intro to differential forms

cephas

let me just check my understanding real quick on the tensor product.

So dx and dy are 1-forms. They are also linear functionals. So how does dx and dy operate on a vector? Is it the same as the derivative? dx(x)=1 or dx(x^2)=2x? I am not quite sure I get that...

And then with the tensor product (which is why I asked the above question) could I go like dxdy(x,y)=1*1=1? or dxdy(y,x)=0? Do I have the right idea here?

lethe

Originally posted by cephas
So dx and dy are 1-forms. They are also linear functionals.

yes, and yes.

So how does dx and dy operate on a vector? Is it the same as the derivative? dx(x)=1 or dx(x^2)=2x? I am not quite sure I get that...

those equations are no good, because x and x^2 are not vectors in the tangent space, and therefore it is invalid to act on them with a differential form. the hardest leap to make in this thread is thinking of vectors as differential operators on functions. d/dx is a good tangent vector that a differential form might eat for breakfast. x^2 is a function that a vector might eat for lunch.

And then with the tensor product (which is why I asked the above question) could I go like dxdy(x,y)=1*1=1? or dxdy(y,x)=0? Do I have the right idea here?

assuming you replace x and y in the arguments of those equations, then yes, you have exactly the right idea.

rick1138

Great thread everyone. Here is a quick example of calculating with forms. In general, the value of a form is the value of the determinant of the matrix formed by the vectors that define it. Consider the wedge sum of two one forms:

α/\β &equiv; 1/2 (α&otimes;β - β&otimes;α)

Hey, does anyone know how I can create sub- and superscripts here, I noticed html is off, I can't finish this post othewise? Any info would be appreciated.

selfAdjoint

Use brackets [ , ] instead of < , >. Otherwise just the same.

rick1138

Thanks, I am going to start again. In general, the value of a form is the value of the determinant of the matrix formed by the vectors that define it. Consider the wedge sum of two one forms:

α/\β &equiv; 1/2 (α&otimes;β - β&otimes;α)

and consider the two one-forms in expanded form:

α &equiv; α₁e₁ + α₂e₂ + α₃e₃

β &equiv; β₁e₁ + β₂e₂ + β₃e₃

and remember these rules for the bases of forms:

e_i /\ e_j = - e_j /\ e_i

(reversing order reverses sign)

e_i /\ e_i = 0

(wedge summing a basis form by itself annihilates it)

so:

α/\β = 1/2 ((α₁e₁ + α₂e₂ + α₃e₃) &otimes; (β₁e₁ + β₂e₂ + β₃e₃) - (β₁e₁ + β₂e₂ + β₃e₃) &otimes; (α₁e₁ + α₂e₂ + α₃e₃)) =
(α₁β₁ e₁ /\ e₁ + α₁β₂ e₁ /\ e₂ +α₁β₃ e₁ /\ e₃ +α₂β₁ e₂ /\ e₁ +α₂β₂ e₂ /\ e₂ +α₂β₃ e₂ /\ e₃ +α₃β₁ e₃ /\ e₁ +α₃β₂ e₃ /\ e₂ +α₃β₃ e₃ /\ e₃) - (β₁α₁ e₁ /\ e₁ + β₁α₂ e₁ /\ e₂ +β₁α₃ e₁ /\ e₃ +β₂α₁ e₂ /\ e₁ +β₂α₂ e₂ /\ e₂ +β₂α₃ e₂ /\ e₃ +β₃α₁ e₃ /\ e₁ +β₃α₂ e₃ /\ e₂ +β₃α₃ e₃ /\ e₃)

which reduces to, by virtue of the rules above:

(α₁β₂ - α₂β₁) e₁ /\ e₂ + (α₂β₃ - α₃β₂) e₂ /\ e₃ + (α₃β₁ - α₁β₃ )e₃ /\ e₁

Which is isomorphic to the cross product, but not quite the same thing - the wedge sum of two one forms produces a two-form, wheras the cross product produces a scalar - the cross product is the Hodge star dual of the wedge sum of two one-forms.

turin

Originally posted by lethe
... [tex]\mathbb{R}^n[/tex] is itself a manifold, albeit a flat one, but we want to extend our idea of a space to include curved spaces, so lemme just give a few examples: a parabola is a curved 1 dimensional manifold, that extends to infinity. a circle is a 1 dimensional manifold that folds back on itself.
...
a manifold is just a space that is not necessarily flat.

Can you give the justification that [tex]\mathbb{R}^n[/tex] is flat and that a parabola is curved (I'm assuming that you mean a parabola to be the 2-D representation of all points (x,y) that exist in the 2-D space which satisfy the coordinate values y = x², or some scaled, translated, or rotated version thereof).

I can see the literal curvature of the parabola, and the literal flatness of the x axis ([tex]\mathbb{R}^1[/tex]) if I view them in the context of the x-y plane, but I don't see how you can characterize such a thing in 1-D, and such characterization seems to be in the spirit of this thread.

