Return to Question

I'm studying root systems and coming up with some observations:

Let $\Phi$ be an irreducible root system and $\Phi^+$ a system of positive roots. Denote by $\Delta=\{\alpha_1, \alpha_2,\ldots,\alpha_n\}$ the corresponding base. Suppose that $\alpha_1, \alpha_2 \in \Delta$ and $\gamma \in \Phi^+$ have the property that $\alpha_1 + \alpha_2$ is not in $\Phi$, but $\gamma + \alpha_1$, $\gamma + \alpha_2$, and $\gamma + \alpha_1 + \alpha_2$ are all in $\Phi$.

Statement 1: In this case, there exists $\alpha_3 \in \Delta\setminus\{\alpha_1,\alpha_2\}$ such that either $\gamma +\alpha_1-\alpha_3 \in \Phi^+$ or $\gamma +\alpha_2-\alpha_3 \in \Phi^+$.

Added: Statement 1 is disproved by Andrei's counter-example.

Statement 2: Either exists $\alpha_k \in \Delta\setminus\{\alpha_1\}$ such that $\gamma +\alpha_1-\alpha_k \in \Phi^+$ or exists $\alpha_l \in \Delta\setminus\{\alpha_2\}$ such that $\gamma +\alpha_2-\alpha_l \in \Phi^+$.

The motivation is to investigate the relationship between a positive root and other lower positive roots. That is to understand the so-called Hasse diagram of a root system.

By a quick glance at lists of all irreducible root systems, I personally think that statement 2 can be correct. There will be a case-by-case proof for the statement thanks to Jim's comment. However, I tried to find a uniform proof. Suppose to the contrary that $\gamma +\alpha_1-\alpha_i \notin \Phi^+$ for all $\alpha_i \in\Delta\setminus\{\alpha_1\}$ and $\gamma +\alpha_2-\alpha_j \notin \Phi^+$ for all $\alpha_j \in\Delta\setminus\{\alpha_2\}$. It follows that the standard inner product $(\cdot,\cdot)$ draws some relations between roots that $(\gamma +\alpha_1,\alpha_i) \le 0$ for all $\alpha_i \in\Delta\setminus\{\alpha_1\}$ and $(\gamma +\alpha_2,\alpha_j) \le 0$ for all $\alpha_j \in\Delta\setminus\{\alpha_2\}$. I'm aware that positive roots forming pairwise obtuse angles must be linearly independent. So we have 2 independent sets having the same cardinality with $\Delta$ are $\{\gamma +\alpha_1,\alpha_2, \ldots,\alpha_n\}$ and $\{\alpha_1,\gamma +\alpha_2, \ldots,\alpha_n\}$.

Update: I can expand a bit by using Carter's Lemma 3 that $r_{\gamma +\alpha_1}r_{\alpha_2}\ldots r_{\alpha_n}$ and $r_{\alpha_1}r_{\gamma +\alpha_2}r_{\alpha_3}\ldots r_{\alpha_n}$ are reduced, where $r_{\theta}$ means the reflection w.r.t $\theta \in \Phi^+$. Can this lead to any contradiction? If not, could you disprove statement 2 by some counter-example?

Any help would be much appreciated.

I'm studying root systems and coming up with an observation:

Let $\Phi$ be an irreducible root system and $\Phi^+$ be a positive root system. Denote by $\Delta=\{\alpha_1, \alpha_2,\ldots,\alpha_n\}$ be the corresponding base. For $\beta, \gamma \in \Phi^+$, $\alpha_1, \alpha_2 \in \Delta$ with the properties that $\beta=\gamma +\alpha_1+\alpha_2$, $\gamma +\alpha_1 \in \Phi^+, \gamma +\alpha_2\in \Phi^+, \alpha_1+ \alpha_2 \notin \Phi^+$ then

Statement: There exists $\alpha_3 \in \Delta\setminus\{\alpha_1,\alpha_2\}$ such that either $\gamma +\alpha_1-\alpha_3 \in \Phi^+$ or $\gamma +\alpha_2-\alpha_3 \in \Phi^+$.

The motivation is to investigate the relationship between a positive positive root and other lower positive roots. That is to understand the so-called Hasse diagram root systems (for instance see definition of Hasse diagram in here)

Any help would be much appreciated.

I'm studying root systems and coming up with some observations:

Let $\Phi$ be an irreducible root system and $\Phi^+$ a system of positive roots. Denote by $\Delta=\{\alpha_1, \alpha_2,\ldots,\alpha_n\}$ the corresponding base. Suppose that $\alpha_1, \alpha_2 \in \Delta$ and $\gamma \in \Phi^+$ have the property that $\alpha_1 + \alpha_2$ is not in $\Phi$, but $\gamma + \alpha_1$, $\gamma + \alpha_2$, and $\gamma + \alpha_1 + \alpha_2$ are all in $\Phi$.

Statement 1: In this case, there exists $\alpha_3 \in \Delta\setminus\{\alpha_1,\alpha_2\}$ such that either $\gamma +\alpha_1-\alpha_3 \in \Phi^+$ or $\gamma +\alpha_2-\alpha_3 \in \Phi^+$.

