My question is on the definition of the covariant derivative w.r.t a connection.

So first the definition of the covariant derivative:

Let $M$ be a smooth manifold with affine connection $\nabla$, and $c:[a,b] \to M$ a smooth curve in $M$. Then the unique map
$\frac{D}{dt}: \Gamma(TM \restriction_{c(t)}) \to \Gamma(TM \restriction_{c(t)})$
that fulfills three properties is called covariant derivative.

Now the first two properties make sense.

iii)If $V$ is induced from a $\mathcal{C}^{\infty}$ vector field $\tilde{V}$ on $M$, in the sense that $V(t) = \tilde{V}_{c(t)}$, then
$\frac{DV}{dt}(t) = \nabla_{c'(t)} \tilde{V}$.

On the left hand side we have a vector-field in $\Gamma(TM\restriction_{c'(t)})$. While on the right hand side we have by definition of the connection one on $\mathfrak{X}(M)$. Does that mean if I restrict the r.h.s to points on $c(t)$ I get the l.h.s ?

  • 2
    $\begingroup$ I would be a little bit more careful with the terminology and notation. For example, you don't have a "vector field in $\Gamma(TM|_{c'(t)})$. Rather, you have a section of the pullback bundle $c^*TM$, or a restriction of a vector field to a curve. This is not a vector field on the curve itself. You also do not have a connection on $\mathfrak{X}(M)$, but a connection on $TM$. Besides those technicalities, yes, that's what it means. $\endgroup$ Aug 25, 2022 at 16:46

1 Answer 1


In short, the answer to your question is yes.

However, perhaps going into a bit more generality would be useful. Let $M$ be a smooth manifold, $\nabla$ an affine connection on $M$, and $c:[0,1]\to M$ a smooth curve.

We may consider the following two things: the pullback bundle $c^*TM$ (which is a bundle over $[0,1]$), and the pullback connection $c^*\nabla$ (which is a connection on the pullback bundle). In your original question you consider the bundle $TM\vert_{c(t)}$, i.e. the restriction of the tangent bundle to the image of the curve. By construction of the pullback bundle, this is the same as $c^*TM$. We will shortly see that $c^*\nabla$ induces $D/dt$.

The pullback connection is characterized as follows. We know it is a map

$$ c^*\nabla:\Gamma(c^*TM)\times\mathcal{X}([0,1])\to\Gamma(c^*TM). $$

Suppose $V\in\mathcal{X}(M)$ is a vector field on $M$ and $X\in\mathcal{X}([0,1])$ is a vector field on $[0,1]$, then $c^*\nabla$ is uniquely determined as the connection on $c^*TM$ that satisfies

$$ (c^*\nabla)_X(c^*V)=c^*(\nabla_{dc(X)}V). $$

In particular, suppose we consider the map $D/dt$ but now defined as

$$ (c^*\nabla)_{\frac{d}{dt}}:\Gamma(c^*TM)\to\Gamma(c^*TM), $$

where $d/dt\in\mathcal{X}([0,1])$ is the standard global coordinate vector field.

Then, as in your question, let $V$ be a vector field along $c(t)$ (in particular, $V\in\Gamma(c^*TM)$) and further suppose $V$ is induced by a vector field $\tilde{V}\in\mathcal{X}(M)$ (in particular, $V=c^*\tilde{V}$). Then by definition of the pullback connection

$$ \dfrac{D}{dt}V=(c^*\nabla)_{\frac{d}{dt}}(c^*\tilde{V})=c^*(\nabla_{c'(t)}\tilde{V}), $$

where we note $c'(t)$ is shorthand notation for $dc(d/dt)$. Now, we note that the RHS $c^*(\nabla_{c'(t)}\tilde{V})$ can be identified as the restriction of $\nabla_{c'(t)}\tilde{V}$ to the image of the curve $c(t)$.


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