One-forms and vectors

1. Sep 30, 2009

NanakiXIII

I'm aware that this may not necessarily be a Relativity question but since GR seems to be a major area of application for these bits of mathematics, I'm going to go ahead and post it on this forum.

I'm trying to understand the fundamental distinction between one-forms and vectors. I thought I understood them until I noticed the following.

$$\nabla f = \frac{df}{dx} \^i + \frac{df}{dy} \^j + \frac{df}{dz} \^k$$

Here we have a one-form (the gradient) which seems to be expressed as a linear combination of vectors (the base Carthesian vectors). Something's wrong. If I write

$$v = x \^i + y \^j + z \^k$$

then v is a vector. But when the components are derivatives, we suddenly get a one-form.

I can think of several things that might be the issue, but I don't know which, if any, is true. First, it may be that the base vectors in the expression for the gradient are not base vectors, but rather one-forms. In Euclidean space, there's not much difference. Second, it may be that somehow the fact that the components are derivatives turns the whole thing into a one-form. I'm not sure, however, how that would work. Last, it may just be that the way I look at it is faulty.

If anyone can offer any insight, I would appreciate it.

2. Sep 30, 2009

dx

When you have a metric, you can change a 1-form into a vector, i.e. there is a unique vector for every 1-form. Since we all learn about the gradient in Euclidean space, we start out thinking of it as a vector. You just have to unlearn this when you move to manifolds without a metric. For a function f on a manifold, the gradient is more naturally thought of as a 1-form:

df = ∂μf dxμ

3. Sep 30, 2009

Rasalhague

I've been struggling with this topic myself. I've been working through the first volume of Bowen & Wang's Introduction to Vectots and Tensors and have found this very helpful in getting a some basic, formal, algebraic definitions of the concepts (vector space, dual space, isomorphism, etc.), but I haven't made much inroad yet into the geometric side of things which I believe is covered more in volume two.

http://repository.tamu.edu/handle/1969.1/2502
http://repository.tamu.edu/handle/1969.1/3609

One thing that puzzled me about hearing that the gradient is the archetypal 1-form is the idea that contravariant vector is synonymous with tangent vector and directional derivative, a tangent vector that points in the direction in which the function is increasing at the greatest rate. But the books on Euclidean vectors that I've read describe the gradient as a special case of a directional derivative. Another thing that puzzles me is dx's notation "df" for a gradient (does this mean a differential?), given that the archetypal vector is said to be a differential displacement, dx, while the archetypal 1-form is the gradient, a kind of derivative (or ordered set of partial derivatives).

This book looks as if it might be a good way of developing some intuition: Geometric Vectors by Gabriel Weinreich.

This review of it reproduces his illustration of the gradient as a set of level curves, similar to how I've seen 1-forms depicted elsewhere, e.g. D. H. Griffel's Linear Algebra and its Applications, Vol. 2, chapter 14 "Duality and its applications".

http://www.maa.org/reviews/vectors.html [Broken]

Weinreich visualises magnitude of a 1-form (linear form, linear functional, cotangent vector, covariant vector, covector) as the density of its level curves, in contrast to the magnitude of a (contravariant) vector which is visualised as the length of an arrow.

Last edited by a moderator: May 4, 2017
4. Sep 30, 2009

dx

This is what doesn't make sense on a space without a metric. Think of a function on a two dimensional real vector space for example. It is easy to see in this case why a tangent vector perpendicular to the contour lines of the function is not a natural way to represent the gradient, because perpendicularity is not covariant under linear transformations. (In a space with a metric, we usually only consider structures compatible with orthogonal transformations, and in that case representing the gradient by a tangent vector is ok.)

5. Sep 30, 2009

Rasalhague

So far, I've only really explored geometrical properties of a Euclidean vector space. I only know a little bit about simpler vector spaces in more abstract, algebraic terms. I think there's something in Bowen & Wang about perpendicularity, so I'll have to read up on that. Hopefully one day it'll be easy for me to see! Is this covariant in the sense invariant (as in covariant derivative), or covariant in the sense of being subject to the same transformation as basis vectors when a new basis is chosen? (I'm guessing the former.)

6. Sep 30, 2009

dx

'Covariant' is a word used in many different ways. My point was simply that the notion of perpendicularity doesnt make sense in a vector space if there is no metric.

7. Sep 30, 2009

dx

When we picture a space or surface in our heads, we always impose intuitively a metric structure on it, even if the space itself has no natural metric defined on it. The best way of avoiding confusion from this, I think, is by imagining performing linear transformations on the space. For example, if you imagine a flat plane peiece of paper, and draw two perpendicular lines on it, when you perform a linear transformation on this space, the lines are no longer perpendicular according to the imagined metric. This indicates that perpendicularity is not a structure compatible with linear transformations.

