449: Vectors Field on Restricted Tangent Vectors Bundle Is C^\infty iff Operation Result on Any C^\infty Function on Super Manifold Is C^\infty on Regular Submanifold

<The previous article in this series | The table of contents of this series | The next article in this series>

A description/proof of that vectors field on restricted tangent vectors bundle is $C^{\mathrm{\infty }}$ iff operation result on any $C^{\mathrm{\infty }}$ function on super manifold is $C^{\mathrm{\infty }}$ on regular submanifold

---

Topics

About: $C^{\mathrm{\infty }}$ manifold

---

The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Description
• 2: Proof
• 3: Note

---

Starting Context

• The reader knows a definition of restricted vectors bundle.
• The reader knows a definition of $C^{\mathrm{\infty }}$ section.
• The reader knows a definition of $C^{\mathrm{\infty }}$ vectors field.
• The reader admits the proposition that for any $C^{\mathrm{\infty }}$ function on any point open neighborhood of any $C^{\mathrm{\infty }}$ manifold, there exists a $C^{\mathrm{\infty }}$ function on the whole manifold that equals the original function on a possibly smaller neighborhood of the point.

---

Target Context

• The reader will have a description and a proof of the proposition that any vectors field on any restricted tangent vectors bundle is $C^{\mathrm{\infty }}$ iff the operation result on any $C^{\mathrm{\infty }}$ function on the super manifold is $C^{\mathrm{\infty }}$ on the regular submanifold.

---

Orientation

There is a list of definitions discussed so far in this site.

There is a list of propositions discussed so far in this site.

---

Main Body

---

1: Description

For any $C^{\mathrm{\infty }}$ manifold, $M^{\prime }$, any regular submanifold, $M\subseteq M^{\prime }$, and any vectors field, $V:M\to T{M}^{\prime }{|}_M$, $V$ is $C^{\mathrm{\infty }}$ if and only if for any $C^{\mathrm{\infty }}$ function, $f\in C^{\mathrm{\infty }}(M^{\prime })$, $Vf$ is $C^{\mathrm{\infty }}$ on $M$.

---

2: Proof

For any point, $p\in M$, there is an adopted chart, $(U_p\subseteq M^{\prime },\varphi )$, of $p$. The corresponding adopting chart is $(U_p\cap M\subseteq M,\pi \circ \varphi {|}_{U_p\cap M})$ where $\pi$ is the projection into the first $n$ components where $n$ is the dimension of $M$.

Let us suppose that $Vf$ is a $C^{\mathrm{\infty }}$ function. Each coordinate function, $x^j:U_p\to \mathbb{R}$, is a $C^{\mathrm{\infty }}$ function on $U_p$, and there is a $C^{\mathrm{\infty }}$ function, $\widetilde{x^j}:M^{\prime }\to \mathbb{R}$, on $M^{\prime }$ that equals $x^j$ on a possibly smaller open neighborhood, $U_p^{\prime }\subseteq U_p$, of $p$, by the proposition that for any $C^{\mathrm{\infty }}$ function on any point open neighborhood of any $C^{\mathrm{\infty }}$ manifold, there exists a $C^{\mathrm{\infty }}$ function on the whole manifold that equals the original function on a possibly smaller neighborhood of the point. On $U_p^{\prime }\cap M$, $V\widetilde{x^j}=V^i\frac{\partial \widetilde{x^j}}{\partial x^i}=V^j$, $C^{\mathrm{\infty }}$ on $U_p^{\prime }\cap M$ by the supposition, and $V$ is $C^{\mathrm{\infty }}$ on $U_p^{\prime }\cap M$. As $V$ is $C^{\mathrm{\infty }}$ on a neighborhood of any point on $M$, $V$ is $C^{\mathrm{\infty }}$ on $M$.

Let us suppose that $V$ is $C^{\mathrm{\infty }}$. $V=V^j\frac{\partial }{\partial x^j}$ on $U_p\cap M$ where $V^j$ is a $C^{\mathrm{\infty }}$ function on $U_p\cap M$. $Vf=V^j\frac{\partial f}{\partial x^j}$ is a $C^{\mathrm{\infty }}$ function on $U_p\cap M$: $\frac{\partial f}{\partial x^j}$ is a $C^{\mathrm{\infty }}$ function on $U_p$, but on $U_p\cap M$, $(\frac{\partial f}{\partial x^j})(x^1,x^2,...,x^n,x^{n+1},x^{n+2},...,x^{n+m})=(\frac{\partial f}{\partial x^j})(x^1,x^2,...,x^n,0,0,...,0)$, which is $C^{\mathrm{\infty }}$ on $U_p\cap M$. As $Vf$ is $C^{\mathrm{\infty }}$ on a neighborhood of any point on $M$, it is $C^{\mathrm{\infty }}$ on $M$.

---

3: Note

$V$ does not operate on any function on $M$, because $V(p)\in Tp{M}^{\prime }$ not in $TpM$.

---

References

<The previous article in this series | The table of contents of this series | The next article in this series>