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Is there some Lie algebra $L(n)$ such that the number of its simple roots grows with the Fibonacci numbers $F(n)$ (or perhaps $F(n+2)$)?

Context: I'm studying the Lucas and Fibonacci cube graphs. Suppose I take the adjacency matrix $A(n)$ of the latter, which is an $F(n) \times F(n)$ symmetric matrix with vanishing elements on the diagonal.

From this, I can construct $C(n) = 2 I(n) - A(n)$, where $I(n)$ is the identity matrix. It turns out that $C$ satisfies all the requirements for being a Cartan matrix. Here is an example for $n=5$ (which is a $5 \times 5$ symmetric matrix)

\begin{equation} C(5) = \left( \begin{array}{ccccc} 2 & -1 & -1 & -1 & 0 \\ -1 & 2 & 0 & 0 & -1 \\ -1 & 0 & 2 & 0 & 0 \\ -1 & 0 & 0 & 2 & -1 \\ 0 & -1 & 0 & -1 & 2 \\ \end{array} \right) \end{equation}

and for $n=6$ (which is an $8 \times 8$ symmetric matrix)

\begin{equation} C(6) = \left( \begin{array}{cccccccc} 2 & -1 & -1 & -1 & 0 & -1 & 0 & 0 \\ -1 & 2 & 0 & 0 & -1 & 0 & -1 & 0 \\ -1 & 0 & 2 & 0 & 0 & 0 & 0 & -1 \\ -1 & 0 & 0 & 2 & -1 & 0 & 0 & 0 \\ 0 & -1 & 0 & -1 & 2 & 0 & 0 & 0 \\ -1 & 0 & 0 & 0 & 0 & 2 & -1 & -1 \\ 0 & -1 & 0 & 0 & 0 & -1 & 2 & 0 \\ 0 & 0 & -1 & 0 & 0 & -1 & 0 & 2 \\ \end{array} \right) \end{equation}

I can also show that the rank of $C(n)$ follows the Fibonacci sequence: Rank $C(n) = F(n)$. Given this, I'd like to understand if it is known whether there exists an algebra such that $C(n)$ is its Cartan matrix. The closest thing I could find were the Hecke algebras, but I haven't seen a clear discussion of their Cartan matrices anywhere.

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A Cartan matrix for a semi-simple Lie algebra has 2's down the diagonal, either 0's in both the $(i,j)$ and $(j,i)$ entries or negative integers. AND a Cartan matrix for a semi-simple Lie algebra is positive definite.

Your "Cartan matrices" are actually generalized Cartan matrices (GCMs). For example, your $C(5)$'s corresponding Dynkin diagram has a loop. No (classical) Cartan matrix has Dynkin diagram with a loop.

Every GCM can be used to build a Kac-Moody Lie algebra. Only Cartan matrices (that are positive definite) yield finite dimensional semi-simple Lie algebras. All others yield infinite dimensional algebras.

Your $C(5)$ is the GCM of a hyperbolic Kac-Moody Lie algebra. Hyperbolic types are essentially one step away from affine types which are one step away from the finite dimensional simple Lie algebras.

Finally, the number of simple roots is the same as the size of your GCM. So a $5 \times 5$ matrix will yield a (possibly infinite dimensional) Kac-Moody Lie algebra with 5 simple roots.

So, yes, your $C(n)$'s correspond to (generally infinite dimensional) Kac-Moody Lie algebras.

Note: Can you create a list of Kac-Moody Lie algebras whose simple roots are numbered according to the Fibonacci sequence? Sure. In a lot of different ways. In fact, we could even make sure they're simple and finite dimensional.

For example, start with [2] (type $A_1$. Then put it together with itself and let's connect the first node of the second copy with the last of the first. This yields $$\begin{bmatrix} 2 & -1 \\ -1 & 2 \end{bmatrix}$$ which is type $A_2$.

Put together $A_k$ and $A_m$ this way and get $A_{k+m}$.

So type A with fibonacci ranks will give such a sequence (corresponding to finite dimensional simple Lie algebras).

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  • $\begingroup$ Thank you for your response, that is rather helpful! However, on the Wikipedia article you linked, it mentions that the rank of hyperbolic Kac-Moody algebras is at most 10. On the other hand, the C(n) I construct have rank F(n). So I'm wondering if this family of Cartan matrices corresponds to some specific algebra. I should mention that I construct C(n) recursively from C(n-1) and C(n-2), if that helps. $\endgroup$
    – Aegon
    Oct 14, 2019 at 2:41
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    $\begingroup$ As an example C5 corresponds to a hyperbolic algebra. As n gets large, each Cn corresponds to a (symmetrizable) Kac-Moody Lie algebra but definitely not finite, affine, or hyperbolic type. So there wouldn't be a whole lot you could say about Cn's in general. $\endgroup$
    – Bill Cook
    Oct 14, 2019 at 2:54

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