Gradient as vector vs differential one-form

[cianfa72]

TL;DR Summary

About the definition of the gradient of a function (map) as a vector or as a differential one-form

It seems to me there is a little of confusion about the definition of gradient.

Take for instance a smooth function $f$ defined on a differentiable manifold. Which is actually its gradient at a given point ?

Someone says gradient is the vector $\nabla f$ defined at each point, whilst others say it is the differential one-form $df$ (i.e. the differential of $f$).

If the manifold is endowed with a metric tensor $g_{\mu \nu}$ then from a vector $v$ assigned at a given point we can calculate the inner product $\nabla f \cdot v$ where the inner product operator actually involves the metric tensor -- i.e. $(\nabla f)^{\mu} v^{\nu} g_{\mu \nu}$ in a given basis on the tangent space.

On the other hand, in a given basis, the action of the one-form $df$ on vector v is defined as $df(v)=\frac {\partial f} {\partial x^i} v^i$.

In other words $df(v)$ is defined for 'free' assigning a differential structure on a manifold, whilst to employ the gradient $\nabla f$ we need the additional structure of inner product on tangent spaces (i.e. the assignment of metric tensor field).

What do you think about ?

 

[fresh_42]

Summary:: About the definition of the gradient of a function (map) as a vector or as a differential one-form

It seems to me there is a little of confusion about the definition of gradient.

Take for instance a smooth function $f$ defined on a differentiable manifold. Which is actually its gradient at a given point ?

Someone says gradient is the vector $\nabla f$ defined at each point, whilst others say it is the differential one-form $df$ (i.e. the differential of $f$).

If the manifold is endowed with a metric tensor $g_{\mu \nu}$ then from a vector $v$ assigned at a given point we can calculate the inner product $\nabla f \cdot v$ where the inner product operator actually involves the metric tensor -- i.e. $(\nabla f)^{\mu} v^{\nu} g_{\mu \nu}$ in a given basis on the tangent space.

On the other hand, in a given basis, the action of the one-form $df$ on vector v is defined as $df(v)=\frac {\partial f} {\partial x^i} v^i$.

In other words $df(v)$ is defined for 'free' assigning a differential structure on a manifold, whilst to employ the gradient $\nabla f$ we need the additional structure of inner product on tangent spaces (i.e. the assignment of metric tensor field).

What do you think about ?

The derivative of a function has many variables: function, direction, point of evaluation, coordinate system. Depending on which you consider fixed or variable you get different notations. E.g. at the beginning of
https://www.physicsforums.com/insights/journey-manifold-su2mathbbc-part/
I have listed 10 possible points of view.

The gradient is defined via coordinates, a differential form is primarily not.

 

[PeterDonis]

Someone says gradient is the vector $\nabla f$ defined at each point, whilst others say it is the differential one-form $df$ (i.e. the differential of $f$).

$\nabla f$ isn't a vector. It's a 1-form. If you consult a source that uses index notation, you will find that the "natural" index position for the $\nabla$ operator is the same as for the partial derivative operator, namely a lower index. A lower index indicates a 1-form. So, for example, we would write $\nabla_\alpha f$ for this form of the gradient of a function $f$.

This means that $\nabla f$ for a scalar function $f$ is actually the same as the differential $df$; just work out the action of $\nabla_\alpha f$ on a vector $v^\alpha$ and you will see that it's the same as what you wrote down for the action of the differential $df$ on the same vector.

Of course if you have a metric on the manifold you can raise the index on the $\nabla$ operator so that it looks like a vector. But you don't need to do that to define the $\nabla$ operator itself.

The actual difference between the $\nabla$ operator and the differential or exterior derivative operator $d$ only shows up when you apply those operators to objects of higher rank than scalars. For the $\nabla$ operator, you have to include terms in the connection coefficients, which are not there when applying it to a scalar function (for a scalar function $\nabla$ is just the partial derivative operator $\partial$). For the exterior derivative, you have to antisymmetrize, and you can only apply it to forms (a scalar function is a 0-form).

 

[PeterDonis]

The gradient is defined via coordinates

It doesn't have to be. You can define the gradient of a scalar function as the 1-form induced by the level surfaces of that function. No coordinates required.

 

[cianfa72]

The gradient is defined via coordinates, a differential form is primarily not.

So basically you are saying gradient makes no sense if there is not a coordinate chart defined, whilst a differential one-form is really a 'geometric' object.

