Inverse of Cartan matrix

Question

The Cartan matrix of the root system $A_n$ looks like, denote it by $A'_n$

$$ A'_n= \begin{bmatrix} 2 & -1 & 0 & 0&\ldots & 0 \\[0.3em] -1 & 2 & -1 & 0 & \ldots & 0 \\[0.3em] 0 &-1 &2 & -1 & \ldots & 0 \\[0.3em] \vdots & & &\ddots & 2& -1 \\[0.3em] 0 & \ldots & & & -1&2 \end{bmatrix}$$

I need to find the inverse matrix explicitly. I know that $\det A'_n=n+1$, using this and the fact that

$$ b_{ij} := (A'_n)_{ij}^{-1} = (-1)^{i+j} \frac{\det (A''_n)_{ij}}{\det (A'_n)_{ij}}$$

where $(A''_n)_{ij}$ is the matrix $A'_n$ with $i$-th row and $j$-th column removed. For the first row I could find that

$$ b_{1k} = \frac{n-k+1}{n+1} $$

But this nice pattern does not continue for other rows. Is there an easier way to approach this problem ? I would appreciate any hints/suggestions/answers for this problem.

Answer 1

The inverse of the Cartan matrix has the following formula:

$(A_n^{-1})_{ij}=\min\{i,j\}-\frac{ij}{n+1}$, for the inverse of other simple Lie algebra, there is a paper https://arxiv.org/pdf/1711.01294.pdf

Answer 2

With a couple of years of delay, I can answer you that on p. 95 of Roselfeld's "Geometry of Lie groups" there are the inverses of the Cartan matrices of simple complex Lie algebras.

EDIT: Up to transposition and without the typos in Rosenfeld, you can also find the same info on p. 295 of Onishchik-Vinberg.

Answer 3

I do not know if this would be helpful, for $A_{6}$ the inverse is the following: $$ \left( \begin{array}{cccccc} \frac{6}{7} & \frac{5}{7} & \frac{4}{7} & \frac{3}{7} & \frac{2}{7} & \frac{1}{7} \\ \frac{5}{7} & \frac{10}{7} & \frac{8}{7} & \frac{6}{7} & \frac{4}{7} & \frac{2}{7} \\ \frac{4}{7} & \frac{8}{7} & \frac{12}{7} & \frac{9}{7} & \frac{6}{7} & \frac{3}{7} \\ \frac{3}{7} & \frac{6}{7} & \frac{9}{7} & \frac{12}{7} & \frac{8}{7} & \frac{4}{7} \\ \frac{2}{7} & \frac{4}{7} & \frac{6}{7} & \frac{8}{7} & \frac{10}{7} & \frac{5}{7} \\ \frac{1}{7} & \frac{2}{7} & \frac{3}{7} & \frac{4}{7} & \frac{5}{7} & \frac{6}{7} \\ \end{array} \right) $$ and for $A_{10}$ is the following: $$ \left( \begin{array}{cccccccccc} \frac{10}{11} & \frac{9}{11} & \frac{8}{11} & \frac{7}{11} & \frac{6}{11} & \frac{5}{11} & \frac{4}{11} & \frac{3}{11} & \frac{2}{11} & \frac{1}{11} \\ \frac{9}{11} & \frac{18}{11} & \frac{16}{11} & \frac{14}{11} & \frac{12}{11} & \frac{10}{11} & \frac{8}{11} & \frac{6}{11} & \frac{4}{11} & \frac{2}{11} \\ \frac{8}{11} & \frac{16}{11} & \frac{24}{11} & \frac{21}{11} & \frac{18}{11} & \frac{15}{11} & \frac{12}{11} & \frac{9}{11} & \frac{6}{11} & \frac{3}{11} \\ \frac{7}{11} & \frac{14}{11} & \frac{21}{11} & \frac{28}{11} & \frac{24}{11} & \frac{20}{11} & \frac{16}{11} & \frac{12}{11} & \frac{8}{11} & \frac{4}{11} \\ \frac{6}{11} & \frac{12}{11} & \frac{18}{11} & \frac{24}{11} & \frac{30}{11} & \frac{25}{11} & \frac{20}{11} & \frac{15}{11} & \frac{10}{11} & \frac{5}{11} \\ \frac{5}{11} & \frac{10}{11} & \frac{15}{11} & \frac{20}{11} & \frac{25}{11} & \frac{30}{11} & \frac{24}{11} & \frac{18}{11} & \frac{12}{11} & \frac{6}{11} \\ \frac{4}{11} & \frac{8}{11} & \frac{12}{11} & \frac{16}{11} & \frac{20}{11} & \frac{24}{11} & \frac{28}{11} & \frac{21}{11} & \frac{14}{11} & \frac{7}{11} \\ \frac{3}{11} & \frac{6}{11} & \frac{9}{11} & \frac{12}{11} & \frac{15}{11} & \frac{18}{11} & \frac{21}{11} & \frac{24}{11} & \frac{16}{11} & \frac{8}{11} \\ \frac{2}{11} & \frac{4}{11} & \frac{6}{11} & \frac{8}{11} & \frac{10}{11} & \frac{12}{11} & \frac{14}{11} & \frac{16}{11} & \frac{18}{11} & \frac{9}{11} \\ \frac{1}{11} & \frac{2}{11} & \frac{3}{11} & \frac{4}{11} & \frac{5}{11} & \frac{6}{11} & \frac{7}{11} & \frac{8}{11} & \frac{9}{11} & \frac{10}{11} \\ \end{array} \right) $$

I think the pattern is quite clear, but I do not know what is a good proof of it.