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  From Manifold to Manifold?

+ 2 like - 0 dislike

Tensor equations are supposed to stay invariant in form wrt coordinate transformations where the metric is preserved. It is important to take note of the fact that invariance in form of the tensor equations is consistent with the fact that the individual components of the tensor may change on passing from one frame to another.[Incidentally,Preservation of the metric implies preservation of norm, angles etc.]

But in General Relativity the tensor equations (examples: the geodesic equation, Maxwell's equations in covariant form) are considered to be invariant in form when we pass from one manifold to another. The metric is not preserved in such situations. Preservation of the value of the line element is consistent with the fact that $g_{\mu\nu}$ may be considered as a covariant tensor of second rank:




Rigorous Calculations:

$ds'^2=g'_{\mu\nu}dx'^\alpha dx'^\beta$

$=g'_{\mu\nu}\frac{\partial x'^\mu}{\partial x^\alpha}{d x^\alpha}\frac{\partial x'^\nu}{\partial x^\beta}{d x^\beta}$

$=g'_{\mu\nu}\frac{\partial x'^\mu}{\partial x^\alpha}\frac{\partial x'^\nu}{\partial x^\beta}{d x^\alpha}{d x^\beta}$

$=>g_{\alpha\beta}=\frac{\partial x'^\mu}{\partial x^\alpha}\frac{\partial x'^\nu}{\partial x^\beta}g'_{\mu\nu}$

Therefore $g_{\mu\nu}$ is a covariant tensor of rank two.

But in the above proof we have assumed the value of $ds^2$ as invariant wrt to our transformation.This not true when different types of manifolds are in consideration.

Non-conservation of the value of $ds^2$ will result in dismissing $g_{\mu\nu}$ as a second rank tensor of covariant type. This will be the situation if we pass from one manifold to another.*It is important to emphasize the fact that the problem will remain even if when we pass from an arbitrary manifold to flat spacetime in particular to the local inertial frame.*Differential considerations are not improving matters as indicated in the above calculation. The very concept of a tensor gets upset by considering different/distinct manifold.

What is the mathematical foundation for the invariance of form of the tensor equations in such applications where we consider different/distinct manifolds?

This post has been migrated from (A51.SE)

asked Nov 15, 2011 in Theoretical Physics by Anamitra Palit (10 points) [ revision history ]
retagged Mar 25, 2014 by dimension10
Although this is probably contained in Luboš's answer below, let me just make the comment that the confusion might be due to the fact that the first sentence in the question is not correct. A tensor equation is such that *all* diffeomorphisms (and not just isometries, as stated) take solutions to solutions. The second paragraph also contains an imprecision. By "when we pass from one manifold to another" I suppose you mean "under a differentiable map from one manifold to another". If so, it is not automatic that a tensor field in one manifold defines a tensor field in the other. (cont'd)

This post has been migrated from (A51.SE)
For example, differential forms pull back under differentiable maps, but unless the map is injective, vector fields do not push forward. So it is probably safe to restrict to diffeomorphisms between manifolds, rather than general differentiable maps.

This post has been migrated from (A51.SE)
The strict answer to your question "What is the mathematical foundation for such applications [as dynamics]" is *naturality*, in the mathematical sense of the word (see "http://mathoverflow.net/questions/56938/what-does-the-adjective-natural-actually-mean"). Many people have favorite textbooks that opened their eyes to the notion of mathematical naturality: one of mine is Jack Lee's *Introduction to Smooth Manifolds*, aka "Smooth Introduction to Manifolds". Generally speaking, the earlier this eye-opening comes to students of math, physics, and engineering, the better! :)

This post has been migrated from (A51.SE)
For clearness I would remark that the injectivity condition on the smooth map $f:M\to N$ is not sufficient to the existence of push-forward $f_{\ast}:\mathcal{X}(M)\to\mathcal{X}(N)$. CConsider for example the inverse of the stereographic map $R^n\to S^n$ or a curve $\mathbb{R}\to\mathbb{T}^2$ whose image is everywhere dense.

This post has been migrated from (A51.SE)

3 Answers

+ 6 like - 0 dislike

The question seems to conflate many different things:

  • the invariance of a mathematical quantity (usually a scalar such as $ds^2$ for the separation of two events in special relativity)
  • covariance of tensors (the values of components of tensors may be calculated from those in another frame but they're not the same thing)
  • universality of equations in different situations (the same equations – defining a theory – have many solutions and different solutions e.g. different shapes of manifolds are generally not related to each other at all)

These things are perhaps related and resemble each other but they're not the same things. In special relativity, some objects such as $p_\mu p^\mu$ for an energy-momentum vector $p^\mu$ are "invariant" which really means that the value of this scalar quantity doesn't change at all if one performs a Lorentz transformation $L$: $$ L(p)^\mu L(p)_\mu = p^\mu p_\mu $$ Then there are tensors which are any objects that transform "covariantly": $$ L(T)_{\alpha\beta\dots \omega} = T_{\alpha' \beta' \dots \omega'} L^{\alpha'}_{\alpha} L^{\beta'}_{\beta} \dots L^{\omega'}_{\omega} $$ which means that they transform as "tensor products of vectors": each index is being contracted with a copy of the Lorentz transformation's matrix.

