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Let $\pi: E \rightarrow M$ be a vector bundle over a manifold $M$, $\mathfrak{X}(M)$ the set of vector fields over $M$, and $\mathcal{E}(M)$ the space of smooth sections of $E$. Then for a given connection $\nabla$, $Y \in \mathcal{E}(M)$, and $X \in \mathfrak{X}(M)$ we call $\nabla_X Y \in \mathcal{E}(M)$ a covariant derivative. If $Y$ is a $(p,q)$ tensor field then so is its covariant derivative $\nabla_X Y$.

Now if $\nabla$ is a linear/affine connection, we may also define the $(p, q+1)$ tensor field $\nabla Y$ as $$\nabla Y(\omega^1, \ldots, \omega^p, V_1, \ldots, V_q, X) = \nabla_XY(\omega^1, \ldots, \omega^p, V_1, \ldots, V_q)$$ which we call the total covariant derivative.

Other than choosing a specific vector field/direction for the covariant derivative, I don't see what is the difference between it and the total covariant derivative. If that is the only difference, what is the point of defining a whole new object? Does $\nabla Y$ geometrically say anything that $\nabla_X Y$ does not?

CBBAM
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    The point is that $\nabla Y$ contains the information of $\nabla_XY$ for every choice of $X$. This allows us to think of a covariant derivative as a map $\nabla: \Gamma(E)\to \Gamma(E\otimes T^*M)$ rather than a map which assigns to each $X$ a map $\Gamma(E)\to \Gamma(E)$. That's really all there is to it. – J.V.Gaiter Dec 19 '23 at 21:11
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    See this related post and the answers there. – jd27 Dec 19 '23 at 21:12
  • @J.V.Gaiter Thank you! – CBBAM Dec 19 '23 at 22:57
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    You can use an analogy from multivariable calculus. Let $f$ be a scalar function f(x,y,z) then $\nabla_\mathbf{v} f$ is $\nabla f\cdot \mathbf{v}$ and the the total derivative is $\nabla f$. The scalar value $\nabla f\cdot\mathbf{v}$ is just a part of the information contained in the three dimensional vector $\nabla f$. – ContraKinta Dec 20 '23 at 22:14

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