不考虑拓扑性质的SU(2)与SO(3)同态的完全纯代数证明

Muhammad Ardhi Khalif, Nur Farida Amalia
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引用次数: 0

摘要

本文的目的是在不考虑两群拓扑的情况下,给出SU(2)和SO(3)之间同态的一个完全的纯代数证明。证明开始于引入一个映射φ: SU(2)→M L(3, C),定义为[φ (U)] i j≡12 tr(σ i U σ j U†)。首先,我们证明了映射φ满足[φ (U 1 U 2)] i j = [φ (U 1)] i k [φ (U 2)] k j,对于每个U 1, U 2∈SU(2)。下一步是证明φ (U)的集合具有正交性质,并且每个φ (U)的行列式为1。之后,我们证明了φ (i2) = i3。最后,为了确定φ确实是同态,而不是同构,我们证明φ(−U) = φ (U),∀U∈SU(2)。
本文章由计算机程序翻译,如有差异,请以英文原文为准。
Complete purely algebraic proof of the homomorphism between SU(2) and SO(3) without concerning their topological properties
The aim of this paper is to provide a complete purely algebraic proof of homo-morphism between SU (2) and SO(3) without concerning the topology of bothgroups. The proof is started by introducing a map ϕ : SU (2) → M L(3, C) de-fined as [ϕ(U )] i j ≡ 12 tr(σ i U σ j U † ). Firstly we proof that the map ϕ satisfies[ϕ(U 1 U 2 )] i j = [ϕ(U 1 )] i k [ϕ(U 2 )] k j , for every U 1 , U 2 ∈ SU (2). The next step is toshow that the collection of ϕ(U ) is having orthogonal property and every ϕ(U ) hasdeterminant of 1. After that, we proof that ϕ(I 2 ) = I 3 . Finally, to make sure thatϕ is indeed a homomorphism, not an isomorphism, we proof that ϕ(−U ) = ϕ(U ),∀U ∈ SU (2).
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