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I previously asked this question on MSE, without success.

By Whitney's embedding theorem, every 2-dimensional manifold (aka. a surface) can be embedded into $\Bbb R^4$. Now, Wikipedia states in this paragraph that we can even embedd into $\Bbb R^3$ if the surface is

  • compact and orientable, or
  • compact and with non-empty boundary.

In the second bullet point, it is clear that I cannot drop either of the conditions. It is not clear to me why I can't drop "compact" in the first bullet point.

Question: Is there an orientable but non-compact surface that does not embedd into $\Bbb R^3$?

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  • $\begingroup$ Total space of canonical bundle is orientable but can't be embedded in $\mathbb{R}^3$. $\endgroup$
    – XT Chen
    Commented Apr 23, 2020 at 15:29
  • $\begingroup$ @XTChen Geat, Thanks! But I checked Wikipedia on this, and I admit, I wasn't able to understand much of it. Can you provide a more accessable explanation as an answer? $\endgroup$
    – M. Rumpy
    Commented Apr 23, 2020 at 15:31
  • $\begingroup$ Sorry. Total space of canonical bundle on $\mathbb{P}^1$ is non orientable. $\endgroup$
    – XT Chen
    Commented Apr 23, 2020 at 15:39
  • $\begingroup$ It is important to specify clearly which definition of "surface" one is using. There is one definition that I believe is at least in the plurality, was what Ian Richards used as his definition: a Hausdorff space that is locally homeomorphic to the plane and has a countable basis for its topology. If instead one considers, e.g., the Prüfer manifold (which has no countable basis but is Hausdorff and locally 2-dimensional), is known not to be metrizable. I doubt that it can be embedded in 3-space. $\endgroup$
    – Daniel Asimov
    Commented Nov 5, 2023 at 23:08

1 Answer 1

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It seems to me that every orientable surface is indeed embeddable in $\mathbb{R}^3$. By Ian Richards' classification theorem (https://www.ams.org/journals/tran/1963-106-02/S0002-9947-1963-0143186-0/S0002-9947-1963-0143186-0.pdf), we know that a non-compact orientable surface is determine by the pair $Y\subset X$ where $X$ is its space of ends (homeomorphic to a compact subset of the Cantor space) and $Y$ is the closed subset of ends with genus.

When genus is finite, i.e. $Y=\varnothing$, the embedding is easily realised as a genus-$g$ surface with a copy of $X$ removed. When genus is infinite, one simply starts with a sphere with a copy of $X$ removed, and adds smaller and smaller handles accumulating to all points of $Y$ (but not to points of $X\setminus Y$).

To ensure this can be done, observe that $Y$ has a countable dense subset. By taking a countable basis of neighborhoods of each of them, we get a countable family $(U_n)_{n\in\mathbb{N}}$ of open subsets of $\mathbb{S}^2\setminus X$ and we want to put one handle in each of them, but no family of handles should approach any point outside $Y$. We put a handle in $U_1$ (i.e. we remove two discs from $U_1$ and glue a handle to them), then inductively add one in the smallest $U_n$ not containing both gluing circle of any previous handle.

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    $\begingroup$ I don't think you need to use the classification of noncompact surfaces --- only the existence of a proper real-valued function (and hence an exhaustion by compact manifolds), and the fact that if $\Sigma$ is a compact surface and we are given any embedding of $\partial \Sigma \hookrightarrow \partial\left([0,1] \times \Bbb R^2\right)$, we may extend it to an embedding $\Sigma \hookrightarrow [0,1] \times \Bbb R^2$. This follows from classification of compact surfaces and isotopy extension. $\endgroup$
    – mme
    Commented Apr 23, 2020 at 16:18

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