On manifolds, you imagine performing diffeomorphisms. Diffeomoprhisms are always linear transformations of the tangent spaces, so tangent vectors are not covariant representations of gradients of functions for the reasons I explained above.

8. Sep 30, 2009

NanakiXIII

So you're saying that

$$\nabla f = \frac{df}{dx} \^i + \frac{df}{dy} \^j + \frac{df}{dz} \^k$$

is simply an incorrect representation, fundamentally? I don't have much trouble understanding the concept of one-forms, though I have trouble picturing them, but I'm trying to reconcile the above notation with the fact that the gradient is a one-form.

9. Sep 30, 2009

dx

There are many ways to see why representing the gradient as a vector is not natural. Personally, I find the geometric picture of what a 1-form is the most insightful, but if you're more algebraically inclined, here's another way:

The gradient of a function is simply the local linear behavior of the function. Given a tangent vector, and the local linear behavior, we can find the directional derivative of the function along the vector. The usual way of representing this operation is to define the gradient as a vector ∇f, and then to take the dot product of this with the vector: ∇f⋅v. The trouble with this is that the dot product is basically the metric, and if you don't have a metric, this expression makes no sense. But obviously, the idea of directional derivative is still meaningful on manifolds without a metric (it just means the rate at which the function is changing along the vector), so we must re-express our concepts so that they are not tied to the metric, and in a form that is more natural in spaces without a metric.

Last edited: Sep 30, 2009
10. Oct 1, 2009

NanakiXIII

I understand that the gradient is not a vector. I'm fine with saying that the gradient of some scalar function is defined as

$$\~df = \{\frac{\partial f}{\partial x},\frac{\partial f}{\partial y},\frac{\partial f}{\partial z}\}$$

and that this expression is a one-form since it transforms covariantly. The thing I'm tripping over is that

$$\frac{\partial f}{\partial x} \^i$$

looks like a vector to me, as do the other two terms for y and z. If they are vectors, then you are just adding up vectors and I don't see how that becomes a one-form. So what I'm really asking is that, knowing that the way we learn of the gradient in Calculus is actually wrong, do we mark the old notation (the one I've quoted in my earlier posts) as wrong, or do we attribute some alternate meaning to it?

11. Oct 1, 2009

dx

That's because they are vectors. The gradient defined as ∇f is a vector.

It doesn't become a 1-form, it is redefined as a 1-form.

It's not wrong. It is just not in a form that easily generalizes to manifolds. When vector calculus was first introduced, it was in the context of 3 dimensional Euclidean space. Classical electrodynamics is usually taught with this formalism, with the gradient defined as a vector, and there is nothing wrong with this. When you try to generalize this to n-dimensional manifolds without metric, it has to be modified.

12. Oct 1, 2009

pervect

Staff Emeritus
My suggestion - go waaayyy back to a linear algebra textbook on vector spaces A vector space is defined abstractly as things that can be multiplied by scalars and added together. The elements of the vector space are called vectors. Vectors have duals, which are sometimes called one-forms. The dual space of a vector space is another vector space, but different from the original space. The duals are formed by taking maps from the vector space to scalars.

The operation is called a duality operation, because the dual of a dual recovers the original space. This theorem is proved in most linear algebra textbooks.

The easiest space to visualize in the context of geometry is the tangent space. If you think of a curved surface, a little section of it that "looks flat" can be thought of as the tangent space where the vectors live, the space of all tangents to curves passing through one point on the surface. The dual of the tangent space is the cotangent space, which is the space of one-forms.

Note that while one-forms are maps from one vector to a scalar, n-forms, in general are maps from n vectors to a scalar.

Hopefully I didn't oversimplify this too much - it's how I think of it.......

13. Oct 2, 2009

Rasalhague

Is there a conventional name for the vector space which people can use when they want to refer unambiguously to the original vector space with respect to which the dual space and the spaces of higher order tensors (which are also vector spaces) are defined. In my notes, I've sometimes called it the "base space" (by analogy with "base field", the field of scalars over which a vector space is defined), just to keep things clear in my head, but is there a real name for it?

Last edited: Oct 2, 2009
14. Oct 2, 2009

NanakiXIII

dx, that's what I was looking for, thank you. My confusion arose from the fact that in my GR course, the teacher's assistant still wrote

$$\nabla f$$,

including the whole vector notation with the Cartesian base vectors, but called that a one-form.

pervect, I had a look, but my Linear Algebra book doesn't seem to cover dual spaces. I think I have a reasonable intuition for what you mean, though. In Linear Algebra, you use column vectors and row vectors. In Quantum Mechanics, you've got your bras and kets. Basically, the dual space is needed to calculate dot products. Of course, you can only do this given a metric, I'm not sure what the use in physics would be when there's no metric.

