# Pondering basis vectors and one forms

Gold Member
So, I've been thinking about this for a while...and I can't seem to resolve it in my head. In this thread I will use a tilde when referring to one forms and a vector sign when referring to vectors and boldface for tensors. It seems to me that if we require the basis vectors and one forms to obey the property that:

$$\tilde{\omega}^j(\vec{e}_i)=\delta_i^j$$

Then, we cannot require the basis vectors and one forms to be "dual" to each other in the sense that we raise and lower their indices with the metric tensor. I.e.:

$$\tilde{\omega}^i=\bf{g}(\vec{e}_i,\quad)$$

Since, unless my metric is the Cartesian metric, I cannot get the first property from the second quantity.

Is this true, or have I messed up somewhere?

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atyy
The "duality" between vector and one forms that you cite does not require a metric. When the vector basis is changed, the dual forms are also changed.

The "duality" between vectors and one forms that can be defined with a metric is different, as the duals do not change if you change basis.

So you are right. The former has nothing to do with raising and lowering indices with a metric (it's defined even without a metric), the latter involves raising and lowering indices with a metric.

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Gold Member
The natural set of one form basis vectors then, is defined by the first property and not the second one then right?

atyy
Yes.

Gold Member
I have another question. If we choose to use the tetrad method to describe our manifold, then the metric is automatically diag(1,1,1,...)? Since we are choosing ortho-normal basis vectors, then it would make sense that that would have to be the case.

I could never figure out Wald's take on the tetrad method...His abstract index notation there makes me really confused what his equations mean...>.>

atyy