Added: Statement 1 is disproved by Andrei's counter-example.

Statement 2: Either exists $\alpha_3 \in \Delta\setminus\{\alpha_1\}$ such that $\gamma +\alpha_1-\alpha_3 \in \Phi^+$ or exists $\alpha_4 \in \Delta\setminus\{\alpha_2\}$ such that $\gamma +\alpha_2-\alpha_4 \in \Phi^+$.

The motivation is to investigate the relationship between a positive root and other lower positive roots. That is to understand the so-called Hasse diagram of a root system.

By a quick glance at lists of all irreducible root systems, I personally think that statement 2 can be correct. There will be a case-by-case proof for the statement thanks to Jim's comment. However, I tried to find a uniform proof. Suppose to the contrary that $\gamma +\alpha_1-\alpha_i \notin \Phi^+$ for all $\alpha_i \in\Delta\setminus\{\alpha_1\}$ and $\gamma +\alpha_2-\alpha_j \notin \Phi^+$ for all $\alpha_j \in\Delta\setminus\{\alpha_2\}$. It follows that the standard inner product $(\cdot,\cdot)$ draws some relations between roots that $(\gamma +\alpha_1,\alpha_i) \le 0$ for all $\alpha_i \in\Delta\setminus\{\alpha_1\}$ and $(\gamma +\alpha_2,\alpha_j) \le 0$ for all $\alpha_j \in\Delta\setminus\{\alpha_2\}$. I'm aware that positive roots forming pairwise obtuse angles must be linearly independent. So we have 2 independent sets having the same cardinality with $\Delta$ are $\{\gamma +\alpha_1,\alpha_2, \ldots,\alpha_n\}$ and $\{\alpha_1,\gamma +\alpha_2, \ldots,\alpha_n\}$. Can this lead to any contradiction? If not, could you disprove statement 2 by some counter-example?

Any help would be much appreciated.

I'm studying root systems and coming up with some observations:

Let $\Phi$ be an irreducible root system and $\Phi^+$ a system of positive roots. Denote by $\Delta=\{\alpha_1, \alpha_2,\ldots,\alpha_n\}$ the corresponding base. Suppose that $\alpha_1, \alpha_2 \in \Delta$ and $\gamma \in \Phi^+$ have the property that $\alpha_1 + \alpha_2$ is not in $\Phi$, but $\gamma + \alpha_1$, $\gamma + \alpha_2$, and $\gamma + \alpha_1 + \alpha_2$ are all in $\Phi$.

Statement 1: In this case, there exists $\alpha_3 \in \Delta\setminus\{\alpha_1,\alpha_2\}$ such that either $\gamma +\alpha_1-\alpha_3 \in \Phi^+$ or $\gamma +\alpha_2-\alpha_3 \in \Phi^+$.

Added: Statement 1 is disproved by Andrei's counter-example.

Statement 2: Either exists $\alpha_k \in \Delta\setminus\{\alpha_1\}$ such that $\gamma +\alpha_1-\alpha_k \in \Phi^+$ or exists $\alpha_l \in \Delta\setminus\{\alpha_2\}$ such that $\gamma +\alpha_2-\alpha_l \in \Phi^+$.

The motivation is to investigate the relationship between a positive root and other lower positive roots. That is to understand the so-called Hasse diagram of a root system.

By a quick glance at lists of all irreducible root systems, I personally think that statement 2 can be correct. There will be a case-by-case proof for the statement thanks to Jim's comment. However, I tried to find a uniform proof. Suppose to the contrary that $\gamma +\alpha_1-\alpha_i \notin \Phi^+$ for all $\alpha_i \in\Delta\setminus\{\alpha_1\}$ and $\gamma +\alpha_2-\alpha_j \notin \Phi^+$ for all $\alpha_j \in\Delta\setminus\{\alpha_2\}$. It follows that the standard inner product $(\cdot,\cdot)$ draws some relations between roots that $(\gamma +\alpha_1,\alpha_i) \le 0$ for all $\alpha_i \in\Delta\setminus\{\alpha_1\}$ and $(\gamma +\alpha_2,\alpha_j) \le 0$ for all $\alpha_j \in\Delta\setminus\{\alpha_2\}$. I'm aware that positive roots forming pairwise obtuse angles must be linearly independent. So we have 2 independent sets having the same cardinality with $\Delta$ are $\{\gamma +\alpha_1,\alpha_2, \ldots,\alpha_n\}$ and $\{\alpha_1,\gamma +\alpha_2, \ldots,\alpha_n\}$.

Update: I can expand a bit by using Carter's Lemma 3 that $r_{\gamma +\alpha_1}r_{\alpha_2}\ldots r_{\alpha_n}$ and $r_{\alpha_1}r_{\gamma +\alpha_2}r_{\alpha_3}\ldots r_{\alpha_n}$ are reduced, where $r_{\theta}$ means the reflection w.r.t $\theta \in \Phi^+$. Can this lead to any contradiction? If not, could you disprove statement 2 by some counter-example?

Any help would be much appreciated.

