parsikoo Posted June 2, 2012 Share Posted June 2, 2012 Are Lorentz Transformation or their products having irreducible representation, and is superposition allowed or social consideration are needed??? Sorry meant special not social Link to comment Share on other sites More sharing options...
elfmotat Posted June 3, 2012 Share Posted June 3, 2012 I'm not sure what you're asking. Trying posing your question in a different way. Link to comment Share on other sites More sharing options...
parsikoo Posted June 3, 2012 Author Share Posted June 3, 2012 I mean considering Lorentz or poincare group, does the product of two transformation is also an irreducible representation, if yes then superposition is valid, as we have irreducible representation?? Link to comment Share on other sites More sharing options...
timo Posted June 3, 2012 Share Posted June 3, 2012 The combination of two (transformation-) group elements again yields a group element (by definition of a "group"). One usually doesn't call that "superposition", though. Link to comment Share on other sites More sharing options...
parsikoo Posted June 3, 2012 Author Share Posted June 3, 2012 agree, but is the product of two group elements will become a irreducible rep.??? if yes then I guess superposition is also valid? Link to comment Share on other sites More sharing options...
ajb Posted June 4, 2012 Share Posted June 4, 2012 I am not sure what you are asking, but the Poincare group does have irreducible representations and these are very important in physics. A representation is a way of describing a group in terms of a vector space and its linear transformations. So elements of the group can be represented by linear transformations acting on some space and the group composition becomes the composition of linear operators. In the finite dimensional case this is just matrices and matrix multiplication. An irreducible representation of a group is a group representation that has no nontrivial invariant subspaces. I do not understand what you mean by is the product of two group elements will become a irreducible rep.??? You pick an irreducible representation and then formulate everything in terms of the linear transformations. You do not leave the specified representation via composition of the linear operations. Any representation of a finite or semisimple Lie group breaks up into a direct sum of irreducible representations. Are you asking about this or something similar, such as taking tensor product of representations? Link to comment Share on other sites More sharing options...
parsikoo Posted June 5, 2012 Author Share Posted June 5, 2012 I am not sure what you are asking, but the Poincare group does have irreducible representations and these are very important in physics. A representation is a way of describing a group in terms of a vector space and its linear transformations. So elements of the group can be represented by linear transformations acting on some space and the group composition becomes the composition of linear operators. In the finite dimensional case this is just matrices and matrix multiplication. An irreducible representation of a group is a group representation that has no nontrivial invariant subspaces. I do not understand what you mean by is the product of two group elements will become a irreducible rep.??? You pick an irreducible representation and then formulate everything in terms of the linear transformations. You do not leave the specified representation via composition of the linear operations. Any representation of a finite or semisimple Lie group breaks up into a direct sum of irreducible representations. Are you asking about this or something similar, such as taking tensor product of representations? The product of two irreducible representations will be also an irreducible representation in specific cases, I just needed a proof if that was the case for two LoT. Link to comment Share on other sites More sharing options...
timo Posted June 5, 2012 Share Posted June 5, 2012 What is the product of two representations? Link to comment Share on other sites More sharing options...
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