# Geometric interpretation of total covariant derivative?

A connection $$\nabla$$ is said to be compatible with riemannian metric $$g$$ if $$\nabla_Z g(X,Y)=g(\nabla_Z X,Y) + g(X,\nabla_Z Y).$$

The total covariant derivative $$(\nabla_Z g)(X,Y)$$ can be calculated as follows: $$(\nabla_Zg)(X,Y)=\nabla_Zg(X,Y)-g(\nabla_ZX,Y) - g(X,\nabla_ZY),$$ where $$\nabla_Zg(X,Y)=Zg(X,Y)$$ is the derivative of smooth function $$g$$ induced by vector $$Z$$.

It is now obvious that compatibility is equivalent to the total covariant derivative being zero, however I want to take a closer look at the term: $$\nabla_Zg(X,Y)=Zg(X,Y)$$, or more generally, $$Zg$$.

In any coordinate chart we can express $$g$$ as $$g=g_{ij}dx_i \otimes dx_j$$.

So would $$Zg=Z^k \frac{\partial g_{ij}}{\partial x_k}dx_i \otimes dx_j$$?

There are a lot of derivatives happening here.

Anyway, in a more general situation, the covariant derivateive of $$(n,m)$$ tensor $$F$$ is defined as:

$$(\nabla_ZF) (w_1,.....,w_n,X_1,...,X_m)=ZF(w_1,.....w_n,X_1,...,X_m)-\Sigma_{i=1}^n(w_1,...\nabla_Zw_i...,w_n,X_1,...X_m)-\Sigma_{i=1}^m(w_1,...w_i...,w_n,X_1,..,\nabla_ZX_i,...X_m)$$

Can anybody give me some sense of what this is measuring? In particular, what does it mean when the total covariant derivative vanishes?? I guess the obvious answer is that it is in some sense compatible with the tensor?? And so in general the total covariant derivative measures how far away a connection is from being compatible with a tensor? Is there any more than this that I should know? Thanks!

• You should really read $Zg(X,Y)$ as $Z(g(X,Y)).$ The mapping $X, Y \mapsto Z(g(X,Y))$ isn't $C^\infty$-linear, so "$Zg$" isn't a tensor. – Anthony Carapetis May 12 at 2:55
• Thanks, yeah the notation was a bit confusing to me. – HaKuNa MaTaTa May 12 at 13:49

I explain the intuitive meaning of $$\nabla g$$ (based on a previous answer of mine). The same idea can be used to understand the covariant derivative $$\nabla F$$ for general tensors $$F$$.

Meaning of $$(\nabla_Z g)(X,Y)$$

It is easiest to see about how the quantities change when we move along a curve. Take a curve $$\gamma$$ on $$M$$ and let $$X$$ and $$Y$$ be parallel vector fields along $$\gamma$$: $$\nabla_{\gamma'}X=0$$ and $$\nabla_{\gamma'}Y=0$$. Then we have \begin{align*} \frac{d}{dt}g(X,Y) &= (\gamma')(g(X,Y)) \\ &= (\nabla_{\gamma'} g)(X,Y) + g(\nabla_{\gamma'}X, Y) + g(X,\nabla_{\gamma'}Y) \\ &= (\nabla_{\gamma'} g)(X,Y). \end{align*} Here $$\gamma'$$ plays the role of $$Z$$. So the quantity $$(\nabla_{\gamma'} g)(X,Y)$$ gives the change of the inner product of $$X$$ and $$Y$$ along $$\gamma$$.

In particular, if $$\nabla g = 0$$, then $$\frac{d}{dt}g(X,Y) = 0.$$ So the inproduct $$g(X,Y)$$ is constant along parallel transport if $$\nabla$$ is compatible with the metric ($$\nabla g=0$$). Said differently: parallel transport w.r.t. the Levi-Civita connection preserves lengths and angles of parallel vector fields.

In general, $$(\nabla_{Z}F)(w_1, \ldots, X_m)$$ measures how much the quantity $$F(w_1,\ldots, X_m)$$ changes when we "walk in the direction $$Z$$".

Some remarks and some bits of intuition

You might ask yourself: Why do we assume that $$X$$ and $$Y$$ are parallel vector fields?

The reason is that we can do so and it simplifies the expression for $$\nabla g$$. Suppose I want to know $$(\nabla_u g)(v,w)(p)$$ at one single point $$p$$. Since $$\nabla g$$ is a tensor, the quantity $$(\nabla_u g)(v,w)(p)$$ only depends on the vectors $$u$$, $$v$$ and $$w$$, not on the values of vector fields around $$p$$. So in order to calculate $$(\nabla_Z g)(X,Y)$$ at $$p$$ we take a curve $$\gamma$$ with $$\gamma(0)=p$$ and $$\gamma'(0) = u$$. Next we take parallel vector fields $$X$$ and $$Y$$ along $$\gamma$$ such that $$X(p)=v$$ and $$Y(p)=w$$.

Second question: Why is $$\nabla g$$ exactly defined as $$(\nabla_Z g)(X,Y) = Z (g(X,Y)) - g(\nabla_Z X,Y) - g(X, \nabla_Z Y) \tag{1}?$$

a. Very loosely speaking $$(\nabla_Z g)(X,Y)$$ measures how $$g(X,Y)$$ changes when we walk in the direction of $$Z$$. But when we walk in the direction of $$Z$$, the vector fields $$X$$ and $$Y$$ also change. In order to take these changes of $$X$$ and $$Y$$ into account, we must substract $$g(\nabla_Z X,Y)$$ and $$g(X, \nabla_Z Y)$$ from $$Z(g(X,Y))$$.

b. Here is another way to interpret the definition. Rewrite eq $$(1)$$ as $$Z (g(X,Y)) = (\nabla_Z g)(X,Y) + g(\nabla_Z X,Y) + g(X, \nabla_Z Y)?$$ In this form, the equation is a kind of Leibniz rule for the derivative. The LHS is the derivative of $$g(X,Y)$$ w.r.t. $$Z$$. The quantity $$g(X,Y)$$ depends on $$g$$, $$X$$ and $$Y$$. So the derivative $$Z(g(X,Y))$$ must depend on how $$g$$, $$X$$ and $$Y$$ change. And indeed, the three terms of the RHS incorporate the changes of $$g$$, $$X$$ and $$Y$$.

• @HaKuNa MaTaTa, there were a lot of questions in your post. I tried to answer them in intuitive way. Feel free to ask for clarification, if you need some. – Ernie060 May 12 at 18:00