1

I was reading Lee's Riemann manifold,in exercise 2.15,which needs to show that:

Let $\tilde{M}$ be n-dim smooth manifold,for any $p\in \tilde{M}$ prove exist some neiborhood around it in $\tilde{M}$ that the local frame adapted to k-submanifold $M\subset \tilde{M}$.

(Where adapted means the local frame has first k tangent vectors at each point as local frame tangent to the submanifold)

I do as follows:

First since $M$ embedded in $\tilde{M}$ which means exist a local frame(taking by slice chart) of $\tilde{M}$ .denoted as $(X_i)_{i=1}^n$ then by slice chart condition we know that first $k$ tangent to $M$.then take Gram-Smidt algorithm on these $(X_i)_{i=1}^n$ where first k vector at each point still span the tangent space for embeded submanifold .

Is my proof correct?There is another question is does this result holds for immersed submanifold?

yi li
  • 5,414
  • I guess this works (for embedded submanifolds) – Didier Mar 08 '21 at 13:28
  • Is my proof correct? I try to use figure eight curve,but seems not contradiction? – yi li Mar 08 '21 at 13:40
  • For immersed figure eight curve, this does not work for the intrsection point because there is no adapted frame of a neighbourhood of $0$ in $\mathbb{R}^2$ such that the first vector at $0$ spans two lines. – Didier Mar 08 '21 at 14:05
  • What do you mean to "span" two lines,the tangent space only 1-dimension? – yi li Mar 08 '21 at 14:22
  • The definition only need the first $k$ component of local frame tangent to submanifold $M\subset \tilde{M}$? – yi li Mar 08 '21 at 14:25
  • "spans" is a synonym of "generates". My point is: can you find a local frame $(X_1,X_2)$ on a neighbourhood of $0$ in $\mathbb{R}^2$ such that the $X_1$ generates the tangent space of the figure eight curve? The answer is no because of the intersection point. Hence one cannot expect use any Gram-Schmidt process. – Didier Mar 08 '21 at 14:33
  • Figure 8 is not a submanifold. Submanifold is always embedded. Figure 8 is an immersed manifold. As that there is an adapted frame along any segment with 2 end points – Deane Mar 08 '21 at 14:36
  • Your answer is correct. – Deane Mar 08 '21 at 14:37
  • Sorry I was still a bit confused why figure-eight curve fails to have adapted frame,for example take $\beta(t) = (sin(t),sin(2t))$ then we may take global frame on $\Bbb{R}^2$ as $E_1 = (1,2)$ and $E_2 = (2,-1)$ with first $(1,2)$ tangent to $M$ (this is all the requirement for the definition on the book).Which is satisfies. expect submanifold means embeded submanifold – yi li Mar 08 '21 at 14:49
  • In Lee's smooth manifold book page 109:"the term smooth submanifold without further qualification means an immersed one" So in this context "submanifold" may means immersed one. – yi li Mar 08 '21 at 15:16
  • And adapted only needs the condition as page 16 of Riemann manifold book shows:" be adapted to M if the first n vector fields are tangent to M" – yi li Mar 08 '21 at 15:21
  • 5
    @yili: You're right that my definition of "submanifold" without qualification means "immersed submanifold." But the exercise you asked about was to prove Proposition 2.14, which said explicitly that $M$ is an embedded submanifold. (By the way, the exercise we're talking about is in my Introduction to Riemannian Manifolds, not my smooth manifolds book.) – Jack Lee Mar 08 '21 at 15:43
  • thank you so much professor,I learnt a lot from your great book,and I still have a lot to learn – yi li Apr 22 '21 at 10:52
  • @yili Hello, can I ask you a question, please? Do the first $k$ vector fields in the adapted frame form a local frame for the submanifold? Thank you. – Boar Dec 14 '23 at 00:53
  • 1
    Hi @Boar, I guess yes? since we can choose the coordinate chart such that the first $k$ coordinates is for the submanifold. – yi li Dec 14 '23 at 14:07
  • @yili Hello, lately I found that we don't actually need a coordinate chart to explain why they form a local frame for the submanifold. It's simply a consequence of the following linear-algebra fact. If a basis for $V$ intersects a $k$-dimensional subspace of $V$ at $k$ basis vectors, then these $k$ vectors will form a basis for the subspace. – Boar Jan 01 '24 at 02:45
  • 1
    Thank you Boar, that's a nice point. – yi li Jan 01 '24 at 03:37
  • 1
    @Boar, a coordinate chart is needed here, unless the submanifold is 0-dimensional. Adapted frame has to be smooth. – Snate May 06 '24 at 22:34
  • 1
    @Boar yes I agree with Snate. I didn't really understand your idea until Snate point it out. To make your idea more clear you can have a look at Prof Lee's introduction to smooth manifold chapter about distribution – yi li May 07 '24 at 04:48
  • thank you for point out Snate – yi li May 07 '24 at 04:56

0 Answers0