Here's my problem: I can't see the fundamental difference between the parabola and the real number line as 1-D manifolds. Both have 1 dimension (another concept I don't quite understand, but I'll defer that until later). I don't see what more you can say without a metric, or at least a coordinatization. If I choose to label the points on the parabola by the arclength along the parabola from the origin (which is, IMO, the most natural way to do it), then how would I know it was curved? Alternatively, how would I know to label points using their x values to show the curvature, when, for the sake of purity, I should not be appealing to any x axis in some x-y plane? In other words, how do I know that the parabola imbeds itself in the x-y plane as a parabola instead of a straight flat line, without already knowing that it was, in fact, a parabola in the x-y plane.

turin

Originally posted by Tom
Vectors v_i (i=1,2,3,...) in Rⁿ are independent iff

a₁v₁+a₂v₂+a₃v₃+...=0

implies that

a₁+a₂+a₃+...=0

Did you mean for:

"a₁+a₂+a₃+...=0"

to be:

"a₁=0, a₂=0, a₃=0, ...?"

turin

Originally posted by lethe
for any vector v, -v is also a vector.

I was just a little uncomfortable with this notation. Do you mean:

-v is defined as (-1)v

where (-1) is a member of the field? Isn't this rather trivial? I'm assuming you mean to require an additive inverse for your vector space (or "abelian group" or whatever you called it). IM very HO, this could be reworded to:

"for every vector, v, there is a vector, v^inv, such that: v + v^inv = 0. This is the existence of an inverse."

Shankar has the more indicative notation, using the kets, to include the minus sign inside the ket, to distinguish it from a literal negative sign as a multiplication by (-1).

turin

Originally posted by lethe
suppose we are given a basis for our original vector space V, {e_μ}. then this induces a natural choice of basis for the dual space V*, {σ^ν}, determined by
[tex]\sigma^\nu(\mathbf{e}_\mu) = \delta^\nu_\mu[/tex]           &nbsp ;               &nbsp ;               &nbsp ;   (1)
...
these linear functionals form a basis of the dual space, that we will have occasion to use.

"This defines THE dual space," as in, "there is ONLY ONE WAY to do it," or, "this defines the dual space," as in, "this is the way we HAPPEN TO do it?" This looks suspiciously like you have sneeked a metric tensor into the discussion under the guise of defining the dual space. Is this the case? Is there some way to define a dual space without using this metric-ish definition? Does this Kronecker-Delta generalize to the metric tensor for general spaces?

turin

Originally posted by jeff
Intrinsic curvature is defined by using the fairly easy to understand idea of "parallel transport". Imagine some closed curve ...

Does this mean that 1-D manifolds can not have intrinsic curvature (how do you make a closed curve on the parabola)?

lethe

Originally posted by turin
Can you give the justification that [tex]\mathbb{R}^n[/tex] is flat and that a parabola is curved (I'm assuming that you mean a parabola to be the 2-D representation of all points (x,y) that exist in the 2-D space which satisfy the coordinate values y = x², or some scaled, translated, or rotated version thereof).

I can see the literal curvature of the parabola, and the literal flatness of the x axis ([tex]\mathbb{R}^1[/tex]) if I view them in the context of the x-y plane, but I don't see how you can characterize such a thing in 1-D, and such characterization seems to be in the spirit of this thread.

i don t really need to be rigorous about the difference between flat and curved in those posts, since i was just mentioning it to give an intuition, and the notion is not actually well defined: the difference between the parabola and the real line is just a different embedding in R², i.e. it is not intrinsic.

the only reason i brought it up was to convince people why the notions we learned in R³ just won t work for a general manifold. R³ is a vector space, and that is what i meant by calling it flat (no metric involved). you can add points in the manifold to each other if it is flat. tangent vectors to the manifold can also be thought of as living in the manifold itself if it is flat.

neither of these things is true if the manifold is not a vector space, and so that s all i meant.

Here's my problem: I can't see the fundamental difference between the parabola and the real number line as 1-D manifolds. Both have 1 dimension (another concept I don't quite understand, but I'll defer that until later). I don't see what more you can say without a metric, or at least a coordinatization. If I choose to label the points on the parabola by the arclength along the parabola from the origin (which is, IMO, the most natural way to do it), then how would I know it was curved? Alternatively, how would I know to label points using their x values to show the curvature, when, for the sake of purity, I should not be appealing to any x axis in some x-y plane? In other words, how do I know that the parabola imbeds itself in the x-y plane as a parabola instead of a straight flat line, without already knowing that it was, in fact, a parabola in the x-y plane.

yes, you are correct. a good observation. instrinsically, all 1D spaces have the same geometry.

lethe

Originally posted by turin
I was just a little uncomfortable with this notation. Do you mean:

-v is defined as (-1)v

where (-1) is a member of the field? Isn't this rather trivial? I'm assuming you mean to require an additive inverse for your vector space (or "abelian group" or whatever you called it). IM very HO, this could be reworded to:

"for every vector, v, there is a vector, v^inv, such that: v + v^inv = 0. This is the existence of an inverse."

