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Silviu

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Thank you!

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- #1

Silviu

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Thank you!

- #2

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In many cases when there exists a non-degenerate bilinear type (0,2) tensor (such as a metric or symplectic form), one will often forego the distinction altogether and colloquially make no difference between a tangent vector and its corresponding one-form as the relation via the type (0,2) tensor is implied.

- #3

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I think that the OP has understood what one-forms are and that the question is more one of nomenclature, i.e., "why are the elements of the dual space not called vectors when the dual space is a vector space?"One forms are maps defined on the tangent space ##TM## of a manifold ##M## that are linear at each point. So at a point of the manifold, the 1 form is just a linear map defined on the tangent plane at that point.

In calculus on manifolds 1 forms and vectors transform differently, one covariantly the other contravariantly. When one has a smooth map between manifolds ##f:M→N## it differential ##df## maps tangent vectors in the tangent space to ##M## at each point ##p## to tangent vectors in the tangent space to ##N## at ##f(p)##. ##df:TM→TN##. This mapping is called covariant. The dual map ##df^{*}## maps the dual tangent space to ##N## to the dual tangent space to ##M##. ##df^{*}:T^{*}N→T^{*}N##. This maps is contra-variant. It maps 1 forms on ##N## to 1 forms on ##M##.

- #4

Geometry_dude

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Is this just a convention, or it is a deeper meaning?

There is a fundamental difference between vectors and covectors (1-forms, even though the term is usually referring to covectorfields), but it is true that one can convert one into another if one has been given a metric. Even then they are different objects, however.

Intuitively, a tangent vector is a vector attached to a point - in physics and depending on context this may or may not be interpreted as the tangent vector of a trajectory, but this is where it comes from.

Interpretations of covectors are not so straightforward, but if you look at the kernel, i.e. the vectors that get mapped onto ##0## by it, then this gives you a hyperplane (plane of one dimension less than the tangent space) in the tangent space. This looks like it's not very useful, but you'd be fundamentally mistaken.

As an example, assume you are given a real-valued function ##f\colon M \to \mathbb R## on your manifold ##M## whose Cartan derivative ##\text{d} f## - a covectorfield - vanishes nowhere. Then, by the preimage theorem, the level sets of ##f## are embedded hypersurfaces and the tangent space of the hypersurface intersecting a given ##p \in M## can be interpreted as the plane given by ##(\text{d} f)_p##.

So geometrically one may interpret covectors at ##p## as giving the linear approximations to a hypersurface intersecting ##p## - but one should be careful with intuitive pictures of that sort.

- #5

Mads Lundwall

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You can understand a vector-field as the orthogonal vectors to the contour lines of one-forms, having a length proportional to the density of theese countour lines in that point: closer lines/higher density -> greater length of corresponding vectors.

- #6

Mads Lundwall

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I got a visual way to se the connection one-forms to vectors (that I have been looking for on the net, but haven't seen this way of explaining it)

You can understand a vector-field as the orthogonal vectors to the contour lines of one-forms, having a length proportional to the density of theese countour lines in that point: closer lines/higher density ->smallerlengt of corresponding vectors.

Sorry INVERSE so it should be smaller length.

- #7

lavinia

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Perhaps the reason is historical since among the first manifolds studied were surfaces in 3 space. On them, tangent vectors lie on tangent lines to the surface.

On manifolds tangent vectors are derivations on smooth functions and do not have this nice geometric picture. But tangent directions are formally equivalent to derivations.

Historically, I do not know why 1 forms are called forms but I imagine that they arose as objects that can be integrated over curves. From this point of view they would have to assign numbers to the tangent vectors to the curve. So they are functionals that act on tangent vectors not tangent vectors themselves.

I do not see how real vectors can be viewed as 1 forms since there is no natural way that I know of to integrate a vector field over a curve. One would have to somehow transform the vector field into a 1 form for instance by using an inner product.

On the other hand, you are right that tangent vectors can be viewed as linear functionals on the space of 1 forms. A vector acts on a 1 form by evaluation. ##v(ω) = ω(v)## and in this way a tangent vector assigns a number to a 1 form at each point of the manifold.

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- #8

victorneto

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There is a fundamental difference between vectors and covectors (1-forms, even though the term is usually referring to covectorfields), but it is true that one can convert one into another if one has been given a metric. Even then they are different objects, however.

A beautiful, intelligible, complete and objective explanation of this important subject can be found on pages 52 the 56 of B.Schutz's Geometric Methods of Mathematical Physics.

- #9

victorneto

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In many cases when there exists a non-degenerate bilinear type (0,2) tensor (such as a metric or symplectic form), one will often forego the distinction altogether and colloquially make no difference between a tangent vector and its corresponding one-form as the relation via the type (0,2) tensor is implied.

The distinction only disappears in Euclidean spaces. In the general case of curved spaces, unidirectional forms and vectors have fundamentally different geometric properties! The clearest example of this is Grad (f), which is often mistaken for a vector, but is not a vector.

