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I'm following the tutorial at this link, where the author states:

These follow from the various way one can iterate covariant derivative

$$\nabla^3_{xyz}s = \nabla^2_{xy}(\nabla_zs) - \nabla_{\nabla^2_{xy}z}s$$ and $$\nabla^3_{xyz}s = \nabla_x (\nabla^2)_{yz}s + \nabla^2_{yz}(\nabla_{x}s)$$

I'm unable to derive these equations myself.

My Attempt

For the second covariant derivative, I seem to be able to derive this. I used the product rule and the fact that we can commute the covariant derivative with contractions to show that:

$$\nabla_x\nabla_y s= \nabla_x C(\nabla s \otimes y) $$ $$\nabla_x\nabla_y s= C(\nabla_x \nabla s \otimes y) + C(\nabla s \otimes \nabla_x y ) $$ $$\nabla_x\nabla_y s= (\nabla_x\nabla s)(y) + \nabla_{\nabla_x y} s $$

I then assume that $(\nabla_x\nabla s)(y)$ is the second covariant derivative, so:

$$ \nabla^2_{xy}s = \nabla_x\nabla_y s - \nabla_{\nabla_x y } s $$

Now at this point we have one expression for the second covariant derivative. To get the next one, I just used the product rule to get:

$$ \nabla_x\nabla_y s = (\nabla_x\nabla)_y s + \nabla_{\nabla_xy}s + \nabla_y(\nabla_x s) $$ $$ \nabla_x\nabla_y s - \nabla_{\nabla_xy}s = (\nabla_x\nabla)_y s + \nabla_y(\nabla_x s) $$

$$ \nabla^2_{xy}s = (\nabla_x\nabla)_y s + \nabla_y(\nabla_x s) $$

But any attempts to do the same thing for the third covariant derivative seem to be failing for me. Is there some straightforward way to get to the results I quoted above from here that I'm not seeing?

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  • $\begingroup$ Your second formula does not look right with $x,x,y$ in $\nabla_x(\nabla^2)_{xy}$. $\endgroup$ Feb 10, 2021 at 13:00
  • $\begingroup$ Oh yes I copied it wrong while I was typing this, easy fix. But the question still stands :P I'll fix that now. Thanks @user10354138 :) $\endgroup$ Feb 10, 2021 at 13:04
  • $\begingroup$ The "cheating" way for higher covariant derivatives in general is to use the fact that the formula is universal, so must hold for the trivial $\mathbb{R}$-bundle. So take $s\in C^\infty(M)$, where you know how to differentiate functions and clean up the non-tensorial bits in $X(Y(Z(s)))$ to get $\nabla^3_{XYZ}s$. $\endgroup$ Feb 10, 2021 at 13:23
  • $\begingroup$ I'd prefer not to reference functions if I can. I'd like a method similar to the one I used for the second covariant derivative if possible (because that is what the author seems to reference if I'm not misunderstanding them) $\endgroup$ Feb 10, 2021 at 13:26
  • $\begingroup$ What's your definition of $\nabla^3 s$? Are you familiar and comfortable with the object referred to as $\nabla s$? The two identities you quote could be taken as definitions, so it's important that you know where to start. $\endgroup$ Sep 6, 2021 at 8:40

2 Answers 2

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My recommendation is to work out your own definitions and notation for everything. When you read someone else's writing, use their notation and proof as a guide to how to write everything including the proof in your own notation. Don't worry about understanding their notation literally.

I find higher covariant derivatives to be very confusing. The way I deal with it is that I view the covariant derivative of a higher order covariant derivative to be just a special case of the covariant derivative of a tensor. For example, the covariant derivative of a second order tensor $T$ is defined to be $$ (\nabla T)(X,Y,Z) = \partial_X(T(Y,Z)) - T(\nabla_XY,Z) - T(Y,\nabla_XZ) $$ So the second order covariant derivative of $T$ is \begin{align*} (\nabla^2T)(X,Y,Z,W) &= (\nabla(\nabla T))(X,Y,Z,W)\\ &=\partial_X(\nabla T(X,Y,Z)) - \nabla T(\nabla_XY,Z,W)\\ &\quad - \nabla T(Y,\nabla_XZ,W) - \nabla T(Y,Z,\nabla_XW) \end{align*} And so on.

