# what is the inner product appeared in front of the integral?

Given a compact manifold with a Riemannian metric $g$, we define the total scalar curvature by $$E(g)=\int_M RdV$$

Let us consider the first variation of $E$ under an arbitrary change of metric. We write $h=\frac{\partial g}{\partial t}$.

One can derive that $$\frac{d}{dt} \int_M RdV=\int_M \langle\frac{R}{2}g-Ric, h\rangle dV$$

Question: In "Lectures On The Ricci Flow" by Peter Topping, $\nabla E(g)=\frac{R}{2}g-Ric$ is concluded by above derivation, where $\nabla E$ is gradient of $E$.

I dont know what is the inner product $\langle .\rangle$ appeared in front of the integral?

Thanks.

• @Jesse Madnick, page 67. – Sepideh Bakhoda Apr 28 '13 at 7:55

The footnote on page 25 states that: "We use $g(\cdot, \cdot)$ and $\langle \cdot, \cdot\rangle$ interchangeably, although with the latter, it is easier to forget any $t$-dependence that $g$ might have."
• what I dont understand is when metric is changing by time, $g(.,.)$ belongs to which time? – Sepideh Bakhoda Apr 28 '13 at 8:04
• I think the time $t$ is variable. That is, it would be more accurate to write $\langle \cdot, \cdot \rangle_t$, but Topping is suppressing $t$ from the notation (even though the dependence is there). I'll edit to include the full quotation. – Jesse Madnick Apr 28 '13 at 8:06
• Oh, but if he's using $\frac{d}{dt}$ to mean $\left.\frac{d}{dt}\right|_{t=0}$ -- which I think he might be doing -- then in this particular case $t = 0$. – Jesse Madnick Apr 28 '13 at 8:08