Past exam of the mathematics course of the University of Cambridge / 2007 / iii / Paper 1 / 2 / b

Solution

R : V^{*} ⟶ \prod_{i \geq 1} V_{i}^{*}, ℓ ⟼ (ℓ ∣_{V_{i}})_{i}

is injective, since the summands span

V

. Given any family

(ℓ_{i})_{i}

ℓ ((v_{i})_{i}) = \sum_{i} ℓ_{i} (v_{i})

. This sum is finite for each vector, regardless of whether the family of functionals has finite support. It defines a linear functional and is inverse to

R

. Hence this is a natural vector-space isomorphism.

(π^{*} (g) ℓ) (v) = ℓ (π (g^{- 1}) v)

. On the

i

th coordinate,

(π^{*} (g) ℓ)_{i} (v_{i}) = ℓ_{i} (χ_{i} (g)^{- 1} v_{i}) = χ_{i} (g)^{- 1} ℓ_{i} (v_{i}) .

Thus

R

intertwines the actions and gives

(V^{*}, π^{*}) ≅ i \geq 1 \prod (V_{i}^{*}, χ_{i}^{- 1}) .

No smoothness restriction has yet been imposed on that product.

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