Therefore, $\nabla^3T = \nabla(\nabla(\nabla T))) = \nabla^2(\nabla T) = \nabla(\nabla^2T)$. Now skew-symmetrization and the Ricci identity should give you what you want.

Note that my personal convention is to never write $\nabla_XT(Y,Z)$. I find that notation difficult to work with, even though I like the way the chain rule identity looks using that notation: $$ \partial_X(T(Y,Z)) = \nabla_XT(Y,Z) + T(\nabla_XY,Z) + T(Y,\nabla_XZ) $$

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  • $\begingroup$ I’m sorry I didn’t respond right away, but I didn’t totally understand how to apply skew-symmetry action and the Ricci identity. Could you elaborate on that a little more? $\endgroup$ Apr 11, 2021 at 4:48
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In order to deal with the third covariant derivative, you need to take two steps:

  1. Understand what the first covariant derivative $\nabla s$ is.
  2. Define $\nabla^3 s$ as $\nabla (\nabla (\nabla s))$.

As you can see, step 2 is quite trivial.


I will assume you already know what $\nabla_X s$ is. If $s$ is a section of $E$ (which might be a tensor product of $TM$'s and $T^*M$'s), $X$ is a tangent field (i.e. a section of $TM$), then $\nabla_X s$ is another section of $E$. Morally, it's $s$ differentiated in the direction of $X$.

So what could $\nabla s$ be? More or less the same thing, but not yet evaluated. This object contains the information about derivatives of $s$ in all directions. It's defined as $(\nabla s)(X) := \nabla_X s$.

Now $\nabla s$ is a $C^\infty(M)$-linear function of $X$, it can be identified with a section of $E \otimes T^*M$. Now we're at the starting point - $\nabla s$ is a section of some vector bundle, and we want to define the covariant derivative $\nabla_X(\nabla s)$ as a section of the same bundle.


This part is already explained in Deane's answer, so let me just add a few words.

Imagine you want second derivatives of $s$, say, in directions $X$ and $Y$. You could consider $\nabla_X (\nabla_Y s)$, but that's just wrong. Please take time to contemplate the most basic example below.

For a moment, consider $s \colon \mathbb{R}^n \to \mathbb{R}$. Computing the directional derivative of the directional derivative $\nabla_X (\nabla_Y s)$, you should see that the derivatives of $Y$ pop out. That's not how the Hessian should work, right?

In short, $\nabla_X (\nabla_Y s)$ contains some information about derivatives of $Y$. This manifests itself in the lack of $C^\infty(M)$-linearity - we have $\nabla_{fX} (\nabla_Y s) = f \cdot \nabla_X (\nabla_Y s)$ but not $\nabla_{X} (\nabla_{fY} s) = f \cdot \nabla_X (\nabla_Y s)$ (if $f$ is a smooth function). Thus, we define $\nabla (\nabla s)(X,Y)$ by subtracting the derivatives of $Y$: $$ \nabla_X (\nabla s) (Y) := \nabla_X (\nabla s (Y)) - (\nabla s) (\nabla_X Y). $$ One can check that the above is $C^\infty(M)$-linear in $Y$, so $\nabla_X (\nabla s)$ is a well-defined section of $E \otimes T^* M$. I've written it this way to emphasize that the same could be done for any section of $E \otimes T^* M$, not just $\nabla s$. But in our case it could be written differently: $$ \nabla_X (\nabla s) (Y) := \nabla_X (\nabla_Y s) - \nabla_{\nabla_X Y} s. $$ Please take time to convince yourself that it's the same thing.

And again, since this is $C^\infty(M)$-linear in $X$, it defines a section $\nabla\nabla s$ (or $\nabla^2 s$) of $E \otimes T^*M \otimes T^*M$.


In the same way one defines $\nabla^3 s = \nabla \nabla \nabla s$: $$ \nabla^3 s (X,Y,Z) = (\nabla_X \nabla^2 s)(Y,Z) = \nabla_X (\nabla^2 s (Y,Z)) - \nabla^2 s(\nabla_X Y, Z) - \nabla^2 s(Y, \nabla_X Z). $$

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  • $\begingroup$ Wow, thanks so much for this! I'll give it a deep read and get back to you very soon :) $\endgroup$ Sep 28, 2021 at 6:15

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