I have that $SL(n,\mathbb{R})$ is an embedded submanifold of dimension $n^2-1$ in $GL(n,\mathbb{R})$, and I know that $T_XGL(n,\mathbb{R})$ is isomorphic to $M(n,\mathbb{R})$ for all $X \in GL(n , \mathbb R)$.
Is there a way I can use this to show that $SL(n,\mathbb{R})$ is a smooth submanifold of $M(n,\mathbb{R})$ and get is dimension? Otherwise, how could I go about it?
Thanks!
There is a general theorem of Cartan that a closed subset of a Lie group that is moreover a subgroup is in fact a closed submanifold, so a Lie subgroup.However, this case can be done with the theorem of implicit functions. $SL(n,\mathbb{R})$ is the set of all matrices $(x_{ij})$ with $\det(x_{ij})=1$. Every point of the hypersurface $\det(x_{ij})=1$ is nonsingular. This is a gemeral fact for level sets $P^{-1}(a)$ of homogenous polynomials $P$ for value $a \ne 0$. If at some point $x$ all the partial derivatives $\frac{\partial P}{\partial x_i}(x)=0$ then, using Euler's identity we get $d \cdot P(x) = \sum x_i \frac{\partial P}{\partial x_i}(x) =0$ so $P(x) \ne a$.
Hence, $SL(n,\mathbb{R})$ is a closed submanifold of $M(n,\mathbb{R})$ of dimension $n^2 -1$.