constructing tensors with specific symmetries

[Shubham Kumar Sinha]

Hi All,

I started using Xact and it is very useful in handling tensor and their derivatives. Now, I want to define a tensor which should be symmetric under indices. I know that if I want to define tensor which should be fully symmetric in indices then i can do the following :

<< xAct`xTensor`;

DefManifold[M, n, IndexRange[i, z]];

DefMetric[-1, met[-i, -j], CD, {";", "\[Del]"},PrintAs -> "\[Delta]"];

DefTensor[H[a,b,c,d], M, Symmetric[{a,b,c,d}]]

But let's suppose if I want to define 4 rank tensor which is symmetric under cyclic operation, that is,

H[a,b,c,d] = H[d,a,b,c]=H[c,d,a,b]=H[b,c,d,a]

and also symmetric under ,

H[a,b,c,d]=H[d,c,b,a]

How to define a tensor which has above symmetry in indices.

It would be helpful if anyone can tell me , if that is possible to construct.

Thanks,

Shubham

[Thomas Bäckdahl]

Hi!

You just need to specify a generating set for your symmetries. The cyclic symmetry is generated by Cycles[{1, 2, 3, 4}], while the other symmetry is generated by Cycles[{1, 4}, {2, 3}].
Therefore, you can define H with
DefTensor[H[a, b, c, d], M, GenSet[Cycles[{1, 2, 3, 4}], Cycles[{1, 4}, {2, 3}]]]

However, observe that using a,b,c,d as indices is incompatible with the line

DefManifold[M, n, IndexRange[i, z]]

Just make sure to only use index symbols in the range used in DefManifold.

If you don't like the Cycles notation, there are other options. See xAct\Documentation\English\xPermDoc.nb section 1.
Internally the generating set is transformed into a strong generating set, but for your application, you probably don't need to think about that.

Regards
Thomas

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