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Let $M$ be a non-compact manifold. Then $\exists$ an embedding $f:\mathbb{R}\to M$.

My attempt: I am trying to show that there is a complete smooth nonvanishing vector field on the manifold $M$ whose integral curve is not a closed curve. I took compact exhaustion and tried defining a vector field, but I seem to run into trouble.

EDIT: Embedding means: Injective immersion which is a proper map.

epsilon_delta
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    Why not just use any coordinate system, which is a diffeomorphism of an open subset with $\mathbb R^n$? By the way, you should probably assume $M$ has positive dimension, otherwise the statement isn't true. – Alex K Apr 15 '21 at 19:54
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    There should always be a smooth embedding from $\mathbb{R}$ to any (positive-dimension) manifold $M$, compact or not. Just embed in a small open ball in $\mathbb{R}^n$ and use a manifold chart to pull the embedding back to $M$. – Neal Apr 15 '21 at 19:54
  • Following the above ideas: $\mathbb{R}$ is homeomorphic (even diffeomorphic) to $(0,1)$. Show that there is always an open intervall $(0,1)$ in a manifold (for example, use a coordinate system). – Didier Apr 15 '21 at 20:03
  • @Didier But That might not give the map to be proper. In particular take $S^1$. Then $ S^1\setminus {N}$ is diffeomorphic to $\mathbb{R}$ but it is not an embedding because the inverse of $S^1$ is whole $R$ which is not compact. In general we need to take a chart whose closure is non-compact, otherwise the preimage of the compct closure, is whole real line which is not compact. – epsilon_delta Apr 15 '21 at 20:16
  • It sounds like you omitted a hypothesis of "proper" in your post. Perhaps you should think carefully about all the hypotheses you want to be satisfied, and edit your post to include them all explicitly. – Lee Mosher Apr 15 '21 at 20:22
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    As it stands, "embedding" appears to mean a topological embedding, i.e. a homeomorphism from $\mathbb R$ onto a subspace of $M$, which is what all the commenters (including myself) took it to mean. – Lee Mosher Apr 15 '21 at 20:23
  • @Didier $\mathbb{R}$ is diffeomorphic to $ (0,1)\times {0}$ but is not an embedding. – epsilon_delta Apr 15 '21 at 20:26
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    @epsilon_delta Your definition of embeding is not standard, as pointed out Lee Mosher. – Didier Apr 15 '21 at 20:27
  • This is the definition is from the book by John M. Le's Intr. to smooth manifolds. I don't know any other definition. – epsilon_delta Apr 15 '21 at 21:06
  • It seems that Lee these "proper embeddings" – Alex K Apr 15 '21 at 21:47
  • @AlexK is it possible to solve the problem in the latest form? Intuitively, it is clear that such an embedding exists but it is hard to write one. – epsilon_delta Apr 15 '21 at 22:28

1 Answers1

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The only proof I can think of for this ends up having a lot of moving parts, but here's a rough sketch:

  • Choose an exhaustion of $M$ by compact submanifolds with boundary $C_0\subset C_1\subset C_2,\cdots\subset M$ (such an exhaustion can be constructed from a Morse exhaustion function).
  • Let $A_i=C_i\setminus\operatorname{int}(C_{i-1})$, and let $A_{i,1}\cdots,A_{i,N_i}$ be the connected components of $A_i$. (This effectively partitions $M$ into compact, connected submanifolds glued together along their common boundaries.)
  • Consider $A_{i_1,j_1}$ and $A_{i_2,j_2}$ to be adjacent if they share a boundary component. This allows one to consider the $A_{i,j}$ to be the vertices of a graph.
  • Show this graph is infinite and connected, and that all vertices have finite degree, implying it contains a ray $A_{i_0,j_0},A_{i_1,j_1},A_{i_2,j_2},\cdots$.
  • Choose and $x_0\in\operatorname{int}(A_{i_0,j_0})$ and $x_k\in\partial A_{i_{k-1},j_{k-1}}\cap\partial A_{i_k,j_k}$.
  • Construct embedded paths $\gamma_k:[0,1]\to A_{i_k,j_k}$ with $\gamma_k(0)=x_k$, $\gamma_k(1)=x_{k+1}$, and $\gamma_k((0,1))\subset\operatorname{int}(A_{i_k,j_k})$, chosen so that $\gamma_k$ and $\gamma_{k+1}$ can be smoothly concatenated.
  • Show the path $\gamma:\mathbb{R}_{\ge 0}\to M$ defined by $\gamma(t)=\gamma_k(t-k)$ for $k\le t< k+1$ is a proper embedding of the ray $\mathbb{R}_{\ge 0}$.
  • Construct a properly embedded closed tubular neighborhood around $\gamma$.
  • Properly embed $\mathbb{R}$ in this neighborhood.
Kajelad
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