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Re: Isomorphism question
Posted:
Jun 12, 1996 1:32 PM


mblanc@skivs.ski.org (Michael Blanc) writes: > : Let SL_2(R) denote the group of 2by2 matrices over the real numbers, > : with a determinant of one. And let SU_2 denote the group of 2by2 > : 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 conjugatelinear inner product on conplex 2space). > : > : 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 nontrivial it is injective on B the group of upper triangular matrices a b 1 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 1dimensional 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 1 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 nontrivial subgroups of finite index and so T must stabilise any such character. An element of T as above acts on U by 2 b > a b
2 and hence f(b) = f(a b). As f is a homomorphism this means that all matrices
1 b'
0 1 2 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 teke@matematik.su.se



