## Submodules

Fancy words: a submodule is a subobject in the category of group representations. What this means is that if and are -modules, and if we have an injective morphism of modules , then we say that is a “submodule” of . And, just to be clear, a -morphism is injective if and only if it’s injective as a linear map; its kernel is zero. We call the “inclusion map” of the submodule.

In practice, we often identify a -submodule with the image of its inclusion map. We know from general principles that since is injective, then is isomorphic to its image, so this isn’t really a big difference. What we can tell, though, is that the action of sends the image back into itself.

That is, let’s say that is the image of some vector . I say that for any group element , acting by on gives us some other vector that’s *also* in the image of . Indeed, we check that

which is again in the image of , as asserted. We say that the image of is “-invariant”.

The flip side of this is that any time we find such a -invariant subspace of , it gives us a submodule. That is, if is a -module, and is a -invariant subspace, then we can define a new representation on by restriction: . The inclusion map that takes any vector and considers it as a vector in clearly intertwines the original action and the restricted action , and its kernel is trivial. Thus constitutes a -submodule.

As an example, let be any finite group, and let be its group algebra, which carries the left regular representation . Now, consider the subspace spanned by the vector

That is, consists of all vectors for which all the coefficients are equal. I say that this subspace is -invariant. Indeed, we calculate

But this last sum runs through all the elements of , just in a different order. That is, , and so carries the one-dimensional trivial representation of . That is, we’ve found a copy of the trivial representation of as a submodule of the left regular representation.

As another example, let be one of the symmetric groups. Again, let carry the left regular representation, but now let be the one-dimensional space spanned by

It’s a straightforward exercise to show that is a one-dimensional submodule carrying a copy of the signum representation.

Every -module contains two obvious submodules: the zero subspace and the entire space itself are both clearly -invariant. We call these submodules “trivial”, and all others “nontrivial”.

[…] We say that a module is “reducible” if it contains a nontrivial submodule. Thus our examples last time show that the left regular representation is always reducible, since […]

Pingback by Reducibility « The Unapologetic Mathematician | September 23, 2010 |

[…] invariant form on our space , then any reducible representation is decomposable. That is, if is a submodule, we can find another submodule so that as […]

Pingback by Invariant Forms « The Unapologetic Mathematician | September 27, 2010 |

[…] know that there’s a copy of the the trivial representation as a submodule of the defining representation. If we use the standard basis of , this submodule is the line […]

Pingback by One Complete Character Table (part 2) « The Unapologetic Mathematician | October 27, 2010 |

[…] in sight has been a vector space over that field. But remember that a representation of is a module over the group algebra , and we can take tensor products over this algebra as […]

Pingback by Tensor Products over Group Algebras « The Unapologetic Mathematician | November 9, 2010 |