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Gln(R) AS a LIE GROUP Contents 1. Matrix Groups Over R, C, and H 1 2

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Gln(R) AS a LIE GROUP Contents 1. Matrix Groups Over R, C, and H 1 2 GLn(R) AS A LIE GROUP MAXWELL LEVINE Abstract. This paper will provide some simple explanation about matrix Lie groups. I intend for the reader to have a background in calculus, point-set topology, group theory, and linear algebra, particularly the Spectral Theorem. I focus on SO(3), the group of rotations in three-dimensional space. Contents 1. Matrix Groups over R, C, and H 1 2. Topology and Matrix Groups 4 3. The Exponential Function 6 4. Tangent Spaces 8 5. Maximal Tori 10 6. Lie Algebras 13 Acknowledgments 16 References 16 1. Matrix Groups over R, C, and H We focus on the matrix groups over the rings R, C, and H, which are all Lie groups. For the most part, we will not discuss abstract Lie groups. However, it is a fact that every compact Lie group is isomorphic to a matrix group (a statement of the theorem can be found in [3]). Hence, the use of matrix groups simply provides a more direct approach to the material. Definition 1.1. For a ring R, the set Mn(R) denotes all n-by-n matrices over R. Mn(R) forms a ring under matrix addition and multiplication, but it has zero 1 0 0 0 0 0 divisors (i.e., 0 0 0 1 = 0 0 ). Furthermore, its additive group is uninteresting 2 because it is simply the additive group of Rn . Therefore, we focus our interest on the multiplicative group of Mn(R). Definition 1.2. For a ring R, the set GLn(R) denotes the multiplicative group of Mn(R), namely the group of invertible matrices over R, and is called the general linear group. If R is commutative, then the determinant function is well-defined. In this case, the set of matrices of determinant 1 is denoted SLn(R) and is called the special linear group. For this paper, we will focus on the case in which R = R, C, H. We know that GLn(R) is a group by virtue of the multiplicative property of determinants: If A, B ∈ GLn(R) and so det A and det B are nonzero, then Date: August 21, 2009. 1 2 MAXWELL LEVINE det(AB) = det(A) det(B) 6= 0, so AB ∈ GLn(R). Similarly, we know that −1 det(A ) = 1/ det(A) 6= 0, so GLn(R) is closed under inverses. Analogous ar- guments can be made to show that SLn(R) is a subgroup of GLn(R). The general linear group of n-by-n matrices is important first of all because it has the group structure that Mn(R) lacks but also because it is all-encompassing in the sense that it contains many important groups. Definition 1.3. For n × n matrices A, O(n, R) = {A : AT A = I} (where AT is the transpose of A) is the orthogonal group, and the subgroup SO(n, R) of O(n, R) of matrices of positive determinant is called the special orthogonal group. Note that it is equivalent to define the orthogonal group as the subgroup of T −1 GLn(R) such that A = A . O(n, R) is a group because A, B ∈ O(n, R) im- plies that (AB)(AB)T = A(BBT )AT = AAT = I and A−1(A−1)T = AT A = I. SO(n, R) is a group because A, B ∈ SO(n, R) implies that det(AB) = det A · det B = 1 · 1 = 1. Note that every matrix n × n can be interpreted as a linear transformation of n-dimensional space. The following proposition elaborates on this notion. Proposition 1.4. Let A ∈ GLn(R) and let hx, yi = x1y1 + ... + xnyn be the Euclidean inner product. The following are equivalent: (1) AAT = I (that is, A ∈ O(n, R)). (2) For entries aij, aji, aijaji = δij. (3) The columns of A are an orthonormal basis, and so are the rows of A. (4) A sends orthonormal bases to orthonormal bases. (5) A preserves the Euclidean norms of vectors under multiplication. The proof is, step by step, more or less immediate. Since the orthogonal group preserves the lengths of all vectors, we can thing of it is representing rotations and reflections in Rn. In this case, the sign of the determinant of a matrix indicates whether it is orientation-preserving. Elements of the special orthogonal group all have positive determinant, which indicates that these transformations are rotations. Definition 1.5. If A = (aij) ∈ GLn(C), then A = (aij) is the matrix of the ∗ T complex conjugates of the entries of A. We define A = A = (aji) to be the conjugate transpose of A. The set U(n) = {A : A∗A = I} is called the unitary group and the subgroup of U(n) of matrices of determinant 1 is the special unitary group. One of the interesting aspects of the general linear group is that GLn(C) is a subgroup of GL2n(R). This can be understood on a conceptual level because an n element in GLn(C) is a transformation from C to itself. Since C is a vector space over R with the basis {1, i}, we can identify Cn with R2n using an injection that a −b sends each entry a + bi of A ∈ GLn(C) to a 2-by-2 section b a of a 2n-by-2n matrix over R. The result is that we reinterpret A as a transformation from R2n to itself. This discussion can be summed up as the following: Proposition 1.6. GLn(C) is isomorphic to a subgroup of GL2n(R). Since we can think of these matrix groups as subgroups of GLn(R), we can conceive of intersections of certain matrix groups, even if they are over different fields. For example, the intersection of O(2n, R) and GLn(C) is U(n). Suppose A GLn(R) AS A LIE GROUP 3 is an element of this intersection. Since GLn(C) consists of all C-linear invertible matrices, we consider A as an element of GL2n(R). Since A ∈ O(2n, R), we know T a −b T that A A = I. Given the isomorphism ϕ, we can think of the matrix b a = a b T −b a as the conjugate of the element a + bi ∈ C. It follows that A corresponds to A∗, so that A ∈ U(n). Definition 1.7. In order to define operations such that R4 is a ring, we write it over the basis {1, i, j, k} and use the relations i2 = j2 = k2 = −1 ij = k ji = −k jk = i kj = −i ki = j ik = −j Under these operations, R4 is a non-commutative division ring which we call the Hamiltonian numbers. This ring contains R in its center and is denoted H. Despite the lack of commutativity, the Hamiltonians are similar to the complex numbers in that they have a conjugation√ operation√ q → q which sends a+bi+cj+dk to a−bi−cj−dk, and a norm |q| = qq = a2 + b2 + c2 + d2, which can be checked. Furthermore, just as multiplication by a complex number is an isometry of C, multiplication by an element of H is an isometry is well, meaning that it preserves distances. This is because for u, v ∈ H, |uv| = |u||v|, so |u(v − w)| = |u||v − w|. Furthermore, we can define an operation on the matrices in GLn(H) that sends ∗ T A = (qij) to A = A = (qji), which is an analog of the conjugate transpose operation. (The question of which is being used should be clear from context). Definition 1.8. The ∗ operation defined is the Hamiltonian conjugate transpose. ∗ The group of matrices Sp(n) = {A ∈ GLn(H): AA = I} is the symplectic group. In fact, there exists an injection of H into GL4(R) given by the action of each of a + bi + cj + dk on the basis elements of H which takes the form a b c d −b a −d c (1.9) a + bi + cj + dk 7→ −c d a −b −d −c b a Thus, we find that GLn(H) ≤ GL4n(R). The property that makes H important in the context of this paper is that the Hamiltonian numbers represent rotations in three-dimensional space. Of particular interest are the Hamiltonians of norm 1, which is denoted S3. S3 is a group under multiplication, and it is isomorphic to SU(2). Furthermore, S1 acts on the space iR + jR + kR of imaginary Hamiltonians, which we denote R3, by rotation. Proposition 1.10. The conjugation operation t 7→ qtq−1 for fixed q ∈ S3 is a rotation of iR + jR + kR, the three-dimensional hyperplane in H. In particular, if we write q = cos θ + u sin θ, then conjugation rotates iR + jR + kR by the angle 2θ around the axis through u. 4 MAXWELL LEVINE Proof. First of all, note that the multiplication of elements in R3 is uv = −u · v + u × v, where · and × are the respective dot and cross products in our standard conception of R3. So if u ∈ iR + jR + kR, then u2 = −1. Since R commutes with all elements of H, t ∈ R means that qtq−1 = t for q 6= 0, so this operation maps R to itself. Furthermore, it is a ring homomorphism and it has the inverse t 7→ q−1tq, so it is a bijection. Now suppose that q(bi+cj+dk)q−1 = a + b0i + c0j + d0k. But then q−1(a + b0i + c0j + d0k)q = a + q−1(b0i + c0j + d0k)q, which contradicts the fact that the conjugation operation has an inverse.
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