Shankar has the more indicative notation, using the kets, to include the minus sign inside the ket, to distinguish it from a literal negative sign as a multiplication by (-1).

yes, i agree with all this. i guess i just can t be bothered with that level of formalism, but i do think that it is very important to see that kind of thing when you first do abstract algebra. it lets you divorce yourself of misconceptions or generalizations that you learned in your high school algebra class.

for abelian groups, i think its pretty harmless. it is trivially easy to show that (-1)v=v^inv

lethe

Originally posted by turin
"This defines THE dual space," as in, "there is ONLY ONE WAY to do it," or, "this defines the dual space," as in, "this is the way we HAPPEN TO do it?" This looks suspiciously like you have sneeked a metric tensor into the discussion under the guise of defining the dual space. Is this the case?

note that i didn t use the kronecker delta to define the dual space, but only to choose a basis. it just happened to be on hand as a way of choosing a basis.

you can choose any basis you like, as long as you can make sure that it is actually a basis (linearly independent, etc). with the choice i made above this was easy to check.

Is there some way to define a dual space without using this metric-ish definition? Does this Kronecker-Delta generalize to the metric tensor for general spaces?

like i said above, the kronecker delta is for choosing a basis, not for defining the dual space. a dual vector acting on a vector gives me a real number. i just have to make a choice for which numbers my basis vectors will give, and i choose 1s and 0s.

this notion does not generalize to other metrics: the metric is not defined between vectors and covectors. some books use an inner product type notation, but i dislike this a lot.

lethe

Originally posted by turin
Does this mean that 1-D manifolds can not have intrinsic curvature (how do you make a closed curve on the parabola)?

you can make a closed curve on the parabola, you just have to be willing to trace back on yourself.

a 1D manifold cannot have any intrinsic curvature, but for other reasons.

turin

Originally posted by lethe
you can add points in the manifold to each other if it is flat.

Can this be done without applying coordinates to the manifold? What does it mean to add point P to point Q?

Originally posted by lethe
instrinsically, all 1D spaces have the same geometry.

What does "geometry" mean? I thought we were discussing pre-geometry manifolds. Does a circle have the same geometry as the real number line?

Originally posted by lethe
... the kronecker delta is for choosing a basis, not for defining the dual space.

I think I understand the distinction here, but I don't understand the significance. If you want to talk about the objects that live in your dual space, then aren't you going to need a basis? Can you give some non-trivial demonstration/identity/proof (not a definition) that does not require a basis?

Originally posted by lethe
a dual vector acting on a vector gives me a real number.

Is this THE definition of a dual vector?

lethe

Originally posted by turin
Can this be done without applying coordinates to the manifold? What does it mean to add point P to point Q?

some manifolds admit algebraic structures, and some don t. linear spaces all do, since it is part of their definition. you do not have to choose coordinates on your manifold to have algebra.

What does "geometry" mean? I thought we were discussing pre-geometry manifolds. Does a circle have the same geometry as the real number line?

in this instance, geometry means curvature. any 1 dimensional manifold has no intrinsic curvature, and thus, locally, all 1D manifolds have the same geometry.

we are discussing differentiable manifolds (pregeometry manifolds, as you say) in this thread. it was you who brought up the issue about the parabola and the line being the same, and so i only mentioned that to make that discussion a little clearer.

I think I understand the distinction here, but I don't understand the significance. If you want to talk about the objects that live in your dual space, then aren't you going to need a basis?

no

Can you give some non-trivial demonstration/identity/proof (not a definition) that does not require a basis?

the dual of the dual of a vector space is canonically isomorphic to the vector space. this theorem can be proved without ever choosing a basis.

the problem with choosing a basis is that there are many equally good bases to pick from, and there is no "best" basis, so sticking to only one is unnatural. but once i have made this unnatural choice, there is a best choice for the basis of the dual space, which i describe above.

Is this THE definition of a dual vector?

yes

chroot

This all strikes me as funny for some reason. A circle is a 1D manifold, and it is curved in the sense that walking along the circle in a constant direction eventually leads you back over your own footsteps. Of course, that definition of "curved" is not mathematically sound.

At the same time, there's no way for a 1D creature who lives on the circle to do any experiments to determine if there is or is not curvature. The only figures he can draw is his 1D space are lines and points, and his lines will always have the same length, no matter which direction he draws them...

I suppose I accept the fact that a 1D curve has no intrinsic curvature, but it bugs me somehow.

- Warren

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