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However, how can I show that those subspaces are themselves irreducible?

Thanks!

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- Thread starter gentsagree
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In summary, the second rank antisymmetric representation of the group SO(4) can be reduced into self-dual and antiself-dual representations. This can be visualized by observing that the projection of the second exterior power of the group's vector space into these subspaces commutes with the group's action. To prove that these subspaces are irreducible, a more detailed framework is needed to show that there are no proper submodules and that the orbit of a basis vector is the entire module.

- #1

- 96

- 1

However, how can I show that those subspaces are themselves irreducible?

Thanks!

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Irreducibility refers to the property of a representation in which it cannot be broken down into smaller, simpler representations. In the context of (anti)self-dual reps, this means that the representation cannot be further decomposed into self-dual or anti-self-dual parts.

The concept of irreducibility is important in understanding the structure of representations and their relationship to each other. It helps identify the building blocks of representations and how they can be combined to form larger, more complex representations.

To determine irreducibility in (anti)self-dual reps, we need to examine the transformation properties of the representation under certain operations. If the representation remains unchanged under these operations, then it is considered irreducible.

No, a representation cannot be both self-dual and anti-self-dual. This is because these two properties are mutually exclusive and a representation can only have one of them.

Understanding irreducibility in (anti)self-dual reps has applications in various fields such as physics, chemistry, and mathematics. It helps in the study of symmetry and conservation laws, as well as in the development of new mathematical models and theories.

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