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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.

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.)

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.

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.

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 (just as $\mathbb{Z}$ is dense in $\mathbb{R}$), 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.

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.