## The Symmetrizer and Antisymmetrizer

Today we’ll introduce two elements of the group algebra of the symmetric group which have some interesting properties. Each one, given a representation of the symmetric group, will give us a subrepresentation of that representation.

The symmetrizer gets its name from the way that it takes an arbitrary vector and turns it into one that the symmetric group will act trivially on. Specifically, we use the element

That is, we take all permutations in the symmetric group and — in a sense — average them out. If we compose this with any permutation we find

But as runs over all the permutations in the group, does as well. So this is just the symmetrizer back again. The upshot is that if we have a representation we find that

Thus every vector in the image of is left unchanged by the action of any permutation. That is, is a subrepresentation on which acts trivially.

The antisymmetrizer, on the other hand, on the other hand, will give us vectors on which the symmetric group acts by the signum representation. We use the group algebra element

Now if we compose this with any permutation we find

Now given a representation we find that

Thus every vector in the image of is multiplied by the signum of any permutation. That is, is a subrepresentation on which acts by the signum representation.

Now, one thing to be careful about: I haven’t said that the subrepresentations are nontrivial. That is, when we (anti)symmetrize a representation, the subrepresentation we get may be zero — maybe no vectors in the representation transform trivially or by the signum representation. In fact, let’s check what happens when we multiply the symmetrizer and antisymmetrizer:

Where the sum comes to zero because we’re just adding up terms where and where , and so everything cancels out. That is, the symmetric part of an antisymmetrized representation is trivial. Similarly, the antisymmetric part of a symmetrized representation is trivial.

[…] and tomorrow I want to take last Friday’s symmetrizer and antisymmetrizer and apply them to the tensor representations of , which we know also carry symmetric group […]

Pingback by Symmetric Tensors « The Unapologetic Mathematician | December 22, 2008 |

[…] of , which actions commute with each other. Our antisymmetric tensors are the image of a certain action from the symmetric group, which is an intertwiner of the action. Thus we have a one-dimensional representation of , which […]

Pingback by The Determinant « The Unapologetic Mathematician | December 31, 2008 |

[…] if its value is unchanged when we swap two of its inputs. Equivalently, it commutes with the symmetrizer. That is, it must kill everything that the symmetrizer kills, and so must really define a linear […]

Pingback by Multilinear Functionals « The Unapologetic Mathematician | October 22, 2009 |

[…] may well not be symmetric itself. Instead, we will take the tensor product of and , and then symmetrize it, to give . This will be bilinear, and it will work with our choice of grading, but will it be […]

Pingback by Tensor and Symmetric Algebras « The Unapologetic Mathematician | October 26, 2009 |

[…] may not be antisymmetric itself. Instead, we will take the tensor product of and , and then antisymmetrize it, to give . This will be bilinear, but will it be […]

Pingback by Exterior Algebras « The Unapologetic Mathematician | October 27, 2009 |

In the development for pi A at the last line there is an extra pi (hat) factor. Should be only:

piA=sgn(pi)A

Comment by Simon | December 28, 2015 |

Thanks; fixed.

Comment by John Armstrong | December 29, 2015 |