Diff. Geom. question

[Ben]

So I have been reviewing my differential geometry notes, and there is an equality I don't quite get.

 

Suppose that [math]M[/math] is a [math]C^{\infty}[/math] manifold. Then for any [math]m \in M[/math] we have a vector space tangent at this point called, reasonably enough, a tangent space. Denote this by [math]T_mM[/math] and suppose that [math]v \in T_mM[/math].

 

Further suppose that the point [math]m \in M[/math] can be defined relative to the set of local coordinate functions [math]\{x^i\}[/math].

 

Then the basis vectors for this space will be the linear operators [math]\frac{\partial}{\partial x^i}[/math], and [math]v \in T_mM[/math] can be written as [math]v=\sum\nolimits _i \alpha^i \frac{\partial}{\partial x^i}[/math] (where the [math]\{\alpha^i\}[/math] is a set of scalars).

 

My question, then, is this: I am told that [math] vx^i = \alpha^i[/math], and I am having trouble extracting this equality.

 

I worked by analogy:

 

Suppose [math]V[/math] is a vanilla vector space with inner products. Let the set [math]\{e_j\} [/math] denote the basis vectors.

 

Now the inner product of an arbitrary vector [math]v = \sum \nolimits_j \alpha^j e_j [/math]with any basis vector will be denoted by [math](v,e_i) = \sum \nolimits_j(a^j e_j, e_i)=\sum \nolimits_j a^j( e_j, e_i)[/math] (since inner products are bilinear) hence [math] (v, e_i) = \sum \nolimits_j a^j(e_j,e_i) = \sum \nolimits_j a^j \delta _{ij} = a^j[/math] where the Kronecker delta is 1 when i = j, and zero otherwise.

 

Obviously, in the present case we may not assume an inner product, and moreover, the [math]\{x^i\}[/math] are coordinate functions, not basis vectors, so my analogy might usually be expected to fail in the present case.

 

I expect I could do this once, but premature senility seems to have set in........

[Ben]

Ah well, it is nothing so romantic as premature senility, rather just a garden-variety stupidity.

 

Doh!

 

So here, after a few minutes of scribbling....

 

If [math]v \in T_mM = \sum \nolimits_j \alpha ^j \frac{\partial}{\partial x^j}[/math] then [math] vx^i = \sum \nolimits_j \alpha^j \frac{\partial}{\partial x^j} x^i = \sum \nolimits_j \alpha^j \delta_{ij} = \alpha^i[/math] (since both coordinate functions and basis vectors are independent by construction).

[Qfwfq]

Gosh I got here almost in time to tell you that! C'mon Ben, you've no idea of the blunders that we can make, I stepped into here just after recognizing one of my very own. :hihi:

[Jay-qu]

I see the maths fairly plainly, but I am struggling to interpret this physically. So v is a tangent to the manifold at a point, and the [math]x^i[/math] are just local coordinates. So then the inner product of the two [math]vx^i[/math] gives you the set of scalars [math]\alpha^i[/math] but what are these scalars?

[Qfwfq]

Concening Ben's thread on differential forms, a scalar function over the manifold is just a 0-form. A (covariant) vecotr field can be defined by the manner he describes 1-forms there and is a differential operator; more precisely the vector turns into a 1-form when applied to a scalar function and contracted with [imath]dx^{\mu}[/imath]. The differential:

 

[math]df(p)=\frac{\partial f(p)}{\partial x^{\mu}}dx^{\mu}[/math]

 

can be regarded as a specific case with the vectors components all being 1 and the [imath]dx^{\mu}[/imath] are the contravariant basis vectors. One can regard the gradient of a function as being a covariant vector that gives an exact 1-form who's integral is the function (up to a constant). Well i'm just blabbering from my vague memories..:D

 

Of course the point [imath]p[/imath] is represented in a given chart by the coordinates so [imath]f(p)[/imath] translates into a function of them. Under a change of chart [imath]f(p)[/imath] becomes a different function, such that the two have the same value for corresponding coordinates, IOW at the same point [imath]p[/imath].

[sanctus]

Jay, the easy interpretation is that [math]{\alpha_i}[/math] are just the coefficients of the decomposition of [math]v[/math] with respect to the basis [math]\frac{\partial}{\partial x^i}[/math], which is a local basis, but still a basis ;-)...

But I guess, you knew that and where more interested in what it is mathematically, described by Qfwfq...

[Qfwfq]

Actually I realize now there seems to be an ambiguity:

So then the inner product of the two [math]vx^i[/math] gives you the set of scalars [math]\alpha^i[/math] but what are these scalars?

yeah, this refers to the vector's components, which are not scalars in the tensorial sense; they are just elements of the field over which the vector spaces are defined. Linear combinations and all that....

[Ben]

So then the inner product of the two [math]vx^i[/math] gives you the set of scalars [math]\alpha^i[/math] but what are these scalars?

Jay, hope the essence of this question has been answered by Q and Sanctus (my thanks to them for illuminating my somewhat opaque contribution).

 

However, I have to say that [math]vx^i[/math] is not an inner product. It not even a set of inner products! I admit that my notation was perhaps confusing, but I cannot find a better one.

 

So, if we recognize that the [math]x^i[/math] are the coordinates curves that pass through the point [math]m \in M[/math], and that the basis vectors are the directional derivatives constrained to "be" on these curves, then by [math]vx^i[/math] I meant that, given this basis and the decomposition that Sanctus mentioned, then when we offer to our generic vector [math]v[/math] a curve, he in return offers us the component of the decomposition relative to that curve.

 

Perhaps a better notation might have been something like [math]v(x^i)[/math] though this is not without its faults; we want the express the above as the evaluation of our vector as a scaling of the basis vectors on this curve.

 

Trouble is this looks like the vector is itself a function, and yet it seems to make no sense to talk about the components as being the image of the coordinates.

 

Or maybe [math]v|_{x^i}[/math]? Which would read "[math]v[/math] evaluated on [math]x^i[/math]".

 

I have a comment or 2 about Q's nice offering, but now I have to do some work.......

[Jay-qu]

That makes sense thanks guys.