Question

Describe the center of the group su(n) and show that it is isomorphic to zn.

Answer 1

Answer:

The centers of the matrix groups U(n) and SU(n)

This note proves an assertion in the hints for one of the Additional Exercises for Chapter 7.

THEOREM. Let U(n) be the group of unitary n×n matrices (the entries are complex numbers,

and the inverse is the conjugate of the transpose), and let SU(n) be the kernel of the determinant

homomorphism U(n) → C − {0}. Then the centers of both subgroups are the matrices of the form

cI, where (as usual) I denotes the identity matrix and |c | = 1. In particular, the center of SU(n)

is a finite cyclic group of order n.

The argument relies heavily on the Spectral Theorem, which implies that for every unitary

matrix A there is a unitary matrix P such that PAP −1

is diagonal.

Proof. We shall first prove the result for U(n). If A lies in the center then for each unitary

matrix P we have A = PAP −1

. Since the Spectral Theorem implies that some matrix PAP −1

is diagonal, it follows that A must be diagonal. We claim that all the diagonal entries of A must

be equal. Suppose that aj,j 6= ak,k. If P is the matrix formed by interchanging the j

th and k

th

columns of the identity matrix, then P is a unitary matrix and B = PAP −1

is a diagonal matrix

with aj,j = bk,k and bj,j = ak,k. But this means that A does not lie in the center of U(n). Therefore

the only matrices which can lie in the center of U(n) have the form cI, and since we are working

with unitary matrices it follows that |c | must be 1.

We now turn to the case of SU(n). If D ⊂ U(n) is the subgroup of diagonal matrices, then

of course D is central and we have U(n) = D · SU(n); in other words, every unitary matrix is the

product of a matrix in SU(n) and a diagonal matrix. Suppose now that A ∈ SU(n) lies in the

center of SU(n); then the observations in the preceding sentence imply that A lies in the center of

U(n), so it follows that the center of SU(n) is equal to D ∩ SU(n). Therefore the determination of

the center reduces to specifying which diagonal matrices cI (where |c | = 1) have determinant equal

to 1. But the determinant of cI is c

n, so the center consists of all matrices cI such that c

n = 1;

i.e., the center consists of all matrices cI such that c is a complex n

th root of 1. Since the set of all

such matrices is isomorphic to Zn, we see that the center of SU(n) is a cyclic group of order n.

Explanation:

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