371 views

Skip to first unread message

Mar 16, 2010, 7:56:45 AM3/16/10

to xAct Tensor Computer Algebra

Hi!

I get problems when I want to generate replacement rules for tensors

or spinors with many indices and large symmetry groups.

MakeRule, does the job, but the number of rules it creates quickly

becomes huge.

For instance, see the example file ProblemLongReplacementRules.nb

There I want to make a replacement rule from

Eq1=\[CapitalOmega][-A, -B, -C, -D, -F] TT[C, D, F, -P, -Q] ==

SS[-A, -B, -P, -Q]

It results in 28 800 rules, and takes a long time to compute.

Does anyone have an idea of how to do a replacement procedure without

having to produce all these rules?

Let us say that we want to substitute Eq1 in

\[CapitalOmega][S, -B, -C, -D, -A] TT[C, D, -H, -S, -Q]

Perhaps it is possible to make a procedure that looks for products of \

[CapitalOmega] and TT with the right number of indices. Then one needs

to make some sort of test that determines if \[CapitalOmega][S, -B, -

C, -D, -A] TT[C, D, -H, -S, -Q] belongs to the right hand side of any

of the 28 800 rules that can be produced from Eq1 withou calculating

all of them. Then one only needs to apply that rule. Unfortunately I

do not know how to do this kind of test.

Any ideas are welcome.

Regards

Thomas

Mar 16, 2010, 1:11:29 PM3/16/10

to xAct Tensor Computer Algebra

Hi Thomas,

Let me start by emphasizing that MakeRule is not supposed to be the

solution for every type of rule you might need. Note that there is no

function in Mathematica which tries to construct general rules. That

would be too difficult. What Mathematica provides is a full language

to construct the rules you might want. xTensor also provides tools to

construct your own rules, and I have designed MakeRule to help users

in more or less simple cases, which can be classified with a few

options. Whenever something becomes big (many indices or many tensors

or many contractions, etc), it is better to construct your own rules

specifically for your problem. There is no hope of having MakeRule

being efficient in all possible cases.

In your particular case, I would start with something like this:

myrule := Omega[oinds__] TT[tinds__] :> With[{

dummies = xAct`xTensor`Private`TakeEPairs[{oinds, tinds}],

frees = xAct`xTensor`Private`TakeFrees[{oinds, tinds}]},

somesign SS @@ frees /;

Length@Intersection[dummies, {oinds}] == 3 &&

Length@Intersection[dummies, {tinds}] == 3

]

This is simple because your three tensors are symmetric, and hence we

do not have to worry about where the indices are. It also assumes that

there are no contractions on either of Omega or TT, because would be

zero by symmetry. You see that in a more general case the rule would

be more complicated. This is the sort of thing that would be difficult

to decide by a MakeRule-type function.

That rule uses some private functions of xTensor, to make it faster,

but could be done with FindIndices, FindFreeIndices, etc. Note that I

haven't bothered to sort out signs, and so you have to find out how to

compute the variable "somesign".

I haven't tested this. Let me know if this works for you. Probably the

conditions at the end should be more careful, but you get the idea.

Cheers,

Jose.

Mar 16, 2010, 3:11:14 PM3/16/10

to xAct Tensor Computer Algebra

Hi!

Thanks for your code.

I managed to extend it to something much more general.

Thanks for your code.

I managed to extend it to something much more general.

Instead of deciding beforehand, all possible patterns of index

combinations,

one could use the canonicalizer to determine if an expression can be

transformed

into the left hand side of Eq1.

I think this idea can be used in many more cases than my specific

example.

See the file ProblemLongReplacementRules2.nb for details.

Is this general enough so it is worth implementing a procedure that

can produce similar rules from other equations?

Regards

Thomas

Mar 16, 2010, 3:42:31 PM3/16/10

to xAct Tensor Computer Algebra

Hi again!

I am sorry for sending too many emails to this group.

I just realized that my method did not work as well as I had hoped.

For instance

\[CapitalOmega][-A, -P, -C, -D, -F] TT[C, D, F, -B, -Q]

will not be treated by the rules because it has a different order of

the free indices.

