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GR191511
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(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).GR191511 said:(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).GR191511 said:(1,1)or(2,0)or(0,2)?
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.GR191511 said:And does a dual vector field have gradient?
They are isomorphic, but not canonically. So, i would say that there is a difference.fresh_42 said:There is mathematically no significant difference since ##V\cong V^*## (except in homological algebra).
That is a completely different question than what the OP is asking.fresh_42 said: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.fresh_42 said: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.
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.WWGD said: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.
What about more than two dimensions?WWGD said: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##.
Ah, yes, I forgot about that. It's only the double-dual . And, yes, that works just for dimension 2.martinbn said: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?
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.WWGD said:Doesn't it matter that the isomorphism isn't natural?
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.WWGD said: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.
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 said: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.
That is a great question. I am not sure how to think about it but maybe we can come up with some ideas.WWGD said: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 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: $$\pi.$$lavinia said:That is a great question. I am not sure how to think about it but maybe we can come up with some ideas.
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 said: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) .
Right. The connection provides the isomorphisms.martinbn said: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.
Well, yes and no. If the connection is not flat then the isomorphism between two tangent spaces becomes path dependent.WWGD said:Right. The connection provides the isomorphisms.
No.martinbn said:I think the point of naturally/canonically isomorphic or not is important for constructions/definitions and not for results.
Why not?S.G. Janssens said:No.
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 said:Well, yes and no. If the connection is not flat then the isomorphism between two tangent spaces becomes path dependent.
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 said: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)##.
I believe WWGD was talking about neighboring points here:Orodruin said:Nobody was talking about neighboring points or points that are infinitesimally close
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?WWGD said:Right. The connection provides the isomorphisms.
Why? There is no indication of that.PrecPoint said:I believe WWGD was talking about neighboring points here:
There are many reasons to do so. It is the entire point of parallel transport, which has numerous applications.PrecPoint said:Why would you try to relate tangent spaces of unrelated points on the manifold?
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 said:And the answer is, the correspondence depends on the existence of a given metric;
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...Orodruin said: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.
So please give some reasons related to the question at hand. The context is the following;Orodruin said:There are many reasons to do so.
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?"fresh_42 said:There is mathematically no significant difference since ##V\cong V^*## (except in homological algebra).
PrecPoint said: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.
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.PrecPoint said:So please give some reasons related to the question at hand. The context is the following;
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.PrecPoint said: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.
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 said:And I really don't understand why we should consider two tangent spaces miles apart.
Let me try to give you the context again. Please pay attention to the bold wordsOrodruin said: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.
martinbn said: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 said:Right. The connection provides the isomorphisms.
Regardless of if the space is flat or not the covariant derivative of a ##(1,0)##-tensor is given by $$X^j_{|h}=\frac{\partial X^j}{\partial x^h}+\Gamma_{l\,\,\,h}^{\,\,\,j}X^l$$Orodruin said:Well, yes and no. If the connection is not flat then the isomorphism between two tangent spaces becomes path dependent.
GR191511 said:(1,1)or(2,0)or(0,2)?And does a dual vector field have gradient?
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).fresh_42 said:There is mathematically no significant difference since ##V\cong V^*## (except in homological algebra).
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.PrecPoint said:(ie a metric tensor field)
Please pay attention to the bold words:PrecPoint said:Please pay attention to the bold words
martinbn said:All tangent spaces to all points on a manifold are isomorphic
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.lavinia said:A natural transformation would seem always to preserve the direction of transformation. @fresh_42 is this correct?
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.fresh_42 said: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
$$
\alpha (W)\circ \varphi^\tau = (V^\tau\stackrel{\varphi^\tau}{\longrightarrow }W^\tau \stackrel{\alpha }{\longrightarrow } W^\dagger) =(V^\tau\stackrel{\alpha }{\longrightarrow }V^\dagger \stackrel{\psi^\dagger}{\longrightarrow }W^\dagger)= \psi^\dagger \circ \alpha (V)
$$
As an example, take a Banach space that is not reflexive but still linearly isomorphic to its second continuous dual space.martinbn said:Why not?