Connectedness of Lie Groups

[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).

 

[Office_Shredder]

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.

 

[dextercioby]

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

 

[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

the maximally connected component of O(n) is SO(n)

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:

the maximally connected component of U(n) is SU(n)

so you need to understand why it is true for O(n) in the first place.

 

[blue_raver22]

I can understand your confusion about the connectedness of Lie groups. The notion of connectedness is a fundamental concept in mathematics and plays a crucial role in the study of Lie groups. Let me try to clarify this by addressing your example.

Firstly, let's define what we mean by a connected group. A connected group is a group where any two elements can be connected by a continuous path within the group. This means that there are no "breaks" or "gaps" in the group - it is one continuous entity. In the context of Lie groups, this means that any two elements can be connected by a continuous path of matrices.

Now, let's look at your example of O(n). As you correctly pointed out, O(n) has two subsets - one with detO=1 (which we call SO(n)) and one with detO=-1. However, these two subsets are not connected by a continuous path within O(n). In other words, there is no way to continuously transform a matrix with detO=1 to a matrix with detO=-1 while staying within O(n). This is why we say that O(n) is not a connected group.

On the other hand, U(n) is a connected group because any two elements in U(n) can be connected by a continuous path of unitary matrices. The two subsets with detU=1 and detU=-1 are connected through the identity matrix, which is in both subsets. This is why we say that U(n) is connected.

In summary, the key difference between O(n) and U(n) is that in O(n) there is no way to continuously connect the two subsets, while in U(n) the subsets are connected through the identity element. This concept of connectedness is important in the study of Lie groups as it helps us understand the structure and properties of these groups.

I hope this explanation has helped clarify the concept of connectedness in Lie groups. If you have any further questions or would like to discuss this further, please do not hesitate to reach out. As scientists, it is important for us to fully understand and clarify any confusion we may have in order to advance our knowledge and understanding of the world.