Rasalhague, I don't have any authority on this, but my guess would be that if you want to refer to your vector space, you would just give it a name. Just call your vector space V, your dual space D, or whatever you like, and you can refer to them all you like, I would think. As long as you are consistent, you might even just call them your 'vector space' and your 'dual space'. I'm afraid that might not fully answer your question, though.

Thank you all for your replies.

15. Oct 2, 2009

Rasalhague

I see that Bowen & Wang

http://repository.tamu.edu/handle/1969.1/2502

define orthognality in terms of the inner product, two vectors being orthogonal to each other if their inner product is zero. The Wikipedia article Inner product space has the same definition. And according to Bowen & Wang's Theorem 12.4, an inner product space is also a metric space, and a metric space is a set M equipped with "a positive real-valued function called the distance function" such that:

(1) $$d(\mathbf{u},\mathbf{v}) \geq 0$$

(2) $$d(\mathbf{u},\mathbf{v}) = 0 \Leftrightarrow \mathbf{u} = \mathbf{v}$$

(3) $$d(\mathbf{u},\mathbf{v}) = d(\mathbf{v},\mathbf{u})$$

(4) $$d(\mathbf{u},\mathbf{w}) \leq d(\mathbf{u},\mathbf{v}) + d(\mathbf{v},\mathbf{w})$$

for all u, v, w in M, which could constitute a vector space, although it doesn't have to.

The distance function for an inner product space is ||u - v||. According to the Wikipedia article Metric, metric can mean distance function, but "In differential geometry, the word metric is also used to refer to a structure defined only on a vector space which is more properly termed a metric tensor (or Riemannian or pseudo-Riemannian metric)." Presumably this is the sense "metric tensor" is the one that we encounter in general relativity? More on this in the section Important cases of generalized metrics:

In differential geometry, one considers metric tensors, which can be thought of as "infinitesimal" metric functions. They are defined as inner products on the tangent space with an appropriate differentiability requirement. While these are not metric functions as defined in this article, they induce metric functions by integration. A manifold with a metric tensor is called a Riemannian manifold. If one drops the positive definiteness requirement of inner product spaces, then one obtains a pseudo-Riemannian metric tensor, which integrates to a pseudo-semimetric. These are used in the geometric study of the theory of relativity, where the tensor is also called the "invariant distance".

http://en.wikipedia.org/wiki/Metric_(mathematics)

Now Bowen & Wang write that if a vector space V has one (and no more than one) inner product, then there's a "canonical isomorphism" (natural isomorphism)

$$G : V \rightarrow V*$$

such that

$$\left\langle G \mathbf{u},\mathbf{v} \right\rangle = \left\langle G \mathbf{v},\mathbf{u} \right\rangle = \mathbf{u} \cdot \mathbf{v}$$

for all u and v in V. Angular brackets here represent what Bowen & Wang call the scalar product, i.e. the action of a one-form (element of V*) on a vector (element of V). They discuss isomorphisms in Ch. 2, Section 6, p. 29, and the idea of a canonical isomorphism in Chapter 7, Section 32, p. 213.

So, if I've understood all this correctly so far, when the vector space has one inner product, there's a unique vector in V corresponding to every one-form (every member of the dual space), and so these one-forms can be treated as (redefined as) vectors, and their action on vectors defined as the inner product.

16. Oct 2, 2009

Rasalhague

Thanks, Nanaki. Yes, this is the approach used by Bowen & Wang. In what I've read so far, the dual space of V is usually denoted V*. I was just wondering if there was a special name for it.

I found it very confusing at first to be told that there's this vector space V and a dual space V*, which is also a vector space, and that V is actually the dual of its own dual, and sometimes elements of V* can be treated as elements of V, so in some sense they're the same, but they're also totally different... I'm still learning the basics, but it started to become clearer after I got the idea that these definitions are relative to V.

17. Oct 2, 2009

NanakiXIII

That's how I understand it, if you're dealing with a space that has a metric, you can link every vector to a one-form counterpart via the metric tensor. I'm not sure, however, how you would go about creating a space that has more than one inner product.

When dealing with general metric spaces, a metric is just a function that defines the distance between any two points in the space. In Euclidean space, the metric is equivalent to applying the Pythagorean theorem to the differences in the coordinates.

In GR, you're usually dealing with the metric tensor. That's not the same thing as a metric, though it is obviously related. The distinction confused me as well, since oftentimes people refer to the metric tensor as just the 'metric', and for Minkowski space, even just the diagonal elements of the metric tensor are referred to as 'the metric' ('it has a -1,1,1,1 metric'), because it is clear in context what is meant.

