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!

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    $\begingroup$ 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. $\endgroup$ – Anthony Carapetis May 12 at 2:55
  • $\begingroup$ Thanks, yeah the notation was a bit confusing to me. $\endgroup$ – 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 $Y$.

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  • $\begingroup$ @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. $\endgroup$ – Ernie060 May 12 at 18:00

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