The Unapologetic Mathematician

Mathematics for the interested outsider

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 ith row and jth 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:


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


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.

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February 16, 2010 - Posted by | Geometry, Root Systems


  1. So cool! I admit that I still cannot see the word “intertwine” without think of Ted “hypertext” Nelson’s mantra: “Everything is profoundly intertwingled!”

    I just spent two delightful full days on the Caltech campus for the Western States Annual Mathematical Physics Meeting that Barry Simon hosts. I always enjoy being the weakest mathematican or physicist in the room, and seeing deep results from fascinating people.

    Comment by Jonathan Vos Post | February 17, 2010 | Reply

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  3. [...] Graphs and Dynkin Diagrams We’ve taken our root system and turned it into a Cartan matrix. Now we’re going to take our Cartan matrix and turn it into a pictorial form that we can [...]

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  4. [...] that for each such root system we can construct a connected Dynkin diagram, which determines a Cartan matrix, which determines the root system itself, up to isomorphism. So what he have to find now is a list [...]

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  6. [...] of and will give us access to . Further, it should be orthogonal to both and , and should have a Cartan integer of with in either order. For this purpose, we pick , which then gives us the last vertex of the [...]

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  7. [...] of the Dynkin diagram of the root system. And for to be an automorphism of , it must preserve the Cartan integers, and thus the numbers of edges between any pair of vertices in the Dynkin diagram. That is, must [...]

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