This is a follow-up to a question I asked previously.
Let ${F_i(\vec{x}),...,F_n(\vec{x})}$, $\vec{x} \in \mathbb{R}^n$ be a set of hypersurfaces. I'm particularly interested in the case where $F_i$ is a polynomial. Assume all $F_i$ are defined such that $F_i(\vec{x}) = 0 \forall \vec{x} \in \mathbb{R}^n$. Therefore, the $F_i$ are the level sets corresponding to the level 0 for some $n+1$ dimensional hypersurfaces. I would like to implicitly define their intersection.
If I am looking for a curve that lies entirely within the intersection, then it is known that such a curve will be mutually perpendicular to the gradient of each function. Under certain conditions, this curve can be defined implicitly as follows:
Let $M = [\bigtriangledown F_1, ..., \bigtriangledown F_n]^T$ and let $\vec{v} \in Null(M)$. Define the curve implicitly to be:
$$\dot{\xi} = C\vec{v(t)}, \xi(0) = \xi_0$$
Where $C$ is an arbitrary constant.
Now, can I do something similar to define the hypersurface of intersection implicitly? Something like a differential equation where I integrate the entire null space? I apologize for the lack of rigour in that question, but I'm not 100% sure how to approach it.
Let us define the map $F : \mathbb{R}^n \to \mathbb{R}^m$ so that $x \mapsto F(x) = \big(F_1(x), F_2(x), ..., F_m(x)\big)$. You have written that $m=n$, but I would not assume that a priori and for me $m \leq n$. Assume these $F_i(x)$ for $i=1,..., m$ are polynomials in $x \in \mathbb{R}^n$ with real coefficients, i.e. $F_i \in \mathbb{R}[x_1,...,x_n]$ where $i=1,...,m$. Then, the differential of the map, aka the tangent map to $F$, is $$dF_x = [\nabla F_1(x)^T,..., \nabla F_m(x)^T]^T$$ where $\nabla F_i(x)$ is the gradient of $F_i$ written as a row-vector, so $\nabla F_i(x)^T$ is its transposed (it is a column-vector). Observe $dF_x$ is an $m \times n$ matrix with polynomial coefficients and so it defines the tangent map $$dF_x : T_x\mathbb{R}^n \to T_x\mathbb{R}^m$$ By convention, all tangent vectors are column-vectors and all dual vectors are row-vectors. Assume that $\text{rank}\big(dF_{x_0}\big) = m$, i.e. the rank of $dF_x$ at a point $x_0$ is maximal. Then, generically, the rank of $dF_x$ is maximal for almost all $x \in \mathbb{R}^n$. In that case, whenever $m=n$ the inverse mapping theorem (aka the inverse function theorem) applies and then $F$ is a local diffeomorphism and so the solutions to $F(x)=0$ are going to be isolated points. So let us assume that $m < n$.
Now if the rank of $dF_x$ is not $m$ generically, but less than $m$, then you need to run some sort of reduction in the ideal $\langle F_1(x),..., F_m(x) \rangle$ which is the polynomial ideal generated by the polynomials $F_i$. In other words, you need to find a minimal set of generator $\langle \tilde{F}_1(x),..., \tilde{F}_k(x) \rangle = \langle F_1(x),..., F_m(x) \rangle$ where $k \leq m$. These minimal set of generators is an independent set of polynomials whose matrix of gradients will have rank $k$ generically. Finding the set of of independent generators can be done by running some sort of Groebner basis method or something else of that sort. I advise you to look that up carefully if you need it as I am a bit rusty on the subject. But let us assume that we have done that beforehand, that $k=m$ and that the $F_1, ... ,F_m$ is an independent set of polynomials so that the rank of $dF_x$ is generically $m$.
