Unitary group

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In mathematics, the unitary group of degree n, denoted U(n), is the group of n×n unitary matrices, with the group operation that of matrix multiplication. The unitary group is a subgroup of the general linear group GL(n, C).

In the simple case n = 1, the group U(1) corresponds to the circle group, consisting of all complex numbers with norm 1 under multiplication. All the unitary groups contain copies of this group.

The unitary group U(n) is a real Lie group of dimension n². The Lie algebra of U(n) consists of complex n×n skew-Hermitian matrices, with the Lie bracket given by the commutator.

The general unitary group (also called the group of unitary similitudes) consists of all matrices $A$ such that $A A *$ is a nonzero multiple of the identity matrix, and is just the product of the unitary group with the group of all positive multiples of the identity matrix.

1 Properties
2 Topology
3 Classifying space
4 See also

[edit] Properties

Since the determinant of a unitary matrix is a complex number with norm 1, the determinant gives a group homomorphism

$\det\colon \mbox{U}(n) \to \mbox{U}(1)$

The kernel of this homomorphism is the set of unitary matrices with unit determinant. This subgroup is called the special unitary group, denoted SU(n). We then have a short exact sequence of Lie groups:

$1\to\mbox{SU}(n)\to\mbox{U}(n)\to\mbox{U}(1)\to 1$

This short exact sequence splits so that U(n) may be written as a semidirect product of SU(n) by U(1). Here the U(1) subgroup of U(n) consists of matrices of the form $\mbox{diag}(e^{i\theta},1,1,\ldots,1)$ .

The unitary group U(n) is nonabelian for n > 1. The center of U(n) is the set of scalar matrices λI with λ ∈ U(1). This follows from Schur's lemma. The center is then isomorphic to U(1). Since the center of U(n) is a 1-dimensional abelian normal subgroup of U(n), the unitary group is not semisimple.

[edit] Topology

The unitary group U(n) is endowed with the relative topology as a subset of M_n(C), the set of all n×n complex matrices, which is itself homeomorphic to a 2n²-dimensional Euclidean space.

As a topological space, U(n) is both compact and connected. The compactness of U(n) follows from the Heine-Borel theorem and the fact that it is a closed and bounded subset of M_n(C). To show that U(n) is connected, recall that any unitary matrix A can be diagonalized by another unitary matrix S. Any diagonal unitary matrix must have complex numbers of absolute value 1 on the main diagonal. We can therefore write

$A = S\,\mbox{diag}(e^{i\theta_1},\dots,e^{i\theta_n})\,S^{-1}.$

A path in U(n) from the identity to A is then given by

$t\mapsto S\,\mbox{diag}(e^{it\theta_1},\dots,e^{it\theta_n})\,S^{-1}.$

Although it is connected, the unitary group is not simply connected. The first unitary group U(1) is topologically a circle, which is well known to have a fundamental group isomorphic to Z. In fact, the fundamental group of U(n) is infinite cyclic for all n:

$\pi_1(U(n)) \cong \mathbb Z.$