6 Representation Theory of the Special Unitary Group SU(N)

6 Representation Theory of the Special Unitary Group SU(N)

6 Representation theory of the special unitary group SU(N) 6.1 Schur-Weyl duality | an overview The Schur-Weyl duality is a powerful tool in representation theory that allows one to put the irreducible representations of the general linear group GL(N) c.f. Definition 6.1) a vector space ⊗n V with dim V = N into 1-to1 correspondence to the irreducible representations of the group Sn on V ⊗n. In particular, it turns out that V ⊗n decomposes as V ⊗n = V S ; λ n ; (6.1) λ ⊗ λ ` Mλ ⊗n where Vλ are the irreducible submodules of GL(N) on V , and Sλ are the so-called Specht modules, which describe the irreducible representations of Sn (c.f., e.g., [4] and other standard ⊗n textbooks). The underlying reason for this is that the actions of GL(N) and Sn on V commute and, even more, the elements of GL(N) are a complete set of actions that commute with those of Sn and vice In these lectures, we will go through the main points of the Schur-Weyl duality, paying particular attention to the role the Young projection operators play in the representation theory of GL(N). We will begin be defining the general linear group GL( N) in section 6.2. • We will then define what we mean by an invariant of GL (N), and show that these invariants are • given by the elements of Sn, section 6.3. In fact, Sn spans the algebra of invariants of GL(N). An important ingredient to seeing this the double commutant theorem, c.f. section 6.3.1. Let v V ⊗n be arbitrary. We will show that, for every Θ , the subspace • 2 2 Yn YΘv (6.2) is invariant and irreducible under the action of Sn. Hence, YΘv is an irreducible C[Sn]- ⊗n submodule and therefore corresponds to an irreducible representation of Sn on V , c.f. section 6.4. Thereafter, we will show that, for every Θ , the subspace • 2 Yn ⊗n YΘV (6.3) is invariant and irreducible under the action of GL( N) | the main ingredient to showing ⊗n this is the fact that Sn spans the algebra of invariants of GL(N ) on V . This shows that the Young projection operators YΘ generate the irreducible ideals (and hence the irreducible representations) of GL(N) on V ⊗n, section 6.5. Finally, in section 6.6, we will argue that the irreducible representations of GL( N) on V ⊗n • are precicely those of the special unitary group SU(N) on V ⊗n. In other words, the Young projection operators on V ⊗n give rise to all the irreducible ideals of SU(N ) on V ⊗n. We will end with an example, constructing the irreducible representations of GL (N) (hence • ⊗n also SU( N)) and Sn on V for n = N = 3, in section 6.7. 58 6.2 Basic definitions Definition 6.1 { General linear group GL(V ) (or GL(N)): Let V be a vector space of dimension N (N not necessarily finite). Consider the subset of End(V ) of all invertible endomorphisms of V . This set forms a group called the general linear group on V , and we denote this group by GL(V ) or sometimes only GL(N) if the vector space V is clear and we want to make the dimension of V explicit. Definition 6.2 { Defining/fundamental representation of GL(V ): Let GL(V ) be the general linear group acting on a vector space V . We can define a representation γ as γ : GL(V ) End(V ) ! (6.6) γ(g) g 7! since there is a well-defined action of GL(V ) on V . The representation γ is referred to as the defining representation of GL(V ) and has dimension dim V = N. 6.3 Invariants of GL(N) Through the defining representation γ of GL(V ) on V , we can define a representation of GL(V ) on V ⊗n via g (v v v ) := γ(g)v γ(g)v γ(g)v = gv gv gv (6.7a) 1 ⊗ 2 ⊗ · · · ⊗ n 1 ⊗ 2 ⊗ · · · ⊗ n 1 ⊗ 2 ⊗ · · · ⊗ n 59 Let be the symmetric group Sn, and let be the general linear group GL(N). Both these groups A B ⊗n have a well-defined action on the vector space V (with dim V = N), and therefore both Sn and GL(N) are subgroups of End (V ⊗n). Furthermore, we have seen in the beginning of section 6.3 that ⊗n the actions of Sn and GL(N) commute on V , such that = S (GL(N))0 = 0 and = GL(N) S0 = 0 A n ⊂ B B ⊂ n A 0 and 0 : (6.18) ) A ⊂ B B ⊂ A ⊗n It follows directly from Maschke's Theorem 3.1 