The following is Exercise 3 of Chapter 2 of my Brazilian edition of do Carmo's Riemannian Geometry:

Let $f: M^n \to \overline M^{n + k}$ be an immersion from a differentiable manifold $M$ to a Riemannian manifold $\overline M$. Assume in $M$ the Riemannian metric induced by $f$: $$ \langle u, v \rangle_p = \langle df_p(u), df_p(v) \rangle_{f(p)}. $$ Let $p \in M$ and $U \subset M$ be a neighborhood of $p$ such that $f(U) \subset \overline M$ be a submanifold of $\overline M$. Let $X, Y$ be vector fields on $f(U)$ and extend then to vector fields $\overline X, \overline Y$ on an open subset of $\overline M$. Define $$ (\nabla_X Y)(p) = \text{tangential component of } \overline \nabla_{\overline X} \overline Y(p), $$ where $\overline \nabla$ is the Riemannian connection of $\overline M$. Prove that $\nabla$ is the Riemannain connection of $M$.

As already discussed in this question, the notation is very confusing and in fact wrong. Proceed as in the answer to the linked question:

if $X, Y$ are vector fields on $U$, we consider the vector fields $df(X), df(Y)$ on $f(U)$ and extend these vector fields to $\overline X, \overline Y$ on an open set of $\overline M$. Recall also that we have the following decomposition of the tangent space at $f(p)$: $$ T_{f(p)}\overline M = df_p(T_pM) \oplus (df_p(T_pM))^\perp, $$ and we call the tangential component of $\overline \nabla_{\overline X} \overline Y$ the terms in $df_p(T_pM)$. So the correct definition for $\nabla$ is $$ \nabla_X Y(p) = (df_p)^{-1}( \text{tangential component of }\overline \nabla_{\overline X} \overline Y(f(p))). $$

First we show that $\nabla$ well-defined, i.e., does not depend on the extensions $\overline X, \overline Y$. Indeed, if $\overline X_1, \overline X_2, \overline Y_1, \overline Y_2$ are two distinct extensions of $df(X), df(Y)$ respectively, then they coincide at $f(p)$. Then their tangential components coincide and $\nabla$ is indeed well-defined.

Now, in order to show that $\nabla$ is a connection, we have to show the three defining properties. Don't we need that $f$ be in fact an embedding, in order to be able to define $g \circ f^{-1}$ to prove, for example, $\nabla_{g X + h Y}Z = g \nabla_X Z + h \nabla_Y Z$?

Also, how to show that this connection is compatible with the Riemannian metric?

Finally, is the following argument for symmetry correct?

Ordering the basis in $T_{f(p)}\overline M$ so that $X_1, \ldots, X_n \in df_p(T_pM)$: \begin{align*} \nabla_{X_i} X_j - \nabla_{X_j} X_i & = (df_p)^{-1}\left(\sum_k\Gamma_{ij}^k X_k\right) - (df_p)^{-1}\left(\sum_k\Gamma_{ji}^k X_k\right) \\ & = (df_p)^{-1} \left(\sum_k(\Gamma_{ij}^k-\Gamma_{ji}^k) X_k \right) \\ & = 0 \end{align*} by the symmetry of $\overline \nabla$. Thus $\nabla$ is symmetric.

  • 2
    $\begingroup$ Although the identification $U \to f(U)$ is confusing when you first see it, this abuse of notation is used by almost everyone. You will have to translate from one to another in your head. $\endgroup$ Aug 7 '20 at 5:54

First, your checking that $\nabla_XY$ is well defined independent of $\overline X, \overline Y$ is unclear: to take an analogy, even if two functions $f_1, f_2$ agree at a point $p$, it does not imply that $f'_1 = f_2'$ at $p$.

To check that $\nabla$ is well-defined, we split into two steps:

  • If $\overline X, \widetilde X$ are both extension of $df(X)$, then for any local vector fields $Z$ on $V\subset \overline M$ and for all $p\in U$, $$ \overline \nabla_{\overline X} Z = \overline \nabla_{\widetilde X} Z\ \ \ \ \ \ \text{ at } f(p).$$ Proof: This follows from the fact that $\overline \nabla$ is $C^\infty$-linear in that component, thus the value $\overline \nabla_{\overline X} Z(f(p))$ depends only on $\overline X(f(p))$.

  • Let $\overline Y, \widetilde Y$ are both extension of $df(Y)$ and $\overline X$ is tangential to $f(U)$, then $$\tag{2} \overline \nabla _{\overline X} \overline Y = \overline \nabla _{\overline X} \widetilde Y\ \ \ \ \ \text{ at }f(p).$$ Proof: this follows from the fact that covariant differentiation can be computed using parallel transport (here): In particular, since $\overline X$ is tangential to $f(U)$, one can find an integral curve of $\overline X$ which lies inside $f(U)$ (For example, let $\gamma : (-\epsilon, \epsilon)\to M$ be an integral curve of $X$. Then $f\circ \gamma$ is an integral curve of $\overline X$ lying inside $f(U)$). Since $\overline Y, \widetilde Y$ agrees on $f(U)$, (2) is shown.

