This is for an optimization class, but I think what I am trying to prove is erring towards differential geometry, if that tag is inappropriate please remove it. My professor left this claim asserted but not proven, I would like to prove it. I found a lower dimensional case here on page 24 that I am trying to adapt.
Here is the background, and the problem.
Let $$g: \mathbb{R}^n \to \mathbb{R}^m,$$ $$g(\mathbf{x}) = [ g_1(\mathbf{x}), \dots , g_m(\mathbf{x})]^{T}$$ where $n \geq m$ and each $g_i$ is a $C^{1}$ map $\mathbb{R}^n \to \mathbb{R}$. Set $k$ to be the codim $n-m$. Define $S$ to be the zero locus of $g$ and for $\mathbf{x}_0 \in S$ define the tangent space $T_{\mathbf{x}_0}S$ to be the set of vectors $\mathbf{v}$ such that there exists a curve $\gamma(t) \subset S$ with $\gamma(0) = \mathbf{x}_0$ and $\gamma'(0) = \mathbf{v}$. Let $Dg(\mathbf{x}_0)$ denote the $m$ x $n$ Jacobian of $g$ at $\mathbf{x}_0$ and assume it has full rank. I would like to prove that
$$\mathrm{Nul}(Dg(\mathbf{x}_0)) \subseteq T_{\mathbf{x}_0} S.$$
Here is what I have so far (almost entirely adapted from the link provided above). Suppose $\mathbf{v} \in \mathrm{Nul}(Dg(\mathbf{x}_0))$, then in particular $\nabla g_i(\mathbf{x}_0) \cdot \mathbf{v} = \mathbf{0}$ and all of the $\nabla g_i(\mathbf{x}_0)$ are linearly independent. After a suitable reordering of the variables, we can apply the Implicit Function Theorem (exactly as stated on the wiki page here, reordering the variables so that $m$ x $m$ matrix is invertible, invoking the full rank of the Jacobian). Now denote the point $\mathbf{x}_0 = (x_1,\dots,x_k,y_1,\dots,y_m)$. Let $\mathbf{a} = (x_1,\dots,x_k)$ and $\mathbf{b} = (y_1, \dots, y_k)$. By invoking the IFT we have some open set $U \subset \mathbb{R}^k$, $\mathbf{a} \in U$, and the existence of a $C^{1}$ map $\phi \colon U \to \mathbb{R}^m$ such that $\phi(\mathbf{a}) = \mathbf{b}$ and $g(\mathbf{x},\phi(\mathbf{x}) = 0$ for all $\mathbf{x} \in U$.
I am attempting to use all of the above to start building a parameterization $\gamma(t)$, but this is where I am stuck. I would like to define $\gamma_i(t) = x_i + v_i t$ for $1 \leq i \leq k$. Since $U$ is open I should be able to choose $\epsilon$ sufficiently small so that, at least for these first $k$ coordinates $\gamma$ stays in $S$ for $t \in (-\epsilon, \epsilon)$. But I don't know how to finish the next $m$ entries of $\gamma$. My assumption is that when the parameterization works out, the derivative portion of the Implicit Function Theorem combined with my hypothesis will allow me to say $\gamma'(0) = \mathbf{v}$.
Thanks!
We have that $T_{x_0} S \subset \mathrm{Nul}(Dg(x_0))$. This follows since given $v \in T_{x_0}S$, we can consider a curve $\gamma \subset S$ such that $\gamma(0)=x_0$ and $\gamma'(0)=v$. Thus, by the chain rule, $$ (g \circ \gamma)'(0)=Dg(x_0) \cdot \gamma'(0)=Dg(x_0) \cdot v.$$ Since the left side is obviously zero as $g$ is constant along $\gamma$, it follows that $v$ is in the kernel of $Dg(x_0)$. But $T_{x_0}S$ has dimension $n-m$, and so thus $\mathrm{Nul}(Dg(x_0))$ by the rank-nullity theorem. Thus, both must coincide.