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I do understand the representation of a vector as: v^{i}∂x_{i}

I also understand the representation of a vector as: v_{i}dx^{i}

So far, so good.

I do understand that when the basis transforms covariantly, the coordinates transform contravariantly, and v.v., etc.

Then, I study this thing called the gradient.

If I work out an indicial notation, I get this:

∇f=(∂f/∂x^{i})dx^{i}

Now comes my trouble.

I can "envision" in my mind, that ∂x_{i}are tangent to the coordinate curves and are, essentially, directions.

But I cannot see the "directions" of dx^{i}

I cannot see them as "basis vectors" as easily as I see ∂x_{i}

I do understand that dx^{i}∂x_{j}= δ^{i}_{j}

And I understand how the direction of, say in 3D space, dx^{1}is perpedicular to the plane formed by ∂x_{2}and ∂x_{3}.

But I cannot easily see the "directions" of the basis of the dual vectors as easily as I can see the basis of the original vectors (as tangent to the coordinate curves).

I cannot make the leap and replace dx^{i}bye,^{1}e,^{2}e, and easily as I can replace ∂x^{3}_{i}withe,_{1}e,_{2}e_{3}

I can begin with the definition of how I construct a dual basis... But I cannot easily make the leap to see the basis as this form dx^{i}. I just don't see "directions" here.

Can someone provide some insight?

Also, given a metric, I can convert the basis of the gradient to a covariant form and the components to a contravariant. So why is the gradient called contravariant, when it can go either way with a metric?

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# A How do I see a dual basis

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