How does the riemann tensor symplify to normal components in the second variation formula?

Question

I have a doubt in the last step of the derivation of the second variation formula for the length functional (Lee's intro to riemannian geometry) (theorem 10.22 see the proof below)

Why is it true that

$Rm(\gamma',\gamma',\cdot,\cdot)=Rm(\cdot,\cdot, \gamma',\gamma')=0$?

The first one seems to be interchange of 1st and 2nd entry with 3rd and 4th entry, but that doesn;t help me to see that it is 0. Furthermore even knowing this...

how does that implies that:

$Rm(V,\gamma',\gamma',V)=Rm(V^\perp,\gamma',\gamma',V^\perp)$ ?

I tried using symmetries of the Riemann tensor, but nothing seems to work. Could you show me how is done?

$Rm(V,\gamma',\gamma',V)=Rm(V^\perp,\gamma',\gamma',V^\perp)$ ?

$Rm(V,\gamma',\gamma',V)=Rm(V^T+V^\perp,\gamma',\gamma',V^T+V^\perp)=$ $Rm(V^\perp,\gamma',\gamma',V^\perp)+Rm(V^T,\gamma',\gamma',V^T)+Rm(V^T,\gamma',\gamma',V^\perp)+Rm(V^\perp,\gamma',\gamma',V^T) $

Somehow the last 3 terms should cancel

As a side thing why is $D_t \gamma'=0?$

Answer 1

• You have skew-symmetry in the first two slots, $R_{abcd}=-R_{bacd}$, and also in the last two slots, $R_{abcd}=-R_{abdc}$, hence the $0$.
• Just plug in $V=V^{\top}+V^{\bot}=c\dot{\gamma}+V^{\bot}$ into $R(V,\dot{\gamma},\dot{\gamma},V)$ and use multilinearity. You’ll get a sum of four terms. One of them will be $R(V^{\bot},\dot{\gamma},\dot{\gamma},V^{\bot})$, and the other three will vanish because each of those terms will have $3$ occurrences of $\dot{\gamma}$, so by the skew-symmetry property mentioned above, will vanish.

Another way of saying this is that $R(\dot{\gamma},\dot{\gamma},\cdot,\cdot)=0$ implies by scalar multiplying in the first slot that $R(V^{\top},\dot{\gamma},\cdot,\cdot)=0$. Similarly, $R(\cdot,\cdot,\dot{\gamma},V^{\top})=0$, so the last three terms vanish.