Vector fields, flows and tensor fields

[TrickyDicky]

Vector fields generate flows, i.e. one-parameter groups of diffeomorphisms, which are profusely used in physics from the streamlines of velocity flows in fluid dynamics to currents as flows of charge in electromagnetism, and when the flows preserve the metric we talk about Killing vector fields and isometries are defined and used i.e. in GR.

How exactly is this generalized to the case of tensor fields?, let's concentrate on rank-2 tensor fields that are most often found in physics. Intuitively one would think they would generate two-parameters groups of diffeomorphisms?. Say for instance if the tensor field is a metric tensor(thinking about Klein's Erlangen program), they would generate isometries in the form of transformations like translations, rotations, reflections involving two directions? For other two-rank tensors that have physical significance , say the stress-energy tensor, what kind of (local)"flow" would it generate?

 

[homeomorphic]

I don't think it does generalize, or, if it does, you wouldn't call it a flow. Flow means you follow the arrows. But in general, tensor fields don't have arrows to follow.

A metric tensor doesn't generate isometries. Things isometries with respect to them. Also, the fact that it's a rank-2 tensor has absolutely nothing to do with having 2-parameters in its isometry group. For example, the isometries of n-dimensional space will have MUCH more than 2-parameters, obviously.

The stress energy tensor is doing something a little like generating a flow, if you interpret it from that viewpoint. It's telling you something about the curvature of space-time. It's a little like time-evolution for PDE's like Maxwell's equations or the wave equation or something. However, calling it a "flow" seems to rely on a well-defined notion of time, and time is sort of tricky to get your hands on in general relativity, so it's a bit different, in general, although I think, in special cases, you can get away with interpreting it that way--telling you how the metric on some space-like surfaces changes over time. The Cauchy problem: I give you the geometry of a space-like surface, you tell me the rest of the geometry. Similar to an initial value problem in ODE (or in PDE, like with the wave equation if I tell you the configuration of a rubber sheet at t = 0, you tell me what happens in the rest of time). If you want to call it a flow, maybe it's sort of a flow in the configuration space of the rubber sheet. And maybe in GR, you might think of it as a flow through the space of geometries for your space-like 3-surface, I guess (don't quote me on it), but I am hesitant to call it a flow.

A more appropriate generalization would be to ask what would happen if you had a field of planes, rather than a field of vectors. Can the planes be realized as the tangent planes to a surface, just vector fields can be realized as the tangent vectors to a curve? Not so easily. This question is addressed by the Frobenius theorem.

Not a particularly good article, but here it is:

http://en.wikipedia.org/wiki/Frobenius_theorem_(differential_topology )

 

[TrickyDicky]

Hi, homeomorphic, thanks for taking a stab at answering this.

I don't think it does generalize, or, if it does, you wouldn't call it a flow. Flow means you follow the arrows. But in general, tensor fields don't have arrows to follow.

Sure, I wouldn't call it flow either, flows as we understand it are one-dimensional, and as you also write below, I had in mind something more like surfaces or planes.

A metric tensor doesn't generate isometries.

Well, to make it clearer, an isometry is any diffeomorphism that preserves the metric tensor, so yes it might be misleading to use the terminology for vector fields in terms of generating flows.

Things isometries with respect to them.

Couldn't understand this sentence, can you rephrase it?

Also, the fact that it's a rank-2 tensor has absolutely nothing to do with having 2-parameters in its isometry group. For example, the isometries of n-dimensional space will have MUCH more than 2-parameters, obviously.

Yes, I was more thinking about the idea of surfaces or planes, in the sense that a tensor of rank-2 may have whatever dimensionality but there is a sense in which it is bidimensional just like n-dimensional vectors are unidimensional.

