# Connectedness of Lie Groups

1. Jan 12, 2014

### gentsagree

I surely am missing something about the notion of connectedness, and I clarify this by means of an example:

O(n), the orthogonal group, has two subsets with detO=1 and detO=-1. Now, the maximally connected component of O(n) is SO(n), which is the subgroup with detO=1 including the Identity, while the other part is simply a coset (and not a subgroup, as it doesn't, of course, contain I).
Thus, O(n) is NOT a connected group.

I do not understand why, on the other hand, U(n) is said to be connected when it has got, in exactly the same way as O(n), two subsets with detU=1 and detU=-1, where we call the former SU(n).

2. Jan 12, 2014

### Office_Shredder

Staff Emeritus
U(n) is a group of complex matrices, so has matrices A with det(A) = z for any complex number |z|=1 (which you will note is a connected set, namely a circle). This doesn't prove the set is connected by itself but does resolve the issue of the determinant.

3. Jan 12, 2014

### dextercioby

Check out C. Chevalley's "Theory of Lie groups", Proposition 3, Page 37.

4. Jan 18, 2014

### Landau

You are focussing on words rather than the meaning of words. I suggest you first try to understand how the facts you state imply that O(n) is not connected.

* connectedness is a property of a topological space, not of a group or a set
* O(n) is partitioned into the 1 and -1 fibers of the determinant, while U(n) is not

Anyway, the key to your 'argument' is
First, it seems to contain a tautology: in my vocabulary a connected component is a maximal connected subset.
Second, do you understand your use of 'the'? It either means 'unique up to homeomorphism', or you forgot to include 'of the identity'. Third, the following statement obtained by substituting words is false:
so you need to understand why it is true for O(n) in the first place.

Last edited: Jan 18, 2014