Derivatives of Finsler metric in the Riemannian case

Question

I'm trying to wrap my head around this article of Zhongmin Shenfor my Master thesis. In Proposition 8.2, he assures that in the Riemannian case in which $L(x,y)=g_{ij}(x)y^iy^j$, given the equality $$ L_{x^ky^l}y^k = 2L_{x^l}, $$ the following is true: $$\frac{\partial g_{il}}{\partial x^k}(x) + \frac{\partial g_{kl}}{\partial x^i}(x) = 2 \frac{\partial g_{ik}}{\partial x^l}(x). $$ I think the Euler theorem ($L_{y^i}y^i = 2L$) must be used somewhere to arrive to the result, but I'm very lost. Can you kindly give some guidance?

Answer 1

Given that $L_{x^ky^l}y^k = 2L_{x^l}$. We have $$L(x,y)=g_{ij}(x)y^iy^j.$$ Differentiating w.r.t. $x^k$ we find that $$L_{x^k}=\frac{\partial g_{ij}}{\partial x^k}y^iy^j \text{, [$x,y$ are omitted for simplicity.]}$$ Differentiating w.r.t. $y^l$ we get $$L_{x^ky^l}=\frac{\partial g_{ij}}{\partial x^k} \frac{\partial y^i}{\partial y^l}y^j+\frac{\partial g_{ij}}{\partial x^k} \frac{\partial y^j}{\partial y^l}y^i$$ $$\implies L_{x^ky^l}y^k=\frac{\partial g_{ij}}{\partial x^k} \frac{\partial y^i}{\partial y^l}y^jy^k+\frac{\partial g_{ij}}{\partial x^k} \frac{\partial y^j}{\partial y^l}y^iy^k$$ $$\implies L_{x^ky^l}y^k=\frac{\partial g_{ij}}{\partial x^k} \delta^i_l y^jy^k+\frac{\partial g_{ij}}{\partial x^k} \delta^j_l y^iy^k,\text{ $\delta^i_j$ being Kronecker's delta}$$ $$\implies L_{x^ky^l}y^k=\frac{\partial g_{lj}}{\partial x^k} y^jy^k+\frac{\partial g_{il}}{\partial x^k} y^iy^k,\text{ $i=l$ on first term $j=l$ on second term}$$ Interchanging $j$ and $k$ in the first term and replacing $j$ by $i$ on that term we get $$L_{x^ky^l}y^k = \frac{\partial g_{kl}}{\partial x^i} + \frac{\partial g_{il}}{\partial x^k}.$$ In similar approach it can be shown that $$2L_{x^l} = 2 \frac{\partial g_{ik}}{\partial x^l}$$ by suitable change of dummy index.

Since $L_{x^ky^l}y^k = 2L_{x^l}$ holds so we have $\displaystyle \frac{\partial g_{il}}{\partial x^k} + \frac{\partial g_{kl}}{\partial x^i} = 2 \frac{\partial g_{ik}}{\partial x^l}$.

Hope it works!