What kind of tensor is the gradient of a vector Field?

[GR191511]

(1,1)or(2,0)or(0,2)?And does a dual vector field have gradient?

 

[Orodruin]

What type of vector field (tangent or dual)? The natural action of the gradient is to add a vector argument (i.e., raise the second number by one) so if you take the gradient of a tangent vector field (type (1,0)) the result would be a type (1,1) tensor. If you take the gradient of a dual vector field, the result would be a type (0,2) tensor.

 

[martinbn]

(1,1)or(2,0)or(0,2)?And does a dual vector field have gradient?

If by gradient of a vector field is meant the covariant differential, then it is (1,1).

 

[fresh_42]

(1,1)or(2,0)or(0,2)?

There is mathematically no significant difference since $V\cong V^*$ (except in homological algebra).

And does a dual vector field have gradient?

I think it makes more sense to ask when a vector field is a gradient field, instead of has. And it is one if it's integrable so that closed curves yield zero.

 

[martinbn]

There is mathematically no significant difference since $V\cong V^*$ (except in homological algebra).

They are isomorphic, but not canonically. So, i would say that there is a difference.

I think it makes more sense to ask when a vector field is a gradient field, instead of has. And it is one if it's integrable so that closed curves yield zero.

That is a completely different question than what the OP is asking.

 

[Ibix]

If you have a vector field $v^a$ and the derivative operator is $\nabla_b$ then the gradient field is $\nabla_b v^a$ which you can tell is a (1,1) tensor from the indices.

 

[GR191511]

Thank you all!

 

[WWGD]

There is mathematically no significant difference since $V\cong V^*$ (except in homological algebra).

I think it makes more sense to ask when a vector field is a gradient field, instead of has. And it is one if it's integrable so that closed curves yield zero.

Doesn't it matter that the isomorphism isn't natural? I mean , we do have, iirc, $V\cong V^{**}$ in a natural, i.e., basis independent way, but the dual itself is isomorphic in a basis independent way.

And, iirc, the answer to the second question is given by Cauchy-Riemann, i. e. , if, given the vector field $U(x,y)+iV(x,y)$, we have $U_x=V_y$.

 

[martinbn]

Doesn't it matter that the isomorphism isn't natural? I mean , we do have, iirc, $V\cong V^{**}$ in a natural, i.e., basis independent way, but the dual itself is isomorphic in a basis independent way.

I know that this thread is about finite dimensional spaces. But if they are not, the the space and its dual are not isomorphic at all, not just non cannonically.

And, iirc, the answer to the second question is given by Cauchy-Riemann, i. e. , if, given the vector field $U(x,y)+iV(x,y)$, we have $U_x=V_y$.

What about more than two dimensions?

 

[WWGD]

I know that this thread is about finite dimensional spaces. But if they are not, the the space and its dual are not isomorphic at all, not just non cannonically.

What about more than two dimensions?

Ah, yes, I forgot about that. It's only the double-dual . And, yes, that works just for dimension 2.

 

[fresh_42]

Doesn't it matter that the isomorphism isn't natural?

I guess I should better have said what I meant. From a mathematical point of view: vector space is vector space. Everything else comes with a goal that has to be achieved. Without context, $V$ and $V^*$ are simply sets of vectors.

 

[lavinia]

One can think of the directional derivative as a function of the direction vector. At each point the value of this function $(∇V)_{x}(s)$ is the directional derivative of the vector field $V$ in the direction of the vector $s$ at the point x

This function is linear in the directions $(∇V)_{x}(as + bt) = a(∇V)_{x}(s) + b(∇V)_{x}(t)$ and so is a vector valued dual vector at x. Letting the point x vary gives a vector valued 1 form and so is a (1,1) type tensor.

Notice that if $f$ is a smooth function then $∇(fV)(s) = df(s)V + f(s)∇V(s)$ .

