Proof of Theorem 1 k-mBIDH G1,G2 ,eˆ is negligible and qS, qH1 are finite.
Proof of Theorem 1. 5: By Proposition 4.16, λ is an algebraic integer, and, thus, all conjugates of λ are algebraic integers, since they satisfy the same monic, integral polynomial. Recall that 1/λ is a conjugate of λ. By Theorem 3.23, algebraic integers form a ring. So, r2 = λ · λ¯ and 1/r2 = 1/λ · 1/λ¯ are also algebraic integers. Since r and 1/r satisfy the polynomials x2 − r2 and x2 − 1/r2, r and 1/r are algebraic integers as well. Furthermore, algebraic integers are closed under multiplication. So, eiθ = 1/r · reiθ is also an algebraic integer. Let σ ∈ G(Nr,eiθ /Q). We will consider all σ that fix r. (The set of all such automorphisms is nonempty, since the identity map is an automorphism that fixes r.) Note that from Theorem 4.16 the conjugates of λ are known. If we assume that r is fixed under some embedding into C, then we can determine the image of eiθ under the embedding as well. Claim: If σ(r) = r, then σ(eiθ) = e±iθ. Suppose by way of contradiction, there exists σ such that σ(r) = r and σ(reiθ) = 1/λ, 1/λ¯, or eiφ. First, assume σ(λ) = 1/λ¯. (If σ(λ) = 1/λ, then compose σ with conjugation. Note that σ(r) will still equal r.) Then, σ(reiθ) = r · σ(eiθ), which implies σ(eiθ) = 1/r2 · eiθ and also σ(e2iθ) = 1/r4 · e2iθ. Thus, e2iθ has a conjugate with norm 1/r4, which is a contradiction by Lemma 6.1. Second, assume σ(reiθ) = eiφ. Then, σ(eiθ) = 1/r · eiφ and, furthermore, σ(e2iθ) = 1/r2 · e2iφ. By Lemma 6.1, the only possible conjugate of e2iθ with norm equal to 1/r2 is precisely 1/r2. But, this implies eiφ = ±1, which is a contradiction.
Proof of Theorem 1. 1 The proof of Theorem 1.2.1 requires a well known result of Xxxxx [Stu80], which says that if a modular form with integer Fourier coefficients is nonvanishing modulo a prime A, then there is a bound on the index of the first coefficient which is nonzero modulo A. To state his theorem, for a rational prime A and a modular form f (z) = Σ∞n=0 a(n)qn ∈ Mk(Γ0(N ), χ) with coefficients in Z, we define ordA(f ) := minn{n : A ‡ a(n)}, and we say ordA(f ) := ∞ if A|a(n) for all n. Σ
Proof of Theorem 1. 4.1 This proof is analogous to the proof of [BO16, Theorem 2.1]. By well known properties of Bessel functions, such as the bounds in (9.8.4) of [AS72], for x ≥ 37.5 the modified Bessel function I1(x) is bounded by N ≤ x2 e−xI1(x) ≤ M where N = 0.394, M = 0.399. First, let 2 ≤ k ≤ 5, and let β := αk . Then by Theorem 5.1.2, for n ≥ 450 we have: 2π . k − 1 Σ1 . x x − 1 + 24n β . 2 −µΣ Σ eµ N − √µ < x
Proof of Theorem 1. 2.2 By Xxxxxxx-Xxxxxxx-Xxxxx theorem, we may assume that A is a sub-algebra of ¥(H) for some (not necessarily separable) Xxxxxxx space H. Since X can be approximated by elements with finite spectra with any precision, we can assume that X has finite spectrum and all its spectral projections of X belong to A, and then apply ap- proximation arguments. Since the original construction of [9] relies only on spectral projections of X, it can be extended to our case with minimal changes. Still, we
Proof of Theorem 1. 3.4 The results of the previous sections reduce the general case to the case of elements whose spectra are contained in sets of the form ε Σ =d=ef (εZ × εZ) ∩ ([—1; 1] × [—1; 1]). It turns out that if σ(T ) C Σε, then we can remove its off-diagonal elements with respect to P in such a way that the element remains normal. The idea is to map the spectrum onto a line segment, then remove off-diagonal elements from the resulting self-adjoint element (it will remain self-adjoint), and then map it back. The choice of particular maps is important since it is the only place where we get a loss in the power of δ. The following lemma describes these maps.
Proof of Theorem 1
