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Let $\Phi$ be a root system in inner product space $E=\mathbb{R}^l$ (inner product $(\cdot,\cdot)$ and $\Delta=\{\alpha,_1,\cdots,\alpha_l\}$ be a fundamental root system.

Consider the root lattice $\Lambda_r=\mathbb{Z}\alpha_1 + \cdots + \mathbb{Z}\alpha_l$ in $E$ and weight lattice $\Lambda$ $$\Lambda =\{\lambda\in E : \frac{2(\lambda,\alpha)}{(\alpha,\alpha)} \in\mathbb{Z} \forall \alpha\in \Delta\}. $$

Let $\Lambda^+ = \{ \lambda\in\Lambda : \frac{2(\lambda,\alpha)}{(\alpha,\alpha)} \geq 0 \forall \alpha\in\Delta\}$.

The miniscules are specific elements of $\Lambda^+$; they are minimal elements of $\Lambda^+$ w.r.t ordering defined as follows: $\lambda_1\leq \lambda_2$ if $\lambda_2-\lambda_1=\sum_{i=1}^l m_i\alpha_i$ with $m_i\geq 0$.

I wanted to see the proof of following fact (seen here this, page 2, top line)

For different minuscules $\lambda,\nu$, the cosets $\lambda+\Lambda_r$ and $\nu+\Lambda_r$ are distinct.

Is this easy to prove? How should we proceed?

I tried a simple case: suppose $\lambda,\nu$ are minuscules with $\lambda=\nu+\alpha-\beta$ where $\alpha,\beta$ are distinct fundamental (i.e. simple) roots. I couldn't get any direction to get contradiction from this.

The actual general problem is that if $\lambda+\Lambda_1=\nu+\Lambda_r$, then $\lambda-\nu$ is integral combination of fundamental roots, so y $\lambda=\nu+m_1\alpha_1 + \cdots + m_l\alpha_l$, so some coefficients could be positive and some negative. The simplest non-trivial case is then exactly one $m_i$ is $+1$, exactly one $m_j$ is $-1$ and rest are zero; I don't get any idea to proceed for contradiction.

Let $\Phi$ be a root system in inner product space $E=\mathbb{R}^l$ (inner product $(\cdot,\cdot)$ and $\Delta=\{\alpha,_1,\cdots,\alpha_l\}$ be a fundamental root system.

Consider the root lattice $\Lambda_r=\mathbb{Z}\alpha_1 + \cdots + \mathbb{Z}\alpha_l$ in $E$ and weight lattice $\Lambda$ $$\Lambda =\{\lambda\in E : \frac{2(\lambda,\alpha)}{(\alpha,\alpha)} \in\mathbb{Z} \forall \alpha\in \Delta\}. $$

Let $\Lambda^+ = \{ \lambda\in\Lambda : \frac{2(\lambda,\alpha)}{(\alpha,\alpha)} \geq 0 \forall \alpha\in\Delta\}$.

The miniscules are specific elements of $\Lambda^+$; they are minimal elements of $\Lambda^+$ w.r.t ordering defined as follows: $\lambda_1\leq \lambda_2$ if $\lambda_2-\lambda_1=\sum_{i=1}^l m_i\alpha_i$ with $m_i\geq 0$.

I wanted to see the proof of following fact (seen here this, page 2, top line)

For different minuscules $\lambda,\nu$, the cosets $\lambda+\Lambda_r$ and $\nu+\Lambda_r$ are distinct.

Is this easy to prove? How should we proceed?

I tried a simple case: suppose $\lambda,\nu$ are minuscules with $\lambda=\nu+\alpha-\beta$ where $\alpha,\beta$ are distinct fundamental (i.e. simple) roots. I couldn't get any direction to get contradiction from this.

The actual general problem is that if $\lambda+\Lambda_r=\nu+\Lambda_r$, then $\lambda-\nu$ is integral combination of fundamental roots, so y $\lambda=\nu+m_1\alpha_1 + \cdots + m_l\alpha_l$, so some coefficients could be positive and some negative. The simplest non-trivial case is then exactly one $m_i$ is $+1$, exactly one $m_j$ is $-1$ and rest are zero; I don't get any idea to proceed for contradiction.

Let $\Phi$ be a root system in inner product space $E=\mathbb{R}^l$ (inner product $(\cdot,\cdot)$ and $\Delta=\{\alpha,_1,\cdots,\alpha_l\}$ be a fundamental root system.

