Lie group and algebras

  • Thread starter ala
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  • #1
ala
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Here is few statements that I proved but I suspect that are incorrect (but I can't find mistake), term group means Lie group same goes for algebra:

1. Noncompact group G doesn't have faithfull (ie. kernel has more that one element) unitary representation.
Proof:
If D(G) is faithfull unitary representation of group G, then D(G) is closed (in topological sense) subgroup of group U(n) (unitary group which is compact and connected, so every it's subgroup is closed), so D(G) is compact (closed subset of compact set is compact), and because D(G) is faithfull we conclude that G is also compact.

2. Kernel of smooth homomorphism is discrete subgroup or whole group.
Proof:
If we denote group with G and kernel with K, K is subgroup. If indentiry component has more than one element than we see that open set containing indentity is represented with identity matrix. Because whole group is genereated with elements from around identiry (my english is bad, but I hope you understand) so whole group G is represented with unitary matrix so K=G. I other case K is discerete subgroup.

And I have few more statements that I doesn't know how to prove nor if they are correct:
3. Semisimple, noncompact group G doesn't have unitary representation.
4. Orthogonal complement (in sense of Killing form) of kernel of representation of algebra doesn't have intersection with kernel (except zero).

If someone sees mistakes here or know for sure that some statements are incorrect, please let me know.
 

Answers and Replies

  • #2
49
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2. I think this is wrong. Take for example G = GL(n,C) (n>2) and H = C (complex numbers). Then the map det(A) - 1 is a smooth map from G to H. But the kernel is a variety of dimension > 0 (because the ideal defining it is not maximal--use Nullstellensatz for example), so is not discrete. I hope what I said is correct.

4. This is probably wrong too. For example, any nilpotent Lie algebra has trivial Killing form. So if the kernel is nontrivial, its complement is the whole Lie algebra so intersects it nontrivially. For an example, take g to be the strictly upper triangular n x n matrices. A representation with nontrivial kernel: just map every matrix to 0.
 
  • #3
ala
22
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First thank you for your reply.
2. Yes, you are right.
Note: Kernel of homomorphism between groups is set of all elements from first group that goes (using homomorphism) to indentity of second group. So mapping could be abs(det(A)) so kernel is unitary group, ie. it is not discrete nor equal to GL(n,C). So now I have modification of statement 2:

2'. Kernel of smooth homomorphism of simple (i.e. doesn't have connected nontrivial subgroups) is discrete subgroup or whole group.
But I don't know how to prove this or to find counterexample. So question is:
Can it be that kernel is nonconnected subgroup that doesn't contain open set in G? (i.e. set that is in topology of G)

4. Yes you are right again, I forgot to write that algebra is semisimple. If you have solvable algebra than orthogonal complement of whole algebra is non trivial so if we have representation that maps elements from this orthogonal complement to zero than statement is not corrent. So is this one correct and how to prove it:

4'. Orthogonal complement (in sense of Killing form) of kernel of representation of semisimple algebra doesn't have intersection with kernel (except zero).

What about statements 1. and 3.?

Best regards...
 
Last edited:
  • #4
ala
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4'. Goes directly from Artin-Cartan theorem. But still I don't know what to do with rest of statements?
 
  • #5
George Jones
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Not all vector space are finite-dimensional.

For which statements in the original post is the above statement relevant?
 
  • #6
ala
22
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Yes, you are right, I forgot to say that they are finite dimensional. (In all statements where representation is used)
 

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