The stress energy tensor is doing something a little like generating a flow, if you interpret it from that viewpoint. It's telling you something about the curvature of space-time. It's a little like time-evolution for PDE's like Maxwell's equations or the wave equation or something. However, calling it a "flow" seems to rely on a well-defined notion of time, and time is sort of tricky to get your hands on in general relativity, so it's a bit different, in general, although I think, in special cases, you can get away with interpreting it that way--telling you how the metric on some space-like surfaces changes over time. The Cauchy problem: I give you the geometry of a space-like surface, you tell me the rest of the geometry. Similar to an initial value problem in ODE (or in PDE, like with the wave equation if I tell you the configuration of a rubber sheet at t = 0, you tell me what happens in the rest of time). If you want to call it a flow, maybe it's sort of a flow in the configuration space of the rubber sheet. And maybe in GR, you might think of it as a flow through the space of geometries for your space-like 3-surface, I guess (don't quote me on it), but I am hesitant to call it a flow.

Well, I was rather thinking that the stress-energy tensor field assures local conservation of momentum-energy, while in contrast other quantities like momentum or energy are quantities that by Noether theorem are conserved by admitting the flow(one-parameter groups of diffeomorfisms) of certain vector fields.

A more appropriate generalization would be to ask what would happen if you had a field of planes, rather than a field of vectors. Can the planes be realized as the tangent planes to a surface, just vector fields can be realized as the tangent vectors to a curve? Not so easily. This question is addressed by the Frobenius theorem.

Yes, this is the kind of visualization I had in mind. Frobenius theorem is quite interesting and pertinent here, it seems to suggest one has to have a space with certain requirements in terms of foliability to make more concrete what I intend with tensor fields.

 

[homeomorphic]

Couldn't understand this sentence, can you rephrase it?

Typo. Things ARE isometries with respect to them.

Well, I was rather thinking that the stress-energy tensor field assures local conservation of momentum-energy, while in contrast other quantities like momentum or energy are quantities that by Noether theorem are conserved by admitting the flow(one-parameter groups of diffeomorfisms) of certain vector fields.

Hmm. Well, I suspect there are analogues of that sort of thing in GR, but I don't know GR well enough.

There might be something interesting along those lines to be found here in Track 2:

http://math.ucr.edu/home/baez/qg-winter2001/

Not sure it's exactly what you are looking for, but it does discuss some things, like what is the analogue of momentum in configuration spaces of fields, like those of GR, so I imagine it might help.

Frobenius theorem is quite interesting and pertinent here, it seems to suggest one has to have a space with certain requirements in terms of foliability to make more concrete what I intend with tensor fields.

Well, the foliability has more to due to with which plane-fields you choose than the space they live in. There's some theorem of Thurston that a compact manifold in which there exists a distribution (plane-field) has a foliation of the same dimension.

 

[Ben Niehoff]

If you do have a foliation by n-surfaces, then you always have an n-parameter group of flows along those n-surfaces. This essentially pops out of the Frobenius theorem, because in order to have a foliation by n-surfaces, you must have n vector fields $X_i$ which are closed under the Lie bracket. These vector fields then generate n flows, each of which preserves the family of n-surfaces.

 

[TrickyDicky]

Thanks for the Baez reference, it is not what I was looking for but it is always refreshing to read.

I'm still interested in knowing if some analogue of integral curves for vector fields is defined for tensor fields, something like (for the case of rank-2 tensors) "integral surfaces".
Anybody?

 

[TrickyDicky]

If you do have a foliation by n-surfaces, then you always have an n-parameter group of flows along those n-surfaces. This essentially pops out of the Frobenius theorem, because in order to have a foliation by n-surfaces, you must have n vector fields $X_i$ which are closed under the Lie bracket. These vector fields then generate n flows, each of which preserves the family of n-surfaces.

Hi Ben I missed this while typing.
So basically one can define such "integral n-surfaces" generated by tensor fields of rank n when that foliation is available?

 

[Matterwave]

Hi Ben I missed this while typing.
So basically one can define such "integral n-surfaces" generated by tensor fields of rank n when that foliation is available?

Not in a direct way. The vector fields under which a tensor field is invariant form a Lie algebra. These vector fields define integral curves which mesh into submanifolds which foliate the manifold.

The latter part of the statement is Frobenius' theorem. The first part of the statement can be seen from direct calculation.

