I often think I have fully understood this, then some question comes up in my mind, and I get confused again (which implies I never understood it in the first place).

We have a co-ordinate basis for vectors [itex]{\partial_\mu}.[/itex] I can think of two ways to get a corresponding basis for covectors.

1. Define basis covectors [itex]{dx^\mu}[/itex] such that

[tex] \partial_\mu dx^\nu = \delta^\nu_\mu.[/tex]

This gives me a problem. Am I allowed to combine vectors and covectors without a metric? This leads me onto the second way...

2. Use the metric so that

[tex]g(\partial_\mu, dx^\nu) = \delta^\nu_\mu[/tex]

but actually... the metric maps from two vectors to a scalar, am I correct? So perhaps what I mean is

[tex]g(\partial_\mu, \partial_\nu) = g^{\mu\nu}\partial_\mu \partial_\nu = \partial_\mu dx^\nu = \delta^\nu_\mu[/tex]

which is in fact the same way as "method 1".

I guess my real question is

Many thanks in advance,

Masud.

P.S. It appears that I have largely "thought aloud" in this post, and may well have answered my own question... but I could do with confirmation.

We have a co-ordinate basis for vectors [itex]{\partial_\mu}.[/itex] I can think of two ways to get a corresponding basis for covectors.

1. Define basis covectors [itex]{dx^\mu}[/itex] such that

[tex] \partial_\mu dx^\nu = \delta^\nu_\mu.[/tex]

This gives me a problem. Am I allowed to combine vectors and covectors without a metric? This leads me onto the second way...

2. Use the metric so that

[tex]g(\partial_\mu, dx^\nu) = \delta^\nu_\mu[/tex]

but actually... the metric maps from two vectors to a scalar, am I correct? So perhaps what I mean is

[tex]g(\partial_\mu, \partial_\nu) = g^{\mu\nu}\partial_\mu \partial_\nu = \partial_\mu dx^\nu = \delta^\nu_\mu[/tex]

which is in fact the same way as "method 1".

I guess my real question is

**can I define covectors without a metric**? This is related to another thread in the forum, but I could do with a quick answer saying if my approach is correct.Many thanks in advance,

Masud.

P.S. It appears that I have largely "thought aloud" in this post, and may well have answered my own question... but I could do with confirmation.

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