Differential geometry and vector fields exercise

[MathNugget]

Given M a smooth manifold, $M=f^{-1}(0)$ for a differentiable function $f: \mathbb{R}^n \rightarrow \mathbb{R}$.
I have to prove that a vector field X on $\mathbb{R}^n$ is tangent in x on M if and only if $X_x(f)=0$.

Suppose $X=\sum_{i=1}^nX_i\frac{\mathcal{d}}{dx_i}$, and $x=(x_1,...,x_n) \in M.$
I know that f(x)=0.
I think $X_x(f)=\sum_{i=1}^n X_i(x)\frac{df}{dx_i}(x)$.

I get stuck here, I am not creative enough to make up something. Do I say there's a neighbourhood around x (probably because $f^{-1}(0)$ is a smooth manifold), where f(x)=0, so the derivative would be 0?

 

[blamocur]

Do I say there's a neighbourhood around x (probably because f−1(0)f^{-1}(0)f−1(0) is a smooth manifold), where f(x)=0, so the derivative would be 0?

f(x) = 0 everywhere on M, isn't it?

I think Xx(f)=∑i=1nXi(x)dfdxi(x)X_x(f)=\sum_{i=1}^n X_i(x)\frac{df}{dx_i}(x)Xx(f)=∑i=1nXi(x)dxidf(x).

Is not that a definition of the derivative along a vector?
It would be helpful if you posted the definition of tangent vector you expected to use.
Since it is given that $M$ is a manifold in $\mathbb R^n$ there must exist a differentiable local map $\phi : \mathbb R^{m} \rightarrow \mathbb R^n$. Can you define tangent vectors in terms of $\phi$?

 

[MathNugget]

f(x) = 0 everywhere on M, isn't it?

sure.

Is not that a definition of the derivative along a vector?
It would be helpful if you posted the definition of tangent vector you expected to use.
Since it is given that $M$ is a manifold in $\mathbb R^n$ there must exist a differentiable local map $\phi : \mathbb R^{m} \rightarrow \mathbb R^n$. Can you define tangent vectors in terms of $\phi$?

Well, $X_x$ is the vector tangent to M in x, that is part of X.
I don't exactly know what $X_x(f)$ is; this example though shows what X(f) is.

Here's the definition I have:
A differential vector field on the open set $U \subset M$ is a differentiable function $X:U\rightarrow TM$ such that $\pi \circ X =Id_U$.

Is this what you're asking? I don't know how I'd define tangent vectors in terms of $\phi$... I thought there already are a bunch of local coordinates, $\frac{\partial}{\partial x^i}$... which I wrote with lower index, because I don't understand it that well, so I prefer writing all index down...

 

[mario99]

I don't exactly know what $X_x(f)$ is

I am pretty sure that $X_x(f)$ means that the derivative of the function $f$ with respect to the vector field $X$ at $x$.

 

[MathNugget]

I am pretty sure that $X_x(f)$ means that the derivative of the function $f$ with respect to the vector field $X$ at $x$.

You are right.


I know how to write the points in terms of $\phi$... we consider a local map $(U, \phi)$, and $f \circ \phi^{-1}: \phi(U)\subset \mathbb{R}^n \rightarrow \mathbb{R}$...
But I dont know how I'd write the tangent vectors, using $\phi$.

Maybe $d_x(\phi^{-1})?$

 

[MathNugget]

I believe now I own the answer:
for a map $(U, \phi)$,
I look at $d(f \circ \phi^{-1})_x(v)=0$? Basing my answer on this. Not totally related, but I think maybe it does the trick?

 

[blamocur]

I believe now I own the answer:
for a map $(U, \phi)$,
I look at $d(f \circ \phi^{-1})_x(v)=0$? Basing my answer on this. Not totally related, but I think maybe it does the trick?

Looks to me like the right way to go. $\phi^{-1}$ is a parametric representation of M and should allow one to reduce the problem to that on $\mathbb R^{n-1}$.

 

[MathNugget]

I'll try to finish this:

Now I know that $d(f \circ \phi^{-1})_x(v)=0, \forall v \in X$.
I suppose this would prove $v \in (f \circ \phi^{-1})(U), \forall (U, \phi), \forall v \in X$, so $X \subset M$.

I have no idea how to get the conclusion.

I think $X_x(f)=\sum_{i=1}^n X_i(x)\frac{df}{dx_i}(x)$.

Maybe I prove that along each coordinate $v=\frac{d}{dx_i}$, $d(f \circ \phi^{-1})_x(\frac{d}{dx_i})=0, \forall v \in X$. So $f \circ \phi^{-1}$ is constant along each vector of the basis of X, so it's overall constant, and then $X_x(f)=0$ indeed...
I can also use that f and phi are bijections, so then f is constant too. Although I think I am making a mistake...or M would be just a single point, somehow.

 

[blamocur]

Now I know that d(f∘ϕ−1)x(v)=0,∀v∈Xd(f \circ \phi^{-1})_x(v)=0, \forall v \in Xd(f∘ϕ−1)x(v)=0,∀v∈X.

