TITLE: $SL(n)$ is a differentiable manifold
QUESTION [9 upvotes]: Prove that $SL(n)=\{A\in \Bbb{R}^{n\times n}:\det(A)=1\}$ is a differentiable submanifold.

The determinant function is smooth since it's a polynomial, and we have $\det^{-1}(1)=SL(n)$. So it suffices to prove that $d \det(A)$ has constant nonzero rank for all $A\in SL(n)$.
By the Laplace's formula, we can write $$\det A = \sum_{j=1}^n (-1)^{i+j} a_{ij} \det (A^{*})$$ where $A^*$ is $A$ with the i-th row and j-th column removed. Then $$\frac{\partial}{\partial a_{ij}} \det A = (-1)^{i+j}\det(A^*).$$
So we need to show that $$\left( (-1)^{i+j}\det(A^*) \right)_{ij}$$ has constant nonzero rank.
How can we proceed?

REPLY [2 votes]: Since $SL(n) = \det^{-1}(1)$ and the determinant $\det : {\mathbb R}^{n^2} \to {\mathbb R}$ is smooth, you just need to show that the gradient $\nabla \det (A)$ is never zero at a unimodular matrix $A$ (ie, $A \in SL(n)$).
As you computed, the $n^2$ entries of $\nabla \det (A)$ are precisely:
$$
\left(\nabla \det (A)\right)_{i,j} = (-1)^{i+j} \det(A^*_{i,j})
$$
Here, $A^*_{i,j}$ is the $(n-1)\times (n-1)$-matrix obtained by deleting row $i$ and column $j$. So, denoting $C_{i,j}:=(-1)^{i+j} \det(A^*_{i,j})$, the matrix $C$ is precisely what is called the cofactor (or "adjugate") matrix of $A$: https://en.wikipedia.org/wiki/Adjugate_matrix, which verifies:
$$
C = (\det A) A^{-1}
$$
This means that $C$ is also unimodular ($\det C=1$). Hence, $\nabla \det (A)$ (with entries $C_{i,j}$) has, at least, one non-zero entry, and can not be the zero vector.