On a smooth manifold $M$, a vector field is a smooth map $X : M \to TM$, where $TM$ is the tangent bundle of $M$. If $\chi(M)$ denotes the space of vector fields on $M$, an affine connection $\nabla$ on $M$ is a mapping $\nabla : \chi(M) \times \chi(M) \to \chi(M)$ which (intuitively) creates a new vector field $\nabla_{X}Y$ given two vector fields $X$,$Y$. If I am correct, given $p \in M$, $\nabla_{X}Y(p)$ can be understood as the derivative of $X$ in the direction of $Y(p)$.

1) Consider that $M \subset \mathbb{R}^n$ to make things "easier. Now, a vector field on $M$ is a smooth map $X : M \to \mathbb{R}^n$ (such that, for all $p$, $X(p) \in T_{p}M$). Given another vector field $Y$ and $p \in M$, the derivative of $X$ in the direction of $Y(p)$ could be (naively) $D_{p}X\big( Y(p) \big)$, where $D_{p}X : T_{p}M \to \mathbb{R}^n$ denotes the differential of $X$ at $p$. But this might not define an affine connection since $D_{p}X\big( Y(p) \big)$ is not necessarily a tangent vector, right ?

2) If $\gamma : I \subset \mathbb{R} \to M$ is a smooth curve and $X$ a vector field along $\gamma$, the covariant derivative $\frac{DX}{dt}$ of $X$ (namely $\nabla_{\frac{d\gamma}{dt}}X$) is defined to be the orthogonal projection on the linear space $T_{\gamma(t)}M$. However, this definition holds only because we implicitly assume that $M$ is equipped with the metric induced by $\mathbb{R}^n$, right ? In other words, if I'm correct and if $g$ is a Riemannian metric (which is not the metric induced by $\mathbb{R}^n$) on $M$, this definition of covariant derivative no longer makes sense.


1 Answer 1


Note that $(\nabla_X Y)(p)$ should be thought of as the derivative of $Y$ at $p$ in the direction of $X(p)$ and not the other way around! (This is manifested in the definition of an affine connection by requiring that $\nabla_X Y$ is $C^{\infty}(M)$-linear in $X$ while only $\mathbb{R}$-linear in $Y$). Regarding your questions:

  1. Up to replacing the role of $X$ and $Y$, you are correct. If however you take the resulting vector $dY|_{p}(X(p))$ (as a vector in $\mathbb{R}^n$) and project it orthogonally onto $T_pM$ (identified as a subspace of $\mathbb{R}^n$) you obtain an affine connection which is precisely the Levi-Civita connection of the Riemannian metric on $M$ induced by $\mathbb{R}^n$.
  2. Once you have a connection on $M$, no matter whether it came from nowhere or from an embedding on $M$ into $\mathbb{R}^n$ and a projection construction as in $(1)$ or from a Riemannian metric, you can define the notion of a covariant derivative of a vector field along a curve. The definition (written informally by $\frac{DX}{dt} = \nabla_{\dot{\gamma}(t)} X$) doesn't involve any orthogonal projection. If $M$ is embedded in $\mathbb{R}^n$ and you work with the Levi-Civita connection of the induced metric (which is the same as the connection coming from the projection construction) then indeed $\frac{DX}{dt} = \operatorname{Proj}(\frac{d}{dt} \left( X(\gamma(t)) \right), T_{\gamma(t)}M)$. If $M$ has a different metric, in doesn't really help you to think of $M$ as a submanifold of $\mathbb{R}^n$ as the embedding is not compatible with the ambient metric and you can't express the connection or the covariant derivative along a curve in terms of the projection construction.
  • $\begingroup$ Thank you for your answer ! I think I get it now !! If $M \subset \mathbb{R}$ is equipped with a Riemannian metric other than the Euclidean one, I should not think about $M$ as a submanifold of $\mathbb{R}^n$ ? Is it more correct to think of it as an abstract manifold, even though it is a subset of $\mathbb{R}^{n}$ ? $\endgroup$
    – Pouteri
    Jul 21, 2016 at 13:01
  • $\begingroup$ Well, you can think of it as a submanifold of $\mathbb{R}^n$ but if the metric $g$ on $M$ is not induced by the Euclidean one, everything related to the geometry of $(M,g)$ (and not the differential manifold properties of $M$) won't be related to the geometry of $M$ as a subset of $\mathbb{R}^n$. For example, nothing prevents you from taking $M$ to be the unit sphere $S^2$ in $\mathbb{R}^3$ with a Riemannian metric whose geometry is that of an ellipsoid and not of a round sphere. $\endgroup$
    – levap
    Jul 21, 2016 at 14:35

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