Fundamental group of PSp(2n,R)

[A. Gama]

What is the fundamental group of the (real) projective symplectic
group, PSp(2n,R). This group is defined as the quotient of the real
symplectic group Sp(2n,R) by its center, Z_2: PSp(2n,R)=Sp(2n,R)/Z_2.
From the homotopy sequence of the fibration 0->Z_2->Sp(2n,R)-
>PSp(2n,R)->0, one sees that the fundamental group of PSp(2n,R) is
either Z or ZxZ_2. But which one is it?

Best regards,
A. Gama

[anton....@googlemail.com]

I think it is Z.
I would argue as follows.
Let n>1. Note that the maximal compact subgroup of PSp(2n,R) is U(n)/
Z_2.
By Iwasawa decompostion, the fundamental group of PSp(2n,R) equals the
one of U(n)/Z_2.
The fibre bundle U(n) -> U(n)/Z_2 induces an exact sequence

1 -> pi_1(U(n)) -> pi_1(U(n)/Z_2) -> Z_2 -> 1

It is known that the inclusion of the center,

U(1) -> U(n)

induces an isomorphism for the fundamental groups.
Using this we can use the same sequence for n=1 and the five-lemma to
deduce

pi_1(U(n)/Z_2) = pi_1(U(1)/Z_2) = Z.

Best,
Anton

[A. Gama]

Hello,

Thank you for your answer!
However, meanwhile, I have reached to a different conclusion as
follows:
As you say, it is enough to compute pi_1(U(n)/Z_2) and I did it by
taking its universal cover and computing the kernel of the projection
(which is isomorphic to the fundamental group of U(n)/Z_2). The
universal cover of U(n)/Z_2 is R\times SU(n) and consider the
projection p:R\times SU(n)->U(n)/Z_2 given by the usual projection p':R
\times SU(n)->U(n), where p'(t,A)=exp(2\pi it)A and composing it with
the projection U(n)->U(n)/Z_2.
So, Ker(p) is the inverse image of {I,-I} under p'.
When you compute this with n odd, it can be seen that it is cyclic,
whereas if you compute it when n is even, it has two generators, one
of which has order 2.
Hence, my conclusion is:
pi_1(U(n)/Z_2)=Z, if n is odd
and
pi_1(U(n)/Z_2)=Z\times Z_2, if n is even.

I think that your argument

"Using this we can use the same sequence for n=1 and the five-lemma to
deduce

pi_1(U(n)/Z_2) = pi_1(U(1)/Z_2) = Z" only works when n is odd.
When n is even you have the sequence
0->SU(n)/Z_2->U(n)/Z_2->U(1)->0, while, for n odd
you have the sequence
0->SU(n)/Z_2->U(n)/Z_2->U(1)/Z_2->0 and this should cause the
difference between the fundamental groups.
Do you agree?

Thank you again,
Best wishes,
AndrŽ Gama