# D/dt g(t) and d/dt D_X Y

1. Apr 27, 2013

### Sajet

Hi!

I'm reading this script* and I fail to understand a rather simple calculation. I assume the problem lies in me not understanding the notation that is used, and I was unable to figure it out or find it in literature.

We have a smooth family of metrics $g = g_t$ on a Riemannian manifold, and we set $h := \frac{\partial}{\partial t}g_t$.

First question:

$\frac{\partial}{\partial t} \nabla_X Y$: Does this mean $\frac{\partial}{\partial t} \nabla_X^t Y$, where $\nabla^t$ is the Levi-Civita connection w.r.t $g_t$?

Second question:

The script says:

$\langle \frac{\partial}{\partial t} \nabla_X Y, Z\rangle = \frac{\partial}{\partial t}g(\nabla_X Y, Z\rangle - h(\nabla_X Y, Z)$

I don't understand this step. Also I don't see the difference between the two terms

$\frac{\partial}{\partial t}g(\nabla_X Y, Z\rangle$ and

$h(\nabla_X Y, Z)$

In class we defined

$\frac{\partial g}{\partial t}(X, Y) := \frac{\partial}{\partial t}(g_t(X, Y))$.

Therefore those two terms seem the same to me.

I would appreciate any help :)

* http://homepages.warwick.ac.uk/~maseq/RFnotes.html , p. 32.

2. Apr 27, 2013

### Fredrik

Staff Emeritus
That's what I would guess too. (I don't see what else it could mean).

I haven't done this sort of thing in a while, but I think that the first term on the right is equal to a sum of three terms (as if we're taking the derivative of a product of three functions), with the d/dt acting on a different "factor" in each term. The term with d/dt acting on the Z is zero.

I interpret the first one as $$\frac{\partial}{\partial t}\left(g(\nabla_X Y, Z)\right)$$ and the second one as
$$\left(\frac{\partial}{\partial t}g\right)(\nabla_X Y, Z).$$

3. Apr 27, 2013

### Sajet

So what exactly is the difference between them?

I would interpret:

$$\frac{\partial}{\partial t}\left(g(\nabla_X Y, Z)\right)$$

as $$\frac{\partial}{\partial t} (t \mapsto \left(g_t(\nabla_X Y, Z)\right))$$

But then, how is the other one different? As I said, we defined:

$\frac{\partial g}{\partial t}(X, Y) := \frac{\partial}{\partial t}(g_t(X, Y))$

Which would lead me to the same equation as the first one?

4. Apr 27, 2013

### Fredrik

Staff Emeritus
$\left(\frac{d}{d t}g\right)(\nabla_X Y, Z)$ is just one of the three terms you get when you compute $\frac{d}{d t}\left(g(\nabla_X Y, Z)\right)$.

I don't understand your definition. Are you sure the left-hand side isn't supposed to be $\frac{d}{dt}g(X,Y)$? That would make sense if it's not really a definition, and your teacher was just trying to explain that the notation $\frac{d}{dt}g(X,Y)$ is to be interpreted as $\frac{d}{dt}\left(g(X,Y)\right)$ and not as $\left(\frac{d}{dt}g\right)(X,Y)$.

5. Apr 27, 2013

### Sajet

Mhh, the definition is definitely the same in my notes. It was supposed to define what exactly is meant when writing $\frac{\partial g}{\partial t}$ in the Ricci Flow-equation.

Would you mind telling me what those three terms are when I compute $\frac{d}{d t}\left(g(\nabla_X Y, Z)\right)$? Maybe my problem is that I don't understand how this derivative is calculated.

6. Apr 27, 2013

### Fredrik

Staff Emeritus
For all vector fields X,Y, we have
$$g(X,Y)=g(X,Y)=g(X^\mu\partial_\mu,Y^\nu\partial_\nu) =g_{\mu\nu}X^\mu Y^\nu,$$ where the $\partial_\mu$ are the tangent vector fields associated with an arbitrary coordinate system. The right-hand side is just a product of three functions, so if we want to compute a derivative of the left-hand side, we can use the product rule on the right-hand side.
$$\frac{d}{dt}g(X,Y) =\left(\frac{d}{dt}g_{\mu\nu}\right) X^\mu Y^\nu + g_{\mu\nu}\left(\frac{d}{dt}X^\mu\right) Y^\nu + g_{\mu\nu}X^\mu \left(\frac{d}{dt}Y^\nu\right).$$

7. Apr 27, 2013

### Sajet

Ah, I didn't think of that. Thanks for all your help!