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Topic: Isomorphism question
Replies: 3   Last Post: Jun 17, 1996 12:20 PM

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Torsten Ekedahl

Posts: 19
Registered: 12/12/04
Re: Isomorphism question
Posted: Jun 12, 1996 1:32 PM
  Click to see the message monospaced in plain text Plain Text   Click to reply to this topic Reply (Michael Blanc) writes:
> : Let SL_2(R) denote the group of 2-by-2 matrices over the real numbers,
> : with a determinant of one. And let SU_2 denote the group of 2-by-2
> : unitary matrices with determinant one ("unitary" meaning that the rows as
> : well as the columns as complex vectors each form an orthonormal basis wrt
> : the standard conjugate-linear inner product on conplex 2-space).
> :
> : I have reason to suspect that these two groups may be isomorphic. Would
> : someone confirm or deny? And when "2" is replaced by "n"?

If I am not mistaken there is a very general result of Borel and Tits
which shows in this particular case that such an isomorphism does not
exist. However, a more direct argument is possible in this particular case
Assume that there is a group homomorphism (of abstract groups)
SL_2(R) -> SU_2
The only normal subgroup of SL_2(R) is the center (of order 2) and
hence if the homomorphism is non-trivial it is injective on B the
group of upper triangular matrices
a b
0 a

of SL_2(R) with a > 0. Now, I claim that any finite dimensional
unitary representation (not necessarily continous!) of B factors
through the homomorphism which to a matrix like that associates a.

The restriction of the representation to the subgroup U

1 b

0 1

is a sum of irreducible representations (being finite dimensional and unitary).
This group is commutative so by Schur's lemma any irreducible
representation is 1-dimensional and hence given by a homomorphism f: U ->
C^* (in fact it lands in the unit circle being unitary).

The set of such f occurring in the representation are permuted
by the action of T :=

a 0
0 a

on U and hence on its homomorphisms to C^*. As the representation is finite
dimensional these are finite in number but as T is a divisible group
it has no non-trivial subgroups of finite index and so T must
stabilise any such character. An element of T as above acts on U by
b -> a b

and hence f(b) = f(a b). As f is a homomorphism this means that
all matrices

1 b'

0 1
with b' = (a - 1)b lie in the kernel of f. Such elements generate U
however so f is trivial and the original representation has U in its kernel.

Torsten Ekedahl

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