Fundamental Group of Matrices

1. Feb 16, 2010

curtdbz

I am reading Munkres and know exactly how to find the fundamental groups of surfaces, using pi_1 and reducing it down to simpler problems. However, I'm completely lost when looking at my final exam it says to find the fundamental groups of matrices! How do you go about doing that! There are no explainations in teh book, or the net (that I can find). The two main ones I'm having trouble with are here, if you could quickly help that would be great.

Find the fundamental groups of
1) The space of 2x2 matrices over C (complex) with 0 determinant with the topology given by the embedding in the vector space of all 2x2 matrices over C.
2) The space of upper triangular matrices of the size 2x2 over C with determinant 1 considered as a closed subspace in the vector space of all matrices size 2x2 over C.

I know the answer to 1, because someone told me it, but I don't know how they came to that conclusion. The answer to one is the integers, Z. Thank you!

2. Feb 17, 2010

quasar987

1 seems hard to me, but the space of question 2 looks contractible: write

a b
c d

for a 2x2 matrix with complex entries. Then the space of such matrices, is, as specified in the question, identified with C^4, and so the space of upper triangular matrices that concerns us is identified with the subspace of C^4 consisting of those quadruples (a,b,0,d) such that ad=1. That is, such that a =/= 0 and d=a^-1=(complex conjugate of a)/|a|². Surely, this space deformation retracts onto, say (1,1,0,1).

3. Feb 17, 2010

wofsy

Det(tA) = (t^n)Det(A) for an nxn matrix.

This gives a strong deformation retraction of the matrices of zero determinant onto the zero matrix.

(A,t) -> tA for t in the unit interval.

Thus the answer to 1 is the trivial group.

For 2 let the matrix a b o d go to the matrix a tb o d. This defines a strong deformation retraction onto the diagonal matrices of determinant 1. These matrices are homeomorphic to the non-zero complex numbers by the map z -> z 0 0 1/z.

The non-zero complex numbers deform onto the complex numbers of norm 1 which is the unit circle in the complex plane. Thus the fundamental group is the fundamental group of the circle which is Z.

The answer to 2 is the integers.

Last edited: Feb 18, 2010
4. Feb 27, 2010

Staff: Mentor

Sounds like you are asking us to do your final take-home exam for you...?