4
$\begingroup$

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 $M$.

$\endgroup$
6
  • 4
    $\begingroup$ 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
  • 1
    $\begingroup$ 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
10
$\begingroup$

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.

$\endgroup$
7
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ 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
  • 2
    $\begingroup$ @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
  • 1
    $\begingroup$ @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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.