# Time derivative of the Levi-Civita Connection (Ricci Flow)

• Apr 27th 2013, 11:29 PM
engmaths
Time derivative of the Levi-Civita Connection (Ricci Flow)
Hello,

I am learning about the Ricci Flow, reading: Lectures on the Ricci flow - Peter Topping (download pdf).

In Proposition 2.3.1 (i.e. page 28) the expression $\frac{\partial}{\partial t}{\nabla}_X Y$ appears (To understand the equation, you may have to read the few lines above Proposition 2.3.1). I don't know how this is defined. Can someone tell me?

In addition, I don't know the definition of $\langle h,\alpha \rangle$ in Proposition 2.3.6.

Regards,
engmaths
• Apr 29th 2013, 08:56 AM
xxp9
Re: Time derivative of the Levi-Civita Connection (Ricci Flow)
In each time t, let V=DxY is a vector field on M.
At time 0, d/dt( V ) = (Vt - V0)/t, this can be defined on each point of M.
<h, a> is the contraction of tensors. The result is a scalar. Just like the contraction between a vector and a co-vector:
<gradf, v>=<df, v> = v(f) = the directional derivative of f on the direction of v.
• May 4th 2013, 11:22 AM
engmaths
Re: Time derivative of the Levi-Civita Connection (Ricci Flow)
Hello xxp9, that helped me, thank you!

I got another quick question that I just want to add here:

Let $\alpha$ be a tensorfield , X a vector field, $\varphi$ a diffeomorphism of a Riemannian Manifold M and L the Lie derivative.

I want to show:
$\varphi^*(L_X\alpha)=L_{(\varphi^{-1})_*X}(\varphi^*\alpha)$

Let b_t be the local flow of X around p. Then $\varphi^{-1}\circ b_t$ is the local flow of $(\varphi^{-1})_*X$ around p. Then
$L_{(\varphi^{-1})_*X}(\varphi^*\alpha)$=d/dt t=0 $((\varphi^{-1}\circ b_t)^*(\varphi^*\alpha))_p$ = d/dt t=0 $((b_t)^*\alpha)_p=(L_X\alpha)_p$
what seems to be wrong. What is wrong?

the flow that induces $(\varphi^{-1})_* X$ is $\varphi^{-1} \circ b_t \circ \varphi$.
Thanks xxp9, I looked up the definition of $(\varphi^{-1})_* X$ and get the right solution now.