 

[Ben Niehoff]

Hi Ben I missed this while typing.
So basically one can define such "integral n-surfaces" generated by tensor fields of rank n when that foliation is available?

No, I'm afraid that idea doesn't make sense. One can talk about a group of flows generated by a collection of vector fields $X_i$. But each flow is still generated by a vector field.

You might be able to make some sense of the notion of "flows" generated by antisymmetric tensors, but I suspect it will still only make sense when those tensors can be written as a wedge product of 1-tensors (i.e. vectors).

The notion of a "flow" is just a very 1-dimensional thing.

 

[TrickyDicky]

Not in a direct way. The vector fields under which a tensor field is invariant form a Lie algebra. These vector fields define integral curves which mesh into submanifolds which foliate the manifold.

The latter part of the statement is Frobenius' theorem. The first part of the statement can be seen from direct calculation.

Yes I can see that , still is there not a way to make it more direct? I'm thinking specifically about rank-2 tensors here, and relying on their multiple role as bilinear map(higher order function) and operator-transformation-local diffeomorphism.

No, I'm afraid that idea doesn't make sense. One can talk about a group of flows generated by a collection of vector fields $X_i$. But each flow is still generated by a vector field.

The notion of a "flow" is just a very 1-dimensional thing.

I already agreed the term flow is not right here just for the reason you mention.
Let's limit the discussion for simplicity to rank-2 tensors.Their role as transformations-mappings-diffeomorphisms as I said above seems to allow a more direct way to relate tensor fields and diffeomorpisms.

You might be able to make some sense of the notion of "flows" generated by antisymmetric tensors, but I suspect it will still only make sense when those tensors can be written as a wedge product of 1-tensors (i.e. vectors).

Yes, well that is basically done in inverse-square Gauss laws relying on the properties of antisymmetric tensors, Euclidean space, Stokes and all that. which allows to ultimately reduce it all to vector fields and their flows. I'm trying to twist it a bit further.

 

[Matterwave]

A general tensor field of rank 2 can be expressed as:

$$T=\sum_{i,j}T^{ij}e_i\otimes e_j$$

How do you propose to turn this into a flow, or a diffeomorphism? Perhaps for tensors of the form:

$$T'=V\otimes W$$

One might be able to use the mesh of integral curves to attempt something, but 1. this equation does not hold in general, and 2. the integral curves will in no way be guaranteed to mesh to form a sub manifold unless $[V,W]=aV+bW$. Even if the integral curves do mesh to form a family of 2 dimensional sub manifolds, what is the flow from one to the other? The sub manifolds, as foliations, are disjoint, and so you'd need a third vector field to define diffeomorphisms from one to the other.

I don't see any advantages to be had from such a narrow definition.

 

[TrickyDicky]

A general tensor field of rank 2 can be expressed as:

$$T=\sum_{i,j}T^{ij}e_i\otimes e_j$$

How do you propose to turn this into a flow, or a diffeomorphism? Perhaps for tensors of the form:

$$T'=V\otimes W$$

One might be able to use the mesh of integral curves to attempt something, but 1. this equation does not hold in general, and 2. the integral curves will in no way be guaranteed to mesh to form a sub manifold unless $[V,W]=aV+bW$. Even if the integral curves do mesh to form a family of 2 dimensional sub manifolds, what is the flow from one to the other? The sub manifolds, as foliations, are disjoint, and so you'd need a third vector field to define diffeomorphisms from one to the other.

I don't see any advantages to be had from such a narrow definition.

Hmmm, I'm afraid we are not talking about the same thing, I'm not saying the same notion of integral curve-flow must be retained in the generalization to tensor fields. It would be the analogue just in the sense of relating fields and diffeomorphisms.So it wouldn't really look like a flow, but then again it is relatively easy to visualize vector fields and their flows, while it is not so easy to visualize even second rank tensor fields.
For instance, think of translations in the Euclidean plane, they are maps R2→R2, and they are two-parameter diffeomorphisms(that are also isometries since they preserve the metric), how would one relate them to a rank-2 tensor field, there appears to be a more direct way than the usual decomposing the group in their x and y directions and from the one-parameter get the vector fields, in the sense that the transformation itself can be seen as a tensor(universal property), but this doesn't exactly capture what I had in mind.
But I realize it is not as easy as I had pictured at first, and I'm not sure it can be fully generalized.