Notes:

- In vector bundle language the gradient is a smooth section of the tensor product bundle $T^{*}⊗T$ where $T$ is the tangent bundle of Euclidean space and $T^{*}$ is the dual bundle

- If one allows vector fields to vary then one gets a mapping from vector fields into smooth sections of this tensor product bundle. This mapping is linear in the sense that it factors over linear combinations of vector fields. If one uses the notation $C^{∞}(E)$ to represent smooth sections of a vector bundle $E$ i.e vector fields then one has a linear mapping

$∇:C^{∞}(T)→C^{∞}(T^{*}⊗T)$

And one has the so called Leibniz rule $∇(fV) = df⊗V + f∇V$

One can now ask the question of what kind of differential operators satisfy this formalism.
These are called connections on the vector bundle.

A connection on the vector bundle $E$ (real or complex) is a linear mapping

$∇:C^{∞}(E)→C^{∞}(T^{*}⊗E)$

that satisfies the Leibniz rule.

- If the vector field is a field of dual vectors i.e. it is a 1 form then one does exactly the same thing. The directional derivative then is a dual vector at each point and the gradient is a dual vector valued dual vector. In this case it is a smooth section of the tensor product bundle $T^{*}⊗T^{*}$ i.e. it is a tensor of type $(0,2)$

 

[lavinia]

Doesn't it matter that the isomorphism isn't natural? I mean , we do have, iirc, $V\cong V^{**}$ in a natural, i.e., basis independent way, but the dual itself is isomorphic in a basis independent way.

In general for two vector spaces of the same finite dimension there is no specific isomorphism between them that comes along for free. Something else is needed to define an isomorphism. For $V$ and its dual one needs for instance a choice of basis for $V$ or a non-degenerate bilinear form. Neither of these come from the definition of a vector space and its dual.

However with the double dual there is a specific isomorphism that is always there. Nothing extra such as an inner product is needed. If $v$ is an element of $V$ then the natural double dual vector is

$v^{**}(w^{*})=w^{*}(v)$ where $w^{*}$ is any dual vector.

 

[WWGD]

In general for two vector spaces of the same finite dimension there is no specific isomorphism between them that comes along for free. Something else is needed to define an isomorphism. For $V$ and its dual one needs for instance a choice of basis for $V$ or a non-degenerate bilinear form. Neither of these come from the definition of a vector space and its dual.

However with the double dual there is a specific isomorphism that is always there. Nothing extra such as an inner product is needed. If $v$ is an element of $V$ then the natural double dual vector is

$v^{**}(w^{*})=w^{*}(v)$ where $w^{*}$ is any dual vector.

Thank you, I'm aware of this in general, as well as the likes of the Musical Isomorphism. I'm just curious as to what property /result follows from vector spaces being naturally isomorphic that doesn't follow when they're just " plain" isomorphic ( in the f.d case, being of the same dimension) .

 

[lavinia]

Thank you, I'm aware of this in general, as well as the likes of the Musical Isomorphism. I'm just curious as to what property /result follows from vector spaces being naturally isomorphic that doesn't follow when they're just " plain" isomorphic ( in the f.d case, being of the same dimension) .

That is a great question. I am not sure how to think about it but maybe we can come up with some ideas.

For starters, the equation $v^{**}(w^{*})=w^{*}(v)$ defines a bilinear pairing $V×V^{*}→F$ ($F$ is the field of scalars) in which for each fixed dual vector one gets a linear map of $V$ into the base field $F$ and for each element of $V$ one gets a linear map of $V^{*}$ into $F$. Each vector in $V$ is both itself and a double dual vector at the same time.

For two arbitrary vector spaces of the same dimension no such pairing automatically exists.

In some sense this pairing says that $V$ and $V^{**}$ are the same vector space. This is more than just an isomorphism.

 

[fresh_42]

That is a great question. I am not sure how to think about it but maybe we can come up with some ideas.

I always had problems distinguishing between natural and canonical. In the end, I was satisfied by the definitions: natural = comes automatically, canonical = done as usual. For e.g. I learned that $\pi\, : \,G\longrightarrow G/N$ is canonical, but not natural. It can't be natural as there is no unique system of representatives. Now, I looked up natural mapping in my homological algebra book, and found: $$\pi.$$
And nLab supports Brunner ...
https://ncatlab.org/nlab/show/natural+transformation
... but apparently does not make a distinction to canonical (redirected to natural):
https://ncatlab.org/nlab/show/core-natural+transformation

However, the translation helps: Canon is a doctrinal or legal tenet of canon, catholic church law. Hence, as it is usually done. I think I'll keep this and will make a difference to natural.

So natural has a precise definition, canonical only a common sense. $V^{**}\cong V$ is natural, $V^*\cong V$ with an ONS canonical.

 

[WWGD]

I think the term ' Natural' does have a precise Category-Theoretical definition.

 

[martinbn]

Thank you, I'm aware of this in general, as well as the likes of the Musical Isomorphism. I'm just curious as to what property /result follows from vector spaces being naturally isomorphic that doesn't follow when they're just " plain" isomorphic ( in the f.d case, being of the same dimension) .

I think the point of naturally/canonically isomorphic or not is important for constructions/definitions and not for results. For example if two vector spaces $V$ and $W$ are isomorphic, because they have the same dimension, then you can add vectors from $V$ and $W$, but that depends on the choice of the isomorphism. If they are canonically isomorphic, then there is a natural way to add vectors from the two spaces. All tangent spaces to all points on a manifold are isomorphic, yet you need more to define a covariant derivative.

 

[WWGD]

I think the point of naturally/canonically isomorphic or not is important for constructions/definitions and not for results. For example if two vector spaces $V$ and $W$ are isomorphic, because they have the same dimension, then you can add vectors from $V$ and $W$, but that depends on the choice of the isomorphism. If they are canonically isomorphic, then there is a natural way to add vectors from the two spaces. All tangent spaces to all points on a manifold are isomorphic, yet you need more to define a covariant derivative.

Right. The connection provides the isomorphisms.

 

[Orodruin]

Right. The connection provides the isomorphisms.

Well, yes and no. If the connection is not flat then the isomorphism between two tangent spaces becomes path dependent.

 

[lavinia]

Continuing with post #15 and using the link in post #16, if I understand it, and some web wandering, it seems that the isomorphism $^{**}$ is "canonical" because it does not involve choosing an extra structure such as a basis. It is "natural" because it describes a natural transformation between the $**$ functor and the identity functor in the category of finite dimensional vector spaces and linear transformations.(see the note below)

The natural transformation - I think - is the commutative square $L^{**} \circ ^{**} = ^{**} \circ L$ for any linear map $L$.

So a natural transformation looks like a homomorphism of functors. In the case of the double dual, $^{**}$ is always an isomorphism on each vector space so it is an isomorphism between the identity functor and the double dual functor. This is stronger seemingly than just a natural transformation. Also the isomorphism is with the identity functor, the functor which sends every object and morphism to itself, so in some sense the isomorphism itself is the identity as suggested in post #15.

Note: The $**$ functor maps each finite dimensional vector space into its double dual and maps linear maps to the induced map on the double dual. The identity functor maps each vector space and linear map to itself.

 

[S.G. Janssens]

I think the point of naturally/canonically isomorphic or not is important for constructions/definitions and not for results.

No.

 

[martinbn]

No.

Why not?

 

[PrecPoint]

Well, yes and no. If the connection is not flat then the isomorphism between two tangent spaces becomes path dependent.

You mean between $T_n(p)$ and $T_n(q)$ of two neighboring points $p,q$ which are not infinitesimally close to each other on the manifold? Because otherwise the affine connection $\Gamma_{h\,\,\,k}^{\,\,\,j}$ is a unique mapping of $T_n(p)$ onto the tangent space $T_n(q)$ of a neighboring point ($dX^j=-\Gamma_{h\,\,\,k}^{\,\,\,j}X^hdx^k)$.

 

[Orodruin]

You mean between $T_n(p)$ and $T_n(q)$ of two neighboring points $p,q$ which are not infinitesimally close to each other on the manifold? Because otherwise the affine connection $\Gamma_{h\,\,\,k}^{\,\,\,j}$ is a unique mapping of $T_n(p)$ onto the tangent space $T_n(q)$ of a neighboring point ($dX^j=-\Gamma_{h\,\,\,k}^{\,\,\,j}X^hdx^h)$.

Nobody was talking about neighboring points or points that are infinitesimally close. The entire point of the connection is to ensure that tangent spaces of infinitesimally close points can be connected. The effects of curvature are quadratic in displacement so they obviously become negligible relative to the linear behaviour as points come closer together.

 

[PrecPoint]

Nobody was talking about neighboring points or points that are infinitesimally close

I believe WWGD was talking about neighboring points here:

Right. The connection provides the isomorphisms.

Relating elements of distinct tangent spaces belonging to neighboring points is the main point of local parallelism. Your answer to WWGD makes no sense. Why would you try to relate tangent spaces of unrelated points on the manifold?

A better question to ask that is actually related to this thread is, when is there a one-to-one correspondence between the tangent space $T_n(p)$ and its dual $T^*_n(p)$ at a point $p$?

And the answer is, the correspondence depends on the existence of a given metric; for an arbitrary differentiable manifold $X_n$ devoid of a metric tensor there is no such relationship between the tangent space and its dual.

 

[Orodruin]

I believe WWGD was talking about neighboring points here:

Why? There is no indication of that.

Why would you try to relate tangent spaces of unrelated points on the manifold?

There are many reasons to do so. It is the entire point of parallel transport, which has numerous applications.

And the answer is, the correspondence depends on the existence of a given metric;

This is only half correct. The correct answer is when the manifold has a non-degenerate bilinear form associated to it. This form may be a metric, a pseudo-metric, symplectic form, etc. depending on the manifold.

 

[PrecPoint]

This is only half correct. The correct answer is when the manifold has a non-degenerate bilinear form associated to it. This form may be a metric, a pseudo-metric, symplectic form, etc. depending on the manifold.

I would just consider those variants of $g_{jh}=\frac{1}{2}\frac{\partial^2F^2(x,\dot{x})}{\partial \dot{x}^j\partial \dot{x}^h}$ for obvious reasons. The word pseudo is just something Riemannian people came up with to distinguish themselves from people using metrics where $ds^2$ is not positive definite. So I don't see your point with this post either. Maybe if you give an example of a bilinear form that is not a tensor...

There are many reasons to do so.

So please give some reasons related to the question at hand. The context is the following;

There is mathematically no significant difference since $V\cong V^*$ (except in homological algebra).

The discussion then followed the classic "$V$ may identified with the dual $(V^*)^*$ of $V^*$" and the usual "but what if the dimension of $V$ is infinite?"

We need a metric tensor field for the statement of fresh_42 to make sense and it has nothing to do with the space being flat or not. And I really don't understand why we should consider two tangent spaces miles apart.

 

[Orodruin]

I would just consider those variants of $g_{jh}=\frac{1}{2}\frac{\partial^2F^2(x,\dot{x})}{\partial \dot{x}^j\partial \dot{x}^h}$ for obvious reasons.

This is clearly incorrect for the case of a symplectic form, which is anti-symmetric.

So please give some reasons related to the question at hand. The context is the following;

No, that is not the context. What you quoted discusses the relation and isomorphy between a vector space V and its dual and double dual. In that context, there is not even a manifold. V can be any finite-dimensional vector space. This relates to the isomorphism between different tangent spaces.

We need a metric tensor field for the statement of fresh_42 to make sense and it has nothing to do with the space being flat or not.

What fresh discussed does not even rely on there being a manifold or a metric. All you need is a non-defenerate map between any vector space V and its dual. This is by definition related to a bilinear form, but you can construct that in any way you wish. For example, given any basis, you can just start mapping the basis vectors to each other in any manner you wish (there are N! possible choices here). It does not need to be symmetric and it does not have to be natural.

And I really don't understand why we should consider two tangent spaces miles apart.

As I said, there are many reasons to consider such isomorphisms. Essentially any application where parallel transport is involved. Are you telling me you do not understand why parallel transport can be relevant?

 

[PrecPoint]

No, that is not the context. What you quoted discusses the relation and isomorphy between a vector space V and its dual and double dual.

Let me try to give you the context again. Please pay attention to the bold words

I think the point of naturally/canonically isomorphic or not is important for constructions/definitions and not for results. For example if two vector spaces $V$ and $W$ are isomorphic, because they have the same dimension, then you can add vectors from $V$ and $W$, but that depends on the choice of the isomorphism. If they are canonically isomorphic, then there is a natural way to add vectors from the two spaces. All tangent spaces to all points on a manifold are isomorphic, yet you need more to define a covariant derivative.

Right. The connection provides the isomorphisms.

Well, yes and no. If the connection is not flat then the isomorphism between two tangent spaces becomes path dependent.

Regardless of if the space is flat or not the covariant derivative of a $(1,0)$-tensor is given by $$X^j_{|h}=\frac{\partial X^j}{\partial x^h}+\Gamma_{l\,\,\,h}^{\,\,\,j}X^l$$
And it has nothing to do with tangent spaces of different points miles apart. Agreed?

(1,1)or(2,0)or(0,2)?And does a dual vector field have gradient?

There is mathematically no significant difference since $V\cong V^*$ (except in homological algebra).

There is a huge mathematical difference between $(1,1)$, $(2,0)$ and $(0,2)$. And we need a one to one correspondence between the tangent space and its dual to raise or lower indices (ie a metric tensor field).

 

[Orodruin]

(ie a metric tensor field)

No, not necessarily a metric tensor field. A non-degenerate bilinear form. This is not the same thing. A metric tensor is a bilinear form but a bilinear form is not necessarily a metric (or even a pseudo-metric). Again, the typical example would be the (fully anti-symmetric) symplectic form in symplectic geometry.

Please pay attention to the bold words

Please pay attention to the bold words:

All tangent spaces to all points on a manifold are isomorphic

 

[lavinia]

A vector space and its dual differ in the way their elements transform under linear maps. If a linear map $L$ maps the vector space $V$ into the vector space $W$ then by definition elements of $V$ are sent to elements of $W$.

However elements of the dual space $V^{*}$ are not sent to elements of $W^{*}$. Instead it is the other way around. Elements of $W^{*}$ are sent to elements of $V^{*}$.

Schematically this transformation difference can be represented by two diagrams:

$L:V→V$ $L^{*}:W^{*}→V^{*}$

This difference in transformation direction distinguishes a vector space and its dual. I would guess that this automatically excludes the possibility of a natural transformation between the identity functor and the "dual"" functor that maps each vector space to its dual and each linear map to its induced map on dual spaces. A natural transformation would seem always to preserve the direction of transformation. @fresh_42 is this correct?

This is a purely algebraic consideration and does not require manifolds or tensors or anything else. However, on a smooth manifold, the differential of a smooth map $f:M→N$ between two manifolds $df:TM→TN$ maps tangent vector in $TM$ to tangent vectors in $TN$ while the dual of the differential $df^{*}:TN^{*}→TM^{*}$ maps $TN^{*}$ back in the opposite direction to $TM^{*}$. The direction of $df$ is called "covariant" while the direction of $df^{*}$ is called "contravariant".

