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Question: I am trying to show that if $M$ and $N$ are smooth manifolds (without boundary), and $$F:M\to N$$ is a smooth embedding, then the differential $$dF:TM\to TN,\quad dF(p,v)=(F(p),dF_p(v))$$ is also a smooth embedding.

In particular, this shows that an embedded submanifold of a smooth manifold gives rise to an embedded submanifold of the tangent bundle in a natural way.

It is not difficult to show that $dF$ is a smooth immersion. Indeed, it has coordinate representations of the form $$dF(x,v)=(F(x),DF(x)v),\quad(x,v)\in \hat{U}\times\mathbb{R}^m\subseteq\mathbb{R}^m\times\mathbb{R}^m$$ so $$D(dF)(x,v)=\begin{pmatrix}DF(x) & 0 \\ \ast & DF(x) \end{pmatrix},$$ which has full rank since $DF(x)$ has full rank. Hence, we at least have that $dF(TM)$ is a immersed submanifold of $TN$.

But now I am stuck in showing that $dF$ is a topological embedding. It is clearly injective, so the inverse $$(dF)^{-1}:dF(TM)\to TM$$ exists. But how do you show it is continuous?

Definitions: Here "smooth" means $C^\infty$. The assumption that $F$ is a smooth embedding means that $F$ is a smooth immersion (i.e. $dF_p:T_pM\to T_{F(p)}N$ is injective at each $p\in M$) and that $F$ is a topological embedding (i.e. $F:M\to F(M)$ is a homeomorphism when $F(M)$ is given the subspace topology inherited from $TN$).

Spenser
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    I think you only need to use the inverse function theorem to prove that $(dF)^-1$ is continuous, you can use it since $DF$ has full rank. – Maxime Scott Jun 01 '15 at 14:33
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    @djvyu72 It is not sufficient. Any smooth immersion has full rank, but there are smooth immersions that are not smooth embeddings. For example, $\beta:(-\pi,\pi)\to\mathbb{R}^2$, $\beta(t)=(\sin 2t,\sin t)$ is not an embedding since its image is compact in $\mathbb{R}^2$. – Spenser Jun 01 '15 at 14:41
  • @Spencer Indeed, my argument doesn't work, to prove your claim, you need to apply the inverse function theorem to $DF.$ I think you it might works if you write down your differential in coordinates (local coordinates) and differentiate it again. If you can prove that it has full rank you are done. – Maxime Scott Jun 01 '15 at 14:52
  • @djvyu72 Thank you very much for your comments, but I am not sure I understand what do you mean. What you say is what I already did: I wrote $dF$ in coordinates, took its differential and showed it has full rank. But this only shows that $dF$ is an immersion (or equivalently that it is a local embedding), but it does not show that $dF$ is a global embedding. – Spenser Jun 01 '15 at 15:01
  • @Spencer You already proved that your map $dF$ was injective, meaning that you have an injective immersion, which is actually an embedding, so I think you are done. – Maxime Scott Jun 01 '15 at 19:36
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    @djvyu72 That is precisely where you are wrong: an injective immersion need not be an embedding. For example, the map $\beta$ that I give above. There are other examples like the well known dense curve on the torus: http://math.stackexchange.com/questions/79051/showing-a-subset-of-the-torus-is-dense – Spenser Jun 01 '15 at 19:42
  • Let us continue this discussion in chat. – Maxime Scott Jun 01 '15 at 20:10
  • @Spenser I don't see why you say that "we at least have that $dF(TM)$ is an immersed submanifold of $TN$." For that to be proved, you would have to exhibit a topology and smooth structure for $dF(TM)$ such that $dF(TM)$ is a topological manifold with respect to the topology and the inclusion map from $dF(TM)$ to $TN$ is a smooth immersion with respect to the smooth structure. This sounds much more plausible in the specific case where $F$ is the inclusion map of $M$ in $N$, since then $F$ and $dF$ would be injective. – Jeff Rubin Jan 14 '24 at 04:48

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Once you know that the image is an immersed submanifold, to show that $F$ is an embedding, it suffices to show that its image is an embedded submanifold; and for that it suffices to show it's embedded in a neighborhood of each point of the image. Let $p\in M$ and $q=F(p)$. Because $F$ is a smooth embedding, it is possible to choose smooth coordinate charts $(U,(x^i))$ containing $p$ and $(V,(y^i))$ containing $q$ such that $F(M)\cap V = F(U)$ and $F|_U$ has a coordinate representation of the form $F(x^1,\dots,x^m) = (x^1,\dots,x^m,0,\dots,0)$. If we let $(x^i, v_i)$ be the associated standard coordinates for $TM$ and $(y^i,w_i)$ those for $TN$, then $dF|_{TU}\colon TU\to TV$ has the coordinate representation $$ dF(x^1,\dots,x^m,v_1,\dots,v_m) = (x^1,\dots,x^m,0,\dots,0,v_1,\dots,v_m,0,\dots,0). $$ The fact that $F(M)\cap V = F(U)$ guarantees that $dF(TM) \cap TV = dF(TU)$, and the coordinate representation above shows that $dF(TU)$ is an embedded $2m$-dimensional submanifold of $TV$.

Jack Lee
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    Thank you very much Professor Lee. I realized this was a way to prove what I wanted, but I didn't think of using the Rank Theorem, so I struggled a lot to show that $dF(TU)$ is embedded. Thanks to your help, I now filled up ever details and have a complete proof. – Spenser Jun 02 '15 at 12:29
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    By the way, I am a huge fan of your book Introduction to Smooth Manifolds, which I am currently self-studying. I wasn't expecting to have an answer from the author himself! Thank you, you are a true inspiration! – Spenser Jun 02 '15 at 12:31
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    @Spenser: You're welcome. And thanks for the kind compliment! – Jack Lee Jun 02 '15 at 14:54
  • Pardon, but why does that coordinate representation show $dF(TU)$ is an embedded $2m$-manifold? – INQUISITOR Nov 23 '20 at 11:20
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    It shows that it satisfies the local slice criterion. See Theorem 5.8 in my Introduction to Smooth Manifolds. – Jack Lee Nov 23 '20 at 16:05
  • @JackLee Is this result still true if $M$ and $N$ have nonempty boundaries? With a different proof of course as slice charts won't work. – nomadicmathematician Feb 28 '24 at 20:32