When you say $d(...)_\mathbf{x}(v) = 0 \; \forall v\in X$ wouldn't you have $v = X_x$ ? What other $v$'s are there at $x$ ?

 

[blamocur]

ds

Maybe I prove...

I doubt you'll get far with timid attempts to guess correct answer. A better strategy would be to post some solution and let others review it. I'll try doing this myself, but I am very rusty on diff. geometry, and I don't have any textbooks, so my sketch of a proof might have holes and errors. For that reason I've decided to break the rules and not wait for mandatory week after the op and post it now. Please feel free to question and ask questions about my write-up.

Let $\phi : U_x \rightarrow \mathbb R^{n-1}$ be a local map at $x$ and $\theta$ its inverse (a.k.a. parametrization of $M$):
$$\theta : V\subset \mathbb R^{n-1} \rightarrow U\subset M \subset \mathbb R^n$$ or
$$\theta : V\subset \mathbb R^{n-1} \rightarrow \mathbb R^n$$Functions $\theta$ and $f$ have derivatives $\nabla\theta$ and $\nabla f$ respectively, where $\nabla \theta$ is a $n \times n-1$ matrix, a.k.a Jacobian, and $\nabla f$, a.k.a. gradient, is a $1\times n$ matrix, a.k.a. row vector. Note: it can be shown that $\nabla \theta$ and $\nabla f$ have ranks $n-1$ and $1$ respectively.

Since $f\circ \theta = 0$ we have $\nabla f \times \nabla \theta = 0$.

Lemma: vector $X\in \mathbb R^n$ is tangent to $M$ if and only if $X = \nabla \theta\times Y$ for some $Y \in \mathbb R^{n-1}$.
The proof of the lemma is left to the reader , but here is the rest of the proof in both directions ("if and only if"):

1. If $X\in \mathbb R^n$ is a tangent vector to M then $$X_x(f) = \nabla f \times X = \nabla f\times \nabla \theta \times Y= 0$$
2. If for some $X\in\mathbb R^n$ we have $\nabla f \times X = 0$ then consider an $n\times n$ matrix $A = (\nabla \theta, X)$, i.e., $\nabla \theta$ with $X$ added as $n$-th column. We now have $\nabla f \times A = 0$, which means that $\det A=0$ and thus the columns of $A$ are linearly dependent, i.e. $X$ belongs to the subspace formed by other columns of $A$, which are tangent vectors to $M$.

 

[MathNugget]

ds

I doubt you'll get far with timid attempts to guess correct answer. A better strategy would be to post some solution and let others review it. I'll try doing this myself, but I am very rusty on diff. geometry, and I don't have any textbooks, so my sketch of a proof might have holes and errors. For that reason I've decided to break the rules and not wait for mandatory week after the op and post it now. Please feel free to question and ask questions about my write-up.

Let $\phi : U_x \rightarrow \mathbb R^{n-1}$ be a local map at $x$ and $\theta$ its inverse (a.k.a. parametrization of $M$):
$$\theta : V\subset \mathbb R^{n-1} \rightarrow U\subset M \subset \mathbb R^n$$ or
$$\theta : V\subset \mathbb R^{n-1} \rightarrow \mathbb R^n$$Functions $\theta$ and $f$ have derivatives $\nabla\theta$ and $\nabla f$ respectively, where $\nabla \theta$ is a $n \times n-1$ matrix, a.k.a Jacobian, and $\nabla f$, a.k.a. gradient, is a $1\times n$ matrix, a.k.a. row vector. Note: it can be shown that $\nabla \theta$ and $\nabla f$ have ranks $n-1$ and $1$ respectively.

Since $f\circ \theta = 0$ we have $\nabla f \times \nabla \theta = 0$.

Lemma: vector $X\in \mathbb R^n$ is tangent to $M$ if and only if $X = \nabla \theta\times Y$ for some $Y \in \mathbb R^{n-1}$.
The proof of the lemma is left to the reader , but here is the rest of the proof in both directions ("if and only if"):

1. If $X\in \mathbb R^n$ is a tangent vector to M then $$X_x(f) = \nabla f \times X = \nabla f\times \nabla \theta \times Y= 0$$
2. If for some $X\in\mathbb R^n$ we have $\nabla f \times X = 0$ then consider an $n\times n$ matrix $A = (\nabla \theta, X)$, i.e., $\nabla \theta$ with $X$ added as $n$-th column. We now have $\nabla f \times A = 0$, which means that $\det A=0$ and thus the columns of $A$ are linearly dependent, i.e. $X$ belongs to the subspace formed by other columns of $A$, which are tangent vectors to $M$.

Thank you.

The timid attempt was *kind of* my idea of proof. It's all I could muster . Frankly, I have no idea what a complete proof is. I usually look up the concepts involved in the exercise (here, differential ecuations on a manifold, vector fieldon manifold etc), make up a story as to how the premises lead to the conclusion, and hope for the best.