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(Vishik's Normal Form) Behavior of a vector field near the boundary of a manifold

Discussion in 'Mathematics' started by Matheus Manzatto, Oct 8, 2018.

  1. I'm trying to demonstrate a special case of Vishik's Normal Form.

    Consider $T^2 := \frac{1}{\sqrt{2}}\cdot\mathbb{T}^2 \subset \mathbb{S}^3$, let $h: \mathbb{\mathbb{S}^3}\to \mathbb{R}$ be the function $h(x_1,x_2,x_3,x_4) = x_1^2+x_2^2 -\frac{1}{2}$, and $\mathbb{S}^+ :=\{x \in \mathbb{S}^3; h(x)\geq 0\}$.

    We say that $X$ is a vector field on $\mathbb{S}^+$ if $$X:\mathbb{S}^+ \to \mathbb{R^4} $$ is a smooth function and $X(p)$ $\in$ $T_p\mathbb{S^3}$, $\forall p \in \mathbb{S}^+$.

    For our purposes we will always consider $X(p) \neq 0$, $\ \forall \ p \in T^2$.

    Notations: $Xh(p)= \nabla h(p) \cdot X(p)$ and $X^2 h(p)= \nabla Xh(p) \cdot X(p).$

    Definition: Let $X$ be a vector field on $\mathbb{S}^+$ and $p$ $\in$ $T^2$. If $Xh(p) = 0$ and $X^2 h(p) \neq 0$, $p$ is called a fold point.

    The theorem that I am trying to demonstrate is as follows:

    Theorem (Vishik's Normal Form) Let $X$ be a vector field on $\mathbb{S}^+$ such that, all the points on $T^2$ that satisfy $Xh(q) =0$ are fold points. Then, if $p$ $\in T^2$ is a fold point, there exists an open set $V_p \subset \mathbb{S}^3$ and a system of coordinates $(x_1, x_2, x_3)$ at $p$ defined in $V_p$ $(x_i(p) = 0, i = 1, 2, 3)$ such that $X|_{V_p}$ is a germ at $V_p\cap T^2$ of the vector field given by: $$\left\{\begin{array}{l} \dot{x}_1=x_2,\\ \dot{x}_2=1,\\ \dot{x}_3=0. \\ \end{array}\right.$$

    I do not have many ideas of how to prove this theorem and I do not know any good reference. Does anyone know how to prove this theorem or can give me some hints (or references)?

    My ideas

    Note that $h$ is a function such that $0$ is regular value, then using a coordinate system $\phi: V \subset \mathbb{R}^3 \to U_p \subset \mathbb{S}^3$ ($\phi(0) = p$), then fuction $h\circ \phi$ satisfies $\text{d}h\circ \phi_0 \neq 0$, so we can suppose that $\frac{\partial}{\partial x}h\circ \phi(0) \neq 0 \Rightarrow$ by local submersion theorem, there is a local coordinate system $\varphi: V' \subset \mathbb{R}^3\to U_p ' \subset \mathbb{S}^3 $ ($\varphi(0)=p$), such that $h\circ \varphi (x,y,z) = x$.

    Defing $Y(q) = D\varphi^{-1}(\varphi(q)) \cdot X(\varphi(q))$, we are able toto study the problem in an open neighborhood of $0$ in the topological space $\mathbb{H}^3 = \{(x,y,z); 0\leq x\}$, with some calculation is possible to show that if $f=h\circ \varphi$, then $Y f(q) = Y_1(q)$ (where $Y=(Y_1,Y_2,Y_3))$, implying by the definition of $X$, that $Yf(p) =0$ and $Y^2 f(p) = Y(p) \cdot \nabla Y_1(p) \neq 0$.

    The points such that $q \in \{0\}\times \mathbb{R}^2$ and $Yf(q) = 0$, form a manifold. To see this, note that $Yf(0,y,z) = Y_1(0,y,z)$, so $Yf(0,0,0) =0$ by hypotesis, and $0\neq Y^2 f(0,0,0) = Y(0,0,0) \cdot \nabla Y_1(0,0,0)$, once $Y(0,0,0)$ $\in$ $\{0\}\times \mathbb{R}^2$, we have that either $$\frac{\partial}{\partial y} Y_1(0,0,0) \neq 0\ \text{or} \ \frac{\partial}{\partial x} Y_1(0,0,0) \neq 0,$$ assuming without loss of generality $\frac{\partial}{\partial y} Y_1(0,0,0)\neq 0$, and using implicit function theorem, there exists $\tau: (-\varepsilon,\varepsilon)\to (-\delta,\delta)$, such that $Y(0,\tau(z),z) =0$. Using that $\Phi (y,z) = (y+\tau(z),z)$, is a local difeomorphism, we can (by changing the coordinates and and shrinking the definition domain of $Y$) assume that $Y = (Y_1,Y_2,Y_3)$, with $Y_1(0,0,z) =0$.

    I do not know very well how to proceed in the demonstration.

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