 

[TrickyDicky]

A general tensor field of rank 2 can be expressed as:

$$T=\sum_{i,j}T^{ij}e_i\otimes e_j$$

How do you propose to turn this into a flow, or a diffeomorphism? Perhaps for tensors of the form:

$$T'=V\otimes W$$

One might be able to use the mesh of integral curves to attempt something, but 1. this equation does not hold in general, and 2. the integral curves will in no way be guaranteed to mesh to form a sub manifold unless $[V,W]=aV+bW$. Even if the integral curves do mesh to form a family of 2 dimensional sub manifolds, what is the flow from one to the other? The sub manifolds, as foliations, are disjoint, and so you'd need a third vector field to define diffeomorphisms from one to the other.

I don't see any advantages to be had from such a narrow definition.

Hmmm, I'm afraid we are not talking about the same thing, I'm not saying the same notion of integral curve-flow must be retained in the generalization to tensor fields. It would be the analogue just in the sense of relating fields and diffeomorphisms.So it wouldn't really look like a flow, but then again it is relatively easy to visualize vector fields and their flows, while it is not so easy to visualize even second rank tensor fields.
For instance, think of translations in the Euclidean plane, they are maps R2→R2, and they are two-parameter diffeomorphisms(that are also isometries since they preserve the metric), how would one relate them to a rank-2 tensor field, there appears to be a more direct way than the usual decomposing the group in their x and y directions and from the one-parameter get the vector fields, in the sense that the transformation itself can be seen as a tensor(universal property), but this doesn't exactly capture what I had in mind.
But I realize it is not as easy as I had pictured at first(wich implied changing the staticity of the integral curves in flows by a more dynamic picture of the curves-thus producing surfaces in their sweeping motion, I thought this could appeal to string theorists as it puts the accent in one dimensional rather than 0-dimensional objects-, and I'm not sure it can be fully generalized.

 

[Matterwave]

How would you use a rank 2 tensor field to map a point P on a manifold to another point Q, maintaining that this is a diffeomorphism? What is your conception of this operation? Because none come to mind to me.

The important property that allows a vector field to define a diffeomorphism is that any point P lies on one and only one congruence (the congruences don't intersect each other). This allows you to define a map from P to Q which lies along the congruence. I see no generalization of this phenomenon to rank 2 tensor fields. What did you have in mind?

 

[TrickyDicky]

How would you use a rank 2 tensor field to map a point P on a manifold to another point Q, maintaining that this is a diffeomorphism? What is your conception of this operation? Because none come to mind to me.

The important property that allows a vector field to define a diffeomorphism is that any point P lies on one and only one congruence (the congruences don't intersect each other). This allows you to define a map from P to Q which lies along the congruence. I see no generalization of this phenomenon to rank 2 tensor fields. What did you have in mind?

No, I'm not trying to map fixed points, the idea is rather mapping curves that lie on parametrized congruences of surfaces instead of curves. This is what I had in mind but I don't know how to make it more explicit or even if it is feasible at all.

 

[Ben Niehoff]

Think about this another way:

Let's say you have a 2-parameter family of diffeomorphisms $\varphi_{st} : M \to M$. By "2-parameter family" we mean that for each $s, t$, we have a unique diffeomorphism. (This will turn out to be very restrictive!)

So, if we hold $s = s_0$ fixed, then we are left with a 1-parameter family of diffeomorphisms $\varphi_{s_0 t}$, which is just a flow along some vector field $X$. Similarly, if we hold $t = t_0$ fixed, we get the 1-parameter family $\varphi_{s t_0}$, which is just a flow along some vector field $Y$.