 

[PeterDonis]

So basically you are saying gradient makes no sense if there is not a coordinate chart defined

As I said in post #4, I'm not sure that is correct. But part of the issue might be different meanings being assigned to the term "gradient".

The $\nabla$ operator and the exterior derivative $d$ operator are different operators, even though they work out to be the same when applied to a scalar function. I am ok with what some have called a physicist's level of rigor here, meaning that I have no problem saying that $\nabla f$ and $df$ are both the "gradient" of $f$, since they both work out to the same result for a scalar function. But at a mathematician's level of rigor, that might not be acceptable, and only the differential $df$ would be properly referred to as the "gradient" of $f$.

A key difference between the $\nabla$ and $d$ operators in general is that the $\nabla$ operator does require some extra structure that the $d$ operator does not, namely a connection on the manifold. (The connection does not have to be derived from or compatible with a metric, although if you also have a metric on the manifold it normally is obtained that way.) But when applied to a scalar function, the connection has no effect; that's why $\nabla f$ and $df$ work out to be the same for a scalar function $f$.

 

[cianfa72]

The actual difference between the $\nabla$ operator and the differential or exterior derivative operator $d$ only shows up when you apply those operators to objects of higher rank than scalars. For the $\nabla$ operator, you have to include terms in the connection coefficients, which are not there when applying it to a scalar function (for a scalar function $\nabla$ is just the partial derivative operator $\partial$). For the exterior derivative, you have to antisymmetrize, and you can only apply it to forms (a scalar function is a 0-form).

So basically you are saying the $\nabla$ operator is the 'covariant derivative operator' and when we apply it to a function (0-form) it 'returns' the same one-form $df$.

 

[PeterDonis]

So basically you are saying the $\nabla$ operator is the 'covariant derivative operator' and when we apply it to a function (0-form) it 'returns' the same one-form $df$.

Yes, because when applied to a scalar function the connection part of the covariant derivative has no effect.

 

[fresh_42]

So basically you are saying gradient makes no sense if there is not a coordinate chart defined, whilst a differential one-form is really a 'geometric' object.

I thought so since I always saw the gradient as a vector of partial differential operators along the Cartesian coordinate directions, i.e. as one half of an inner product. However, see @PeterDonis 's post #4.

 

[PeterDonis]

I always saw the gradient as a vector of partial differential operators along the Cartesian coordinate directions, i.e. as one half of an inner product.

As I said in post #2, if there is a metric on the manifold, you can of course raise the index on the $\nabla$ operator and treat it as a vector. However, if there is no metric on the manifold, but only a connection, you cannot do this.

A "vector of partial differential operators" is not the same as the $\nabla$ operator treated as a vector (i.e., with its index raised), because there are no terms in the connection present. As I have said, the connection terms have no effect when applying the operator to a scalar function, so you can sort of get away with this "vector of partial differential operators" in this case. However, AFAIK this "vector" does not actually transform as a vector in general.

 

[cianfa72]

I thought so since I always saw the gradient as a vector of partial differential operators along the Cartesian coordinate directions, i.e. as one half of an inner product. However, see @PeterDonis 's post #4.

ok, you said 'one-half' of an inner product since picking Cartesian orthonormal coordinates we are actually picking a tensor metric $g_{\mu \nu}$ which in that coordinate system is the identity.

 

[fresh_42]

ok, you said 'one-half' of an inner product since picking Cartesian orthonormal coordinates we are actually picking a tensor metric $g_{\mu \nu}$ which in that coordinate system is the identity.

We are typing in the wrong order. See post #10.

 

[fresh_42]

My misunderstanding was that I always took gradient as a term for Euclidean spaces, and other notations like $\nabla$ if general manifolds are the topic.

 

[Infrared]

The convention I've always seen is that $\nabla f$ is the vector corresponding to the $1$-form $df$ via a metric. It seems that Peter is using $\nabla f$ and $df$ interchangeably?

 

[cianfa72]

ok, so in case of 0-form $f$ we get $\nabla f = df$. We said $\nabla$ is a covector operator hence it has a lower index $\nabla_{\mu}$ and components $\frac {\partial} {\partial x^{\mu}}$ in any dual basis $\{dx^{\mu}\}$.

What about $\nabla^{\mu}$ obtained raising the lower index of $\nabla_{\mu}$ ?

