Alphabeta Math
ExampleConstruction: Literature-sourcedVerification: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-16
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The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

  • Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
  • AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
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These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Frobenius reciprocity for group representations without tensor products

Example

Let HG be groups and let k be a field. For a left k-linear H-representation V, define IndHG(V) to be the functions f:GV such that

f(gh)=h1f(g)

for gG and hH, and such that the left cosets on which f is nonzero form a finite set. With (af)(g)=f(a1g), this is a G-representation and

IndHGResHG.

Facts & Assumptions

Given: A subgroup HG, a field k, an H-representation V, and a G-representation W.

[F1]

A left group action satisfies ex=x and (gh)x=g(hx) (Left group actions, transitive actions, and faithful actions).

[F2]

A vector space is an abelian group under addition with scalar laws λ(u+v)=λu+λv, (λ+μ)v=λv+μv, (λμ)v=λ(μv), and 1v=v (Vector space over a field).

[F3]

A map T:VW is linear exactly when T(au+bv)=aT(u)+bT(v) for all scalars and vectors (Linear map between vector spaces over the same field).

[F4]

A subgroup contains the identity and is closed under products and inverses (Subgroup).

[F5]

The left coset of H represented by g is gH={gh:hH}, and the right coset is Hg={hg:hH} (Left and right cosets gH and Hg of a subgroup).

[F6]

For sets A,B, the functions AB form the set BA (The set BA of all functions AB).

[F7]

A set is finite when it is in bijection with some natural number (The cardinality A of a finite set).

[F8]

For a finite set S and a function into a commutative monoid, sSas is the enumeration-independent finite sum, and the sum over the empty set is 0 (A finite sum in a commutative monoid indexed by an arbitrary finite set).

[F9]

Finite sums are invariant under bijective reindexing, split over disjoint unions, and satisfy the two finite Fubini formulas (Finite commutative-monoid sums are invariant under bijective reindexing, split over disjoint unions, and satisfy the finite Fubini rule).

Verification

technique · direct
1.1

The defining equations cut out a vector subspace of the function set VG from [F6]. Pointwise operations preserve the covariance equation, and finite unions preserve finite coset support.

F2F3F5F6F7algebra
2.1

The formula (af)(g)=f(a1g) preserves the covariance equation and finite support. The equations in [F1] show that it is a left G-action, and pointwise operations show that the action is linear. Hence IndHG(V) is a G-representation.

step 1.1F1F2F4algebra
2.2

For an H-equivariant linear map u:VResHG(W), set u^(f):=gHG/Hgu(f(g)), where zero summands are omitted. If g is replaced by gh, the summand becomes ghu(h1f(g))=gu(f(g)), so it depends only on the coset. The nonzero index set is finite, and [F8] makes the empty-support value zero.

step 1.1F1F3F5F7F8algebra
3.1

Define jV:VResHGIndHG(V) by jV(v)(h)=h1v for hH and jV(v)(g)=0 for gH. The subgroup axioms make this well-defined with support in H, and direct calculation gives jV(av)=ajV(v) for aH; it is linear by [F3].

step 2.1F1F3F4algebra
3.2

Linearity follows termwise from [F3]. Left multiplication bijects the relevant coset sets, so reindexing with [F9] gives u^(af)=au^(f). Thus u^ is a G-equivariant linear map, naturally in V and W.

step 2.1step 2.2F1F3F9algebra
4.1

Every f has the finite decomposition f=gHgjV(f(g)): at any xG, only the coset xH contributes, and its contribution is f(x). Therefore a G-map T satisfies (TjV)^(f)=T(f).

step 2.1step 3.1step 2.2F8F9
5.1

The function jV(v) is supported on the single coset H, and its value at 1 is v, so (u^)jV=u. Together with step 4.1, the assignments uu^ and TTjV are inverse natural bijections, proving IndHGResHG.

step 3.1step 2.2step 3.2step 4.1F4F8

Depends on

Used by

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Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 68 results over 23 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources