$SL(n)$ is a differentiable manifold
First, we need a theorem:
Suppose $U\subset \mathbb{R}^{n+m}$ is an open set and $f \colon U \to \mathbb{R}^m$ is a $\mathcal{C}^{\infty}$ map. Let $q \in \mathbb{R}^m$ and $M = f^{-1}(q)$. If $\mathrm{D}f(x)$ has rank $m$ for all $x \in M$, then $M$ is an $n$-dim. submanifold of $\mathbb{R}^{n+m}$.
Statement:
Let $SL_n(\mathbb{R}) = \{ A \in \mathrm{Mat}_{n \times n}\ \lvert \mathrm{det} A = 1\}$. Then:
- If $A \in SL_n(\mathbb{R})$, then $\mathrm{D \ det}(A)$ has rank $1$.
- $SL_n(\mathbb{R})$ is an $(n^2 -1) -$ manifold.
Proof:
Note, that $\mathrm{det} \colon \mathbb{R}^{n^2} \to \mathbb{R}$ is polynomial and so $\mathcal{C}^{\infty}$. To show $(1)$ it is only necessary to show that some directional derivative $\mathrm{D_{B}det}(A)$ is nonzero. We compute for $A \in SL_n(\mathbb{R})$,
$$ \mathrm{D_{A}det}(A) = \frac{d}{dt} \mathrm{det}(A + tA)\lvert_{t=0} = \frac{d}{dt} (1 + t)^n \mathrm{det}(A)\lvert_{t=0} = \frac{d}{dt}(1+t)^n \lvert_{t=0} = n. $$ Hence, $(1)$ is shown. To see $(2)$, we use the above theorem. The set $GL_{n}(\mathbb{R}) \subset \mathbb{R}^{n^2 -1} \times \mathbb{R}$ is an open set and $\mathrm{det}\colon GL_{n}(\mathbb{R}) \to \mathbb{R}$ is a $\mathcal{C}^{\infty}$- map. The set $\mathrm{det}^{-1}(1) = SL_{n}(\mathbb{R})$, and $\mathrm{D \ det}(A)$ has rank $1$ for each $A \in SL_n(\mathbb{R})$. Therefore, by the above theorem, $SL_{n}(\mathbb{R})$ is an $(n^2 - 1)$-submanifold.