Is the Contracted Metric Equal to the Dimension of the Manifold?

[Decimal]

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!

 

[TeethWhitener]

Don't forget the implied summation:

https://en.wikipedia.org/wiki/Einstein_notation

 

[Decimal]

Oh wow, I completely forgot that that applies at the end as well. Thanks a lot!