0
$\begingroup$

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?

$\endgroup$
  • 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 Levy May 23 '17 at 11:11
  • 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 Humphreys May 23 '17 at 15:50
  • $\begingroup$ 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). $\endgroup$ – Vladimir May 23 '17 at 19:32
  • 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 Humphreys May 23 '17 at 20:37
  • 1
    $\begingroup$ @JimHumphreys, the paper is available at Lusztig's web site: www-math.mit.edu/%7Egyuri/papers/car.ps . $\endgroup$ – LSpice May 24 '17 at 1:04
3
$\begingroup$

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 $D_{2n}$.

$\endgroup$
  • $\begingroup$ Paul, thank you very much. (a) Indeed, $2A^{-1}$ for $D_{2n}$ is integer valued. $\endgroup$ – Vladimir May 23 '17 at 23:40
  • 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. $\endgroup$ – Jim Humphreys May 24 '17 at 0:08
  • $\begingroup$ I agree with the latest comment of Prof. Humphreys (many thanks!). I have edited my original post. $\endgroup$ – Vladimir May 24 '17 at 0:13
  • $\begingroup$ 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. $\endgroup$ – Vladimir May 24 '17 at 12:45
  • $\begingroup$ In case (b) we obtain: (c) $n_i = \sum_{j =1}^{r} B_{i j}$, $i = 1, \dots, r$, where $B = A^{-1}(I + P)$. $\endgroup$ – Vladimir May 24 '17 at 12:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.