Let $f: M \to N$ be an immersion of a differentiable manifold $M$ into a Riemannian manifold $N$. Assume that $M$ has the Riemannian metric induced by $f$. Let $p \in M$ and let $U \subset М$ be a neighborhood of $p$ such that $f(U) \subset N$ is a submanifold of $M$. Further, suppose that $X, Y$ are differentiable vector fields on $f(U)$ which extend to differentiable vector fields $X^*, Y^*$ on an open set of $N$. Define $$(\nabla_x Y)(p) =\text{ tangential component of }(\overline{\nabla}_{x^*} Y^*)(p),$$ where $\overline{\nabla}$ is the Riemannian connection of $N$. Prove that $\nabla$ is the Riemannian connection of $M$.

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    $\begingroup$ In the future, please cite where you got this problem, and what your motivation is for asking. For instance, this is Exercise 3 in Chapter 2 of do Carmo's "Riemannian Geometry." My own professor actually just assigned this problem to us as homework last week. (Also, welcome to math.stackexchange!) $\endgroup$ Oct 1 '11 at 19:07
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    $\begingroup$ At any rate, you know the Riemannaian connection on $M$ is unique, so you only need to check that $\nabla$ is metric compatible and symmetric. $\endgroup$ Oct 1 '11 at 19:09
  • $\begingroup$ Excuse me, is my first time $\endgroup$ Oct 1 '11 at 19:32
  • $\begingroup$ One more thing: if possible, try to use Latex syntax when writing in order to make your text more legible. Your next texts will appear in a very nice fashion. $\endgroup$
    – matgaio
    Apr 21 '12 at 23:47
  • $\begingroup$ @JesseMadnick I would appreciate if you take a look at my comments. I do not know why I cannot call you from your answer, so I call you from here. Thank you. $\endgroup$ Mar 24 '15 at 10:08

As in the problem statement, we let $\overline{\nabla}$ denote the Riemannian connection on $N$, and define $(\nabla_XY)(p) = \pi^\top[\overline{\nabla}_{X^*}Y^*(p)]$, where $\pi^\top\colon TN|_{f(U)} \to TM$ denotes the tangential projection. We aim to show that $\nabla$ is the Riemannian connection on $M$.

By the uniqueness of the Riemannian connection on $M$, it suffices to show that (1) $\nabla$ is a connection, (2) $\nabla$ is compatible with the metric, and (3) $\nabla$ is symmetric.

(1) $\nabla$ is a connection

Let $X_1, X_2, Y_1, Y_2$ be differentiable vector fields on $f(U)$ that extend to differentiable vector fields on an open subset of $N$. We note that $$(\nabla_{X_1 + X_2}Y)(p) = (\overline{\nabla}_{X_1^* + X_2^*}Y^*)(p) = (\overline{\nabla}_{X_1^*}Y^* + \overline{\nabla}_{X_2^*}Y^*)(p)^\top = (\nabla_{X_1}Y)(p) + (\nabla_{X_2}Y)(p)$$ and $$(\nabla_X(Y_1 + Y_2))(p) = (\overline{\nabla}_{X^*}Y_1^* + \overline{\nabla}_{X^*}Y_2^*)(p)^\top = (\nabla_XY_1)(p) + (\nabla_XY_2)(p)$$ and if $h\in C^\infty(f(U))$ is any differentiable function on $f(U)$, then $$\begin{align*} (\nabla_X(hY))(p) & = (\overline{\nabla}_{X^*}(hY^*))(p)^\top \\ & = \left[(\overline{\nabla}_{X^*}h)(p)Y^*(p) + h(p)(\overline{\nabla}_{X^*}Y^*)(p) \right]^\top \\ & = \left[ (Xh)(p)\,Y^*(p)\right]^\top + h(p)(\overline{\nabla}_{X^*}Y^*)(p)^\top \\ & = (Xh)(p)\,Y(p) + h(p)\,(\nabla_XY)(p), \end{align*}$$ which shows that $\nabla$ is a connection.

(2) $\nabla$ is compatible with the metric.

