# The Unapologetic Mathematician

## 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