Second, we show that $\nabla$ is indeed a connection. To begin with, we show

(1) For any local vector fields $X, Y$ on $U$ and local smooth functions $\varphi:U \to \mathbb R$, we have $$\nabla_{\varphi X} Y (p) = \varphi(p) \nabla_X Y(p), \ \ \ \forall p\in U.$$ Proof: let $\overline \varphi$ be a smooth function on $V\subset \overline M$ which extends $\varphi\circ f^{-1} : f(U) \to \mathbb R$. That is, for all $f(p) \in f(U)$ we have $$ \varphi (p) = \overline \varphi (f(p)).$$ Then $\overline \varphi \overline X$ is an extension of $df (\varphi X)$. So \begin{align*} \nabla _{\varphi X} Y(p) &= df^{-1} \bigg(\text{tangential component of } \overline \nabla_{\overline\varphi \overline X} \overline Y(f(p))\bigg) \\ &= df^{-1} \bigg(\text{tangential component of }\ \overline\varphi (f(p)) \overline \nabla_{\overline X} \overline Y(f(p))\bigg) \\ &= \varphi (p) df^{-1} \bigg(\text{tangential component of }\overline \nabla_{\overline X} \overline Y(f(p))\bigg) \\ &= \varphi (p) \nabla_X Y (p). \end{align*}

(2) We show also that $\nabla$ is compatible with the pullback metric $g = f^*\bar g$, let $X, Y, Z$ be vector fields. Then by definition,

$$ X g(Y, Z)(p) = \frac{d}{dt}\bigg|_{t=0} g_{\gamma(t)}( Y(\gamma(t)), Z(\gamma(t))),$$

where $\gamma : (-\epsilon, \epsilon) \to M$ is any curve with $\gamma(0) = p$, $\gamma'(0) = X(p)$. Using the definition of pullback metric,

$$ g_{\gamma(t)}( Y(\gamma(t)), Z(\gamma(t))) = \bar g_{f(\gamma(t))} (df_{\gamma(t)} Y(\gamma(t)), df_{\gamma(t)} Z(\gamma(t))).$$

Since $f\circ \gamma$ is a curve in $\overline M$ with $f\circ \gamma (0) = f(p)$, $(f\circ \gamma)'(t) = df_{\gamma(t)} X(\gamma(t))$, we have

\begin{align*} \frac{d}{dt}\bigg|_{t=0} g_{\gamma(t)}( Y(\gamma(t)), Z(\gamma(t)))&= \overline X \bar g (\overline Y, \overline Z) f(p)\\ &= \bar g(\overline \nabla _{\overline X} \overline Y , \overline Z ) + \bar g(\overline Y , \overline \nabla _{\overline Y} \overline Z) \ \ \ \ \ \text{ at } f(p)\\ &= \bar g(df (\nabla _{X} Y) , df ( Z) ) + \bar g(df(Y) , df(\nabla _{Y} Z) \\ &= g(\nabla_X Y, Z) + g(Y, \nabla_XZ) \end{align*} at $p$. Note we used that $\overline Y, \overline Z$ are tangential to $f(U)$, so that we have $$ \bar g (\overline \nabla_{\overline X} \overline Y, \overline Z) = \bar g ((\overline \nabla_{\overline X} \overline Y)^\top, \overline Z),$$ where $(\cdot)^\top$ denotes the tangential part of a vector.

Finally, in your checking of the symmetry of $\nabla$ you used $\Gamma_{ij}^k = \Gamma_{ji}^k$, which a priori you do not know yet. Indeed the symmetry of $\nabla$ is equivalent to the symmetry of $\Gamma$.

To give a correct proof we, just like all the others properties we proved, pushforward everything to $\overline M$, prove the property there and then pullback: by definition,

\begin{align*} \nabla_X Y- \nabla_Y X &= df^{-1} \left( \overline\nabla_{\overline X} \overline Y - \overline\nabla _{\overline Y} \overline X\right)^\top \\ &= df^{-1} ([\overline X, \overline Y]^\top). \end{align*}

Since $f(U)$ is a submanifold and $\overline X, \overline Y$ are tangential to $f(U)$,

$$ [\overline X, \overline Y]^\top = [\overline X, \overline Y] = [df (X), df(Y)]$$ (this can be check directly, assuming that $f(U)$ is a plane $\mathbb R^n \subset \mathbb R^{n+k}$. The Riemannian structure is not used here). Then by this, we have $$\nabla_X Y- \nabla_Y X = [X, Y].$$

  • $\begingroup$ Thank you very much for the most complete and clear answer (and also for all the help you've given me in the past days). However, I have some doubts. To begin with, don't we need some argument to say that $\overline \varphi \overline X$ extends $df( \varphi X)$? Also, why does $\overline \varphi$ "comes out" of $(df)^{-1}$ as $\varphi$? $\endgroup$ Aug 7 '20 at 17:11
  • $\begingroup$ Regarding the first question, is it a pointwise argument? Something along these lines: for each $p \in U$ we have $\varphi(p) \in \mathbb R$, so $df(\varphi(p) X_p) = \varphi(p) df(X_p)$, implying that the value of the vector field $df(\varphi X)$ at $q \in f(U)$ is $\varphi(f^{-1}(q)) df(X)_q = \overline \varphi(f(p)) df(X)_q$? $\endgroup$ Aug 7 '20 at 17:15
  • $\begingroup$ How do we know that there is an integral curve of $\overline X$ in $f(U)$, when showing that $\nabla$ is well-defined? $\endgroup$ Aug 7 '20 at 17:49
  • $\begingroup$ A more fundamental question: In Lee's Intro to Smooth Manifolds, page 182, he says that for a given smooth map $F: M \to N$ and $X \in \mathfrak X(M)$, there may not be any vector field on $N$ that is $F$-related to $X$. But of course, if $df(X)$ is a vector field on $N$, it is trivially $f$-related to $X$. So how can we know that this is the case? $\endgroup$ Aug 7 '20 at 18:43
  • 1
    $\begingroup$ Concerning your question, (1) yes it is just a pointwise argument that you can pull out $\varphi (p)$. (2) I've added something about finding the integral curve (3) $df(X)$ is not a vector field on $N$. it is defined only on $f(U)$ and we need to extend it to a (local) vector field on $N$. @DaniloGregorin $\endgroup$ Aug 8 '20 at 17:49

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.