 

[PeterDonis]

The convention I've always seen is that $\nabla f$ is the vector corresponding to the $1$-form $df$ via a metric. It seems that Peter is using $\nabla f$ and $df$ interchangeably?

The operator $\nabla$ in every GR textbook I've seen is the covariant derivative operator, which, as I said, has a lower index, just as the partial derivative operator does. When operating on a scalar function, as I've said, $\nabla f$ gives the same result as $df$; but that's only the case for a scalar function.

Can you give any references that show the usage you describe?

 

[PeterDonis]

ok, so in case of 0-form $f$ we get $\nabla f = df$.

Yes.

We said $\nabla$ is a covector operator hence it has a lower index $\nabla_{\mu}$ and components $\frac {\partial} {\partial x^{\mu}}$ in any dual basis $\{dx^{\mu}\}$.

No, its components are $\partial_\mu$ plus correction terms in the connection coefficients; what the correction terms are depends on what kind of object $\nabla$ is operating on. When operating on a scalar function $f$, there are no correction terms; but there are when operating on any object of higher rank.

What about $\nabla^{\mu}$ obtained raising the lower index of $\nabla_{\mu}$ ?

Just raise the index with the inverse metric.

 

[cianfa72]

No, its components are $\partial_\mu$ plus correction terms in the connection coefficients; what the correction terms are depends on what kind of object $\nabla$ is operating on. When operating on a scalar function $f$, there are no correction terms; but there are when operating on any object of higher rank.

ok, I was thinking about operator $\nabla_{\mu}$ acting just on 0-forms.

Just raise the index with the inverse metric.

$g^{\mu \nu} \nabla_{\nu}$, so as above are its components $g^{\mu \nu} \partial_\nu$ in the tangent space $\{ \partial_\mu\}$ basis ?

 

[Infrared]

Can you give any references that show the usage you describe?

See https://en.wikipedia.org/wiki/Gradient#Riemannian_manifolds

I think we're just using different notation- I mean $\nabla$ just as the gradient operator and you mean it as the connection. I wasn't reading carefully earlier.

 

[PeterDonis]

$g^{\mu \nu} \nabla_{\nu}$, so as above are its components $g^{\mu \nu} \partial_\nu$ in the tangent space $\{ \partial_\mu\}$ basis ?

When acting on a scalar function, yes. When acting on higher rank objects, there will be additional terms corresponding to the connection coefficient terms with an index raised.

 

[PeterDonis]

I think we're just using different notation- I mean $\nabla$ just as the gradient operator and you mean it as the connection.

Not the connection, the covariant derivative operator, which includes terms in the connection (when operating on an object of rank 1 or higher).

I note that the Wikipedia page you linked to derives the notation of $\nabla$ as a vector operator from vector calculus. In the section you referenced, on the gradient in differential geometry, it is clear that $\nabla$ as a vector operator is actually obtained by raising an index on $\nabla$ as a covector operator. Also, that section only discusses applying the operator to a scalar function $f$; elsewhere in the article it is clear that when operating on objects of higher rank, additional terms in the connection are present (and again, treating the $\nabla$ operator as a vector requires raising an index).

 

[pervect]

Using the graphical techniques of MTW, a vector can be regarded graphically as a little arrow with a tail and a tip, while a gradient in 3 dimensions can be regarded as a stack of parallel plates. In 2 dimensions, the gradient would be represented by a stack of parallel lines, rather than a stack of parallel plates.

The "arrow" representation of vectors is a pretty standard representation. The stacks-of-plates definition arises from the fact that the number of plates the "arrow" pierces is a scalar, and a one-form is formally defined as a map from a [edit] vector to a scalar. So if one take a stack of plates, and uses it to operate on a vector by the "number of piercings" mechanic, the result is a scalar, making the graphical definition match the formal one.

An example of a gradient would be a contour map, which is a 2d map that represents a scalar "height" at every point on the 2d map, by plotting the surfaces of constant height on the 2d map. Magnifying the map to consider just a small section of it and considering only a small neighborhood around some particular point, the contour map becomes a set of parallel lines, the 2d representation of the one-form.

The gradient of a road would be defined as the change in height of the road per unit length of travel around the road. If we represent segment of a road by a vector (usually a unit-length vector), then the gradient is a one-form, because it takes in the vector representing the roads direction, and outputs a change in the height, which is a scalar.