Next, consider the system $$dF_x (v) = 0 \,\,\, \text{ for } \,\,\, v \in T_x\mathbb{R}^n$$ Now, the tangent field of $n-m$ planes $$\ker(dF)_x = \{v \in T_x\mathbb{R}^n \, | \, dF_x (v) = 0 \}$$ is an integrable distribution of the tangent spaces to the leaves of the foliation $$F^{-1}(y) = \{x \in \mathbb{R}^n \, | \, F(x) = y \}$$ Observe you have a system of the form $dF_x(v) = 0$ where the matrix $dF_x$ is a matrix with polynomial coefficients. Thus you can solve the system as a system with polynomial coefficients, i.e. $dF_x$ is an $m \times n$ matrix with coefficients (entries, or elements) in the ring of polynomials $\mathbb{R}[x]$. This ring is a subring of the field of rational functions $\mathbb{R}(x) = \mathbb{R}(x_1,...,x_n)$, where the elements of the latter look like $\frac{P(x)}{Q(x)}$ with $P, Q \in \mathbb{R}[x]$. Now treat the linear system $dF_x(v_x) = 0$ as a linear system in the field $\mathbb{R}(x) $, since a lot of the linear algebra works in an arbitrary field, you can find a set of $n-m\,\,\ $ $\mathbb{R}(x)$-linearly independent vector-solutions $X_1(x), X_2(x), ... X_{n-m}(x)$, so that each $X_i$ is a vector column of dimension $n$ with rational coefficients (entries that are rational functions of $n$ variables). So if we define $X(x) = [X_1(x),..,X_{n-m}(x)]$, an $n \times (n-m)$ matrix with rational coefficients from $\mathbb{R}(x)$, then $$dF_x \, X = 0$$ Just like in the case of equations with integer coefficients, solved in the field of rational numbers, you can even find an appropriate polynomial such that if you multiply all the $X_i$ by it, you end up with $\mathbb{R}(x)$ linearly independent vector-columns $X_1(x), ... , X_{n-m}(x)$ with polynomial coefficients. Thus you have ended up obtaining a set of polynomial vector fields $X_1(x), ... , X_{n-m}(x)$ that span the distribution $\ker(dF_x)$, i.e. they span the tangent spaces of the leaves of the foliation $F^{-1}(y)$.
These guys $X_1(x), ... , X_{n-m}(x)$ are exactly the analog of what you are referring to in your original posting. If you take the phase flows $x(t_i) = \Phi_i^{t_i}(x)$ of the vector fields $X_i(x)$, then $$\frac{d}{dt_i} \Phi_i^{t_i}(x) = X_i\big(\Phi_i^{t_i}(x)\big), \,\,\,\,\, \Phi_i^{0}(x) = x$$ and then the map $$\Psi_{x_0} \, : \, (t_1, t_2, ..., t_{n-m}) \, \mapsto \,\Phi_1^{t_1} \circ \Phi_2^{t_2} \circ ... \circ \Phi_{n-m}^{t_{n-m}}(x_0)$$ is a local chart for the leaf $M_{x_0} = \{x \in \mathbb{R}^n \,\,|\,\,F(x) = F(x_0)\}$ passing through the point $x_0$.
Edit. Assume you are given a smooth vector field $X(x)$ in $\mathbb{R}^n$. Then component-wise, it can be written like a vector of $n$ smooth functions (in your case, even polynomial functions), depending on the $n$ vector $x=(x_1,...,x_n)$ $$ X(x) = \begin{bmatrix} \, X^1(x_1, ..., x_n)\\ X^2(x_1, ..., x_n)\\ \vdots\\ X^n(x_1, ..., x_n) \, \end{bmatrix}$$ Each vector field has a family of curves that are tangent to the vector field and through each point in space, there is exactly one such curve passing through that point, which means no two such curves intersect. These are the so called integral curves, also called phase curves, of the vector field $X(x)$. They are obtained as the solutions to the system of ODEs \begin{align} \frac{dx_1}{dt} &= X^1(x_1, ..., x_n)\\ \frac{dx_2}{dt} &= X^2(x_1, ..., x_n)\\ \vdots\\ \frac{dx_n}{dt} &= X^n(x_1, ..., x_n) \end{align} which in vector form is written as $$\frac{dx}{dt} = X(x)$$ Now pick an arbitrary point $x_0$ in $\mathbb{R}^n$ and try to solve the initial value problem \begin{align} \frac{dx}{dt} &= X(x)\\ x(0) &= x_0 \end{align} which means you are trying to find a curve $x=x(t)$ in $\mathbb{R}^n$ that is tangent to the vector field $X(s)$ and passes through the point $x_0$, i.e. $$\frac{dx}{dt}(t) = X\big(x(t)\big) \,\,\, \text{ for all } \,\, t \in (-\varepsilon, \varepsilon) $$ and $x(0) = x_0$. By the classical theorems of ODEs, there exists exactly one such curve (existence and uniqueness of solutions of systems of smooth ODEs). Now, denote by $$x(t) = \Phi^t(x_0 )$$ for $t \in (-\varepsilon, \varepsilon)$ this curve, a notation which says that the curve depends on the parameter $t$ and on the initial condition $x_0$. When I change $t$, I move along the curve, but when I change $x_0$ I move from one curve tangent to the vector field $X$ to another. So technically we have a map $$\Phi \, : \, (-\varepsilon, \varepsilon) \times \mathbb{R}^n \, \to \, \mathbb{R}^n$$ By another result of ODE theory, this map is a smooth map (in your case it is even real analytic, because your vector field is polynomial, which means real analytic). If you fix the parameter $t$, you get a diffeomorphism (bijection, smooth in both directions) $$\Phi^t \, : \, \mathbb{R}^n \, \to \, \mathbb{R}^n$$ It is a bijection because of the existence and uniqueness result for ODEs. For a fixed $x_0$ what you have is that $$\frac{d}{dt} \Phi^t(x_0) = X\big(\Phi^t(x_0)\big)$$ $$\Phi^0(x_0) = x_0$$ This map $\Phi$ is called the phase flow map (or simply the phase flow) of the vector field. Again due to the existence and uniqueness of solutions of smooth ODEs, for any $x \in \mathbb{R}^n$ and any $t, s \in (-\varepsilon, \varepsilon)$ the phase flow has the properties $$\Phi^t\Big(\Phi^s(x)\Big) = \Phi^{t+s}(x)$$ $$\Phi^0(x) = x$$ In your case you have a (family of) $n-m$ hypersurface $F^{-1}(y_0) = \{x \in \mathbb{R}^n \, | \, F(x) = y_0\}$ where $y_0 \in \mathbb{R}^m$, which basically says that $x = (x_1,...,x_n) \in F^{-1}(y_0)$ if and only if \begin{align} &F_1(x_1, ..., x_n) = y_{1,0}\\ &F_2(x_1, ..., x_n) = y_{2,0}\\ & \,\,\,\, \vdots \\ &F_m(x_1, ..., x_n) = y_{m,0}\\ \end{align} Moreover, you have $n-m$ such vector fields $X_1, ..., X_{n-m}$ all tangent to $F^{-1}(y_0)$. This means that you have $n-m$ phase flow maps $\Phi_1^{t_1}, ...., \Phi_{n-m}^{t_{n-m}}$ for depending on $n-m$ different independent parameters $(t_1, t_2,..., t_{n-m}) \in \mathbb{R}^{n-m}$. The fact that the vector fields are tangent is equivalent to the fact that for each $j = 1,...,n-m$ $$F\big(\Phi_j^{t_j}(x)\big) = F(x)$$ or in coordinates $$F_i\big(\Phi_j^{t_j}(x)\big) = F_i(x)$$ for $i=1,..., m$. If you compose the maps $$\Phi^{t_j}_j \, : \, \mathbb{R}^n \, \to \, \mathbb{R}^n$$ for $j=1,...,n-m$ you get again a map $$\Psi(t_1,...,t_{n-m},x) = \Phi^{t_1}_1\Big(...\Phi^{t_{n-m}}_{n-m}(x)\Big) = \Phi^{t_1}_1\circ ... \circ\Phi^{t_{n-m}}_{n-m}(x)$$ where $$\Psi \, : \, \mathbb{R}^{n-m} \times \mathbb{R}^n \, \to \, \mathbb{R}^n$$ with the property that $$F\Big(\Psi(t_1,...,t_{n-m},x)\Big) = F\Big(\Phi^{t_1}_1\big(...\Phi^{t_{n-m}}_{n-m}(x)\big)\Big) = F\Big(\Phi^{t_1}_1\circ ... \circ\Phi^{t_{n-m}}_{n-m}(x)\Big) = F(x)$$ so if you fix a point $x_0 \in \mathbb{R}^n$ on the surface $F(x) = y_0$, meaning $F(x_0) = y_0$, you obtain a map $$\Psi_{x_0}(t_1,...,t_{n-m}) = \Psi(t_1,...,t_{n-m},x_0)= \Phi^{t_1}_1\Big(...\Phi^{t_{n-m}}_{n-m}(x_0)\Big) = \Phi^{t_1}_1\circ ... \circ\Phi^{t_{n-m}}_{n-m}(x_0)$$ where $$\Psi_{x_0} \, : \, \mathbb{R}^{n-m} \, \to \, \mathbb{R}^n$$ which parametrizes the $n-m$ hypersurface $F(x) = y_0$.