- #1
Decimal
- 75
- 7
Homework Statement
[/B]
For any metric ##g_{ab}## show that ##g_{ab} g^{ab} = N## where ##N## is the dimension of the manifold.
Homework Equations
$$ g_{ab} = \mathbf {e_a} \cdot \mathbf {e_b} $$
$$ g^{ab} = \mathbf {e^a} \cdot \mathbf {e^b} $$
The Attempt at a Solution
When I substitute the above equations I get the following: $$ g_{ab} g^{ab} = (\mathbf {e_a} \cdot \mathbf {e_b}) \cdot (\mathbf {e^a} \cdot \mathbf {e^b})$$ $$ g_{ab} g^{ab} =(\mathbf {e_a} \cdot \mathbf {e^a}) \cdot (\mathbf {e_b} \cdot \mathbf {e^b}) = 1 $$
So I arrive at a scalar 1 when contracting the tensor, which I think is to be expected, since the covariant and contravariant metric are each others inverses. However I don't see how this is then supposed to equal the dimension of the manifold. Any help would be greatly appreciated!
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