The Unapologetic Mathematician

Cartan Matrices

As we move towards our goal of classifying root systems, we find new ways of encoding the information contained in a root system $\Phi$. First comes the Cartan matrix.

Pick a base $\Delta\subseteq\Phi$ of the root system. Since $V$ is finite-dimensional and $\Delta$ is a basis of $V$, $\Delta$ must be finite, and so there’s no difficulty in picking some fixed order on the simple roots. That is, we write $\Delta=\{\alpha_1,\alpha_2,\dots,\alpha_n\}$ where $n=\dim(V)$. Now we can define the “Cartan matrix” as the $n\times n$ matrix whose entry in the $i$th row and $j$th column is $\alpha_i\rtimes\alpha_j$. These entries are called “Cartan integers”.

The matrix we get, depends on the particular ordering of the base we chose, of course, so the Cartan matrix isn’t quite uniquely determined by the root system. This is relatively unimportant, actually. More to the point is the other direction: the Cartan matrix determines the root system up to isomorphism!

That is, let’s say $\Delta'\subseteq\Phi'\subseteq V'$ is another root system $\Phi'$ in another vector space $V$ with another identified base $\Delta'=\{\alpha'_1,\dots,\alpha'_n\}$. Further, assume that for all $1\leq i,j\leq n$ we have $\alpha'_i\rtimes\alpha'_j=\alpha_i\rtimes\alpha_j$, so the Cartan matrix determined by $\Delta'$ is equal to the Cartan matrix determined by $\Delta$. I say that the bijection $\alpha_i\mapsto\alpha'_i$ extends to an isomorphism $\phi:V\rightarrow V'$ that sends $\Phi$ onto $\Phi'$ and satisfies $\phi(\alpha)\rtimes\phi(\beta)=\alpha\rtimes\beta$ for all roots $\alpha,\beta\in\Phi$.

The unique extension to $\phi$ is trivial. Indeed, since $\Delta$ is a basis for $V$ all we have to do is specify all the images $\phi(\alpha_i)$ and there is a unique linear transformation $\phi:V\rightarrow V'$ extending the mapping on basis vectors. And it’s an isomorphism, since the image of our basis of $V$ is itself a basis of $V'$, so we can turn around and reverse everything.

Now our hypothesis that the bases give rise to the same Cartan matrix allows us to calculate for simple roots $\alpha,\beta\in\Delta$:

\displaystyle\begin{aligned}\sigma_{\phi(\alpha)}(\phi(\beta))&=\phi(\beta)-(\phi(\beta)-\phi(\alpha)\phi(\alpha)\\&=\phi(\beta)-(\beta\rtimes\alpha)\phi(\alpha)\\&=\phi(\beta-(\beta\rtimes\alpha)\alpha)\\&=\phi(\sigma_\alpha(\beta))\end{aligned}

That is, $\phi$ intertwines the actions of each of the simple reflections $\sigma_\alpha$. But we know that the simple reflections with respect to any given base generate the Weyl group!

And so $\phi$ must intertwine the actions of the Weyl groups $\mathcal{W}$ and $\mathcal{W}'$. That is, the mapping $\sigma\mapsto\phi\circ\sigma\circ\phi^{-1}$ is an isomorphism $\mathcal{W}\rightarrow\mathcal{W}'$ which sends $\sigma_\alpha$ to $\sigma_{\phi(\alpha)}$ for all $\alpha\in\Delta$.

We can go further. Each root $\beta$ is in the $\mathcal{W}$-orbit of some simple root $\alpha$. Say $\beta=\sigma(\alpha)$ for $\alpha\in\Delta$. Then we find

$\displaystyle\phi(\beta)=\phi(\sigma(\alpha))=(\phi\circ\sigma\circ\phi^{-1})(\phi(\alpha))\in\Phi'$

And so $\phi$ must send $\Phi$ to $\Phi'$. A straightforward calculation (unwinding the one before) shows that $\phi$ must then preserve the Cartan integers for any roots $\alpha$ and $\beta$.

February 16, 2010 Posted by | Geometry, Root Systems | 8 Comments