# An orientable surface that cannot be embedded into $\Bbb R^3$? [duplicate]

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 $$\Bbb R^3$$?

• Total space of canonical bundle is orientable but can't be embedded in $\mathbb{R}^3$. – XT Chen Apr 23 at 15:29
• @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? – M. Rumpy Apr 23 at 15:31
• Sorry. Total space of canonical bundle on $\mathbb{P}^1$ is non orientable. – XT Chen Apr 23 at 15:39

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.
• 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. – Mike Miller Apr 23 at 16:18