# Question about differential of embedding

For any $C^{\infty}$ manifold $M$, the tangent bundle $TM$ of $M$ is also a $C^{\infty}$ manifold.

Hence we can think about the differential $df:TM\rightarrow TN$ of maps $f:M\rightarrow N$ between smooth manifolds as a smooth map between two smooth manifolds $TM$ and $TN$.

When I come up with this, I'm totally convinced that for any (smooth) embedding $\iota:N\rightarrow M$ between smooth manifolds, the differential $d\iota:TN \rightarrow TM$ also becomes an embedding (It must be in common sense; embedding is subobject!), but I cannot prove this in detail.

In particular, I'm stuck on thinking about the differential of the differential $d(d\iota)$. It's quite hard to imagine for me.

Is this true? In that case, I want to see a detailed proof for this one.

• Can you prove this when $\iota$ is the embedding of $\mathbb{R}^n$ into $\mathbb{R}^m$, with $n\leq m$? That's the best place to start, I think. – froggie Dec 4 '12 at 16:12
• to write down $d(d\iota)$ explicitely at some tangent vector $X\in TM$, you only need to write down $d\iota$ in local oordinates and differentiate it, try it for yourself, it's not hard. – Olivier Bégassat Dec 4 '12 at 16:18
• I think it is possible to get around explicit calculations. It is clear that $d\iota$ is bijective, has full rank along the zero section and is linear restricted to fibers. Maybe with the help of local trivializations it should become clear – Wolfgang Spindeler Dec 4 '12 at 16:45
• Thanks everyone, in particular froggie. I can use the result on $\mathbb{R}^n$ with the normal chart for embedded manifold. – cjackal Dec 4 '12 at 17:21
• The result is also true when $i:N\to M$ is merely an immersion, see my answer in math.stackexchange.com/questions/3233772/… – Zero May 26 '19 at 0:17

Take normal charts $\left\{ (U_\alpha , \psi_\alpha)\right\}$ for $N$, that is, charts of $M$ s.t $\psi_\alpha \circ \iota \circ(\psi_\alpha \mid_{U_\alpha \cap N})^{-1}(a^1,\cdots , a^n)=(a^1,\cdots,a^n,0,\cdots,0)$.

($\psi_\alpha \mid_{U_\alpha\cap N}:U_\alpha\cap N \rightarrow \mathbb{R}^n$ : restriction onto first $n$ coordinates)

Then $(\hat{U_\alpha},\hat{\psi_\alpha})$ is a chart for $TM$, where $\hat{U_\alpha}=\pi^{-1} (U_\alpha), \hat{\psi_\alpha}(p,X)=(\psi_\alpha (p),d\psi_\alpha^1(X),\cdots,d\psi_\alpha^m(X))$.

($\pi:TM\rightarrow M$:natural projection)

Same for $TN$ : $(\widehat{U_\alpha\cap N}, \widehat{\psi_\alpha \mid_{U_\alpha \cap N}})$ is a chart for $TN$.

Identify $\mathbb{R}^n \cong \mathbb{R}^n \times \left\{0,0,\cdots,0\right\}\subset \mathbb{R}^m$, hence tangent vector $X$ at $p\in T_p N$ looks locally like $\sum_{i=1}^n a^i \frac{\partial}{\partial x^i}$ where $x^i = (\psi_\alpha\mid_{U_\alpha \cap N})^i=\psi_\alpha^i$

Since $d\iota (p,X)=(p,d\iota_p X)$, $\hat{\psi_\alpha}\circ d\iota (p,X)=(\psi_\alpha(p), d\psi_\alpha^1 (d\iota_p X),\cdots, s\psi_\alpha^m (d\iota_p X))=(\psi_\alpha(p),a^1,\cdots,a^n,0,\cdots ,0).$

$\therefore \hat{\psi_\alpha}\circ d\iota \circ \widehat{\psi_\alpha \mid_{U_\alpha\cap N}}^{-1}(p^1,\cdots, p^n,a^1,\cdots , a^n)=(p^1,\cdots,p^n,0,\cdots,0,a^1,\cdots,a^n,0,\cdots,0)$ : embedding.