# 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.

## 6 Comments	»

1. [...] is the identity matrix. The matrix commutes with for every complex scalar , and so Schur’s lemma will apply to all of [...]

Pingback by | October 1, 2010 | Reply

2. [...] we can apply Schur’s lemma to all of them. In the middle two equations, we see that both and must be either be invertible or [...]

Pingback by | October 4, 2010 | Reply

3. [...] time, Schur’s lemma tells us that each is an intertwinor between and itself. And so we conclude that each of the [...]

Pingback by | October 5, 2010 | Reply

4. [...] What I do want to get into right now, is calculating the center of the matrix algebra . The answer is reminiscent of Schur’s lemma: [...]

Pingback by | October 6, 2010 | Reply

5. [...] this point, Schur’s lemma kicks in to tell us that if then is the zero matrix, while if then is a scalar times the [...]

Pingback by | October 22, 2010 | Reply

6. [...] straightforward concept: it’s a module such that its only submodules are and . As usual, Schur’s lemma tells us that any morphism between two irreducible modules is either or an isomorphism. And, as [...]

Pingback by | September 15, 2012 | Reply

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## About this weblog

This is mainly an expository blath, with occasional high-level excursions, humorous observations, rants, and musings. The main-line exposition should be accessible to the “Generally Interested Lay Audience”, as long as you trace the links back towards the basics. Check the sidebar for specific topics (under “Categories”).

I’m in the process of tweaking some aspects of the site to make it easier to refer back to older topics, so try to make the best of it for now.
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