Centres of GL_n and SL_n, and discrete normal subgroups
Statement
- Compute the centres of and .
- Show that a discrete normal subgroup of a connected topological group is central.
Hint
For part 1, test commutation with the matrix units or with . For part 2, look at the map for fixed .
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Solution
1. Centre of . Let commute with all invertible matrices. Then for it commutes with , hence with . Now has as its -th column and zeros elsewhere, while has the -th row of as its -th row and zeros elsewhere. Comparing entries: the entry for gives , the entry for gives , and the entry gives . As are arbitrary, is diagonal with equal diagonal entries. Thus is a scalar matrix: (conversely scalar matrices are central).
Centre of . The same argument applies since for . So the centre consists of the scalar matrices of determinant : with . If is odd, and . If is even, and .
2. Discrete normal subgroups. Let be connected, discrete (every point of is isolated in , with the subspace topology). Fix and define , . It is well defined since is normal and continuous (into , with values in ). Its image is a connected subset of contained in the discrete space (in the subspace topology, a connected subset of a discrete space is a point). So is constant, equal to : for all . Hence is central.
Application. In (connected) the subgroup is discrete, normal and central, as predicted. Connectedness is needed: in the non-connected group the cyclic group generated by is finite (so discrete) and normal (conjugating by a reflection gives and conjugating by a rotation changes nothing), but it is not central, since a reflection does not commute with .
Remark. This is the first step in showing that a covering group's kernel is central: for a connected covering of topological groups, the kernel is discrete normal, hence central, hence abelian.