# Compatibility of covariant derivative with the connection

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 ?

• 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. Aug 25, 2022 at 16:46

## 1 Answer

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 $$c(t)$$.