Re: Please help Representation theory questions, GL(n;R), GL(n;C)

On 17-10-2005 12:43, James wrote (without pressing the ENTER key too
often :-( ):

As before, we have the adjoint action Ad_GL(n) : gl(n) ---> gl(n), but
now I will write it as a representation, Ad_GL(n) : GL(n) ---> GL(gl(n)),
which is of dimension n^2 since gl(n) = M(n;R).
Then, there is an equivalence of representations Ad_GL(n) = f_n (x) (f_n)^*
where ^* denotes the dual representation.  This equivalence of
representations is obtained by showing that Hom(V,V) is isomorphic to
V (x) V^* and the isomorphism is actually a G-isomorphism with respect to
some G-actions.

Now, my question : The following appeared in the last lecture of my class: Let U(n) is the unitary n x n matrices. Every matrix in GL(n;C) can be
written as 1/2 (A + A*) + 1/2 (A - A*), the sum of a symmetric and
antisymmetric matrix. Also, A = 1/2 (A + A*) - i*1/2*(A-A*), so any A in
M(n;C) = gl(n;C) can be thought of as the copmlexification of the symmetric
matrices, where gl(n;C) is the Lie algebra of GL(n;C). So,
gl(n;C) = (Symmetric matrices) (x) C. So, for U(n), we have Ad_U(n) (x) C =
u_n (x) u_n^*, where u_n is the natural representation of U(n) on C^n, or
you can think of the representation as the restriction of the representation
of GL(n;C) on C^n.

I understood only the first two sentences.  Can anyone make sense of the rest
and please please give me your thoughts?  What does Ad (x) C mean in general,
because I am seeing this in my textbook?  (C is the complex numbers)>


First of all, if you write A as 1/2 (A + A*) + 1/2 (A - A*), this is
*not* a way of writing it "the sum of a symmetric and antisymmetric
matrix", because, in general, A + A* is not symmetric; it is hermitian.

So should I have written that every matrix A in GL(n;C) can be written
as the sum of a hermitian and anti-hermitian matrix? 

Yes. It is also true that every matrix A in GL(n,C) can be written as
the sum of a symmetric and an anti-symmetric matrix, but that's
irrelevant here.

Then, should I have written gl(n;C) = (Hermitian matrices) (x) C
instead of gl(n;C) = (Symmetric matrices) (x) C?  I'm not quite sure
what is meant by gl(n;C) = (Symmetric matrices) (x) C.


If you don't know what it means, then you shouldn't write, whether true
or false. But no, the statement gl(n;C) = (Hermitian matrices) (x) C is
not exactly true, although it is close to a true statement. What is true
is that the natural map

(Hermitian matrices) (x) C ----> gl(n,C)
                  A + i B   |->  A + i B

is a vector space isomorphism and even an isomorphism of representations
of the unitary group.

OTOH, there's an obvious error at the beginning of the third sentence:
it should read:

   A = 1/2 (A + A*) + i (A - A*)/(2i).

Then, yes, you do get A as a B + i C, where both B and C are /hermitian/
(not symmetric) matrices.

In general, if _r_ is a representation of a group G on a real vector
space V, then r (x) C is the representation of G on V (x) C defined by

((r (x) C)(g))(v + i w) = (r(g))(v) + i (r(g))(w).

Thank you. Before, I stated that Ad_GL(n) = f_n (x) (f_n)^*. Is there
an easy way to see that Ad_U(n) (x) C = u_n (x) u_n^* ? It must have
something to do with gl(n;C) = (Symmetric matrices) (x) C (or if Symmetric
must be replaced by Hermitian, sorry I still don't know). 

It's easy to prove that if _f_, _g_, and _h_ are real representations of
some group G and if _f_ is isomorphic to g (x) h, then the
complexification of _f_ (i. e., f (x) C) is isomorphic to the tensor
product of the complexifications of _g_ and _h_. So, yes, you do have
Ad_U(n) (x) C = u_n (x) u_n^*.

Now that I think about it, I'm not even sure which vector space U(n) acts
on in the representation Ad_U(n).  Is it its Lie algebra?  You stated
above what r (x) C meant, for a general real representation r, but Ad_U(n)
I assume is not a real representation (is it?).


Yes, Ad_U(n) is a the adjoint representation of U(n), in which U(n) acts
on its Lie algebra, which is the Lie algebra of the anti-hermitian
matrices. And this is a real vector space but not a complex one (in
general, if A is anti-hermitian, i A is not).

Best regards,

Jose Carlos Santos
.

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