Finally, since we claim that each $\varphi_{st}$ corresponds to exactly one diffeomorphism, we must in fact have that $X$ and $Y$ commute!

$$[X,Y] = 0$$
So even if you start with the idea of a "2-parameter family of diffeomorphisms", it turns out that this idea is even more restrictive than the idea of a group of flows on the leaves of a foliation. Because in order to have a foliation, $[X,Y]$ need only be closed into $a X + b Y$; it needn't be zero.

 

[TrickyDicky]

Yes, I can see what you mean, Ben. Thanks for the clarifying post.
The idea is quite restrictive rather than a generalization as I thought. Only under very specific conditions it can be achieved.
I'll think some more about it just in case I see if there is any potential situation where this could be applicable or if I think up some other question.

 

[TrickyDicky]

I was thinking about the two restrictive conditions you wrote above and since I already mentioned the stress-energy tensor as a possible example I was wondering if being a both symmetric and divergenceless tensor field would meet those criteria or at least get close to it(I guess the dimensionality of the space would also matter by the Frobenius theorem?).

 

[Matterwave]

No, I'm not trying to map fixed points, the idea is rather mapping curves that lie on parametrized congruences of surfaces instead of curves. This is what I had in mind but I don't know how to make it more explicit or even if it is feasible at all.

A diffeomorphism is always mapping points to points. If you map a line to a line, you are still mapping points on that line to another point on another line. Point P to point Q is the most basic and most general diffeomorphism map you can obtain.

As regards to your questions about the stress-energy tensor "meeting the requirements" put forth by Ben, Ben's restrictions are on the vector fields generating each specific flow. The stress energy tensor cannot be expressed as the simple direct product of two vectors. Two 4-vectors has a total of 8 independent components, while the Stress-energy tensor has 10 independent components in general.

 

[TrickyDicky]

A diffeomorphism is always mapping points to points. If you map a line to a line, you are still mapping points on that line to another point on another line. Point P to point Q is the most basic and most general diffeomorphism map you can obtain.

If you map a vector in one direction to a vector in another direction with a tensor I guess you are in some sense also mapping points but not fixed points in the sense I think you mean by P and Q.

As regards to your questions about the stress-energy tensor "meeting the requirements" put forth by Ben, Ben's restrictions are on the vector fields generating each specific flow.

The vanishing commutator involves both vector fields

The stress energy tensor cannot be expressed as the simple direct product of two vectors. Two 4-vectors has a total of 8 independent components, while the Stress-energy tensor has 10 independent components in general.

The tensor product of two 4-vectors would have 16 components, being symmetric they are reduced to 10.

 

[Matterwave]

If you map a vector in one direction to a vector in another direction with a tensor I guess you are in some sense also mapping points but not fixed points in the sense I think you mean by P and Q.

Of course you can change the "direction" of the vector. This is the whole idea behind Lie dragging and Lie derivatives. The flow generated by a vector field can do the same thing (in fact, in general, you WILL change the "direction" of the vector, unless the vector is Lie dragged by your flow)! You certainly don't need a tensor to do this. If I map points to points, I also map lines to lines, by definition. A diffeomorphic map of points to points maps curves to curves, and surfaces to surfaces, and volumes to volumes.

The vanishing commutator involves both vector fields

I'm not sure where you're going with this.

The tensor product of two 4-vectors would have 16 components, being symmetric they are reduced to 10.

The tensor product of two 4-vectors have 16 components, but at most 8 of them can be independent. Since you started with 8 total components, you can't just magically get more independent components out of thin air. This is the reason why not every rank 2 tensor can be expressed as the direct product of 2 vectors. Like I mentioned in my previous post, not every rank two tensor can be expressed as:

$$T=V\otimes W$$

Only a small subset of tensors can be expressed this way. Most other tensors must be expressed as sums of direct products of vectors.

 

[TrickyDicky]

The tensor product of two 4-vectors have 16 components, but at most 8 of them can be independent. Since you started with 8 total components, you can't just magically get more independent components out of thin air. This is the reason why not every rank 2 tensor can be expressed as the direct product of 2 vectors. Like I mentioned in my previous post, not every rank two tensor can be expressed as:

$$T=V\otimes W$$

Only a small subset of tensors can be expressed this way. Most other tensors must be expressed as sums of direct products of vectors.