This means that one might need to produce the same number of rules

as the number of permutations of the free indices (reduced by the

symmetries). 6 in this case. (Times 2 for the different signs.)

This is more difficult to write, but it could work in for many cases

where MakeRule is not the prefered choice.

Regards

Thomas

Mar 17, 2010, 7:50:24 PM3/17/10

to xAct Tensor Computer Algebra

Hi Thomas,

I like your idea, but it is not clear how well it scales for more

complicated problems, because in this example we use heavily the fact

that you work with symmetric spinors.

Concerning the problem you mention, it is easy to solve: This is a

small modification of your code, also introducing a Which construct to

avoid having two rules:

myrule := Omega[oinds__] TT[tinds__] :>

With[{

dummies = xAct`xTensor`Private`TakeEPairs[{oinds, tinds}],

frees = xAct`xTensor`Private`TakeFrees[{oinds, tinds}] },

With[{LHS = (TT[C, D, F, ##] & @@ Complement[{tinds}, dummies])

(Omega[-C, -D, -F, ##] & @@ Complement[{oinds},

dummies]) },

Which[

ToCanonical[ Omega[oinds] TT[tinds] - LHS ] === 0, SS @@

frees,

ToCanonical[ Omega[oinds] TT[tinds] + LHS ] === 0, - SS @@

frees

]

] /;

Length@Intersection[dummies, {oinds}] == 3 &&

Length@Intersection[dummies, {tinds}] == 3

]

Cheers,

Jose.

Mar 18, 2010, 8:40:55 AM3/18/10

to xAct Tensor Computer Algebra

Hi!

This is probably the best way for the particular example, but I am

more interested in a way to generare this kind of rules from quite

general equations.

This is probably the best way for the particular example, but I am

more interested in a way to generare this kind of rules from quite

general equations.

I managed to slightly modify EqualExpressionsQ to do the missing

parts. See the file ProblemLongReplacementRules3b.nb for details.

I think this idea is general enough to work as an alternative to

MakeRule

in the cases with many symmetries and dummy indices.

I think it does the minimal ammount of tests, except that it uses all

permutations of the free indices. This is still quick for my examples,

but

it could be improved for more complicated ones. Many permutations will

be

equivalent due to the symmetries, but I do not know how to compute

this.

Do you know how to compute all permutations that are not equivalent

with

respect to the symmetries of the free indices of a tensor?

I think this can be related to the ToCanonical[%/.Sym-

>ImposeSymmetry]

problem we discussed earlier.

I tried to write a procedure that can renerate rules like this for

arbitrary products of two tensors. This can also be found in

ProblemLongReplacementRules3b.nb.

Do you like the idea?

Do you think it is worth implementing something as general as MakeRule

for this method?

Regards

Thomas

Mar 23, 2010, 12:08:41 PM3/23/10

to xAct Tensor Computer Algebra

Hi!

I have tried to find a good and quick way to generate a list of all

permutations of the free indices that are not equivalent with respect

to the symmetries of the free indices.

Unfortunately I have not succeeded.

I can find the group of all permutations of the free indices (G1), and

the group of symmetries of the expression that only involves the free

indices (G2).

Say that one want to do this for expr1:

inds1 = FindIndices@Evaluate@expr1

dummies1 = xAct`xTensor`Private`TakeEPairs@inds1

dummiespos1 = List @@ (First@First@Position[inds1, #, 1] & /@

dummies1)

G1 = Last@Last@SymmetryOf@expr1

G2 = Stabilizer[dummiespos1, G1]

The next problem I suppose is to generate a list of all cosets of G2

in the group G1. Is there a way to do this without generating a list

of all elements in G1 and then elliminate the ones that are not

needed?

The symmetries of the free indices can probably be larger than G2 due

to some strange symmetry that could for instance interchange two free

indices at the same time as two pairs of dymmy indices are

interchanged. It would be complicated to compute this, and for this

purpose one would not gain much, if anything at all.

Another thing:

I found that MakeRule does not care about symmetries of covariant

derivatives with several indices. When working with space spinors one

will have covariant derivatives with two indices that are symmetric,

so there are real cases when one needs this.

Regards

Thomas

Mar 23, 2010, 12:14:19 PM3/23/10

to xAct Tensor Computer Algebra

I am sorry I mixed up the code.

