Chapter 19: Problem 2

Show that the special orthogonal group \(S O(n)\), the pseudo-orthogonal groups \(O(p \cdot q)\) and the symplectic group \(S p(n)\) arc all closed subgroups of \(G L(n, \mathbb{R})\). (a) Show that the complex groups \(S L(n, \mathbb{C}), O(n, C), U(n), S U(n)\) are closed subgroups of \(G L(n, C)\) (b) Show that the unitary groups \(U(n)\) and \(S U(n)\) are compact groups.

Short Answer

Expert verified

All groups mentioned in the exercise are closed subgroups of GL(n, R) or GL(n, C), as they are themselves groups containing matrices that are part of the parent group, and also contain all their limit points. The unitary groups U(n) and SU(n) are compact, as they are both closed and bounded.

Step by step solution

Definitions

First, we need to understand the groups we are dealing with. Special orthogonal, pseudo-orthogonal, and symplectic groups, among others, are defined by certain constraints on matrices. For example, the special orthogonal group \(S O(n)\) consists of all n by n orthogonal matrices with determinant 1. They preserve a Euclidean metric, such as the length of vectors and angle between vectors.

Subgroups of GL(n, R) or GL(n, C)

To show that these groups are subgroups of GL(n, R) or GL(n, C), we must show that they satisfy two conditions: they must themselves be groups, and they must be contained within the parent group. The group GL(n, R) or GL(n, C) contains all n by n invertible matrices with real or complex entries respectively, so it's clear that all the matrices in the groups mentioned in the problem are contained within GL(n, R) or GL(n, C).

Show Closure

To show that a set is closed, we show that it contains all of its limit points. This means that if we have a sequence of matrices from this group that converges to a limit, that limit matrix must also be in the group. This can be shown for each of the groups in the exercise, often by using the definitions of the groups and the operations of matrix multiplication and taking inverses.

Compactness of Unitary Groups

To show that the unitary groups U(n) and SU(n) are compact, we need to show that they are both closed and bounded. It has already been shown that they are closed. To show that they are bounded, we note that every element of these groups is a unitary matrix, meaning its entries are complex numbers of absolute value 1. This restricts how large the entries can be, and thus bounds the matrices.