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I am currently learning about the covariant derivative on smooth manifolds with the following definition:

Definition. A covariant derivative of vector fields is an operation that associates to two vector fields $X$ and $Y$ a new vector field $\nabla_XY$ and satisfies the following properties.

  1. $\mathcal{C}^{\infty}(M)$-linearity with respect to $X$: $$ \nabla_{X_1+X_2}Y = \nabla_{X_1}Y + \nabla_{X_2}Y,\quad \nabla_{fX}Y = f\nabla_XY; $$
  2. additivity and the Leibniz rule with respect to $Y$: $$ \nabla_X(Y_1+y_2) = \nabla_XY_1+\nabla_XY_2,\quad \nabla_X(fY) = D_Xf\cdot Y + f\nabla_XY. $$

Then there is this theorem, for which I do not understand the proof. enter image description here

How am I supposed to understand the transfer the vectorfield via $\varphi$ part? Where does the covariant derivative on $U_i$ come from? Can you please explain me, what the first part with the chart means in exact terms? Thank you

Didier
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QED
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1 Answers1

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Presumably, the author means that $\nabla^i$ is the pullback through $\varphi_i$ of the covariant derivative $\bar \nabla$ on $\Bbb R^n$ defined by taking the directional derivatives in the following way. If $\bar X= \sum_j \bar X^j\partial_j$ and $\bar Y = \sum_k \bar Y^k \partial_k$ are vectors fields on $\Bbb R^n$, then $\bar \nabla_{\bar X} \bar Y = \sum_{j,k} \bar X^j \frac{\partial \bar Y^k}{\partial x^j} \partial_k$.

Let $X$ and $Y$ be vector fields on $M$. The pullback covariant derivative $\nabla^i = {\varphi_i}^*\bar\nabla$ is given by the formula $\nabla^i_XY= {\varphi_i}^*\left(\bar \nabla_{{\varphi_i}_*X}{\varphi_i}_*Y\right)$, and therefore, the covariant derivative $\nabla$ on $M$ is given by

$$ \nabla_XY = \left(\sum_i\lambda_i\nabla^i\right)_XY= \sum_i\lambda_i {\varphi_i}^*\left(\bar\nabla_{{\varphi_i}_*X}{\varphi_i}_*Y\right). $$ You can check that it indeed defines a covariant derivative since the pullback $\varphi_i^*$ and the pushforward ${\varphi_i}_*$ are $\mathcal{C}^{\infty}(M;\Bbb R)$-linear.

In addition, I would like to say that this construction works on all manifolds, not only compact ones (as suggests the text OP has linked). Indeed, we only need to have a partition of unity associated to a locally finite open cover. Since all manifolds possess such a partition of unity, the same proof works.

Finally, let us mention that there is a quicker but more conceptual way to prove the existence of a covariant derivative on any manifold. By the Whitney's embedding Theorem, any manifold embeds into some euclidean space $\Bbb R^N$. The orthogonal projection of the directional derivative of vector fields on the tangent space yields such a covariant derivative. It is the famous Levi-Civita connection, which is in fact associated to the Riemannian metric induced by the embedding.

In case your question is more about the definition of the pushforward or the pullback through a diffeomorphism $\varphi\colon U\to V$, they are defined by $$ (\varphi_*X)(q) = d\varphi(\varphi^{-1}(q)) X(\varphi^{-1}(q)) $$ for $q\in V$ and $X$ a vector field on $U$, and $$ (\varphi^* Y)(p) = [d\varphi(p)]^{-1} Y(\varphi(p)) $$ for $p\in U$ and $Y$ a vector field on $V$.

Didier
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    So $\bar{X}$ is a Vector field in $\mathbb{R}^n$ right? What do you mean with $\partial_i$? – QED Oct 02 '22 at 00:29
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    What is the covariant derivative on $\mathbb{R}^n$? I know this for surfaces, by taking the tangential part of the directional derivative, but in $\mathbb{R}^n$ everything is tangent to this seems odd – QED Oct 02 '22 at 00:35
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    @QED Yes, $\bar X$ is a vector field on $\Bbb R^n$. $\partial_i$ is the $i$th vector of the canonical basis of $\Bbb R^n$, this is standard notation in geometry. If you read carefully what I wrote, you will see that I have already defined the covariant derivative $\bar \nabla$ on $\Bbb R^n$. The one for surfaces you know is the Levi-Civita connection for embedded surfaces which is just a particular case. Here you are studying the full generality apparently. – Didier Oct 02 '22 at 07:56
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    But what is the product $\partial_i \partial_j$? – QED Oct 02 '22 at 10:19
  • @QED I have ever written such a thing, did I? – Didier Oct 02 '22 at 10:30
  • Ah I see, you changed it now. Thank you for the clear answer – QED Oct 02 '22 at 10:32
  • @QED Yes. $X^i$ is a function, $Y^j$ also, $\partial_i Y^j$ is its $i$th partial derivative, and you have the right to take their product. I have edited it to make it clearer that $\partial_iY^j$ is a partial derivative of a function – Didier Oct 02 '22 at 10:32