My book is An Introduction to Manifolds by Loring W. Tu. Pictured below is the last example from Section 22, Manifolds with Boundary.

enter image description here

I have been trying to wrap my head around this for about 2 hours (3.5 hours, if you include the 1.5 hours spent on typing up this question).

My questions are:

  1. Should the given "$c_{*,p}$" be $c_{*,p}: T_p([a,b]) \to T_{\color{red}{c(p)}}C$ ?

  2. Is the given "$c_{*,p}: T_p([a,b]) \to \{\text{see (1) for range}\}$" actually $(j \circ c)_{*,p} = j_{*,c(p)} \circ c_{*,p}$ where

    • $c_{*,p}: T_p([a,b]) \to T_{c(p)}M$
    • $j: C \to M$ and $j_{*,c(p)}: T_{c(p)}C \to T_{c(p)}M$, both are inclusion,
    • so the given "$c_{*,p}$" is an "induced" differential, where "induced" refers to restricting range like in Subsection 11.4 ?

    • 2.1. Is the given "$c_{*,p}$" then an isomorphism and thus $c$ is a local diffeomorphism by Remark 8.12 on the Inverse Function Theorem? How is this relevant? I think this answers question (6) below.

  3. It's not stated as to what $M$ is, but I think $M$ is a smooth oriented n-manifold with boundary. Is this relevant, and why or why not?

    • 3.1. Must $n=1$ in this example?
  4. What exactly is the orientation on $C$? I think the orientation on $[a,b]$ is given by smooth vector field $\frac{d}{dx}$ on $(a,b)$, smooth outward-pointing vector field $\frac{d}{dx}$ at $x=b$ and smooth outward-pointing vector field $-\frac{d}{dx}$ at $x=a$ and orientation form $dx$ on all of $[a,b]$ (I think it's the same form for each boundary point and for the interior unlike with the vector field), so for $C$, I think the smooth outward-pointing vector field is $c_{*,p}[\frac{d}{dx}\mid_p]$ and something to do with $c$ and $dx$ like $c^{*}(dx)$, $d(c \circ x)$ or $c \circ (dx)$.

    • 4.1. Also I seem to have only a local orientation at $p$, namely, $c_{*,p}[\frac{d}{dx}\mid_p]$. What's the original orientation supposed to be? We can define the pushforward $c_{*}[\frac{d}{dx}]$ if $c$ is injective (Subsection 14.5), but how do we know $c$ is injective?

      • 4.1.1. There might be other ways to define the pushforward. Hopefully at least one of those pushforward definitions is smooth. I'm about to read more here.
  5. Where do we use injectivity of $c_{*,p}$, either the original or the given "$c_{*,p}$" (whose injectivity follows from composition of injections is an injection)?

  6. How do we know $\partial (c[a,b]) = c (\partial [a,b])$ and $ (c[a,b])^o = c ([a,b]^o)$?

    • I think this would follow from Proposition 22.4 if $c$ were injective, but (see question $(4.1)$).

    • I think this would follow from Proposition 22.4 if $c$ were a local diffeomorphism, which I think follows from a "yes" to question $(2.1)$ or if $c$ were a local diffeomorphism onto its image (which I think means that the restricted range $c$, $c: [a,b] \to c[a,b]$ is a local diffeomorphism)

  7. Are "sections" relevant? I think even if $c$ is not injective, $c$ can have sections even if $c$ has no inverse or something.

  • 1
    $\begingroup$ Oh goodness, what a lot of questions. I admit I'm a bit perplexed by the wording of the example. After all, an immersion of $[a,b]$ into $M$ makes me think of things like a figure eight in a plane: locally injective, but a figure 8 is not a manifold (with or without boundary)... $\endgroup$ – Rylee Lyman Mar 13 '19 at 1:39
  • $\begingroup$ @RyleeLyman But it's an immersed submanifold even if it's not a manifold? Thanks for saying you're perplexed by the wording! $\endgroup$ – user636532 Mar 13 '19 at 1:39
  • 1
    $\begingroup$ Yeah, that's what was tripping me up. In that case, I assume that somehow the immersion is actually an embedding, in which case why not say so? Maybe something else is meant... In any case, all the example seems to be trying to say is that $[a,b]$ has a "usual" orientation in the sense that $a < b$, and that this carries through to its image? $\endgroup$ – Rylee Lyman Mar 13 '19 at 1:45
  • 4
    $\begingroup$ Actually, the example is incorrect as stated. For example, the map $c\colon [0,2\pi]\to \mathbb R^2$ given by $c(x) = (\cos x,\sin x)$ is a $C^\infty$ immersion, and its image is the unit circle, which is a manifold with boundary (whose boundary happens to be empty). But $c(0)$ and $c(2\pi)$ are not boundary points, so it doesn't make sense to talk about boundary orientations there. I imagine Tu had in mind that $c$ should be injective. Since an injective smooth immersion with compact domain is automatically an embedding, it follows then that $c[a,b]$ is diffeomorphic to $[a,b]$. $\endgroup$ – Jack Lee Mar 15 '19 at 18:32
  • 1
    $\begingroup$ If you want to ask a new question, you should post it as a separate question. MSE frowns on back-and-forth exchanges in the comments. $\endgroup$ – Jack Lee Jun 20 '19 at 19:23
  1. Yes, we have $p \in [a,b]$, not in $M$.
  2. I think that somehow immersed submanifolds should be included under consideration by this example, otherwise the wording is needlessly cumbersome. In this case I think your intuition is right, but If $M$ has dimension two, say, $c$ should not be a local diffeomorphism. Well, perhaps onto its image, but that's exactly what an immersion is already.
  3. In this example we should consider $M$ that have orientation and those that do not, boundary or not, and the dimension should be arbitrary.
  4. The orientation on the image of $C$ is more or less as you describe. Think of the vector field version of it as the tangent line to the curve, pointing in the direction of the orientation on $[a,b]$, i.e. if $M$ is $\mathbb{R}^n$, exactly the derivative of $c$ at $p$, thought of as a vector in the tangent space of $\mathbb{R}^n$ at $c(p)$.
  5. Actually here I think all we use is that $c$ is a smooth function. The immersion condition also probably tells us that at any points $p$ and $q$ with $c(p) = c(q)$, the tangent vectors point in different directions?
  6. Here I think being locally injective should be good enough? Take some small neighborhood and see if 22.4 gets you there?
  7. I don't think so for this example? I don't think about differential geometry every day, so I could be wrong.

Your Answer

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