that V is a completely reducible Sn-module. That V ⊗n is also a completely reducible GL(N)-module follows from the Peter-Weyl Theorem [16] (which we will state without proof): Note 6.1: Peter-Weyl Theorem As we will explain in Note 6.2, GL(N) is a Lie group, which means that, in particular, it is a differentiable manifold. A manifold is said to be compact if it is compact as a topological space, that is every open cover has a finite subcover, c.f., e.g., [17]. (Very losely speaking, you may think of a cover of a manifold M as another manifold M 0 enclosing it. The requirement that every subcover is finite can be thought of that every submanifold N 0 M 0 that is also ⊂ a cover for M is finite, implying that M was finite to start off with. As an example, the unit sphere is a compact manifold, but an infinite plane would not be.) It turns out that the representations of such compact Lie groups are completely reducible: Theorem 6.2 { Reducible carrier spaces of compact groups (Peter-Weyl [16]): Let ' be a unitary representation of a compact group G on a complex Hilbert space H. Then H splits into an orthogonal direct sum of irreducible finite-dimensional unitary representations of G. Now, the group GL(N) is not compact (as a manifold). However, as we will see in the later section 6.6, its subgroup SU(N) is compact. Furthermore, as we will argue in section section 6.6, the irreducible representations of GL(N) are precicely those of SU(N) and vice versa (c.f. Theorem 6.5) by means of the so-called unitarian trick, it follows that the Peter- Weyl Theorem 6.2 does indeed apply to the group GL(N) as well. If all the things said in this note do not quite make sense to you yet, try re-reading this note after you have read section 6.6 | this should clear things up for you. ⊗n Therefore V is completely reducible as an Sn-module, as well as as a GL(N)-module. Therefore, the Double commutant theorem 6.1 asserts that = S = 00 and = GL(N) = 00 A n A B B = 00 and = 00 : (6.19) )A A B B 62 6.6 The unitarian trick: irreducible representations of SU(N) from GL(N) Having established the connection between the irreducible representations of Sn and the irreducible representations of GL(N) on V ⊗n, Definition 6.4 { Special unitary group SU(N): Let GL(N) be the general linear group on a vector space V with dim(V ) = N. We define SU(N) to be the subset of matrices in GL(N) that are unitary (with respect to the canonical scalar product on V ) and have determinant 1, SU(N) = U GL(N) UU y = 1 and det U = 1 : (6.41) 2 n o It can be shown that SU( N) is in fact a group ( c.f. ) and we call it the special unitary group on V . (The term special refers to the fact that the elements of SU (N) are unimodular, i.e. have determi- nant 1, and unitary referes to the property that UU y = 1 for every U SU(N).) 2 The following intermediate result will turn out to be quite useful when establishing the correspon- dence between the irreducible representations of GL( N) and the irreducible representations of SU(N) on V ⊗n: Proposition 6.3 { Square matrix decomposition: Let M n×n be the space of all n n matrices with entries in C, and let Hn×n and An×n be the spaces × of all Hermitian, respectively, anti-Hermitian n n matrices with entries in C. Then, × M n×n = Hn×n An×n : (6.44) ⊕ Proof of Proposition 6.3. It is clear that Hn×n + An×n M n×n since a sum of a Hermitian and ⊂ an anti-Hermtitian n n matrix will still yield an n n matrix. Conversely, let m M n×n. Then, × × 2 we can write 1 1 m = (m + my) + (m my) ; (6.45) 2 2 − =:m+ =:m − | {z } | {z } 66 y where m is the Hermitian conjugate of m. Notice that m+ is Hermitian and m− is anti-Hermitian. Thus, we also hace that M n×n Hn×n + An×n, implying that ⊂ M n×n = Hn×n + An×n (6.46) It remains to show that Hn×n An×n = 0 to obtain the desired result: Let m Hn×n An×n. Since \ f g 2 \ m Hn×n, m is Hermitian. Furthermore, since also m An×n, m is anti-Hermitian. Therefore, 2 2 2 n n 2 n n m =======m H × my =======m A × m m = m : (6.47) − ) − However, since all entries in m are elements of C, m = m only holds for m being the zero matrix.

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