The Unapologetic Mathematician

Mathematics for the interested outsider

Schur’s Lemma

Now that we know that images and kernels of G-morphisms between G-modules are G-modules as well, we can bring in a very general result.

Remember that we call a G-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 G-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 f:V\to W is a morphism between two simple objects, then either f is an isomorphism or it’s the zero morphism from V to W. Indeed, since V is simple it has no nontrivial subobjects. The kernel of f is a subobject of V, so it must either be V itself, or the zero object. Similarly, the image of f must either be W itself or the zero object. If either \mathrm{Ker}(f)=V or \mathrm{Im}(f)=\mathbf{0} then f is the zero morphism. On the other hand, if \mathrm{Ker}(f)=\mathbf{0} and \mathrm{Im}(f)=W we have an isomorphism.

To see how this works in the case of G-modules, every time I say “object” in the preceding paragraph replace it by “G-module”. Morphisms are G-morphisms, the zero morphism is the linear map sending every vector to 0, and the zero object is the trivial vector space \mathbf{0}. If it feels more comfortable, walk through the preceding proof making the required substitutions to see how it works for G-modules.

In terms of matrix representations, let’s say X and Y are two irreducible matrix representations of G, and let T be any matrix so that TX(g)=Y(g)T for all g\in G. Then Schur’s lemma tells us that either T is invertible — it’s the matrix of an isomorphism — or it’s the zero matrix.

September 30, 2010 Posted by | Algebra, Category theory, Group theory, Representation Theory | 6 Comments

   

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