Let $A$ be a Cartan matrix of some finite-dimensional simple Lie algebra $g$ and $A^{-1}$ is the inverse matrix. Is it correct that that: (a) $2 A^{-1}$ is integer-valued matrix only if $g$ is in the following list: $B_n$, $C_n$, $D_{2k}$, $G_2$, $F_4$, $E_7$ and $E_8$; (b) for $g$ belonging to the set: $A_n$, $D_{2k +1}$, $E_6$, the matrix $ A^{-1} (I + P)$ is integer-valued, where $I$ is the unit matrix and $P$ is the matrix of permutation, corresponding to the generator of $Z_2$-group - the symmetry group of the Dynkin diagram?
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2$\begingroup$ ${\rm det}\, A$ equals the order of the centre of the corresponding simply-connected group, so the answer to (a) is almost yes by case-checking - except in type $D_4$ the symmetry group is $\mathfrak{S}_3$. $\endgroup$– Paul LevyMay 23, 2017 at 11:11
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1$\begingroup$ Can you provide any motivation for the question(s), or at least a couple of typical examples? What you've written doesn't quite make sense to me. Note too that $A^{-1}$ already has integer entries for types $E_8, F_4, G_2$ Meanwhile maybe it's useful to suggest two references: section 13.1 of my 1972 textbook on Lie algebras, along with a short (but somewhat hard-to-find) paper by Lusztig and Tits ams.org/mathscinet-getitem?mr=1329156. $\endgroup$– Jim HumphreysMay 23, 2017 at 15:50
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$\begingroup$ The question is motivated by certain physical problem, where some integer valued matrix $B$ appears. For the case (a): $B = 2A^{-1}$ and for the case (b): $B = A^{-1} (I + P)$. As we expected $(a)$ is a well-known fact. The conjecture (b) was verified for $E_6$. For classical series $A_r$, $D_r$ the (b) was verified by MATHEMATICA for some ranks $r$ ($r > 4$ for $D$-series). $\endgroup$– VladimirMay 23, 2017 at 19:32
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1$\begingroup$ @Vladimir: Your comments are helpful, but the case of $D_4$ still needs more discussion as indicated by Paul Levy in his comment. For (a) this case is certainly not "a well-known fact" (what is the inverse Cartan matrix here?). I'm not familiar with your formulation, which seems to work much of the time though I can't see what would make it true in general. $\endgroup$– Jim HumphreysMay 23, 2017 at 20:37
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1$\begingroup$ @JimHumphreys, the paper is available at Lusztig's web site: www-math.mit.edu/%7Egyuri/papers/car.ps . $\endgroup$– LSpiceMay 24, 2017 at 1:04
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1 Answer
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For (b), the rows of $A^{-1}$ correspond to something like the fundamental weights, expressed in terms of the simple roots. (Perhaps in general you have to take co-weights and co-roots and/or the columns of $A^{-1}$, but in case (b) the root system is always simply-laced so this essentially changes nothing anyway.) Then you are asking whether $\varpi_i+\gamma(\varpi_i)$ is an integer linear combination of simple roots, where $\varpi_i$ is a fundamental weight and $\gamma$ is the involution generating $Z_2$ (or rather, the automorphism of the weight lattice which is induced by a generator of $Z_2$).
Then your statement (b) is incorrect in general. In type $D_{n}$, the diagram involution swaps $\alpha_{n-1}$ and $\alpha_n$, and by a straightforward calculation we have:
$\varpi_{n-1}+\varpi_n = \alpha_1+2\alpha_2+\ldots + (n-2)\alpha_{n-2} + \frac{n-1}{2}(\alpha_{n-1}+\alpha_n).$
In particular, if $n$ is even then the last two coefficients are not in ${\mathbb Z}$.
I would note also (something I missed in my earlier comment) that the "only if" statement in (a) is wrong for type $D_{2n}$.
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$\begingroup$ Paul, thank you very much. (a) Indeed, $2A^{-1}$ for $D_{2n}$ is integer valued. $\endgroup$– VladimirMay 23, 2017 at 23:40
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1$\begingroup$ @Paul: As Vladimir points out, and your answer emphasizes, the cases for $D_n$ differ when $n$ is even or odd. This reflects the fact that the fundamental group for $n$ even is $\mathbb{Z}_2 \times \mathbb{Z}_2$, whereas for $n$ odd it's cyclic of order 4. A basic problem I was overlooking is that Vladimir's formulation needs to place $D_n$ for $n$ even in one case and for $n$ odd in the other case. So the formulation needs to be more precise to be correct; the symmetry group of the graph can't distinguish the two cases. $\endgroup$ May 24, 2017 at 0:08
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$\begingroup$ I agree with the latest comment of Prof. Humphreys (many thanks!). I have edited my original post. $\endgroup$– VladimirMay 24, 2017 at 0:13
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$\begingroup$ The sums $n_i = 2 \sum_{j =1}^{r} A^{-1}_{i j}$ ($i = 1,\dots, r$, where $r$ is rank of the Lie algebra) are integer numbers, which are components of twice dual Weyl vector in the basis of simple co-roots. $\endgroup$– VladimirMay 24, 2017 at 12:45
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$\begingroup$ In case (b) we obtain: (c) $n_i = \sum_{j =1}^{r} B_{i j}$, $i = 1, \dots, r$, where $B = A^{-1}(I + P)$. $\endgroup$– VladimirMay 24, 2017 at 12:51