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!
Best Answer
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 $Y$.