 

[checksix]

pervect, as usual you give such a simple and readable explanation of a concept. Thanks from a quiet lurker here.

 

[PeterDonis]

a one-form is formally defined as a map from a scalar to a vector.

Actually it's the other way around: it's a map (more precisely a bounded linear map) from vectors to scalars.

 

[robphy]

a one-form is formally defined as a map from a scalar to a vector.

Actually it's the other way around: it's a map (more precisely a bounded linear map) from vectors to scalars.

Indeed... the rest of the paragraph is

So if one take(s) a stack of plates, and uses it to operate on a vector by the "number of piercings" mechanic, the result is a scalar, making the graphical definition match the formal one.

 

[PeterDonis]

Indeed... the rest of the paragraph is

Yes, the rest of the paragraph is fine; I just thought the particular wording I quoted was backwards.

 

[jbergman]

Lee's Smooth manifolds has a nice exposition on this subject.

In $\mathbb{R}^n$ the gradient of a smooth function $f$ is defined as

$$grad \, f = \sum_{i=1}^n \frac{\partial f}{\partial x^i} e_i$$

Or as Lee writes tangent vectors,
$$grad \, f = \sum_{i=1}^n \frac{\partial f}{\partial x^i} \frac{\partial}{\partial x^i}$$

This does not make sense independent of coordinates. To see this look at the example of $f(x,y)=x^2$ on $\mathbb{R}^2$.

In polar coordinates,

$$ grad f \ne \frac{\partial f}{\partial r}\frac{\partial}{\partial r} +
\frac{\partial f}{\partial \theta}\frac{\partial}{\partial \theta}$$

Instead, as Peter stated, it's better to think of the gradient as a 1-form and we get a coordinate independent definition, using Einstein notation.

$$df = \frac{\partial f}{\partial x^i} dx^i$$

And as already stated you can again recover a vector field from the gradient covector field or differential if you have a metric.

i.e.

$$ grad \, f = g^{ij} \frac{\partial f}{\partial x^i} \frac{\partial }{\partial x^i}$$

 

[cianfa72]

When acting on a scalar function, yes. When acting on higher rank objects, there will be additional terms corresponding to the connection coefficient terms with an index raised.

ok, so when acting on scalar function alone we can form the 'inner product' between two vectors, namely $\nabla^{\mu} f$ and the vector $v$ of components $\{ v ^{\mu}\}$. The inner product in the given basis involves the metric tensor $g_{\mu \nu}$ so we get

$(\nabla^{\mu} f) g_{\mu \alpha} v ^{\alpha} = g^{\nu \mu} (\partial_{\nu} f) g_{\mu \alpha} v ^{\alpha} = (\partial_{\nu} f) g^{\nu \mu} g_{\mu \alpha} v^{\alpha}= (\partial_{\nu} f) \delta ^{\nu}_{~\alpha} v^{\alpha}= (\partial_{\nu} f) v^{\nu}$

 

[vanhees71]

$\nabla f$ isn't a vector. It's a 1-form. If you consult a source that uses index notation, you will find that the "natural" index position for the $\nabla$ operator is the same as for the partial derivative operator, namely a lower index. A lower index indicates a 1-form. So, for example, we would write $\nabla_\alpha f$ for this form of the gradient of a function $f$.

This means that $\nabla f$ for a scalar function $f$ is actually the same as the differential $df$; just work out the action of $\nabla_\alpha f$ on a vector $v^\alpha$ and you will see that it's the same as what you wrote down for the action of the differential $df$ on the same vector.

Of course if you have a metric on the manifold you can raise the index on the $\nabla$ operator so that it looks like a vector. But you don't need to do that to define the $\nabla$ operator itself.

The actual difference between the $\nabla$ operator and the differential or exterior derivative operator $d$ only shows up when you apply those operators to objects of higher rank than scalars. For the $\nabla$ operator, you have to include terms in the connection coefficients, which are not there when applying it to a scalar function (for a scalar function $\nabla$ is just the partial derivative operator $\partial$). For the exterior derivative, you have to antisymmetrize, and you can only apply it to forms (a scalar function is a 0-form).

In other words only with a fundamental form as an additional element of your manifold you have a canonical, i.e., basis/coordinate independent one-to-one mapping between one-forms and vectors.

 

[vanhees71]

Lee's Smooth manifolds has a nice exposition on this subject.