I agree that not every rank-2 tensor field can be the product of 2 vector fields, but I'm obviously referring to those that indeed can be expressed that way.

I don't know where you got the part about the 8 independent components for the product of two 4-vector fields. As I said this number will depend basically on the symmetries of the tensor. As an inmediate counterexample an antisymmetric rank-2 tensor from the product of two vectors in 4 dim has 6 independent components instead of 8 as you claim.

 

[Matterwave]

I agree that not every rank-2 tensor field can be the product of 2 vector fields, but I'm obviously referring to those that indeed can be expressed that way.

I don't know where you got the part about the 8 independent components for the product of two 4-vector fields. As I said this number will depend basically on the symmetries of the tensor. As an inmediate counterexample an antisymmetric rank-2 tensor from the product of two vectors in 4 dim has 6 independent components instead of 8 as you claim.

I claim that 8 is the maximum (in 4 dimensions), because you have 4 components in the first vector, 4 in the second. Any combination of the two can AT MOST produce 8 independent components.

I'm using this argument to argue that the Stress-energy tensor, which has 10 independent components, CANNOT be expressed in this way, and therefore would not be good for your purposes.

 

[TrickyDicky]

Let's use the dyadic terminology that here might clarify things:a general second order tensor is called a dyadic. And certainly it doesn't have to be result of the product of two vectors. A second order tensor that is the direct product of two vectors is a dyad: ad dyadic tensor of rank one, using the term rank with a different meaning than in previous posts where it was synonimous of order . Now every dyadic can be expressed as a linear combination of dyads.

Now the stress-energy tensor is not a dyad so I can see it is not good for my purpose, but not for the reason you mention as in general the dyadic product of two vectors has $n^2$ independent components (with n being the dimension), that may or not be reduced in function of the symmetries of the resultant tensor.

Matterwave, it would be interesting if you could provide some reference that shows that the number of independent components after the tensor product operation is performed on two vectors is the sum n+n instead of $n^2$, I find it difficult since each component of a dyadic product is obtained by multiplication of the components of each vector and except some of the vector components are zero one gets $n^2$ independent components before possible reductions due to symmetry.
Note that there is something that sometimes is referred to simply as dyad, that is a combination of two n-vectors juxtaposed that doesn't involve any multiplication, that is simply a linear transformation matrix and that indeed has just n+n independent components at most.

 

[homeomorphic]

Tricky is right. If you tensor two vector spaces, the dimension multiplies. So, 4 dimensions tensor 4 dimensions would give you 16 components. Unless, you are looking at some subspace of the tensor product, such as the anti-symmetric ones. I'm pretty sure it's true that the stress-energy tensor can't be expressed as a tensor product of two guys, in general, though.

The flow I was talking about at the beginning, involving the stress-energy tensor was basically what people do as a toy example in quantum gravity sometimes, where the space-time is taken to be a 3-manifold cross R, where R represents time. Then, you do a flow in what I would gather is some space of connections on the 3-manifold, (modulo gauge tranformations, maybe)--or in other words, the space of possible geometries on it. That's the only flow I can see coming up in relation to the stress energy tensor. Not a flow in space-time, but a flow in that moduli space, which I think is infinite-dimensional (though possibly not, once you mod out by gauge equivalence, but I don't really know this stuff, so I'm trying to fudge it as best I can). Of course, in general, you wouldn't want to cross with R to get your space-time, so this picture breaks down.

 

[Matterwave]

Let's use the dyadic terminology that here might clarify things:a general second order tensor is called a dyadic. And certainly it doesn't have to be result of the product of two vectors. A second order tensor that is the direct product of two vectors is a dyad: ad dyadic tensor of rank one, using the term rank with a different meaning than in previous posts where it was synonimous of order . Now every dyadic can be expressed as a linear combination of dyads.