Field equations in special relativity are covariant: they (after all terms are moved to the left hand side and the right hand side vanishes) transform as tensors which means that if they vanish (hold) in one reference frame, they also do in another. However, the particular numerical values of the components of a (covariant) tensor do depend on the reference frame. They're not "invariant" (unchanging); instead, they're just "covariant" (they change together with the coordinates, according to a universal tensor rule).

In general relativity, the fields such as the Ricci tensor are functions of the spacetime coordinates. At each point, the objects transform as tensors (as explained above) under coordinate transformations that reduce to Lorentz transformations in the vicinity of the given point (up to a certain approximation). In fact, the tensor transformation rule above may be generalized and should be generalized from $SO(3,1)$ to $GL(4,R)$. This is also useful to write down how the tensor fields transform under general diffeomorphisms i.e. not necessarily linear coordinate transformations. For general coordinate transformations, the definition of a "tensor" is more constraining: for example, partial derivatives of vectors no longer transform as tensors.

With this more constraining definition, general relativity dictates field equations that have the form "tensor field vanishes". For a given theory, the equations of motion have the universal form – e.g. the Maxwell-Einstein equations, to be specific. The well-definedness and the uniqueness of the equations of motion is what we mean by having a single theory. However, a single theory or a single set of equations in physics always has many solutions. In special relativity, one may produce new (mathematically but not physically new) solutions by Lorentz transformations from a given one; in general relativity, one may obtain new (mathematically but not physically new) solutions by any diffeomorphisms applied on a given solution.

The tensor fields transform covariantly but they are not invariant and they depend on the situation – on the shape of the manifold etc. At any rate, at this point, you should understand why your question makes no sense. The mathematical foundation of "such application" is elementary linear algebra, differential geometry, special relativity, or general relativity, depending on what you're exactly asking about. However, you're not exactly asking about anything so your question can't be answered. Be sure that there doesn't exist any contradiction along the lines that you apparently wanted to propose in the wording of your question.

This post has been migrated from (A51.SE)
answered Nov 15, 2011 by Luboš Motl (10,278 points) [ no revision ]
+ 6 like - 0 dislike

The tensor equations you mention are not invariant, they are covariant. Big difference. Both are differential equations, which transform linearly under nonlinear transformations from one manifold to another because they are differential equations at a point. The nonlinear transformation from one manifold to another induces a linear transformation of the tangent space at each point of one manifold to the tangent space at the corresponding point on the other.

The metric tensor $g_{\mu\nu}$ is linearly covariant under transformations. Given that there is a metric tensor on both manifolds, and that one is the image of the other under a diffeomorphism, the distance from point $A$ to point $B$, for example, is invariant under that transformation. Loosely, a tensor object is invariant when all the indices are summed (typically under the Einstein summation convention, which hides many linear algebra theorems in the formalism).

Luboš posted a more extensive Answer as I was finishing this, so consider this a relatively simple-minded counterpoint to that.

I suggest you need a decent Math-Physics crossover book that discusses manifolds and differential geometry at an intermediate level, of which there are a number. I've been happy with Nakahara, http://www.amazon.com/Geometry-Topology-Physics-Graduate-Student/dp/0750306068.

This post has been migrated from (A51.SE)
answered Nov 15, 2011 by Peter Morgan (1,230 points) [ no revision ]
+ 3 like - 0 dislike

This depends on what you mean by "pass from one manifold to another". In General Relativity one generally considers a single manifold $\mathcal{M}$ and diffeomorphisms $\phi: \mathcal{M} \rightarrow \mathcal{M}$. I think the idea you are trying to get at is that if you consider a geometry on $\mathcal{M}$, that is a pair $(\mathcal{M} , g)$ where $g$ is a metric field (a smooth 2-form field which is non-degenerate and has the right signature) which satisfies then Einstein Field Equations on $\mathcal{M}$ then the geometry, given by $(\mathcal{M} , (\phi^{-1})^*g)$, where $(\phi^{-1})^*g$ is the pullback of $g$ along the inverse of $\phi$, also satisfies the Einstein Field Equations.

This is the origin of the notion that GR has a "gauge redundancy" which is its invariance under active diffeomorphisms and is quite different (although related to) the fact that the tensor equations of GR are covariant under coordinate transformations (that is switching from one coordinate chart to another).

It is only in the above sense that when you "pass from one manifold to another" the "metric is preserved".

This post has been migrated from (A51.SE)
answered Nov 20, 2011 by Kyle (335 points) [ no revision ]

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