# Lie Subgroup - Cartan's Theorem

1. Nov 18, 2008

### malawi_glenn

Hi, I was reading Cartan's Theorem:

A Group H is a Lie Subgroup to Lie Group G if H is a closed subgroup to G.

Now first of all, is this a definition of Lie Subgroup?

Second, what does it mean that the subgroup is "closed"? I thought all groups where closed under group multiplication.. :/ Help?

2. Nov 18, 2008

A group is a set G with elements (possible infinite and/or uncountable) such as

$$G= \left\{g_1,g_2,\cdots g_n\right\}$$

with the following

1) a multiplication rule $$g_i*g_j=g_k$$ for some i,j and k. (closure)
2) an identity element I, st $$I*g_i=g_i*I=g_i$$
3) every element has an inverse, i.e $$g_i*g_i^{-1}=I$$

A SUBSET H is a subgroup if $$h_i*h_j=h_k$$ for every element h in the subset H

3. Nov 18, 2008

### malawi_glenn

yes i know, but what is meant by a CLOSED subgroup?!

you forgot associativity by the way....

and that the inverse to h is in H ...

4. Nov 18, 2008

I don't understand what the problem is.

All subgroups are closed otherwise it's not a subgroup it's just merely some subset of G

5. Nov 18, 2008

### malawi_glenn

closed with respect to what? group multiplication?

Why is the theorem stressing CLOSED subgroup, the word has no power if there are no subgroups which are NOT CLOSED (open)..

6. Nov 18, 2008

yes group multiplication.

I don't know why it's specifying closed subgroup to be honest. The word is redundent.

Yes I didn't bother with associativity but at this time in the evening I couldn't be arsed lol

7. Nov 18, 2008

### d_leet

I'm fairly certain it means closed in the topological sense as a subset of the Lie group G.

8. Nov 18, 2008

### cristo

Staff Emeritus
I don't think it means closed as in closed under group multiplication since (as both of you say) that is always true in the definition of a group. What the statement probably means is that H is a Lie subgroup of G if H is a (topologically) closed subgroup of G.

9. Nov 18, 2008

### malawi_glenn

Ok, so a subgroup of G wich is also a closed subset of G?

10. Nov 18, 2008

### malawi_glenn

So how for instance do one determine if a group is topologically closed? For example SO(2), is that a closed group?

11. Nov 18, 2008

### samalkhaiat

The word Lie need not be there.

A subgroup H of a group G is a subset {h} of elements of G which closes with respect to the multiplication already defined by G ( h.k in H for all h,k in H) and which contains the inverse of each of its elements h, and the identity e. If the subset {h} does not close, i.e., if

$$h.k \in G \ \mbox{not in H}, \ h,k \in H$$

then the set {h} does form a group.

Subgroups arise by imposing a restriction on the original group, e.g. restricting the 3-dimensional group of rotations to rotations about one axis, or to discrete rotations.
The most interesting subgroups are the so-called invariant subgroups; an invariant subgroup H of G is one which is invariant with respect to conjugation with G, i.e.,

$$ghg^{-1} \in H \ \ \mbox{for all} \ \ h \in H, \ g \in G$$

The translation subgroups of the spacetime groups are invariant subgroups because they are transformed into themselves by rotations.
Also, when H is an invariant subgroup, the coset manifold G/H forms a group called the quotient group.

regards

sam

Last edited: Nov 18, 2008
12. Nov 18, 2008

### malawi_glenn

But I am asking about What a Lie Subgroup is. What is it?

13. Nov 18, 2008

### samalkhaiat

14. Nov 18, 2008

### malawi_glenn

so,

a subgroup to a Lie group is a "Lie subgroup" ?

Now SO(2) is a liegroup, is U(1) a Lie Sub group to SO(2)?

15. Nov 18, 2008

### samalkhaiat

Yes.

No. they are isomorphic.

U(1) is a subgroup of SU(2) and SO(2) is a subgroup of SO(3).

16. Nov 18, 2008

### malawi_glenn

ok, and since SO(3) is a Lie Group, SO(2) is a Lie Subgroup to SO(3)?

17. Nov 18, 2008

### samalkhaiat

the group G = {e} of the identity, is a subgroup of any group A. If A is a Lie group, would we call {e} a Lie subgroup? Yes, it is a trivial Lie group.

18. Nov 19, 2008

### malawi_glenn

Ok I think I got it!

Really hard for a physicsists to get involved into abstract algebra, but I try hard to understand.

Thanx for all replies!

19. Nov 19, 2008

### morphism

No. Presumably we want a "Lie subgroup" to be a Lie group itself. Now, any subgroup H of a Lie group G can inherit the topological structure of G (as a subspace of G) - so at least there is a topology on H that we can talk about - but H may not inherit the manifold structure associated to G. In other words, H may not be a submanifold of G. Example: the real numbers R form a Lie group under addition, the usual topology, and the usual smooth structure. However, the subgroup Q of rational numbers is not a manifold, so it's not a Lie group. (Note that Q is not topologically closed in R.) The actual definition of a Lie subgroup of a Lie group varies a bit, but a common one is the following: a subgroup H of a Lie group G is a Lie subgroup of G if H is an embedded submanifold of G. Check your text to see what definition the author is using.

Now, it's a standard result that a Lie subgroup of a Lie group G is topologically closed in G. Cartan's theorem states that the converse of this result is true (at least for real Lie groups!).

20. Nov 19, 2008

### malawi_glenn

Ok I have no book but is using different internet resources.

So let me get this straight, A Lie Subgroup is a subgroup to a Lie Group which is a Lie Group itself?