In the language of what is sometimes called "abstract nonsense" the differential is a covariant functor from the category of smooth manifolds and smooth maps to the category of smooth vector bundles and smooth vector bundle morphisms while the dual of the differential defines a contravariant functor to the category of smooth vector bundles and smooth vector bundle morphisms. This is the origin of the terms "push forward" and "pullback".

Notes:
-The coordinates of a vector with respect to a basis are dual vectors when they are viewed as linear functionals that assign a coordinate to each vector. Similarly, the coordinates of dual vectors are dual dual vectors and under the canonical isomorphism are identified with vectors.

 

[fresh_42]

A natural transformation would seem always to preserve the direction of transformation. @fresh_42 is this correct?

If I read nLab correctly, and there is no better online source I am aware of when it comes to homological algebra then there is no distinction between covariant or contravariant functors in the definition of a natural transformation between categories, i.e. it is valid in both cases.

Assume we would have two different functors $\tau,\dagger \, : \,\text{Vec}\longrightarrow \text{Vec}^*.$ A transformation $\alpha \, : \, \tau \longmapsto \dagger$ is natural if it is compatible with the morphisms: Say $\varphi : V^\tau \longrightarrow W^\tau$ and $\psi : V^\dagger \longrightarrow W^\dagger$ then we demand that
$$
\alpha (W)\circ \varphi^\tau = (V^\tau\stackrel{\varphi^\tau}{\longrightarrow }W^\tau \stackrel{\alpha }{\longrightarrow } W^\dagger) =(V^\tau\stackrel{\alpha }{\longrightarrow }V^\dagger \stackrel{\psi^\dagger}{\longrightarrow }W^\dagger)= \psi^\dagger \circ \alpha (V)
$$
Hence being natural says something about the mapping $\tau \longmapsto \dagger$ rather than the direction of $\varphi$ or $\psi.$ I think that makes sense as we are interested how $V^*\cong V^\tau$ and $V^*\cong V^\dagger$ are related, and not how morphisms in either category are defined. If I'm not mistaken, then $\varphi,\psi \, : \,W\longrightarrow V$ is not ruled out.

There is another interesting note on that page:
A transformation which is natural only relative to isomorphisms may be called a canonical transformation.

At least this explains the difficulty to distinguish the terms with the example $\text{Vec}$ and $\text{Vec}^*.$

 

[lavinia]

If I read nLab correctly, and there is no better online source I am aware of when it comes to homological algebra then there is no distinction between covariant or contravariant functors in the definition of a natural transformation between categories, i.e. it is valid in both cases.

Assume we would have two different functors $\tau,\dagger \, : \,\text{Vec}\longrightarrow \text{Vec}^*.$ A transformation $\alpha \, : \, \tau \longmapsto \dagger$ is natural if it is compatible with the morphisms: Say $\varphi : V^\tau \longrightarrow W^\tau$ and $\psi : V^\dagger \longrightarrow W^\dagger$ then we demand that
$$
\alpha (W)\circ \varphi^\tau = (V^\tau\stackrel{\varphi^\tau}{\longrightarrow }W^\tau \stackrel{\alpha }{\longrightarrow } W^\dagger) =(V^\tau\stackrel{\alpha }{\longrightarrow }V^\dagger \stackrel{\psi^\dagger}{\longrightarrow }W^\dagger)= \psi^\dagger \circ \alpha (V)
$$

Right. What I meant to say more clearly was there is no idea of a natural transformation between a covariant functor and a contravariant functor.

The thinking was whether one could choose isomorphisms $J_{V}$ between each vector space $V$ an its dual so that for every linear map $L:V→W$, $L^{*} \circ J_{W} \circ L =J_{V}$. This cannot be done by choosing a basis for each vector space.

This seems to illustrate a difference between the idea of a natural/canonical isomorphism and a natural transformation.

 

[S.G. Janssens]

Why not?

As an example, take a Banach space that is not reflexive but still linearly isomorphic to its second continuous dual space.