Chapter 19: Problem 3

Show that \(G L(n, C)\) and \(S L(n, \mathbb{C})\) are connected Lie groups. Is \(U(n)\) a connected group?

Short Answer

Expert verified

Yes, both \(G L(n, C)\) and \(S L(n, \mathbb{C})\) are connected Lie groups, and \(U(n)\) is also a connected group.

Step by step solution

Understand the Groups

We should remember that \(G L(n, C)\) is the general linear group of \(n \times n\) complex matrices, \(S L(n, \mathbb{C})\) is the special linear group of \(n \times n\) complex matrices with determinant 1, and \(U(n)\) is the unitary group consisting of \(n \times n\) unitary matrices.

Show That GL(n,C) and SL(n,C) are Connected

To prove that \(G L(n, C)\) is connected, it's sufficient to show that any element in \(G L(n, C)\) can be connected to the identity. This is true because any \(n \times n\) matrix \(A\) can be connected to the identity via the straight line path \(f(t) = (1 - t) I + t A\), where \(t \in [0,1]\). The determinant function \(det: G L(n, C) \rightarrow \mathbb{C}^{*}\) is a polynomial map and hence continuous. Since the image of the continuous function on a connected space is still connected, the special linear group \(S L(n, \mathbb{C})\) which is the pre-image of \(1 \in \mathbb{C}\) under determinant function is also connected.

Is U(n) a Connected Group?

Yes, \(U(n)\) is a connected group. This can be proven by showing that any unitary matrix can be connected to the identity on a continuous path. Any unitary matrix \(U\) can be diagonalized by a unitary matrix \(V\). This means we can write \(U = VDV^{-1}\), where \(D\) is a diagonal matrix with eigenvalues on the unit circle. Now, we can move continuously from \(D\) to \(I\) along the unit circle and use \(V\) to carry this movement back to the full matrix space. This argument can be made more precise by the use of eigenvector cycles and polar coordinates.