What Does the Metric Tensor Imply in This Context?

[redstone]

I've read that the metric tensor is defined as
{{g}^{ab}}={{e}^{a}}\cdot {{e}^{b}}

so does that imply that?
{{g}^{ab}}{{g}_{cd}}={{e}^{a}}{{e}^{b}}{{e}_{c}}{{e}_{d}}={{e}^{a}}{{e}_{c}}{{e}^{b}}{{e}_{d}}=g_{c}^{a}g_{d}^{b}

 

[TSny]

No, that can't be correct. For example, let $a = 1, b = 2, c = 1, d = 2$ and assume we're dealing with the standard Minkowski metric.

Then $g^{ab}g_{cd} = g^{12}g_{12} = 0\cdot 0 = 0$

But, $g^a_cg^b_d = g^1_1g^2_2 = 1\cdot 1 = 1$

 

[Dick]

I've read that the metric tensor is defined as
{{g}^{ab}}={{e}^{a}}\cdot {{e}^{b}}

so does that imply that?
{{g}^{ab}}{{g}_{cd}}={{e}^{a}}{{e}^{b}}{{e}_{c}}{{e}_{d}}={{e}^{a}}{{e}_{c}}{{e}^{b}}{{e}_{d}}=g_{c}^{a}g_{d}^{b}

Nope. You lost the dot product completely when you went to the next expression. If your metric is diagonal is g^{01}g_{01}=g^0_0 g^1_1?

 

[Chestermiller]

The easiest way to understand the metric tensor is to use dyadic notation:

I = (e_i\cdote_j)eⁱe^j = g_ij eⁱe^j = (eⁱ\cdote^j)e_ie_j = g^ij e_ie_j

Any vector or tensor dotted with the metric tensor returns that vector or tensor unchanged. Thus, the metric tensor can be regarded as the identity tensor.

 

[redstone]

Makes sense. thanks for all the help