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 $M$.

My questions are:

  • $X$ and $Y$ are vector fields on $f(U) \subset M$. Then $\nabla_X Y(p)$ does not make sense. Shouldn't it be $\nabla_X Y(f(p))$? Also, $\nabla$ is not a connection on $M$, but on $f(U)$, isn't it? So what does the problem want us to prove?
  • What does "tangential component" mean?
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    $\begingroup$ Yes. It's sloppy writing by identify $U$ as $f(U)$. Since $\overline{M}$ is a Riemannian manifold, each tangent space at $f(U)$ split into tangent space of the submanifold and its orthogonal complement. Tangential component obviously the projection onto tangent space of submanifold. $\endgroup$ Jul 31, 2020 at 20:18

1 Answer 1


There is a bit of abusing notation here. The goal is to define $\nabla $, which is a connection on the tangent bundle of $M$. Thus the goal IS to define $\nabla _X Y(p)$, where $p\in M$ and $X, Y$ are local vector fields of $M$ around $p$. (I think it is very important to know that they are not defining something at $f(p)$: there might be $p\neq q$ so that $f(p) = f(q)$)

The way to do so is

(1) push-forward the local vector fields $X, Y$ to $\mathrm df(X), \mathrm df(Y)$ respectively, if $X, Y$ are on $U$, then $\mathrm df(X), \mathrm df(Y)$ are on $f(U)$ (they abuse notations here, identifying $X, Y$ with $\mathrm df(X), \mathrm df(Y)$)

(2) extend $\mathrm df(X), \mathrm df(Y)$ to a local vector fields $\overline X, \overline Y$ respectively on $\overline M$ around $f(p)$, and

(3) define $\nabla_X Y(p) := \text{tangential component of }\overline\nabla _{\overline X} \overline Y (f(p))$ (Note that it is $f(p)$ on the right hand side. I guess it is a typo) As suggested in the comment, the tangent space at $T_{f(p)} \overline M$ split into $df (T_pM)$ and $(df (T_pM))^\perp$, the orthogonal complement. The tangential component are taken with respect to this decomposition. Thus the more precise definition should be $$\nabla_X Y(p) :=(\mathrm df)^{-1} \bigg( \text{tangential component of }\overline\nabla _{\overline X} \overline Y (f(p))\bigg)$$

I suppose they went on to show that $\nabla$ is well defined, independent of the extension $\overline X, \overline Y$. Indeed $\nabla $ is the Levi-Civita connection on $M$ with respect to the pullback metric $$\langle u, v \rangle_p := \langle df_p(u), df_p(v) \rangle_{f(p)}.$$

  • $\begingroup$ I guess the choice of $U$ would make $f|_U$ embedding no ? Therefore $U$ and $f(U)$ diffeomorphic. $\endgroup$ Jul 31, 2020 at 20:57
  • $\begingroup$ Yes (I think I don't understand your comment, can you be more direct/specific?) @SiKucing $\endgroup$ Jul 31, 2020 at 20:59
  • $\begingroup$ It is said in the question that $U$ is a nbd of $p$ such that $f(U)$ is a submanifold of $\overline{M}$, which i interpret as embedded submanifold. With this choice, $f|U$ is embedding and therefore we can't have $f(p)=f(q)$ for $p\neq q$ isn't it ? $\endgroup$ Jul 31, 2020 at 21:04
  • $\begingroup$ Ar I see. My point is that there could be $q\notin U$ so that $f(q) = f(p)$. Think of the immersion $f: \mathbb R \to \mathbb R^2$, $f(t) = (\cos t, \sin t)$. It is important to know that $\nabla$ is defined on $\mathbb R$, but not on the image $\mathbb S^1$. @SiKucing $\endgroup$ Jul 31, 2020 at 21:07
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    $\begingroup$ Did the answer here help? @DaniloGregorin $\endgroup$ Aug 6, 2020 at 17:56

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