This is my definition of a smooth manifold I am supposed to work with:
Let $\mathcal{M} \subseteq \mathbb{R}^{n}$. The set $\mathcal{M}$ is a $k$-dimensional smooth submanifold of $\mathbb{R}^{n}$ if:
$\mathcal{M}$ is given locally as the zero set of a function $F\colon W \to \mathbb{R}^{n-k}$, where $W \subseteq \mathbb{R}^{n}$ is open in $\mathbb{R}^{n}$, $F$ is of class $C^{\infty}$, and so that $\mathcal{M} \cap W = \{\, x \in W \mid F(x)=0\, \}$. My definition also requires that the Jacobian $\mathcal{J}_{F}(x)$ has rank $n-k$ for every point $x \in \mathcal{M} \cap W$.
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The special linear group is the set $\mathrm{SL}(n,\mathbb{R}) = \{\, A \in M_{n \times n}(\mathbb{R}) \mid \det(A)=1\, \}$. I am trying to prove that this set is a smooth submanifold. From my reading online, I have understood that this submanifold has dimension $n^2-1$.
I'm going to make the identification $\mathbb{R}^{n^2}\cong M_{n \times n}(\mathbb{R})$. My thoughts on proving this are to define a function $G\colon M_{n \times n}(\mathbb{R}) \to \mathbb{R}$ given by $G(A)=\det(A)-1$.
We have obviously $M_{n \times n}(\mathbb{R}) \cap \mathrm{SL}(n,\mathbb{R}) = \mathrm{SL}(n,\mathbb{R})$ where $M_{n \times n}(\mathbb{R}) \cong \mathbb{R}^{n^2}$ is open, and we know that $G$ is smooth since $\det$ is smooth. Furthermore, $\mathrm{SL}(n,\mathbb{R}) = \{\, A \in M_{n \times n}(\mathbb{R}) \mid G(A)=0\, \}$.
Finally, the Jacobian of the function $\mathcal{J}_{G}$ is going to be a column vector of length $n^2$, and I will need to show that it has rank $1$. But since its a column vector, isn't this an immeadiate conclusion?
So to me it seems like this proof is finished. However, it seems a little too easy. Have I made any errors or omissions?
To show that the Jacobian really is of rank $1$ (i.e., not equal to the zero vector), I suggest you expand $\det A$ by minors along the first column, say: $$\det A = \sum_{i=1}^n (-1)^{i+1}a_{i,1}M_{i,1}.$$ Of course, $(-1)^{i+1}M_{i,1}$ is in fact an entry of the Jacobian. So the Jacobian can only be zero, if also $\det A=0$. But we are looking at the Jacobian for points $A\in SL_n(\Bbb R)$!