$SL(3,\mathbb{R})$ is a smooth manifold? How do you show $SL(3,\mathbb{R})$ is a smooth manifold? I am thinking to use the preimage theorem, but what kind of thing I need to show first before I can apply the theorem?
 A: As suggested in the comments consider the function
$$F:\quad{\mathbb R}^{3\times 3}\to {\mathbb R},\qquad A\mapsto\det(A)\ .$$
Then $$M:=SL(3,{\mathbb R})=\bigl\{A\in {\mathbb R}^{3\times 3}\>\bigm|\>F(A)=1\bigr\}\ .$$ In order to show that $M$ is an eight-dimensional smooth manifold it is enough to show that for $A\in M$ one has $dF(A)\ne0$. To show the latter it is sufficient to present a single vector $X\in {\mathbb R}^{3\times 3}$ with $dF(A).X\ne0$. We choose $X:=A$ and then have
$$dF(A).A=\lim_{\epsilon\to0+}{F(A+\epsilon A)-F(A)\over\epsilon}=\lim_{\epsilon\to0+}{(1+\epsilon)^n -1\over\epsilon}=n\ne0\ .$$
A: Another approach (depending on the assumptions you're allowed to make):


*

*The general linear group $\mathrm{GL}_3 \mathbb{R}$ is a smooth manifold, since it's an open subset of $\mathbb{R}^9$.

*In fact, since the multiplication and inversion maps are smooth, $\mathrm{GL}_3 \mathbb{R}$ is a Lie group.  (It's actually enough to show that multiplication is smooth to conclude this.)

*The determinant function is smooth, since it's a polynomial in the entries of its inputs.  It's also a homomorphism of Lie groups.

*$\mathrm{SL}_3 \mathbb{R}$ is the kernel of the determinant, and is thus a closed subgroup of $\mathrm{GL}_3 \mathbb{R}$.

*Closed subgroups of Lie groups are themselves Lie groups, so you're done!
Admittedly, there's a lot going on here, and you're likely working from more basic principles.  Still, it's good to see the larger context, even as you're struggling with the foundations.
