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I know that there is a very formal definition for "manifold" out there. However, I'm working within the confines of a first course in multivariable calculus. I was wondering if the following statement is true:

Let

$$x = f(q_1, q_2) \in \mathbb{R}$$ and $$y = g(q_1,q_2)\in \mathbb{R}$$

be functions such that:

  1. $q_i \in \mathbb{R}$ and

2.

$$J = \begin{bmatrix}\frac{\partial f}{\partial q_1} & \frac{\partial f}{\partial q_2}\\\frac{\partial g}{\partial q_1} & \frac{\partial g}{\partial q_2}\end{bmatrix}$$

is invertible everywhere. Then these two equations describe a manifold.

Is that correct? If so, I suspect that it describes an extremely limited subset of all manifolds, but I'll take what I can get. If it is not correct, can someone please let me know why?

Michael Stachowsky
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    Do you mean that the graph $F:=(f,g)$ is a surface if $J(F) \neq 0 $? If so, then the inverse function gives you a local diffeomorphism between an open ball $B((x,y);r)$ about $(x,y)$ and $f(x,y)$, which is a patch for a surface. Then you need to work on the overlaps, and show that they are smooth, or at least continuous if you just want a topological manifold. – MSIS Jul 04 '17 at 19:49
  • I see. If I require that $f$, $g$ be invertible, I assume that I then don't need the Jacobian condition, is that right? – Michael Stachowsky Jul 04 '17 at 19:53
  • I don't know if the two are equivalent, or if this is sufficient; let me think. Do you know the result that the level set of a non-critical point is a manifold? – MSIS Jul 04 '17 at 19:55
  • The general result is that if a is a regular value of F, then $F^{-1}(a)$ is a submanifold of the ambient space, by the inverse/implicit function theorem. – MSIS Jul 04 '17 at 20:00
  • I see. So then, if I let $x - f(q_1,q_2) = 0$ and $y - g(q_1,q_2) = 0$ and assumed that the inverses of these new functions exist, then I have a manifold? – Michael Stachowsky Jul 04 '17 at 20:08
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    In each case, yes, if 0 is a critical value, then each map ( the set of points describing the inverse) separately defines a manifold. What I am not sure about is for non-scalar maps, i.e., vector valued maps ( meaning $\mathbb R^k; k=2 $ or higher), this is true if J(f) is invertible, not sure it is enough for invertibility of functions alone. You can see, e.g., for $F(x,y)=x^2+y^2-1 $, computing its differential, that 0 is not a critical value. This implies, by the theorem, that $F^{-1}(0)$ is a manifold , the circle. I am just not sure if this is true for maps into $\mathbb R^2$ or higher. – MSIS Jul 04 '17 at 20:14
  • Ah, I see. So both f and g separately would define manifolds with this definition, but it may be the case that the points (x,y) do not lie on a manifold? However, isn't it true that the Cartesian product of two manifolds is a manifold? – Michael Stachowsky Jul 04 '17 at 20:21
  • Yes, but the set {$(f(q_1,q_2), g(q_1, q_2))$} is not the same as the Cartesian product of the two functions. – MSIS Jul 04 '17 at 20:24
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    It is an interesting question, will get to it when I have time from work. – MSIS Jul 04 '17 at 20:25
  • Thanks. All I really need is a reasonable definition that describes some class of manifolds that is accessible to second year students as a brief introduction to the concept. It can be extremely restricted. – Michael Stachowsky Jul 04 '17 at 20:29
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    A way of seeing that makes sense to me is that you're gluing together pieces, say open sets of Euclidean space using "nice-enough" maps. Any such gluing produces a manifold. – MSIS Jul 04 '17 at 20:30
  • Assuming $f$ and $g$ are continuously-differentiable, those two equations (interpreted as conditions on Cartesian $4$-space with coordinates $(q_{1}, q_{2}, x, y)$) describe the graph of a regular mapping (or "immersion") $F:\mathbf{R}^{2} \to \mathbf{R}^{4}$. This set is not generally a manifold (e.g., there can be self-crossings, as in a figure-8), but when restricted to sufficiently small pieces of the domain, the result is a smooth $2$-manifold. 2. Invertibility of $f$ and $g$ do not guarantee the Jacobian condition; think of $(x, y) = (q_{1}^{3}, q_{2}^{3})$ at the origin.
    • – Andrew D. Hwang Jul 04 '17 at 20:45
  • @AndrewD.Hwang: is there any way to restrict my functions' domains so that I can be assured that I get a manifold? For example, could I say "let $f$ and $g$ be defined on some domain $q_1 \in [a, b]$, $q_2 \in [c,d]$ such that $f$ and $g$ are bijective on that domain"? Or is there a lot more to it and I just can't get the type of definition I'm looking for? – Michael Stachowsky Jul 04 '17 at 21:02
  • @MSIS Logically, then, if I sufficiently restrict my domain I to avoid needing to glue anything together, would that work? I know I'm looking for an absurdly small subset of manifolds, so do let me know if such a set is so trivial as to be not worth mentioning... – Michael Stachowsky Jul 04 '17 at 21:03
  • @AndrewD.Hwang: Can I say that the inverse image ( level set) of a regular value of a map into $\mathbb R^k $ is a manifold, or does this hold only for k=1? – MSIS Jul 04 '17 at 21:06
  • @MSIS: Briefly, "yes in general". In detail, if $F:\mathbf{R}^{n+k} \to \mathbf{R}^{k}$ is continuously differentiable, and if for some $y$ in $F(\mathbf{R}^{n+k})$ the derivative $DF(x)$ has full rank $k$ for every point $x$ in $F^{-1}(y)$, then the level set ${x : F(x) = y}$ is a smooth $n$-manifold. (The key technical tool is the implicit function theorem.) – Andrew D. Hwang Jul 04 '17 at 21:44
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    @Michael: My apology: My first comment was incorrect. If $F:\mathbf{R} \to \mathbf{R}^{2}$ is continuously differentiable, then the graph defined by the equations $(x, y) = F(q_{1}, q_{2})$ in $\mathbf{R}^{4}$ is a smooth $2$-manifold. (I was wrongly thinking you'd defined a general immersion from $\mathbf{R}^{2}$ to $\mathbf{R}^{4}$; a graph immersion is special enough that technical snags I mentioned do not arise.) – Andrew D. Hwang Jul 04 '17 at 21:52
  • Great. So then is my original statement correct, provided f and g are continuously differentiable? Or is it sufficient for them to be differentiable and I don't need any restrictions on my Jacobian? – Michael Stachowsky Jul 04 '17 at 22:43
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    @MichaelStachowsky: Yes, using the Jacobian condition, since we can set up $F:=(f,g) $ as $F^{-1}(0,0) $ and , by construction, $(0,0)$ is a regular value ( any value is) , so its inverse image the graph of , $F:=(f,g)$, describes a manifold. – MSIS Jul 05 '17 at 03:44