Connected components of aLie group consists of block matrices

Question

I and my classmate discussed about the number of the connected components of a Lie group consists of matrices like

\begin{pmatrix} A & B & C \\ 0 & E & F \\ 0 & 0 & I \end{pmatrix}

Where A, B, ..., I are matrices of order n. I believe this Lie group has 2 connected components, since the determinants of the matrices are nonzero, but my classmate said that there are 8 connected components, since the determinants of each block matrices are nonzero. I wonder who is correct and why the other one is wrong?

Answer 1

To test your two hypotheses against each other, check what happens for $n=1$; that will refute one. Then try to prove the remaining hypothesis in general.

Answer 2

As the determinant of this block triangular matrix is

$$\det(A)\det(E)\det(I)$$

I see indeed, as your classmate, $8$ equivalence classes according to the sign triples of $(\det(A),\det(E),\det(I))$ :

$$(-,-,-),(-,-,+),\cdots (+,+,+)$$

For example, any arc connecting a $(-,-,-)$ case to a $(-,-,+)$ case would cross a $(-,-,0)$ case by continuity of function $\det$.

Remark : the other components play no role.

An exemple : in the case of $n=1$, imagine space $\mathbb{R}^3$ with coordinate axes labelled by $a,e,i$ : the equivalence classes would be materialized by the 8 (open) octants.