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Thread: Lebesgue measure of a manifold embedded in the euclidean space

  1. #1
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    Lebesgue measure of a manifold embedded in the euclidean space

    Let M be a m-dimensional manifold embedded in a n-dimensional euclidean space. If m is strictly less than n, is its Lebesgue measure zero?

    Can anyone give me a reference which can prove/disprove this?

    Thanks
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Re: Lebesgue measure of a manifold embedded in the euclidean space

    Quote Originally Posted by uasac View Post
    Let M be a m-dimensional manifold embedded in a n-dimensional euclidean space. If m is strictly less than n, is its Lebesgue measure zero?

    Can anyone give me a reference which can prove/disprove this?

    Thanks
    Here's how I would think about doing it. Since $\displaystyle M$ can be embedded into $\displaystyle \mathbb{R}^n$ and subspaces of Lindelof spaces are Lindelof we know that $\displaystyle M$ is Lindelof and so we can find a countable atlas $\displaystyle \{(U_k,\varphi_k)\}_{k\in\mathbb{N}}$ for $\displaystyle M$. Since the countable union of null sets is null it suffices to show that each $\displaystyle U_k$ has measure zero. To do this consider that by considering the map $\displaystyle \mathbb{R}^m\xrightarrow{\iota\circ\varphi_k^{-1}}\mathbb{R}^n$ (where $\displaystyle \iota$ is the embedding $\displaystyle M\hookrigtharrow \mathbb{R}^n$) you get a smooth map from $\displaystyle \mathbb{R}^m$ into $\displaystyle \mathbb{R}^n$ and since $\displaystyle m<n$ we know (by lemma 10.3 in Lee) the image of $\displaystyle \iota\circ\varphi_k^{-1}$ has measure zero. And, since measure zero sets are preserved under diffeomorphisms we can pull back along $\displaystyle \iota$ to get that $\displaystyle U_k$ has measure zero.
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  3. #3
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    Re: Lebesgue measure of a manifold embedded in the euclidean space

    I'm just starting to learn differential geometry, but your arguments seem sound to me. Thank you.
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