# Expressing third covariant derivatives in terms of second covariant derivatives

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?

• Your second formula does not look right with $x,x,y$ in $\nabla_x(\nabla^2)_{xy}$. Feb 10, 2021 at 13:00
• 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 :) Feb 10, 2021 at 13:04
• 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 $\nabla^3_{XYZ}s$. Feb 10, 2021 at 13:23
• 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) Feb 10, 2021 at 13:26
• 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. Sep 6, 2021 at 8:40

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: $$\partial_X(T(Y,Z)) = \nabla_XT(Y,Z) + T(\nabla_XY,Z) + T(Y,\nabla_XZ)$$

• 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? Apr 11, 2021 at 4:48

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.

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 $$\nabla^3 s = \nabla \nabla \nabla s$$: $$\nabla^3 s (X,Y,Z) = (\nabla_X \nabla^2 s)(Y,Z) = \nabla_X (\nabla^2 s (Y,Z)) - \nabla^2 s(\nabla_X Y, Z) - \nabla^2 s(Y, \nabla_X Z).$$