# I Difference between vectors and one-forms

1. Aug 30, 2017

### Silviu

Hello! I am reading some introductory differential geometry and they define the vector space associated to a point of a manifold as the tangent plane at that point. Intuitively it makes sense to call these vectors (just as the speed is the tangent to the trajectory), but why are those called vectors and the one in the dual vector space one-forms. Is this just a convention, or it is a deeper meaning? Like the dual vector space is still a vector space, so its members can be considered vectors, while the real vectors can be considered one forms, right?
Thank you!

2. Aug 30, 2017

### Orodruin

Staff Emeritus
This is just a matter of nomenclature. Of course, one-forms are also vectors, but tangent vectors are more often referred to just as "vectors". If you would want to be more specific you would say "tangent vectors" and "one-forms" (or "dual vectors" or "covectors").

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. Aug 30, 2017

### Orodruin

Staff Emeritus
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?"

4. Aug 31, 2017

### Geometry_dude

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. Sep 12, 2018

I got a visual way to se the connection one-forms to vectors (that I have been looking for on the net, but havent 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 lenght proportional to the density of theese countour lines in that point: closer lines/higher density -> greater length of corresponding vectors.

6. Sep 15, 2018

Sorry INVERSE so it should be smaller length.

7. Sep 15, 2018

### lavinia

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.

Last edited: Sep 16, 2018
8. Sep 16, 2018

### victorneto

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. Sep 16, 2018

### victorneto

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. Sep 16, 2018

### Orodruin

Staff Emeritus
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!

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. Sep 16, 2018

### stevendaryl

Staff Emeritus
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. Sep 16, 2018

### Orodruin

Staff Emeritus
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.

13. Sep 16, 2018

### stevendaryl

Staff Emeritus
When dealing with Euclidean space, it's sort of hard to get a feel for the distinction between vectors and one-forms, and why the distinction is important. I think it might help to think about a "space" where there is no sensible notion of turning vectors into one-forms. One example might be the state space of thermodynamics. A point in this space can be characterized by the numbers $P,T$ (pressure and temperature) for a system with a fixed number of particles. Another thermodynamic variable, say volume $V$, is a function of these two: $V(P,T)$. The state space can be thought of as a two-dimensional space, and as the system changes, it traces out a curve in this space.

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

1. 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".
2. 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 $\frac{\partial V}{\partial T}$.
Both of these objects are clearly vector-like. But unlike in Euclidean 2-D space, there is no sensible metric. It doesn't make any sense at all to try to determine the "length" of a velocity vector like $(\frac{dP}{dt},\frac{dT}{dt})$. What are you going to do? It certainly can't be $\sqrt{(\frac{dP}{dt})^2 + (\frac{dT}{dt})^2}$---that doesn't even have consistent units. Similarly for differentials---there is no good notion of length.

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. Sep 16, 2018

### lavinia

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.

Last edited: Sep 16, 2018
15. Sep 16, 2018

### stevendaryl

Staff Emeritus
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. Sep 16, 2018

### lavinia

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 $x^{i}(x_{j}) =δ_{ij}$

Last edited: Sep 16, 2018
17. Sep 16, 2018

### Orodruin

Staff Emeritus
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.

18. Sep 16, 2018

### lavinia

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

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. Sep 16, 2018

### Orodruin

Staff Emeritus
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.)

20. Sep 16, 2018

### lavinia

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.

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 is naturally isomorphic to its double dual.

Last edited: Sep 16, 2018