Second order derivative of a contrast function gives Reimanian metric:

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Let $M$ be a $C^{\infty}$ manifold. We can define a contrast function $\rho:M\times M\to \mathbb{R}$ satisfies $\rho(x,y)\geq0$ for all $x,y\in M$ and equality holds iff $x=y$.

In Eguchi's 1991 paper "Geometry of minimum contrast",he defines a Riemannian metric on M by $G(X,Y)=-\rho(X|Y)$ where $X,Y$ are vector fields on M. The notation $\rho(X|Y):=X_{x}Y_{y}\rho(x,y)$ where $ X_{x}=X^{i}_{x}\bigg(\frac{\partial}{\partial x_{i}}\bigg)_{x},Y_{y}=Y^{i}_{y}\bigg(\frac{\partial}{\partial x_{i}}\bigg)_{y}.$

I am trying to prove G as a Riemannian metric for that I need to prove

$\bullet$ G is bilinear. Which can be prove from the definition.

$\bullet$G(X,X)>0 for all $X\neq 0 .$

$\bullet$G(X,Y)=G(Y,X).

To prove positive-definiteness and symmetry properties we need the following two results

$\bullet \rho(Y| \cdot)=0$ for any vector field $Y$.

$\bullet$ $\rho(XY|\cdot)=-\rho(X|Y).$

I am unable to get these two and following is what I am trying:

Since $\rho (x,y)$ has minimum value $0$ at $x=y$ therefore we have

$\bigg(\frac{\partial}{\partial x_{i}}\bigg)_{x}\rho(x,y)\bigg|_{y=x}=0$ and $\bigg(\frac{\partial}{\partial x_{i}}\bigg)_{y}\rho(x,y)\bigg|_{y=x}=0$

Let $X,Y$ be two any vector fields over M.

Then $\rho(Y|X)=Y_{x}X_{y}\rho(x,y)\bigg|_{y=x}=0$.

That is $\rho(Y| \cdot)=0$ for any vector field $Y$.

Applying $X$ to $\rho(Y| \cdot)=0$ ,we have

$X\bigg(\rho(Y| \cdot)\bigg)=0$

$\implies X(Y_{x}\rho(x,y)\bigg|_{y=x})=0.$

The followings are from the paper

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How should I go to get the results?Is my way correct or am I mising something? I will be thankful if anyone can give me some hints on it.