In $\mathbb{R}^n$ the gradient of a smooth function $f$ is defined as

$$grad \, f = \sum_{i=1}^n \frac{\partial f}{\partial x^i} e_i$$

Or as Lee writes tangent vectors,
$$grad \, f = \sum_{i=1}^n \frac{\partial f}{\partial x^i} \frac{\partial}{\partial x^i}$$

This does not make sense independent of coordinates. To see this look at the example of $f(x,y)=x^2$ on $\mathbb{R}^2$.

In polar coordinates,

$$ grad f \ne \frac{\partial f}{\partial r}\frac{\partial}{\partial r} +
\frac{\partial f}{\partial \theta}\frac{\partial}{\partial \theta}$$

Instead, as Peter stated, it's better to think of the gradient as a 1-form and we get a coordinate independent definition, using Einstein notation.

$$df = \frac{\partial f}{\partial x^i} dx^i$$

And as already stated you can again recover a vector field from the gradient covector field or differential if you have a metric.

i.e.

$$ grad \, f = g^{ij} \frac{\partial f}{\partial x^i} \frac{\partial }{\partial x^i}$$

This is highly misleading, because it only works out using Cartesian coordinates for the vector space of an Euclidean affine space.

The natural general definition of a bare differentiable manifold is to define (alternating) differential forms as derivative operators on alternating tensor fields. Then the gradient of a scalar field naturally occurs as a one-form, i.e., a co-vector field,
$$\mathrm{d} \Phi=\mathrm{d}q^j \partial_j \Phi.$$

 

[jbergman]

This is highly misleading, because it only works out using Cartesian coordinates for the vector space of an Euclidean affine space.

The natural general definition of a bare differentiable manifold is to define (alternating) differential forms as derivative operators on alternating tensor fields. Then the gradient of a scalar field naturally occurs as a one-form, i.e., a co-vector field,
$$\mathrm{d} \Phi=\mathrm{d}q^j \partial_j \Phi.$$

That's the exact same thing I wrote.

$$df = \frac{\partial f}{\partial x^i} dx^i$$

is a covector field.

 

[PeterDonis]

This is highly misleading

I'm as confused by this statement as @jbergman since what you wrote is the same as what he wrote (in the latter, coordinate independent part of his post). The only difference is that he also wrote an additional step where you use the metric to raise the index of the gradient covector field, which works if you have a metric on the manifold; you've already agreed that that's possible.

 

[vanhees71]

What is misleading is the first equation reading
$$\text{grad} f= \frac{\partial f}{\partial x^i} \vec{e}_i=(\partial_i f) \vec{e}_i,$$
which obviously is not an invariant vector, because you shouldn't sum over two lower indices. What's correct is
$$\text{grad} f= (\partial_i f)\vec{e}^i,$$
which clearly shows that it is a co-vector (1-form) not a vector.

Only for a Cartesian basis you can identify $\vec{e}_i$ with $\vec{e}^i$, because lowering and raising indices in this and only this case goes with the Kronecker $\delta_{ij}=\delta^{ij}$.

 

[PeterDonis]

What is misleading is the first equation reading
$$\text{grad} f= \frac{\partial f}{\partial x^i} \vec{e}_i=(\partial_i f) \vec{e}_i,$$
which obviously is not an invariant vector

And @jbergman explicitly said that this only works in $\mathbb{R}^n$ and does not work in the absence of coordinates.

Only for a Cartesian basis you can identify $\vec{e}_i$ with $\vec{e}^i$

And that part of @jbergman's post, as he explicitly said, was restricted to this case. He then went on, in the latter part of his post, to give a better formulation that is not dependent on any particular choice of coordinates. So he already gave the clarification that you are trying to give here. You're just repeating what he said in different words.

 

[jbergman]

What is misleading is the first equation reading
$$\text{grad} f= \frac{\partial f}{\partial x^i} \vec{e}_i=(\partial_i f) \vec{e}_i,$$
which obviously is not an invariant vector, because you shouldn't sum over two lower indices. What's correct is
$$\text{grad} f= (\partial_i f)\vec{e}^i,$$
which clearly shows that it is a co-vector (1-form) not a vector.

Only for a Cartesian basis you can identify $\vec{e}_i$ with $\vec{e}^i$, because lowering and raising indices in this and only this case goes with the Kronecker $\delta_{ij}=\delta^{ij}$.

That was the whole point of the post, that the definition that is often presented to students in multivariable calculus is not coordinate free.