- #10

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This is only partially true. Yes, they are different vector spaces, but a non-degenerate form introduces a bijection between those two vector spaces, thus introducing an equivalence. The distinction between tangent vectors and 1-forms a priori also exists in Euclidean spaces!The distinction only disappears in Euclidean spaces. In the general case of curved spaces, unidirectional forms and vectors have fundamentally different geometric properties!

The clearest example of this is Grad (f), which is often mistaken for a vector, but is not a vector.

This depends on what you mean by grad(f). The typical intended meaning of grad(f) is actually a tangent vector, related to the 1-form ##df## through the inverse metric, but nomenclature vanishes and it is also common to see ##df## called "gradient".

- #11

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This depends on what you mean by grad(f). The typical intended meaning of grad(f) is actually a tangent vector, related to the 1-form ##df## through the inverse metric, but nomenclature vanishes and it is also common to see ##df## called "gradient".

Defining grad(f) to be a vector is weird, though, even if common. What you typically do with a gradient is to take its scalar product with a vector to get a directional derivative. So the use of the metric "cancels out" (you use the inverse metric to create a vector, and then the metric to combine two vectors).

- #12

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I agree, the natural object is ##df##, not the tangent vector related to it, since ##df(V)## is just the directional derivative of ##f## in the direction of ##V##. Still, it is the most common usage of "gradient" that I have encountered.Defining grad(f) to be a vector is weird, though, even if common. What you typically do with a gradient is to take its scalar product with a vector to get a directional derivative. So the use of the metric "cancels out" (you use the inverse metric to create a vector, and then the metric to combine two vectors).

- #13

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There are two different types of vector-like quantities that you can define for this space.

- You can pick a parametrized curve through state space (for example, the state of a system as as function of time). The tangent vector of this curve has components (in the ##P,T## coordinate system) ##\frac{dP}{dt}## and ##\frac{dT}{dt}##. We could call this the "velocity vector".
- You can pick a scalar function (for example, volume ##V##) and consider how that function changes as you move about the state space. The differential (or whatever you want to call it) of this function has components ##\frac{\partial V}{\partial P}## and ##\frac{\partial V}{\partial T}##.

It also doesn't make any sense to try to take the scalar product of two velocity vectors. In the same way, any way you try to do it is going to lead to an expression that has inconsistent units. It similarly doesn't make any sense to take the scalar product of two differentials.

However, you can certainly take a kind of scalar product of a velocity vector and a differential: If you have a velocity vector with components ##(\frac{dP}{dt},\frac{dT}{dt})## and a differential with components ##\frac{\partial V}{\partial P}, \frac{\partial V}{\partial T}##, you can contract them to get the scalar:

##\frac{dV}{dt} = \frac{dP}{dt} \frac{\partial V}{\partial P} + \frac{dT}{dt} \frac{\partial V}{\partial T}##

This quantity has consistent units of (volume)/(time), and has a ready interpretation of the rate of change of volume.

- #14

lavinia

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I

However, you can certainly take a kind of scalar product of a velocity vector and a differential: If you have a velocity vector with components ##(\frac{dP}{dt},\frac{dT}{dt})## and a differential with components ##\frac{\partial V}{\partial P}, \frac{\partial V}{\partial T}##, you can contract them to get the scalar:

##\frac{dV}{dt} = \frac{dP}{dt} \frac{\partial V}{\partial P} + \frac{dT}{dt} \frac{\partial V}{\partial T}##

Here you are choosing an inner product by a choice of a basis for the phase space. One way to define an isomorphism between a vector space and its dual space is to choose a basis.

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- #15

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Here you are choosing an inner product by a choice of a basis for the phase space. One way to define an isomorphism between a vector space and its dual space is to choose a basis.

It's not an "inner product", because it doesn't take a pair of vectors, it takes one vector and one one-form. And the definition of product is independent of the basis. (In the sense that every choice of basis ##x^\mu## gives the same result for the combination ## \sum_j \frac{\partial f}{\partial x^j} \frac{dx^j}{dt}##).

I don't see how choosing a basis sets up an isomorphism between a vector and its dual space.

- #16

lavinia

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It's not an "inner product", because it doesn't take a pair of vectors, it takes one vector and one one-form. And the definition of product is independent of the basis. (In the sense that every choice of basis ##x^\mu## gives the same result for the combination ## \sum_j \frac{\partial f}{\partial x^j} \frac{dx^j}{dt}##).

Your basis is the vectors ##∂/∂P## and ##∂/∂T##. The inner product sets these vectors to be orthonormal. With this inner product the gradient vector of the function ##V## is the vector ##(∂V/∂P)∂/∂P + (∂V/∂T)∂/∂T##.

A basis for a vector space always determines an inner product by setting the basis to be orthonormal. Conversely, an inner product determines a set of orthonormal bases each of which gives you the inner product back again.

