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Let $M$ be the Moebius vector bundle over $S^1$.

Is it possible to embedd the total space of $M\oplus (S^1\times \mathbb{R^1})$ over $S^1$ into $\mathbb{R}^3$?

I suppose this isn't possible but I don't know an argument.

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It's not possible, since the Whitney sum would be a nonorientable $3$-manifold, which cannot be embedded in $\mathbb R^3$.

  • 0
    What's the reason that a nonorientable $3$-manifold cannot be embedded in $\mathbb{R}^3$?2012-02-13
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    An orientation reversing path in $M$ would also reverse orientation of the ambient space.2012-02-13
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    I'm using that an $n$-dimensional manifold embedded in an $n$-dimensional manifold sits as an open subset, which follows from "invariance of dimension."2012-02-13
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    Slightly differently: If there existed an embedding, the image would have to be open by invariance of domain, and open subsets of $\mathbb R^3$ are orientable.2012-02-13
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    There you go :D2012-02-13
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    Invariance of domain, that's what I meant!2012-02-13