Show that $SL(n, \mathbb{R})$ is a $(n^2 -1)$ smooth submanifold of $M(n,\mathbb{R})$ I need to show for $n=3$ that $SL(n,\mathbb{R})=\{A \in M(n, \mathbb{R}) : detA=1 \}$ is a $(n^2 -1)$ dimensional smooth submanifold of the vector space $M(n,\mathbb{R})$ of all real $n \times n$ matrices. I would assume that I need to use the regular value theorem and use the determinant map to get the result, but I'm a bit unsure on how to set this up correctly. Any help would be appreciated.
 A: Ben West's solution is extremely clean and simple, and it should be accepted as the answer. 
This is the solution that I was thinking about.
$GL_n = \det^{-1}(\mathbb{R}\setminus\{0\})$
Since $\det$ is continuous, $GL_n$ is an open subset of $M_n$, meaning it has dimension $n^2$.
Restrict $\det$ to $GL_n$. Then, for any $A\in GL_n$ and any $B\in M_n \sim T_A GL_n$, we have:
\begin{align}
d(\det)_A(B) &= \lim_{t\to0} \frac{\det(A+tB) - \det(A)}{t} \\
& =\det(A) \lim_{t\to0} \frac{\det(I+tA^{-1}B)-1}{t}\\
&= \det(A)\mathrm{tr}(A^{-1}B)
\end{align}
Since $\det$ maps into a 1-dimensional space, we just need one $B$ that makes $\mathrm{tr}(A^{-1}B)≠0$. So, take $B=A$.
This shows that every non-zero number is a regular value of $\det$.
A: You want to consider the smooth map $\det\colon M_n(\mathbb{R})\to\mathbb{R}$. If you can show $1$ is a regular value of $\det$, then $SL(n,\mathbb{R})=\det^{-1}(1)$ is a smooth manifold of dimension $$\dim(M_n(\mathbb{R}))-\dim(\mathbb{R})=n^2-1$$
by the regular value theorem. 
To show $1$ is a regular value of $\det$, first note that the domain and codomain are vector spaces, so they may be identified with their own tangent spaces. 
If $A\in M_n(\mathbb{R})$, then $d(\det)_A(A)$ can be computed as
$$\lim_{t\to 0}\frac{\det(A+tA)-\det(A)}{t}=\lim_{t\to 0}\frac{(1+t)^n\det(A)-\det(A)}{t}=n\cdot\det(A)
$$
This shows $d(\det)_A$ is nonzero linear functional, hence surjective, for invertible $A$. This implies in particular that $1$ is a regular value of $\det$. 
