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This is the point of my confusion: Let $M$ be a manifold and $K$ an open subset of $M$, then $K$ is a submanifold of $M$. Submanifold are manifolds themselves, so we can speak about dimensions of $M$ and $K$ respectively. I need: do $M$ and $K$ have the same dimension? Some examples show that they do: let's say interval is not open in $R^2$, so all open subsets of $R^2$ are two-dimensional.

Why I am confused is how we can prove the existence of outward normal, which exists it is enough to say about $\Omega \subset R^n$. I then know that $\Omega$ has the same local vectors as $R^n$, and outward normal is locally normal to all vectors, to all $n$ independent vectors which is impossible in $R^n$. But, outward normal does exist... Where do I make a mistake??

nikola
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    Converse: https://math.stackexchange.com/questions/84577/open-subsets-in-a-manifold-as-submanifold-of-the-same-dimension –  Jul 21 '19 at 09:19

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We put outward normal on $\partial \Omega$, which has one dimension less, allowing that to happen. And them $\Omega$ MUST be the same dimension, or dimension $n$, because $\partial \Omega$ has a dimension exactly one less then $\Omega$, and if $\Omega$ had dimension $k < n$ then $\partial \Omega$ would have dimension $k-1$, allowing to have normal space of dimension $n-k+1$, which is more than 1, which means we have that outward normal isn't well defined.

nikola
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  • Sometimes trying to explain your question to a friend who knows no mathematics at all solves problem for you, I saw a solution when he told me "tangent is on boundary of circle, not on circle". He could be a hell of a geometrist.... – nikola Feb 14 '17 at 19:41
  • Yes, this is the idea. $\Omega$ cannot be an arbitrary open set, though - the topological boundary of an open submanifold is not necessarily a submanifold. We typically would require $\Omega$ to be a "domain with $C^1$ boundary". – Anthony Carapetis Feb 14 '17 at 19:43