Now the stress-energy tensor is not a dyad so I can see it is not good for my purpose, but not for the reason you mention as in general the dyadic product of two vectors has $n^2$ independent components (with n being the dimension), that may or not be reduced in function of the symmetries of the resultant tensor.

Matterwave, it would be interesting if you could provide some reference that shows that the number of independent components after the tensor product operation is performed on two vectors is the sum n+n instead of $n^2$, I find it difficult since each component of a dyadic product is obtained by multiplication of the components of each vector and except some of the vector components are zero one gets $n^2$ independent components before possible reductions due to symmetry.
Note that there is something that sometimes is referred to simply as dyad, that is a combination of two n-vectors juxtaposed that doesn't involve any multiplication, that is simply a linear transformation matrix and that indeed has just n+n independent components at most.

In fact this counting scheme is the proof that there is no way to express the general rank 2 tensor $T$ as $T=V\otimes W$. $V$ and $W$ between them have $2n$ components, while a tensor of rank 2 has $n^2$ components. Therefore, there is not enough freedom in 2 vectors to specify a general rank 2 tensor.

If you want a source, you can look at Bernard Schutz Geometrical methods of mathematical physics. Chapter 2, exercise 2.5 actually asks the student to prove this themselves (for the solution, see the appendix, wherein he says exactly what I said).

 

[homeomorphic]

Ok, I see what you were saying, now. I don't know about Tricky.

 

[Matterwave]

Tricky is right. If you tensor two vector spaces, the dimension multiplies. So, 4 dimensions tensor 4 dimensions would give you 16 components. Unless, you are looking at some subspace of the tensor product, such as the anti-symmetric ones. I'm pretty sure it's true that the stress-energy tensor can't be expressed as a tensor product of two guys, in general, though.

The flow I was talking about at the beginning, involving the stress-energy tensor was basically what people do as a toy example in quantum gravity sometimes, where the space-time is taken to be a 3-manifold cross R, where R represents time. Then, you do a flow in what I would gather is some space of connections on the 3-manifold, (modulo gauge tranformations, maybe)--or in other words, the space of possible geometries on it. That's the only flow I can see coming up in relation to the stress energy tensor. Not a flow in space-time, but a flow in that moduli space, which I think is infinite-dimensional (though possibly not, once you mod out by gauge equivalence, but I don't really know this stuff, so I'm trying to fudge it as best I can). Of course, in general, you wouldn't want to cross with R to get your space-time, so this picture breaks down.

The "generator" of this "flow" is highly non-trivial. To move from one Cauchy hyper-surface (geometric 3-volume slice of space time) to the next, one defines lapse and shift functions. But these are consequences of the coordinate systems, and the method of splicing, and not necessarily on true dynamical evolution.

The statement "modulo gauge transformations" sounds simple enough, but is not simple to put in practice. The 3-metric and associated conjugate momenta do not really create a good phase space due to all the possible diffeomorphisms of a space-time (read: possible coordinate transformations). Even taking the configuration space to be Wheeler's superspace (the space of all 3-geometries) there is still the arbitrariness in the particular slicing of space-time that one chooses. And I know of no way to get rid of this arbitrariness in any 3-slicing of space-time.

What I'm saying is, the "flow" you mentioned associated to this concept is probably not going to be "generated" by the stress-energy tensor in the sense that the time translations in a classical system are generated by the Hamiltonian. The Hamiltonian in the ADM formalism is actually just a sum of constraint equations, and therefore must equal 0.

 

[TrickyDicky]

In fact this counting scheme is the proof that there is no way to express the general rank 2 tensor $T$ as $T=V\otimes W$. $V$ and $W$ between them have $2n$ components, while a tensor of rank 2 has $n^2$ components. Therefore, there is not enough freedom in 2 vectors to specify a general rank 2 tensor.

If you want a source, you can look at Bernard Schutz Geometrical methods of mathematical physics. Chapter 2, exercise 2.5 actually asks the student to prove this themselves (for the solution, see the appendix, wherein he says exactly what I said).

Right, I knew this proof, that is why I said that I agreed a general order two tensor is not necessarily the tensor product of two vectors, a general second order tensor can be constructed with the sum of $n^2$ tensor products.

