By Louis Auslander

Complaints of the yank Mathematical Society

Vol. sixteen, No. 6 (Dec., 1965), pp. 1230-1236

Published by way of: American Mathematical Society

DOI: 10.2307/2035904

Stable URL: http://www.jstor.org/stable/2035904

Page count number: 7

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**Extra resources for An Account of the Theory of Crystallographic Groups**

**Example text**

18. Define a representation π of G as follows: H(π) = L2 (G/H, μ). For x ∈ G and f ∈ L2 (G/H, μ), define π(x)f (ω) = σ (x −1 , ω)f (x −1 · ω), 22 Basics for all ω ∈ G/H . To see that π is a representation, note first that π (x)f ∈ L2 (G/H, μ) and π (x)f 2 2 |f (x −1 · ω)|2 σ (x −1 , ω) dμ(ω) = G/H = |f (ω)|2 dμ(ω) = f 2 2. 18, one checks that π (x)π(y) = π(xy), for any x, y ∈ G. Clearly, π(e) is the identity operator. Thus, π is a homomorphism of G into the group of unitary operators on L2 (G/H, μ).

41 is often referred to as a separating vector for the algebra π (G) . When ξ is a separating vector for π(G) , the map T → T ξ is an injective linear map from π(G) into H(π ). So π(G) cannot be too large if π is a cyclic representation. A general representation can be decomposed into a sum of cyclic representations. 42 Let π be a representation of G. Then π is equivalent to the direct sum of cyclic representations of G. 5 Representations of L1 (G) and functions of positive type 27 We saw that one can add representations, in a sense, by taking the direct sum which acts on the direct sum of the respective Hilbert spaces.

81 When G = R, suppose R is parametrized by {χγ : γ ∈ R}, where χγ (t) = e2πiγ t , for t ∈ R. Then we can write f (γ ) = f (t)e2πiγ t dt, R for γ ∈ R, where we are writing γ in place of χγ . Then R |f (γ )|2 dγ = 2 1 2 R |f (t)| dt, for all f ∈ L (R) ∩ L (R). Therefore, both R and R can be equipped with unscaled Lebesgue measure and the Plancherel equality holds. 80. Thus, if we define the modulation representation MG of G on L2 (G) by MG (x)ξ (χ) = χ (x)ξ (χ ), for all x ∈ G, ξ ∈ L2 (G), and almost all χ ∈ G, then λG MG .

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