CHAPTER 4

Connectedness

BY

Dr. Dibyendu De

Associate Professor Department of University Of Kalyani West Bengal, India E-mail : [email protected]

1 Module-5: Matrix Lie groups

To conclude the discussion on connected spaces we examine connectedness of some

matrix Lie groups, where by a matrix Lie groups we mean closed subgroups of Gl(n, R) and Gl(n, C).

Definition 1. The general linear over the real numbers, denoted Gl(n, R), is the group of all n × n invertible matrices with real entries. The over

the complex numbers, denoted Gl(n, C), is the group of all n × n invertible matrices with complex entries.

Henceforth we shall denote Mn(C) as the space of all n × n matrices with complex entries.

n2 The topology of Mn(C) is the subspace topology induced from R . The convergency in the space Mn(C) can be described by the following result.

Proposition 1. Let Am be a sequence of complex matrices in Mn(C). We say that Am converges to a matrix A if each entry of Am converges (as m → ∞) to the corresponding entry of A

Definition 2. A subgroup G of Gl(n, C) is said to be matrix if it is closed in 2 Gl(n, C), where Gl(n, C) is equipped with the subspace topology induced from Rn .

Example 1. The general linear groups (over R or C) are themselves matrix Lie groups. Of course, Gl(n, C) is a closed subgroup of itself. Moreover, Gl(n, R) is a subgroup of

Gl(n, C), and if Am ∈ Gl(n, R) and Am converges to A, then the entries of A are real. Thus, either A is not invertible or A ∈ Gl(n, R).

Example 2. The general linear groups (over R or C) is the group of n × n invertible matrices (with real or complex entries) having determinant one are denoted by Sl(n, R) and Sl(n, C). It is easy to observe that both these are subgroups of Gl(n, C). Furthermore,

2 if An is a sequence of matrices with determinant one and An converges to A, then A also has determinant one, because the determinant is a continuous function. Thus, Sl(n, R) and Sl(n, C) are matrix Lie groups.

The orthogonal and special orthogonal groups, O(n) and SO(n). A matrix A is orthogonal if its transpose is it’s inverse. Hence, detA = ±1, for all orthogonal matrices A. Equivalently A is orthogonal if it preserves the inner product,

namely if hx, yi = hAx, Ayi for all vectors x, y ∈ Rn. The set O(n) of all n × n real orthogonal matrices is a subgroup is a subgroup of

Gl(n, C). The limit of a sequence of orthogonal matrices is orthogonal, because the relation A · Atr = I is preserved under taking limits. Thus, O(n) is a matrix Lie group.

The set of n × n orthogonal matrices with determinant one is the special orthogonal

group SO(n). Clearly, this is a subgroup of O(n), and hence of Gl(n, C). Moreover, both orthogonality and the property of having determinant one are preserved under limits, and so SO(n) is a matrix Lie group. The unitary and special unitary groups, U(n) and SU(n) A matrix A with complex entries is unitary if it preserves the inner product, namely

if hx, yi = hAx, Ayi for all vectors x, y ∈ Cn. Equivalently A matrix A is unitary if A · A∗ = I, i.e., if A∗ = A−1. Since detA∗ = detA, we see that if A is unitary, then det (A∗ · A) = 1. for all unitary matrices A.

This, in particular, shows that every unitary matrix is invertible. The same argument as for the shows that the set of unitary matrices forms a closed sub- group of Gl(n, C) and denoted by U(n).

The set of unitary matrices with determinant one is the special unitary group SU(n). It is easy to check that SU(n) is a matrix Lie group. Component of matrix Lie group The next proposition shows that components, which is a topological object, becomes an algebraic object in matrix Lie group.

Proposition 2. If G is a matrix Lie group, then the component of G containing the identity is a subgroup of G.

3 Proof. Let X be the component of G. and A, B belong to X. Then there exist continuous paths A(t) and B(t) with A(0) = B(0) = I, A(1) = A, and B(1) = B. Then, A(t)B(t) is a continuous path starting at I and ending at AB so that AB ∈ X. Furthermore, A(t)−1 is a continuous path starting at I and ending at A−1, and so the inverse of any element lying in the identity component is again in the identity component. Thus, the identity component is a subgroup.

Disconnectivity of Gl(n, R)

Example 3. Gl(n, R) is not connected.

Proof. Observe that Gl(n, R) = Gl+(n, R)∪G−(n, R) where Gl+(n, R) = {A ∈ Gl(n, R): |A| > 0} and Gl−(n, R) = {A ∈ Gl(n, R): |A| < 0}. It remains to observe that determinant func- tion from Gl(n, R) to R is a continuous function (Justify yourself about the above state- ment).

Connectivity of Gl(n, C) On the other hand if we consider Gl(n, C), then it will be connected. One can imagine the case n = 1. Then clearly Gl(n, R) is disconnected where as Gl(n, C) is in fact punctured complex plane, which is clearly pathconnected. This idea will be applied in the following proof.

Theorem 1. Gl(n, C) is path connected.

Proof. Let A and B be two matrices in Gl(n, C). We wish to find a path from A to B in GL(n, C). We look for this path among the matrices of the form (1 − z)A + zB, where

z ∈ C. These matrices form a plane, parameterized by the complex coordinate z, and the plane includes A at z = 0 and B at z = 1. Any element of our required path from A to B should not belong to the {(1−z)A+zB : |(1 − z)A + zB| = 0}. Now (1 − z)A + zB, is an n × n complex matrix whose entries are linear terms in z. Its determinant is therefore a polynomial of degree at most n in z and so, by the fundamental theorem of algebra, the above equation has at most n roots. These roots represent n points in the plane of matrices (1 − z)A + zB, not including the points A and B. This allows us to find a path, from A to B in the plane, avoiding

the points with determinant zero, as required. Thus Gl(n, C) is path-connected.

Connectivity of U(n) and SU(n)

4 Theorem 2. The groups U(n) and SU(n) are connected, for all n ≥ 1.

Proof. We know that every unitary matrix has an orthonormal basis of eigenvectors, with eigenvalues of the form eiθ.Therefore any unitary matrix U can be decomposed as   eiθ1 0     −1 U = U1  ...  U1   0 eiθn with U1 unitary matrix and θi ∈ R. Trivially every matrix of the above form is unitary as all of the factors are unitary. Now, define a path f : I → U(n) by the formula

  ei(1−t)θ1 0     −1 f(t) = U1  ...  U1 .   0 ei(1−t)θn Then f(0) = U and f(1) = I. Thus, any two elements U and V of U(n) can be connected to each other by a continuous path that runs from U to I and then from I to V . This proves that U(n) is path connected. By a similar argument, we can prove that SU(n) is connected.

We end this module introducing the famous Hisenberg group.

Definition 3. The set of all 3 × 3 real matrices A of the form   1 a b      0 1 c    0 0 1

where a, b, c are arbitrary real numbers, is called Heisenberg group.

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