# Riemann Manifold equipped with Euclidean metric

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?

• It is not clear what is to he proved. It seems that 1 is just stating the definition, and 2 is an property of the connection and there is no 3. – Arctic Char Jul 20 '20 at 13:25
• Thanks for the response. I was wondering If someone could prove 2. and if it is possible to relate it with the differentiation of the inner product in the given link. PS: Sorry about 3. mistake. – Thoth Jul 20 '20 at 13:36
• Do you know the definition of $\bar \nabla _{X_p} Y_p$? – Arctic Char Jul 20 '20 at 16:09
• $\bar{\nabla}_{X_p} Y_p=X_p(Y_p)=\sum_{i,j=1}^{n}\mathcal{X}^i(p)\frac{\partial\mathcal{Y}^j}{\partial x_i} \Bigr|_{p}\frac{\partial}{\partial x^j}\Bigr|_{p}$, where $\mathcal{X}_i,\mathcal{Y}_j \in \mathbb{R}$ the coordinate coefficients of vector fields $X_p,Y_p$, respectively. Is there any specific property of $g_p$ which can be used along with $\bar{\nabla}_{X_p}Y_p$ definition? – Thoth Jul 20 '20 at 17:48
• Yes you've got the correct definition. Then (2) is really a direct calculations. – Arctic Char Jul 20 '20 at 17:59

## 1 Answer

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.