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?

  • $\begingroup$ 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. $\endgroup$ – Arctic Char Jul 20 '20 at 13:25
  • $\begingroup$ 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. $\endgroup$ – Thoth Jul 20 '20 at 13:36
  • $\begingroup$ Do you know the definition of $\bar \nabla _{X_p} Y_p$? $\endgroup$ – Arctic Char Jul 20 '20 at 16:09
  • $\begingroup$ $\bar{\nabla}_{X_p} Y_p=X_p(Y_p)=\sum_{i,j=1}^{n}\mathcal{X}^i(p)\frac{\partial\mathcal{Y}^j}{\partial x_i} \Bigr|_{p}\frac{\partial}{\partial x^j}\Bigr|_{p}$, 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? $\endgroup$ – Thoth Jul 20 '20 at 17:48
  • $\begingroup$ Yes you've got the correct definition. Then (2) is really a direct calculations. $\endgroup$ – Arctic Char Jul 20 '20 at 17:59

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.


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