0
$\begingroup$

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!

$\endgroup$
  • 1
    $\begingroup$ 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
1
$\begingroup$

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 $Y$.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ @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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.