I have a paperback 3rd edition and on page 65 you can find Proposition 5.8. My question is about part (c):

If $A$ is of indefinite type, then $$ \overline{X} = \{ h \in \{ \frak h_{\mathbb{R}} \}| \left \langle \alpha, h \right \rangle \geq 0 \text{ for all } \alpha \in \Delta_+^{im} \}, $$ where $\overline{X}$ denotes the closure of $X$ in the metric topology of $\frak h_{\mathbb{R}}$.

Some context: $A$ is a generalized Cartan matrix, $\frak h_{\mathbb{R}}$ is a real form of the Cartan subalgebra in the Kac-Moody algebra associated to $A$ and finally $X$ is the Tits cone.

Question: if one reads the proof given by Kac, it is written for $X$ and not $\overline{X}$. Is there a mistake or am I missing something here?

  • $\begingroup$ I glanced at that passage in the 3rd edition, which gets convoluted by the time you reach part (c) treating the case of an indefinite matrix A. (The finite and affine cases, which come up more often in applications, are a little more straightforward.) Keep in mind that the Tits cone C is defined in general by non-strict inequalities (so it serves as a fundamental domain). But the presence of imaginary roots complicates life, so you have to refer back to earlier sections and assemble steps in (c) very carefully. Probably "mistake" is too strong, but it's not easy to follow. $\endgroup$ Jul 27, 2012 at 13:34
  • $\begingroup$ But if you read the proof he says let X^{\prime} be the right hand side and then shows that it is equal to X itself! The closure in metric topology is not mentioned even once! $\endgroup$
    – Najdorf
    Jul 27, 2012 at 13:43
  • $\begingroup$ I used the word "mistake" because I think the statement is false and you should replace the closure with the cone. Of course I might be mistaken myself, hence "possible mistake". $\endgroup$
    – Najdorf
    Jul 27, 2012 at 13:45
  • $\begingroup$ @Najdorf: [Note I called C the "Tits cone", but that label belongs to the W -saturated set X. Definite "mistake".] Since X' is closed and is shown to contain X, it also contains the closure of X. The proof isn't written down carefully, but I think if you assemble the various steps they do add up to a correct proof of the reverse inclusion for the closure of X. ["Misprint"?] This book is known for its rough spots, even the 3rd edition, but it's an essential source. The only similar textbooks are those by Moody-Pianzola and Carter, which have different emphases. Good luck $\endgroup$ Jul 27, 2012 at 17:52

1 Answer 1


After taking a closer look at the proof by Kac of Prop. 5.8 c), I can see that it's too sketchy to be followed easily. Here the generalized Cartan matrix is assumed to be of indefinite type, which I have little experience with. But the basic steps in the proof might be organized as follows:

In the background is the infinite root system $\Delta= \Delta_+ \cup \Delta_-$ along with a further partition $\Delta_+ = \Delta_+^{re} \cup \Delta_+^{im}$. Here $\Delta_+$ consists of $\mathbb{Z}^+$-linear combinations of the fixed simple roots $\alpha_1, \dots, \alpha_n$. The $\alpha_i$ are linearly independent and lie in the dual space of $\mathfrak{h}_\mathbb{R}$ where $\mathfrak{h}$ is the finite dimensional Cartan subalgebra. The Weyl group $W$ is generated by all reflections $r_\alpha$ with $\alpha \in \Delta_+$.

The Tits cone $X$ is the image under $W$ of $C:= \{ h \in \mathfrak{h}_\mathbb{R} | \langle \alpha_i, h \rangle \geq 0 \text{ for all } i\}$. The candidate for its metric closure is $X' := \{h \in \mathfrak{h}_\mathbb{R} | \langle \beta, h \rangle \geq 0 \text{ for all } \beta \in \Delta_+^{im}\}$. Being defined by inequalities, $X'$ is closed.

By Prop. 5.2 a), $\Delta_+^{im}$ is $W$-invariant (so $X'$ is). Obviously $X' \supset C $, so $X' \supset \overline{X}$.

In the reverse direction, consider just those $h \in X'$ for which $\langle \alpha_i, h \rangle \in \mathbb{Z}$ for all $i$. These elements of $X'$ are dense in the metric topology, so it's enough to show they all lie in $X$ (where they will form a dense subset of $\overline{X}$).

Use Thm. 5.6 c) to find $\beta \in \Delta_+^{im}$ such that the "Cartan integers" $\langle \beta, \alpha_i^\vee \rangle <0$ for all $i$. (So $\beta= \sum_i b_i \alpha_i$ with all $b_i>0$.)

In turn, for all $\gamma \in \Delta_+^{re}$, $$r_\gamma (\beta) = \beta - \langle \beta, \gamma^\vee \rangle \gamma = \beta + s \gamma$$ with $s$ larger than the sum of coefficients of $\gamma$ because $\langle \beta, \alpha_i^\vee \rangle <0$ for all $i$. Thanks to Prop. 5.2 c), all $\beta +s \gamma \in \Delta_+^{im}$. Since $h \in X'$, it follows that $\langle \beta + s\gamma, h \rangle \in \mathbb{Z}^+$. In particular, only finitely many such $\gamma$ exist with $\langle \gamma, h \rangle \leq -1$.

But Prop. 3.12 c) characterizes $X$ as the set of all $h$ for which only finitely many $\gamma \in \Delta_+$ satisfy $\langle \gamma, h \rangle <0$. Combined with the special choice of $h$, we get $h \in X$ as desired.

ADDED: To fill in details of the argument I've automatically tended to think in terms of Zariski-density and Zariski-closure, but something else must be going on here to deal with the metric topology. This is the point at which I'm doubtful about the strategy used by Kac. But given the brevity of the argument it's probably necessary to look further into the surrounding material for some kind of insight. (Maybe it's just a question of pointing to the fact that both $X$ and $X'$ are cones? It would be easier if the author of the book told us what he was thinking about.)

  • $\begingroup$ Why are all integral points of X^{\prime} (defined as points where all simple roots give us integers) dense in the metric topology? I think it would be far from dense and actually discrete. $\endgroup$
    – Najdorf
    Jul 31, 2012 at 18:11
  • $\begingroup$ @Najdorf: See my added paragraph. I think I expanded correctly the steps sketched in the book, but then my attempt to invoke "density" was oversimplified. Since both $X$ and $X'$ are cones, maybe one can tweak my argument by adding that all positive real multiples of the chosen $h$ are also in $X$. Anyway, the argument given in the book is too sketchy even if it turns out to be basically correct (modulo a misprint). And there may not be any alternative sources to consult for c). Maybe it's time to consult Victor Kac directly? $\endgroup$ Aug 1, 2012 at 17:11
  • $\begingroup$ Does Kac answer email? Actually I found what seems to be a full proof in "Introduction to Kac-Moody Algebra" by Zhexian Wan (Proposition 5.6, page 98). $\endgroup$
    – Najdorf
    Aug 14, 2012 at 7:54
  • 1
    $\begingroup$ @Najdorf: Your question is a good one, which I can't answer. It's helpful to have a reference to Wan's book, which I hadn't thought of (though I think our library does own a copy). Please update your question if you get better insight than I got from reading Kac. $\endgroup$ Aug 14, 2012 at 18:08

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