Editing Special linear group (section)

==Topology==
Any invertible matrix can be uniquely represented according to the [[polar decomposition]] as the product of a [[unitary matrix]] and a [[Hermitian matrix]] with positive [[eigenvalue]]s. The [[determinant]] of the unitary matrix is on the [[unit circle]], while that of the Hermitian matrix is real and positive. Since in the case of a matrix from the special linear group the product of these two determinants must be 1, then each of them must be 1. Therefore, a special linear matrix can be written as the product of a [[special unitary matrix]] (or [[special orthogonal matrix]] in the real case) and a [[Positive-definite matrix|positive definite]] Hermitian matrix (or [[symmetric matrix]] in the real case) having determinant 1.

It follows that the topology of the group <math>\operatorname{SL}(n,\C)</math> is the [[product topology|product]] of the topology of <math>\operatorname{SU}(n)</math> and the topology of the group of Hermitian matrices of unit determinant with positive eigenvalues. A Hermitian matrix of unit determinant and having positive eigenvalues can be uniquely expressed as the [[matrix exponential|exponential]] of a [[traceless]] Hermitian matrix, and therefore the topology of this is that of <math>(n^2-1)</math>-dimensional [[Euclidean space]].<ref>{{harvnb|Hall|2015}} Section 2.5</ref> Since <math>\operatorname{SU}(n)</math> is [[simply connected]],<ref>{{harvnb|Hall|2015}} Proposition 13.11</ref> then <math>\operatorname{SL}(n,\C)</math> is also simply connected, for all <math>n\geq 2</math>.

The topology of <math>\operatorname{SL}(n,\R)</math> is the product of the topology of [[special orthogonal matrix|SO]](''n'') and the topology of the group of symmetric matrices with positive eigenvalues and unit determinant. Since the latter matrices can be uniquely expressed as the exponential of symmetric traceless matrices, then this latter topology is that of {{nowrap|(''n'' + 2)(''n'' − 1)/2}}-dimensional Euclidean space. Thus, the group <math>\operatorname{SL}(n,\R)</math> has the same [[fundamental group]] as <math>\operatorname{SO}(n)</math>; that is, <math>\Z</math> for <math>n=2</math> and <math>\Z_2</math> for <math>n>2</math>.<ref>{{harvnb|Hall|2015}} Sections 13.2 and 13.3</ref> In particular this means that <math>\operatorname{SL}(n,\R)</math>, unlike <math>\operatorname{SL}(n,\C)</math>, is not simply connected, for <math>n>1</math>.