This is what I did mean to send:

This is what I did mean to send:

inds1 = FindIndices@Evaluate@expr1

frees1 = FindFreeIndices@Evaluate@expr1

dummies1 = xAct`xTensor`Private`TakeEPairs@inds1

dummiespos1 = List @@ (First@First@Position[inds1, #, 1] & /@

dummies1)

G1=xAct`xTensor`Private`symperms[inds1, frees1]

G2 = Stabilizer[dummiespos1, Last@Last@SymmetryOf@expr1]

Regards

Thomas

Mar 23, 2010, 1:14:18 PM3/23/10

to xAct Tensor Computer Algebra

Hi Thomas,

> I have tried to find a good and quick way to generate a list of all

> permutations of the free indices that are not equivalent with respect

> to the symmetries of the free indices.

> Unfortunately I have not succeeded.

It is not a trivial thing.

> I can find the group of all permutations of the free indices (G1), and

> the group of symmetries of the expression that only involves the free

> indices (G2).

> Say that one want to do this for expr1:

(copied here the code from your second email)

>

> inds1 = FindIndices@Evaluate@expr1

> frees1 = FindFreeIndices@Evaluate@expr1

> dummies1 = xAct`xTensor`Private`TakeEPairs@inds1

> dummiespos1 = List @@ (First@First@Position[inds1, #, 1] & /@ dummies1)

> G1=xAct`xTensor`Private`symperms[inds1, frees1]

> G2 = Stabilizer[dummiespos1, Last@Last@SymmetryOf@expr1]

OK.

> The next problem I suppose is to generate a list of all cosets of G2

> in the group G1. Is there a way to do this without generating a list

> of all elements in G1 and then elliminate the ones that are not

> needed?

A list of representatives of the cosets is called a "transversal", and

can be computed using the strong-generating-set description of the

groups, though not very efficiently. Unfortunately xPerm does not yet

have that algorithm. I'll add it at some point. The need for this

algorithm has appeared already, for different reasons, in other

discussions with Alfonso Garcia-Parrado and Guillaume Faye.

> The symmetries of the free indices can probably be larger than G2 due

> to some strange symmetry that could for instance interchange two free

> indices at the same time as two pairs of dymmy indices are

> interchanged. It would be complicated to compute this, and for this

> purpose one would not gain much, if anything at all.

Yes. You use a pointwise stabilizer to compute G2, but what is needed

is a setwise stabilizer, another algorithm not yet present in xPerm.

> Another thing:

> I found that MakeRule does not care about symmetries of covariant

> derivatives with several indices. When working with space spinors one

> will have covariant derivatives with two indices that are symmetric,

> so there are real cases when one needs this.

I've just had a look at the corresponding part of xTensor.nb,

subsection 10.2. Copied from there, in relation to the function

SymmetryEquivalentsOfTensor, which is the function actually called by

MakeRule :

"It would be very simple to generalize this function to arbitrary

expressions (change SymmetryGroupOfTensor to SymmetryOf and compute

indices with FindIndices)"

Perhaps you can modify SymmetryEquivalentsOfTensor to do what you

want. I don't have the time to do this myself.

Cheers,

Jose.

Apr 7, 2010, 6:51:56 PM4/7/10

to xAct Tensor Computer Algebra

Hi!

I managed to write a code that generates replacement rules for all

cases where the left hand side is a product of tensors.

See the file ProblemLongReplacementRules7.nb

I managed to write a code that generates replacement rules for all

cases where the left hand side is a product of tensors.

See the file ProblemLongReplacementRules7.nb

The permutations of the free indices are computed using the pointwise

stabilizer of the symmetry group of the left hand side.

This can be extended to a setwise stabilizer when that is implemented.

The transversal was already implemented in xPerm, so I used that.

There might be better algorithms, but at least it seems to work.

I do not know if my way of coding is safe and efficient, so if anyone

has any comments on this - let me know.

The code is not properly tested yet.

Regards

Thomas Bäckdahl

> ...

>

> read more »

Reply all

Reply to author

Forward

0 new messages

Search

Clear search

Close search

Google apps

Main menu