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.
Differential Geometry – Show That SL(n, R) is a (n^2 -1) Smooth Submanifold
differential-geometryfunctional-analysismanifoldssmooth-manifolds
Best Answer
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 $\det$.