# Vector fields, flows and tensor fields

Matterwave
Gold Member
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).

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

Matterwave
Gold Member
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.

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.

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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
Gold Member
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?

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.

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