# do Carmo Riemannian Geometry Exercise 2.3: definition of $\nabla$ for an immersion - Part II

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.

• 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. 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 $$\nabla_X Y- \nabla_Y X = [X, Y].$$

• 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 $\varphi$? Aug 7 '20 at 17:11
• 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 $\varphi(f^{-1}(q)) df(X)_q = \overline \varphi(f(p)) df(X)_q$? Aug 7 '20 at 17:15
• How do we know that there is an integral curve of $\overline X$ in $f(U)$, when showing that $\nabla$ is well-defined? Aug 7 '20 at 17:49
• 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? Aug 7 '20 at 18:43
• 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 Aug 8 '20 at 17:49