If ##x_{i}## is a basis then then the dual basis are the linear maps ##x^{i}## such that ##x^{i}(x_{j}) =δ_{ij}##

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- #17

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The one-form he is referring to is ##dV = (\partial_P V) dP + (\partial_T V) dT##, which is what makes sense with the chain rule for ##dV##. There is no natural metric on thermodynamic state space and therefore no gradient (in the sense of a tangent vector), but that does not mean that one cannot contract one-forms with tangent vectors (in fact, it is what one-forms are by definition), which is what is going on here.Your basis is the vectors ##∂/∂P## and ##∂/∂T##. The inner product sets these vectors to be orthonormal. With this inner product the gradient vector of the function ##V## is the vector ##(∂V/∂P)∂/∂P + (∂V/∂T)∂/∂T##.

Choosing a basis for a vector space is equivalent to choosing an inner product.

- #18

lavinia

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Right. I am just pointing out that one is implicitly defining an inner product.The one-form he is referring to is ##dV = (\partial_P V) dP + (\partial_T V) dT##, which is what makes sense with the chain rule for ##dV##. There is no natural metric on thermodynamic state space and therefore no gradient (in the sense of a tangent vector), but that does not mean that one cannot contract one-forms with tangent vectors (in fact, it is what one-forms are by definition), which is what is going on here.

To get an isomorphism between a vector space an its dual one needs a choice of metric or what is equivalent a choice of basis.

- #19

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I don't see why you would want to do that. Such a metric holds absolutely no physical relevance (even the units would be completely messed up if you define coordinates such as ##T## and ##P## to be orthonormal) and thermodynamic state space is the prime example of a manifold used in physics where there is no metric or even relevant bilinear form needed for most of the physics. (The prime example of a manifold used in physics with a (pseudo) metric of course being spacetime and the prime example of a manifold with an anti-symmetric bilinear form being phase space = a symplectic manifold.)Right. I am just pointing out that one is implicitly defining a metric. The one form ##dV## by itself is metric/basis independent.

- #20

lavinia

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This is a mathematical point not a discussion of physics. The OP's question was whether there is a difference between a tangent space and its dual and the point is that they are isomorphic but not naturally isomorphic. One needs to choose an inner product or equivalently choose a basis.I don't see why you would want to do that. Such a metric holds absolutely no physical relevance (even the units would be completely messed up if you define coordinates such as ##T## and ##P## to be orthonormal) and thermodynamic state space is the prime example of a manifold used in physics where there is no metric or even relevant bilinear form needed for most of the physics. (The prime example of a manifold used in physics with a (pseudo) metric of course being spacetime and the prime example of a manifold with an anti-symmetric bilinear form being phase space = a symplectic manifold.)

The OP asked whether a tangent space/bundle and its dual/1 forms are interchangeable. Part of the answer to this is that at each point they form isomorphic vector spaces but the isomorphism is not natural. An isomorphism requires a metric and this is the equivalent to choosing a basis. This point is true whether or not the resulting metric has physical meaning.

As opposed to this, a vector space

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- #21

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@stevendaryl certainly intended his example to be based on physics or he would not have chosen that example. Sure, you can define an isomorphism (they are vector spaces of the same dimension so they are obviously both isomorphic to ##\mathbb R^n##), but in this particular example there is no reason for actually doing so and the concept of a "direction of maximal increase" is irrelevant.This is a mathematical point not a discussion of physics.

- #22

lavinia

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Right. I understand what Stevendaryl was trying to say. But the mathematical point remains and I think this was what the OP was after.@stevendaryl certainly intended his example to be based on physics or he would not have chosen that example. Sure, you can define an isomorphism (they are vector spaces of the same dimension so they are obviously both isomorphic to ##\mathbb R^n##), but in this particular example there is no reason for actually doing so and the concept of a "direction of maximal increase" is irrelevant.

- #23

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Sure, but every time someone introduces a metric on thermodynamic state space, a puppy dies.Right. I understand what Stevendaryl was trying to say. But the mathematical point remains.

- #24

lavinia

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Ok. My pet coyote ate the puppy.Sure, but every time someone introduces a metric on thermodynamic state space, a puppy dies.

- #25

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Your basis is the vectors ##∂/∂P## and ##∂/∂T##. The inner product sets these vectors to be orthonormal. With this inner product the gradient vector of the function ##V## is the vector ##(∂V/∂P)∂/∂P + (∂V/∂T)∂/∂T##.

I don't see why you would want to say that the basis vectors are orthonormal. I certainly didn't specify that, and I avoided mentioning an inner product. There is no good reason to have an inner product.

- #26

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Right. I am just pointing out that one is implicitly defining an inner product.

I don't see why choosing a basis implies anything about an inner product. A basis is just a choice of independent vectors, in the sense that no vector is equal to a linear combination of the others. Choosing a basis does not imply anything about an inner product. If you call it an "orthonormal basis", then that certainly implies something about the inner product, but not all bases are orthonormal.

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