13
$\begingroup$

Given a function $f: \mathbb{R}^n \rightarrow \mathbb{R}^m $ which is smooth, show that $$\operatorname{graph}(f) = \{(x,f(x)) \in \mathbb{R}^{n+m} : x \in \mathbb{R}^n\}$$ is a smooth submanifold of $\mathbb{R}^{n+m}$.

I'm honestly completely unsure of where or how to begin this problem. I am interested in definitions and perhaps hints that can lead me in the right direction.

$\endgroup$
21
$\begingroup$

This is an application of the Implicit Function Theorem. You have a function $f : \mathbb{R}^n \to \mathbb{R}^m$ and you construct the graph given by: $\{ (x,y) \in \mathbb{R}^n \times \mathbb{R}^m : y = f(x) \}.$ Let me define a new map, say, $G : \mathbb{R}^{n+m} \to \mathbb{R}^m$ given by $G : (x,y) \mapsto y-f(x).$ I have defined this the way I have so that the graph of $f$ is the zero-level set of $G$, i.e. the graph of $f$ is the set of $(x,y) \in \mathbb{R}^n \times \mathbb{R}^m$ such that $G(x,y) = 0.$

In brutal detail this map is really:

$$G : (x_1,\ldots,x_n,y_1,\ldots,y_m) \mapsto (y_1-f_1(x_1,\ldots,x_n),\ldots,y_m-f_m(x_1,\ldots,x_n)) \, .$$

We need to calculate the Jacobian Matrix of $G$. A quick calculation will show you that:

$$ J_G = \left[\begin{array}{c|c} -J_f & I_m \end{array}\right] ,$$

where $J_f$ is the $m \times n$ Jacobian matrix of $f$ and $I_m$ is the $m \times m$ identity matrix. The matrix $J_G$ is an $m \times (m+n)$ matrix.

To be able to apply the IFT, we need to show that $0$ is a regular value of $G$. (After all, the graph of $f$ is $G^{-1}(0).$) We can do this by showing that none of the critical points get sent to 0 by $G$. Notice that $G$ has no critical points because $J_G$ always has maximal rank, i.e. $m$. This is clearly true since the identity matrix $I_m$ has rank $m$.

It follows that the graph of $f$ is a smooth, parametrisable $(n+m)-m=n$ dimensional manifold in a neighbourhood of each of its points.

$\endgroup$
  • $\begingroup$ I'm having trouble understanding your definition of G. There are multiple functions $f_1,\ldots , f_m$ because of the values $y_1,\ldots,y_m$ to be equal to those, and if that is the case, wouldn't each of the coordinates of the range be equal to 0? $\endgroup$ – ackshooairy Nov 26 '12 at 19:55
  • $\begingroup$ The function $f$ goes from $\mathbb{R}^n$ to $\mathbb{R}^m$, i.e. you give it $n$ numbers and it gives you $m$ numbers back. The numbers you give it are $x_1,\ldots,x_n$ while the numbers it gives you are $f_1,\ldots,f_m.$ In longhand: $$f(x) = f(x_1,\ldots,x_n) = (f_1(x_1,\ldots,x_n), \ldots, f_m(x_1,\ldots,x_n)).$$ Does this help? $\endgroup$ – Fly by Night Nov 26 '12 at 20:12
  • 1
    $\begingroup$ The the graph of $f$ is given by the equation $G(x,y)=0$. As you said: the coordinates of the range of the graph will all be zero. This is what the IFT does: It takes a system of equations and tells you if you get a smooth, parametrisable manifold as the solution space. The trick is to find a system of equations whose solution gives you what you're interested in. $\endgroup$ – Fly by Night Nov 26 '12 at 20:23
  • 2
    $\begingroup$ But applying the IFT surely is an overkill... You have a global chart $\mathbb{R}^n \to \operatorname{graph}(f) \subseteq \mathbb{R}^{n+m}$ given by $x \mapsto (x,f(x))$ which is a smooth bijection onto its image and its inverse is given by the restriction of the projection $(x,y) \mapsto x$ to the graph. $\endgroup$ – Martin Apr 23 '13 at 1:01
  • 1
    $\begingroup$ Actually, applying the IFT seems to be circular in this case. The proof of the fact that the pre-image of a regular value is a submanifold proceeds by showing that you get charts by writing the preimage locally as graph of a function. $\endgroup$ – Martin Apr 23 '13 at 1:10
3
$\begingroup$

The map $\mathbb R^n\mapsto \mathbb R^{n+m}$ given by $t\mapsto (t, f(t))$ has the Jacobi matrix $\begin{pmatrix}I_n\\f'(t)\end{pmatrix}$, which has a full rank $n$ for all $t$ (because of the identity submatrix). This means that its value range is a manifold. Is there anything unclear about it?

How is this a proof that it is a manifold?

A manifold of rank $n$ is such set $X$ that for each $x\in X$ there exists a neighborhood $H_x\subset X$ such that $H_x$ is isomorphic to an open subset of $\mathbb R^n$. In this case, the whole $X=graph(f)$ is isomophic to $\mathbb R^n$. The definition of a manifold differs, often it is required for the isomophism to be diffeomophism, which is true here as well.

Think of it this way: A manifold $X$ of rank $2$ is something, in which: wherever someone makes a dot there by a pen, I can cut a piece of $X$ and say to this person: "See, my piece is almost like a piece of paper, it's just a bit curvy.

The definition of manifold might seems strage here because here you can take the neighborhood as the whole $X$. This is not always the case: A sphere is a manifold as well, but a whole sphere is not isomorphic to $\mathbb R^2$, you have to take only some cut-out of it.

$\endgroup$
  • $\begingroup$ Let me try to reiterate. This matrix comes from the derivative of my graph function where $I_n$ is the derivative of $t$ and $f'(t)$ is the derivative of $f(t)$. That I believe I understand. I'm not sure about how rank $n$ means that the range is a manifold or even exactly what it is to be a submanifold. Could you please elaborate a little on that? $\endgroup$ – ackshooairy Nov 26 '12 at 19:38
  • $\begingroup$ It's been long since I had Calculus, so I'm not sure I'll give the exact explanation, but I'll try. $\endgroup$ – yo' Nov 26 '12 at 19:40
  • $\begingroup$ Your further explanation as well as the other answer were very helpful. +1 Thank you. $\endgroup$ – ackshooairy Nov 26 '12 at 20:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.