But I think you might have misunderstood what this proof implies from what Schutz write in the Appendix. You seemed to imply that the tensor product of two vectors has at most 2n independent components, and therefore by counting the independent components of a tensor one might deduce if it is the result of a tensor product or not. This is not correct. Just do the calculation, multiply two vectors using the outer product, you get a second order tensor with $n^2$ independent components. Granted you have only used 2n components to build it and that is why a general tensor is not necessarily the outer product of two vectors, but still after the product the $n^2$ components formed from the 2n are considered to be independent.

 

[TrickyDicky]

I'm pretty sure it's true that the stress-energy tensor can't be expressed as a tensor product of two guys, in general, though.

Certainly, I just found its decomposition as a sum of outer products. But an example of a tensor that is a dyad is the spatial stress tensor(and in fact several others also used in engineering and physics).

The flow I was talking about at the beginning, involving the stress-energy tensor was basically what people do as a toy example in quantum gravity sometimes, where the space-time is taken to be a 3-manifold cross R, where R represents time. Then, you do a flow in what I would gather is some space of connections on the 3-manifold, (modulo gauge tranformations, maybe)--or in other words, the space of possible geometries on it. That's the only flow I can see coming up in relation to the stress energy tensor. Not a flow in space-time, but a flow in that moduli space, which I think is infinite-dimensional (though possibly not, once you mod out by gauge equivalence, but I don't really know this stuff, so I'm trying to fudge it as best I can). Of course, in general, you wouldn't want to cross with R to get your space-time, so this picture breaks down.

As commented by matterwave it is only in very specific instances that the 3+1 decomposition is possible in GR(for instance in static spacetimes...), so it wouldn't work in general.

My idea was more radical than this though, and it is not something that can be generalized either, as Ben showed it needs some requisites that very few tensor fields(if any, I'm yet to find a single example) would have.

 

[Matterwave]

Right, I knew this proof, that is why I said that I agreed a general order two tensor is not necessarily the tensor product of two vectors, a general second order tensor can be constructed with the sum of $n^2$ tensor products.

But I think you might have misunderstood what this proof implies from what Schutz write in the Appendix. You seemed to imply that the tensor product of two vectors has at most 2n independent components, and therefore by counting the independent components of a tensor one might deduce if it is the result of a tensor product or not. This is not correct. Just do the calculation, multiply two vectors using the outer product, you get a second order tensor with $n^2$ independent components. Granted you have only used 2n components to build it and that is why a general tensor is not necessarily the outer product of two vectors, but still after the product the $n^2$ components formed from the 2n are considered to be independent.

This doesn't make sense to me...If I can use the counting to prove that a general rank 2 tensor cannot be expressed as a dyad, why can't I use this counting to prove that a specific rank 2 tensor cannot be expressed as a dyad?

Can you come up with an example where my reasoning fails? A rank two tensor with more than 8 independent components that can be expressed as the direct product of 2 vectors alone?

 

[TrickyDicky]

This doesn't make sense to me...If I can use the counting to prove that a general rank 2 tensor cannot be expressed as a dyad, why can't I use this counting to prove that a specific rank 2 tensor cannot be expressed as a dyad?

You can. What you can't is take an arbitrary rank 2 tensor and decide whether it is a dyad or a sum of dyads just based on the number of independent components of the tensor, this is using the proof backwards.

Can you come up with an example where my reasoning fails? A rank two tensor with more than 8 independent components that can be expressed as the direct product of 2 vectors alone?

The best way is to compute it as I said, and realize that the fact that you use only 8 initial components to construct 16 components doesn't mean that just 8 of the final 16 are independent, if this was so you should be able to say which 8 are the independent ones, but you can't.

 

[TrickyDicky]

I guess it is true the whole thing doesn't make much sense with tensor fields, certainly not at the level of smooth manifolds where the usual flows for vector fields are defined. Perhaps after introducing a metric and curvature it could make more sense for certain geometries' isometry groups.

 

[Matterwave]

I don't see any way to make it work...