0\cdot f)(g)=f(g_0^{-1}g). \]
The covariance relation is preserved by left translation, so this is a \(G\)-representation.
3. Choose representatives \(T\subseteq G\) for the left cosets \(G/H\). Then both models are naturally isomorphic, as vector spaces, to\[ \bigoplus_{t\in T} V. \]
For the tensor model, \[ k[G]=\bigoplus_{t\in T} t\,k[H] \] as a right \(k[H]\)-module, hence \[ k[G]\otimes_{k[H]}V \cong \bigoplus_{t\in T}V. \]
For the function model, the covariance relation determines \(f\) uniquely from the values \(f(t)\) on the representatives \(t\in T\). Thus evaluation on \(T\) gives the same direct-sum vector space.
4. The two identifications in 3 intertwine the \(G\)-actions.
For each \(g\in G\) and \(t\in T\), write uniquely \[ gt=t'h \qquad(t'\in T,\ h\in H). \] In the tensor model, \[ g\cdot(t\otimes v)=t'h\otimes v=t'\otimes hv. \] In the function model, left translation produces exactly the same permutation of cosets together with the same \(H\)-action on the corresponding \(V\)-coordinate. Hence the two \(G\)-modules are isomorphic.
Therefore the tensor-product and equivariant-function constructions are equivalent definitions of induction.