18. Oct 2, 2009

Hurkyl

Staff Emeritus
Every symmetric bilinear form gives you an inner product.

19. Oct 2, 2009

NanakiXIII

I thought the inner product was defined using the metric tensor, but I suppose I don't really know that for sure. Or perhaps I'm mixing things up.

20. Oct 2, 2009

dx

The metric tensor is a symmetric bilinear form on the tangent space.

21. Oct 2, 2009

Rasalhague

I haven't really got to grips with the metric tensor yet, but Bowen and Wang define the inner product without reference to it. For them, an inner product space is a vector space V equipped with a function f : V x V --> C, where C denotes the field of scalars over which the vector space is defined, e.g. the complex numbers, and V x V is the cartesian product, such that:

(1) $$f\left(\mathbf{u},\mathbf{v} \right) = \overline{f\left(\mathbf{v},\mathbf{u} \right)}$$

(2) $$\lambda f\left(\mathbf{u},\mathbf{v} \right) = f\left(\lambda \mathbf{u},\mathbf{v} \right)$$

(3) $$f\left(\mathbf{u}_{1} + \mathbf{u}_{2},\mathbf{v} \right) = f\left(\mathbf{u}_{1},\mathbf{v} \right) + f\left(\mathbf{u}_{2},\mathbf{v} \right)$$

(4) $$f\left(\mathbf{u},\mathbf{u} \right) \geq 0$$

(5) $$f\left(\mathbf{u},\mathbf{u} \right) = 0 \Leftrightarrow \mathbf{u} = \mathbf{0}$$

for all u, v, u_1, u_2 in V, and all lambda in C. The overline in (1) denotes the complex conjugate, not an issue when the scalars are the reals, since every real number is its own complex conjugate.

22. Oct 2, 2009

DrGreg

The point is that an inner product space is a vector space with an inner product. In that context it is understood there is just one inner product, the one used to define the space in the first place. There's no reason why you can't take the same vector space and give it a different inner product, and thus define a different inner product space. As this discussion is taking place in the relativity forum rather than the linear algebra or topology forums, in the case of relativity the "inner product", or should that be "pseudo-inner product", is given by the metric tensor which, in any scenario, is determined by the energy-momentum-stress distribution, so you don't have any choice.

And, yes, in relativity, the word "metric" is always a shorthand for "metric tensor", and not the metric of a mathematical metric space.

$$\mathbf{\nabla} f = \frac{\partial f}{\partial x} \textbf{i} + \frac{\partial f}{\partial y} \textbf{j} + \frac{\partial f}{\partial z} \textbf{k}$$
$$df = \frac{\partial f}{\partial x}dx + \frac{\partial f}{\partial y}dy + \frac{\partial f}{\partial z}dz$$​

The first defines a vector. The second is shorthand for a 1-form, i.e. a function

$$df(\textbf{i}dx + \textbf{j}dy + \textbf{k}dz) = \frac{\partial f}{\partial x}dx + \frac{\partial f}{\partial y}dy + \frac{\partial f}{\partial z}dz$$​

which could also be written as

$$df(u_x\textbf{i} + u_y\textbf{j} + u_z\textbf{k}) = \frac{\partial f}{\partial x}u_x + \frac{\partial f}{\partial y}u_y + \frac{\partial f}{\partial z}u_z$$​

The two definitions are related by

$$df(\textbf{u}) = \textbf{u} \cdot \mathbf{\nabla} f$$​

All the above works fine in a Euclidean space (the "dot" in the last equation is the Euclidean dot product). But with a different metric tensor it all goes wrong, as you no longer have a Euclidean dot product. Instead you have to convert from the 1-form's covariant components into the associated vector's contravariant components via the metric tensor (or vice versa).

$$V^a = g^{ab}V_b$$
$$V_a = g_{ab}V^b$$​

23. Oct 2, 2009

DrGreg

A metric tensor is a generalisation of an inner product. Conditions (4) and (5) no longer hold. (In the context of relativity, everything is real so you don't need to worry about complex numbers.)

24. Oct 2, 2009

Rasalhague

The Wikipedia article Metric (mathematics) says, "A manifold with a metric tensor is called a Riemannian manifold. If one drops the positive definiteness requirement of inner product spaces, then one obtains a pseudo-Riemannian metric tensor, which integrates to a pseudo-semimetric." Does condition (4) define positiveness, and (5) definiteness?

25. Oct 2, 2009

DrGreg

Yes.

(I've only ever heard of "positive" (4) (sometimes called "positive semi-definite") and "positive-definite" (4)+(5), I don't think "definite" is ever used on its own. But I could be wrong. If, in a finite dimensional vector space, you represent the form by a matrix, f(x,y) = yHGx, the same terminology can be applied to the matrix itself.)