# Interval and a general tensor

1. Apr 7, 2013

### redstone

I'm trying to understand what kind of relation the metric can have with a general tensor B.

$$d{{x}^{a}}d{{x}^{b}}{{g}_{ab}}=d{{s}^{2}}$$
$$\frac{d{{x}^{a}}d{{x}^{b}}{{g}_{ab}}}{d{{s}^{2}}}=1$$
$$\frac{d{{x}^{a}}d{{x}^{b}}{{g}_{ab}}}{d{{s}^{2}}}=\frac{1}{D}g_{a}^{a}$$
$$\frac{d{{x}^{a}}d{{x}^{b}}{{g}_{db}}g_{a}^{d}}{d{{s}^{2}}}=\frac{1}{D}g_{d}^{a}g_{a}^{d}$$
$$\frac{d{{x}^{a}}d{{x}^{b}}{{g}_{db}}g_{a}^{d}}{d{{s}^{2}}}A_{m}^{n}= \frac{1}{D}g_{d}^{a}g_{a}^{d}A_{m}^{n}$$
Define: $$B_{am}^{dn}=g_{a}^{d}A_{m}^{n}$$
substitute in
$$\frac{d{{x}^{a}}d{{x}^{b}}{{g}_{db}}}{d{{s}^{2}}}B_{am}^{dn}=\frac{1}{D}g_{d}^{a}B_{am}^{dn}$$
$$\frac{d{{x}^{a}}d{{x}^{b}}{{g}_{db}}}{d{{s}^{2}}}B_{am}^{dn}=\frac{1}{D}B_{am}^{an}$$

It all looks Ok to me. Does all of the following look reasonable, or is there a problem somewhere?