This is a computation:

$$\begin{align*} X_p\langle Y_p, Z_p \rangle_g & = X^*_p\langle Y^*_p, Z^*_p \rangle_{\overline{g}} \\ & = \overline{\nabla}_{X^*}\langle Y^*, Z^* \rangle_{\overline{g}}(p) \\ & = \langle (\overline{\nabla}_{X^*}Y^*)(p), Z^*_p\rangle_{\overline{g}} + \langle Y^*_p, (\overline{\nabla}_{X^*}Z^*)(p)\rangle_{\overline{g}} \\ & = \langle (\overline{\nabla}_{X^*}Y^*)(p)^\top + (\overline{\nabla}_{X^*}Y^*)(p)^\perp, Z_p \rangle_{\overline{g}} + \langle Y_p, (\overline{\nabla}_{X^*}Z^*)(p)^\top + (\overline{\nabla}_{X^*}Z^*)(p)^\perp \rangle_{\overline{g}} \\ & = \langle (\nabla_XY)(p), Z_p \rangle_g + \langle Y_p, (\nabla_XZ)(p)\rangle_g, \end{align*}$$ where I've switched to subscripts $X_p = X(p)$ for a slight ease of notation. (Also, $\perp$ denotes the normal projection.)

(3) $\nabla$ is symmetric.

Finally, symmetry follows from noting that

$$\begin{align*} (\nabla_XY)(p) - (\nabla_YX)(p) & = (\overline{\nabla}_{X^*}Y^*)(p)^\top - (\overline{\nabla}_{Y^*}X^*)(p)^\top \\ & = \pi^\top\!\left( (\overline{\nabla}_{X^*}Y^*)(p) - (\overline{\nabla}_{Y^*}X^*)(p) \right) \\ & = \pi^\top([X^*, Y^*]_p) \\ & = \pi^\top([X, Y]_p) \\ & = [X, Y]_p \end{align*}$$ since $[X,Y]$ is tangent to $M$ whenever $X$ and $Y$ are.

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    $\begingroup$ I have some questions about your answer. We know that $\overline{\nabla}:\mathfrak{X}(\overline{M})\times\mathfrak{X}(\overline{M}) \rightarrow\mathfrak{X}(\overline{M})$. If $\overline{X}$ and $\overline{Y}$ are defined in an open subset of $\overline{M}$ we can extend them to $\overline{M}$, so $\overline{\nabla}_{\overline{X}}\overline{Y}$ makes sense. Now, $h\in\mathcal{C}^{\infty}(f(U))$ is not necessarily defined in an open subset of $\overline{M}$ so, what does $\overline{\nabla}_\overline{X}(h\overline{Y})$ mean? $\endgroup$ Mar 23 '15 at 16:18
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    $\begingroup$ In addition, since $\nabla:\mathfrak{X}(f(U))\times\mathfrak{X}(f(U)) \rightarrow \mathfrak{X}(f(U))$, it would only be a connection if every vector field of $f(U)$ was extendable, because we have proven the connection axioms for this kind of vector fields. (Excuse my ignorance, I am just trying to understand) $\endgroup$ Mar 24 '15 at 8:39
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    $\begingroup$ In my humble opinion we should follow the next steps:1)Every vector field of $M$ has an associated vector field in $f(U)$. 2)Every vector field of $f(U)$ that comes from one of $M$ is extendable to $N$. 3)Now $\nabla:\mathfrak{X}(M)\times\mathfrak{X}(M)\rightarrow\mathfrak{X}(M)$, so we can proove those three properties. $\endgroup$ Mar 24 '15 at 10:31
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    $\begingroup$ @SrinivasaGranujan Since $\nabla$ is a local operator, we can choose $U$ sufficiently small such that $f(U)$ is a regular submanifold of $N$. For any smooth vector field $X$, we just consider $X|_U$, and it's easy to extend $f_* (X|_U)$. Hence, $\nabla$ is the LC connection on $M$. $\endgroup$
    – user360777
    May 1 '17 at 12:56
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    $\begingroup$ @rmdmc89 the fact that f is an immersion implies its a local embedding. So for small enough $U$, $f : U \to f(U)$ is an embedding of $f(U)$ in $N$. $\endgroup$ Dec 22 '18 at 14:54

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