Dual vectors, 1-forms & other fun stuff

[Ben]

the [imath]dx_i[/imath] are not the basis vectors so [imath]df[/imath] is not the gradient, it's the differential. The former is a vector and the latter isn't.

Then I take it that you disagree with virtually everything I said in this thread?

 

Was I not at pains to point out that they are BOTH differential 1-forms, and that the [math]dx^i[/math] are indeed a basis for the cotangent space [math]T_p^{\ast}M[/math] for some arbitrary differentiable manifold [math]M[/math]?

 

In the context of differentiable manifolds, I think you may have to go a long way to find a source that denies that the gradient may be identified with a 1-form.

 

Since this is essentially the point of my thread (call it's creation ill-judged if you must), I rather wish you had voiced your disagreement earlier........

[Qfwfq]

...and that the [math]dx^i[/math] are indeed a basis for the cotangent space [math]T_p^{\ast}M[/math]

I know they are a basis of the cotangent space, but not of the tangent space. I thought covariant components as being those for a basis in the tangent space. :confused:

 

Well, my memories are a bit distant but I wouldn't have said the 1-form is the gradient; I see the nexus but wouldn't have called them the same thing. I would have identified the 1-form with the differential and not with the gradient. Sure, the two are related, but they're not the same thing.:shrug:

[Ben]

Well, let's see what you mean by is.

 

Suppose we are working in a vector space with an inner product defined - it's a metric space. Suppose further that the gradient of a function is given by [math]\nabla f[/math] and that it is a vector.

 

Then we know that the following must be true: for [math] g \in V^{\ast} \otimes V^{\ast}[/math] that [math]g(\nabla f, v) = \alpha \in \mathbb{R},\,\,\, v \in V[/math].

 

We must also have some [math]\varphi \in V^{\ast}[/math] such that [math]g(\nabla f, v) = \varphi(v) =\alpha[/math]. The object [math]\varphi[/math] is by construction a 1-form when [math]V^{\ast} = T^{\ast}_pM[/math].

 

Also that [math]df \in T^{\ast}M[/math]. Since the symbol is arbitrary, set [math]\varphi = df[/math], thus [math]g(\nabla f, v) = df(v)[/math].

 

Then, without argument (math joke!), now we will have that [math]g(\nabla

f,-) = df \in V^{\ast}[/math].

 

So we have the following situation: we have [math]\nabla f \in T_pM,\,\,\, g(\nabla f,- )= df \in T^{\ast}_pM[/math]; in the absence of an inner product, the LHS of this equality doesn't exist, so here we are FORCED to use the covariant form of the gradient, which is [math]df[/math].

 

This is just an example of the ugly truth I referred to earlier, that what we think of as being an identity all too often turns out to be so only up to isomorphism, in this case given by [math]T^{\ast}_pM \simeq T_pM[/math]

[Qfwfq]

Then, without argument (math joke!), now we will have that [math]g(\nabla

f,-) = df \in V^{\ast}[/math].

Which was kinda what I meant by saying that [imath]\nabla f[/imath] is related to [imath]df[/imath]. So I guess that's your meaning of is, as I suspected. ;) :D

 

As I said in another thread, it takes me a bit of effort, despite having seen these things already, only long ago. You can expect it to be much harder for others.