Riemann Manifold equipped with Euclidean metric

Question

I am trying to understand the following equations about the Euclidean metric:

1. $g_p = dx^1 \otimes dx^1 + \dots +dx^n \otimes dx^n$
2. $X_p\cdot g_p (Y_p,Z_p) = g_p(\bar{\nabla}_{X_p}Y_p,Z_p) + g(Y_p, \bar{\nabla}_{X_p}Z_p)$

About 1. I thing I got it if you thought $g_p$ as an element of the space defined by the tensor product of the cotangent spaces $T^*M \otimes_{M} T^*M = \bigcup\limits_{p \in M} T^*_pM \otimes T^*_pM$. 2. seems line differentiating inner product but how can we prove it with $X_p$ applied as a derivation to a smooth manifold?

Thanks. Could you please help to prove 2. analytically?

Answer 1

Let $(x^1, \dots, x^n)$ a local coordinate system and $$X_p = \sum_i \mathcal{X}_{i=1}^n \frac{\partial}{\partial x^i}\Biggr|_p,$$ $$Y_p = \sum_i \mathcal{Y}_{i=1}^n \frac{\partial}{\partial x^i}\Biggr|_p,$$ $$Z_p = \sum_i \mathcal{Z}_{i=1}^n \frac{\partial}{\partial x^i}\Biggr|_p.$$ vector fields. Moreover, for a Riemann metric $g$ we have

$$g(Y_p,Z_p) = \sum_{ij} \mathcal{Y}^i_p\mathcal{Z}^j_p g_{ij}\left(\frac{\partial}{\partial x^i}\Biggr|_p, \frac{\partial}{\partial x^j}\Biggr|_p\right).$$.

In addition, for two vector fields $Y_p,Z_p$ the Euclidean connection $\bar{\nabla}_{X_p}Y_p$ is given by

$$\bar{\nabla}_{X_p}Y_p = \sum_j X_p(\mathcal{Z}^j)\frac{\partial}{\partial x^j}\Biggr|_p$$

Now apply $X_p$ as derivation on $g$ and we obtain

$$X_p \cdot g(Y_p,Z_p) = \sum_{i,j}X_p(\mathcal{Y}^i\mathcal{Z}^j)g_{ij}\left(\frac{\partial}{\partial x^i}\Biggr|_p, \frac{\partial}{\partial x^j}\Biggr|_p\right) = \sum_{i,j}[(X_p\mathcal{Y}^i\mathcal{Z}^j) +\mathcal{Y}^i X_p\mathcal{Z}^j) ]g_{ij}\left(\frac{\partial}{\partial x^i}\Biggr|_p, \frac{\partial}{\partial x^j}\Biggr|_p\right) = g(\bar{\nabla}_{X_p}Y_p,Z_p) + g(Y_p,\bar{\nabla}_{X_p} Z_p).$$

^{Thank @Arctic Char for useful comments.}