I'm studying root systems and coming up with an observation:

Let $\Phi$ be an irreducible root system and $\Phi^+$ a system of positive roots. Denote by $\Delta=\{\alpha_1, \alpha_2,\ldots,\alpha_n\}$ the corresponding base. Suppose that $\alpha_1, \alpha_2 \in \Delta$ and $\gamma \in \Phi^+$ have the property that $\alpha_1 + \alpha_2$ is not in $\Phi$, but $\gamma + \alpha_1$, $\gamma + \alpha_2$, and $\gamma + \alpha_1 + \alpha_2$ are all in $\Phi$.

Statement: In this case, there exists $\alpha_3 \in \Delta\setminus\{\alpha_1,\alpha_2\}$ such that either $\gamma +\alpha_1-\alpha_3 \in \Phi^+$ or $\gamma +\alpha_2-\alpha_3 \in \Phi^+$.

The motivation is to investigate the relationship between a positive positive root and other lower positive roots. That is to understand the so-called Hasse diagram of a root system.

By a quick glance at lists of all irreducible root systems, I personally think that the above statement can be correct. There will be a case-by-case proof for the statement thanks to Jim's comment. However, I tried to find a uniform proof. Suppose to the contrary that $\gamma +\alpha_1-\alpha_i \notin \Phi^+$ for all $\alpha_i \in\Delta\setminus\{\alpha_1\}$ and $\gamma +\alpha_2-\alpha_j \notin \Phi^+$ for all $\alpha_j \in\Delta\setminus\{\alpha_2\}$. It follows that the standard inner product $(\cdot,\cdot)$ draws some relations between roots that $(\gamma +\alpha_1,\alpha_i) \le 0$ for all $\alpha_i \in\Delta\setminus\{\alpha_1\}$ and $(\gamma +\alpha_2,\alpha_j) \le 0$ for all $\alpha_j \in\Delta\setminus\{\alpha_2\}$. I'm aware that positive roots forming pairwise obtuse angles must be linearly independent. So we have 2 independent sets having the same cardinality with $\Delta$ are $\{\gamma +\alpha_1,\alpha_2, \ldots,\alpha_n\}$ and $\{\alpha_1,\gamma +\alpha_2, \ldots,\alpha_n\}$. Can this lead to any contradiction? If not, could you disprove the statement by some counter-example?

Any help would be much appreciated.

I'm studying root systems and coming up with some observations:

Let $\Phi$ be an irreducible root system and $\Phi^+$ a system of positive roots. Denote by $\Delta=\{\alpha_1, \alpha_2,\ldots,\alpha_n\}$ the corresponding base. Suppose that $\alpha_1, \alpha_2 \in \Delta$ and $\gamma \in \Phi^+$ have the property that $\alpha_1 + \alpha_2$ is not in $\Phi$, but $\gamma + \alpha_1$, $\gamma + \alpha_2$, and $\gamma + \alpha_1 + \alpha_2$ are all in $\Phi$.

Statement 1: In this case, there exists $\alpha_3 \in \Delta\setminus\{\alpha_1,\alpha_2\}$ such that either $\gamma +\alpha_1-\alpha_3 \in \Phi^+$ or $\gamma +\alpha_2-\alpha_3 \in \Phi^+$.

Added: Statement 1 is disproved by Andrei's counter-example.

Statement 2: Either exists $\alpha_3 \in \Delta\setminus\{\alpha_1\}$ such that $\gamma +\alpha_1-\alpha_3 \in \Phi^+$ or exists $\alpha_4 \in \Delta\setminus\{\alpha_2\}$ such that $\gamma +\alpha_2-\alpha_4 \in \Phi^+$.

The motivation is to investigate the relationship between a positive root and other lower positive roots. That is to understand the so-called Hasse diagram of a root system.

By a quick glance at lists of all irreducible root systems, I personally think that statement 2 can be correct. There will be a case-by-case proof for the statement thanks to Jim's comment. However, I tried to find a uniform proof. Suppose to the contrary that $\gamma +\alpha_1-\alpha_i \notin \Phi^+$ for all $\alpha_i \in\Delta\setminus\{\alpha_1\}$ and $\gamma +\alpha_2-\alpha_j \notin \Phi^+$ for all $\alpha_j \in\Delta\setminus\{\alpha_2\}$. It follows that the standard inner product $(\cdot,\cdot)$ draws some relations between roots that $(\gamma +\alpha_1,\alpha_i) \le 0$ for all $\alpha_i \in\Delta\setminus\{\alpha_1\}$ and $(\gamma +\alpha_2,\alpha_j) \le 0$ for all $\alpha_j \in\Delta\setminus\{\alpha_2\}$. I'm aware that positive roots forming pairwise obtuse angles must be linearly independent. So we have 2 independent sets having the same cardinality with $\Delta$ are $\{\gamma +\alpha_1,\alpha_2, \ldots,\alpha_n\}$ and $\{\alpha_1,\gamma +\alpha_2, \ldots,\alpha_n\}$. Can this lead to any contradiction? If not, could you disprove statement 2 by some counter-example?

Any help would be much appreciated.