Schur’s Lemma
Now that we know that images and kernels of 
–morphisms between –modules are -modules as well, we can bring in a very general result.
Remember that we call a -module irreducible or “simple” if it has no nontrivial submodules. In general, an object in any category is simple if it has no nontrivial subobjects. If a morphism in a category has a kernel and an image — as we’ve seen all -morphisms do — then these are subobjects of the source and target objects.
So now we have everything we need to state and prove Schur’s lemma. Working in a category where every morphism has both a kernel and an image, if 
is a morphism between two simple objects, then either 
is an isomorphism or it’s the zero morphism from 
to 
. Indeed, since is simple it has no nontrivial subobjects. The kernel of is a subobject of , so it must either be itself, or the zero object. Similarly, the image of must either be itself or the zero object. If either 
or 
then is the zero morphism. On the other hand, if 
and 
we have an isomorphism.
To see how this works in the case of -modules, every time I say “object” in the preceding paragraph replace it by “-module”. Morphisms are -morphisms, the zero morphism is the linear map sending every vector to 
, and the zero object is the trivial vector space 
. If it feels more comfortable, walk through the preceding proof making the required substitutions to see how it works for -modules.
In terms of matrix representations, let’s say 
and 
are two irreducible matrix representations of , and let 
be any matrix so that 
for all 
. Then Schur’s lemma tells us that either is invertible — it’s the matrix of an isomorphism — or it’s the zero matrix.
September 30, 2010 -
Posted by John Armstrong |
Algebra, Category theory, Group theory, Representation Theory
The Unapologetic Mathematician 
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