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Annihilators of closed subgroups of Euclidean space

Example

Let n≥0 and 0≤k≤n be integers. Coordinates are indexed by 0≤i<n (The standard list e:n→Fn with ei(i)=1F and ei(j)=0F for j≠i is an ordered basis of Fn; hence dim⁡FFn=n, and F0 is the zero space with basis ∅ and dimension 0). Work in G=Rn with the dual identified with Rn by ξ↦(x↦e2πi ξ⋅x), computed below from the classification of characters of the line and the product-dual identification (Continuous characters of the real line are exponentials, Duals of finite products and of discrete direct sums). No choice principle is used by the computations below; the general quotient-dual and closed-subgroup identifications cited in the dependency list give the context of clause (2), whose coordinate content is computed directly here.

(1) If V≤Rn is a linear subspace, then V⊥={ξ:ξ⋅v∈Z for all v∈V}; when V=Rk×{0}n−k this is {0}k×Rn−k, the coarse orthogonal complement, whereas for the lattice H=Zk×{0}n−k it is Zk×Rn−k. For k=n the annihilator of the full-rank lattice Zn is again the lattice Zn, while for k<n the annihilator contains the line R ek and is not discrete; so an annihilator is not in general an orthogonal complement.

(2) For H=Zk×{0}n−k the quotient Rn/H is identified with (R/Z)k×Rn−k, and its characters are computed on the standard coordinates by the discrete Fourier pairing: its characters are exactly the maps x+H↦e2πi ξ⋅x with ξ∈Zk×Rn−k, whose pullbacks are precisely the characters of Rn trivial on H.

(3) If A is an invertible n×n real matrix and H=AZn, then H⊥=A−TZn with A−T:=(A−1)T=(AT)−1.

Facts & Assumptions

Given: The group Rn with its standard topology and the dual identification constructed from the character classification of the line and the product-dual identification (Continuous characters of the real line are exponentials, Duals of finite products and of discrete direct sums).

[F1]

Every continuous homomorphism φ:Rn→T is φ(x)=e2πi ξ⋅x for a unique ξ∈Rn, and ξ↦φξ is an isomorphism of topological groups Rn→Rn^: the classification of characters of the line supplies the algebraic bijection, and finite-product duality reduces the topology check to the line. On the line, compact K is bounded, say ∣t∣≤M; continuity of s↦e2πis at 0 makes e2πiξt uniformly close to e2πiξ0t on K when ∣ξ−ξ0∣ is small. Conversely, if ∣ξ−ξ0∣≥ϵ, then t=1/(2∣ξ−ξ0∣)∈[−1/ϵ,1/ϵ] gives ∣e2πi(ξ−ξ0)t−1∣=2, so the uniform ball of radius 1 on this compact interval forces ∣ξ−ξ0∣<ϵ. Uniform balls are compact-open neighbourhoods by The compact-open character group is a Hausdorff topological abelian group, and compactness and boundedness of these line sets follow from Heine-Borel in Rn: with the Euclidean metric a subset of Rn is compact if and only if it is closed and bounded, and the proof by bisection uses no choice principle; the same holds on the real line. The value at ±πi is −1 by exp⁡(x+iy)=ex(cos⁡y+isin⁡y), ∣exp⁡(x+iy)∣=ex, and eiπ+1=0. At n=0 both groups are trivial. (Continuous characters of the real line are exponentials, Duals of finite products and of discrete direct sums, The Pontryagin dual with the compact-open topology)

[F2]

H⊥={γ:γ(h)=1 for every h∈H}, and e2πit=1 exactly when t∈Z, while e2πi(t+k)=e2πit for k∈Z. (The annihilator of a subgroup, ker⁡(exp⁡)=2πiZ, and exp⁡z=exp⁡w exactly when z−w∈2πiZ, The integers as equivalence classes of pairs of naturals)

[F3]

The quotient Rn/H for H=Zk×{0}n−k is (R/Z)k×Rn−k with the product topology: each projection R→R/Z is continuous and open, since the saturation of an open set is the union of its integer translates. The finite product of these maps and identity maps is therefore continuous, open and surjective, with kernel H, and the induced quotient bijection is continuous and open. A continuous character on Rn constant on the cosets of H factors through the quotient map, uniquely and continuously; conversely characters of the quotient pull back to characters of Rn that are trivial on H. (The quotient group G/N and coset product (gN)(hN)=ghN, The discrete, indiscrete, cofinite, cocountable, particular-point and Sierpinski topologies, For a quotient map q:X→Y, a map out of Y is continuous iff its composite with q is; a continuous map on X constant on the fibres of q factors uniquely through q; and a composite of quotient maps is a quotient map, Homeomorphism, open map, closed map, embedding, and what it means for a property to be topological)

[F4]

For a real matrix A, a vector ξ and z∈Zn one has (ATξ)⋅z=ξ⋅(Az) and A−T=(AT)−1; the standard basis vectors e0,…,en−1 lie in Zn, and y∈Rn satisfies y⋅z∈Z for all z∈Zn exactly when y∈Zn. (Transpose is linear and involutive, and (AB)T=BTAT, The transpose AT of a matrix, Invertible matrices and the general linear group GL⁡n(F), The standard list e:n→Fn with ei(i)=1F and ei(j)=0F for j≠i is an ordered basis of Fn; hence dim⁡FFn=n, and F0 is the zero space with basis ∅ and dimension 0, The integers as equivalence classes of pairs of naturals)

Verification

1.1F1F2F4

For a linear subspace V≤Rn, a character γξ lies in V⊥ exactly when ξ⋅v∈Z for every v∈V, by [F1] and [F2]. When V=Rk×{0}n−k and v=t ei for 0≤i<k and t∈R, the condition t ξi∈Z for all real t forces ξi=0 (otherwise take t=1/(2ξi)); the remaining coordinates are unconstrained, so V⊥={0}k×Rn−k.

1.2F1F2F4

For H=Zk×{0}n−k, the condition ξ⋅h∈Z for all h∈H reads ∑0≤i<kξihi∈Z for all integers h0,…,hk−1. Taking h=ei gives ξi∈Z for 0≤i<k, and the remaining coordinates are unconstrained; hence H⊥=Zk×Rn−k. In particular H⊥ is not discrete when k<n, since the nonzero vectors m−1ek lie in it and converge to 0 as positive integers m→∞, and the full-rank case k=n gives (Zn)⊥=Zn.

1.3F1F2F4

For H=AZn with A invertible, γξ∈H⊥ if and only if ξ⋅(Az)∈Z for all z∈Zn, that is (ATξ)⋅z∈Z for all z∈Zn, which by [F4] holds exactly when ATξ∈Zn, that is ξ∈(AT)−1Zn=A−TZn.

2.1F2F3step 1.2

For H=Zk×{0}n−k, the quotient identified in [F3] has the standard coordinates (x0,…,xk−1) mod Z and x′∈Rn−k. A character γξ with ξ∈Zk×Rn−k satisfies γξ(h)=e2πi∑0≤i<kξihi=1 for all h∈H, so it is constant on cosets and factors through the quotient, where it takes the value e2πi(∑0≤i<kξixi+ξ′⋅x′) on the class of x; these are exactly the characters of the quotient, which is the stated discrete Fourier pairing on the torus factor.

3.1step 1.1step 1.2step 1.3step 2.1∎

Clauses (1), (2) and (3) are proved in steps 1.1 and 1.2, step 2.1 and step 1.3; the example also records that "the annihilator of a lattice is a lattice" holds only in the full-rank case of clause (3), not for the degenerate subgroups Zk×{0}n−k with k<n.

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