Quantum Mechanics for Mathematicians

so the dual representation is given by conjugating all elements of the matrix.
The same calculation as in the real case here gives

det(L−^1 ) det(L) = det(L†) det(L) =det(L) det(L) =|det(L)|^2 = 1

so det(L) is a complex number of modulus one. The map

L∈U(n)→det(L)∈U(1)

is a group homomorphism.
We have already seen the examplesU(1),U(2) and SU(2). For general
values ofn, the study ofU(n) can be split into that of its determinant, which
lies inU(1) so is easy to deal with, followed by the subgroupSU(n), which is a
much more complicated story.

Digression.Note that it is not quite true that the groupU(n)is the product
groupSU(n)×U(1). If one tries to identify theU(1)as the subgroup ofU(n)
of elements of the formeiθ 1 , then matrices of the form

ei

mn 2 π

1

forman integer will lie in bothSU(n)andU(1), soU(n)is not a product
of those two groups (it is an example of a semi-direct product, these will be
discussed in chapter 18).
We saw at the end of section 3.1.2 thatSU(2)can be identified with the three-
sphereS^3 , since an arbitrary group element can be constructed by specifying one
row (or one column), which must be a vector of length one inC^2. For the case
n= 3, the same sort of construction starts by picking a row of length one inC^3 ,
which will be a point inS^5. The second row must be orthonormal, and it can be
shown that the possibilities lie in a three-sphereS^3. Once the first two rows are
specified, the third row is uniquely determined. So as a manifold,SU(3)is eight
dimensional, and one might think it could be identified withS^5 ×S^3. It turns out
that this is not the case, since theS^3 varies in a topologically non-trivial way
as one varies the point inS^5. As spaces, theSU(n)are topologically “twisted”
products of odd dimensional spheres, providing some of the basic examples of
quite non-trivial topological manifolds.

4.7 Eigenvalues and eigenvectors

We have seen that the matrix for a linear transformationLof a vector spaceV
changes by conjugation when we change our choice of basis ofV. To get basis-
independent information aboutL, one considers the eigenvalues of the matrix.
Complex matrices behave in a much simpler fashion than real matrices, since in
the complex case the eigenvalue equation

det(λ 1 −L) = 0 (4.5)