Consider the root lattice $\Lambda_r=\mathbb{Z}\alpha_1 + \cdots + \mathbb{Z}\alpha_l$ in $E$ and weight lattice $\Lambda$ $$\Lambda =\{\lambda\in E : \frac{2(\lambda,\alpha)}{(\alpha,\alpha)} \in\mathbb{Z} \forall \alpha\in \Delta\}. $$

Let $\Lambda^+ = \{ \lambda\in\Lambda : \frac{2(\lambda,\alpha)}{(\alpha,\alpha)} \geq 0 \forall \alpha\in\Delta\}$.

The miniscules are specific elements of $\Lambda^+$; they are minimal elements of $\Lambda^+$ w.r.t ordering defined as follows: $\lambda_1\leq \lambda_2$ if $\lambda_2-\lambda_1=\sum_{i=1}^l m_i\alpha_i$ with $m_i\geq 0$.

I wanted to see the proof of following fact (seen here this, page 2, top line)

For different minuscules $\lambda,\nu$, the cosets $\lambda+\Lambda_r$ and $\nu+\Lambda_r$ are distinct.

Is this easy to prove? How should we proceed?

Let $\Phi$ be a root system in inner product space $E=\mathbb{R}^l$ (inner product $(\cdot,\cdot)$ and $\Delta=\{\alpha,_1,\cdots,\alpha_l\}$ be a fundamental root system.

Consider the root lattice $\Lambda_r=\mathbb{Z}\alpha_1 + \cdots + \mathbb{Z}\alpha_l$ in $E$ and weight lattice $\Lambda$ $$\Lambda =\{\lambda\in E : \frac{2(\lambda,\alpha)}{(\alpha,\alpha)} \in\mathbb{Z} \forall \alpha\in \Delta\}. $$

Let $\Lambda^+ = \{ \lambda\in\Lambda : \frac{2(\lambda,\alpha)}{(\alpha,\alpha)} \geq 0 \forall \alpha\in\Delta\}$.

The miniscules are specific elements of $\Lambda^+$; they are minimal elements of $\Lambda^+$ w.r.t ordering defined as follows: $\lambda_1\leq \lambda_2$ if $\lambda_2-\lambda_1=\sum_{i=1}^l m_i\alpha_i$ with $m_i\geq 0$.

I wanted to see the proof of following fact (seen here this, page 2, top line)

For different minuscules $\lambda,\nu$, the cosets $\lambda+\Lambda_r$ and $\nu+\Lambda_r$ are distinct.

Is this easy to prove? How should we proceed?

I tried a simple case: suppose $\lambda,\nu$ are minuscules with $\lambda=\nu+\alpha-\beta$ where $\alpha,\beta$ are distinct fundamental (i.e. simple) roots. I couldn't get any direction to get contradiction from this.

The actual general problem is that if $\lambda+\Lambda_1=\nu+\Lambda_r$, then $\lambda-\nu$ is integral combination of fundamental roots, so y $\lambda=\nu+m_1\alpha_1 + \cdots + m_l\alpha_l$, so some coefficients could be positive and some negative. The simplest non-trivial case is then exactly one $m_i$ is $+1$, exactly one $m_j$ is $-1$ and rest are zero; I don't get any idea to proceed for contradiction.

Let $\Phi$ be a root system in $E=\mathbb{R}^l$ and $\Delta=\{\alpha,_1,\cdots,\alpha_l\}$ be a fundamental root system.

Consider the root lattice $\Lambda_r=\mathbb{Z}\alpha_1 + \cdots + \mathbb{Z}\alpha_l$ in $E$ and weight lattice $\Lambda$ $$\Lambda =\{\lambda\in \mathbb{R}^l : \frac{2(\lambda,\alpha)}{(\alpha,\alpha)} \in\mathbb{Z} \forall \alpha\in \Delta\}. $$

Let $\Lambda^+ = \{ \lambda\in\Lambda : \frac{2(\lambda,\alpha)}{(\alpha,\alpha)} \geq 0 \forall \alpha\in\Delta\}$.

The miniscules are specific elements of $\Lambda^+$; they are minimal elements of $\Lambda^+$ w.r.t ordering defined as follows: $\lambda_1\leq \lambda_2$ if $\lambda_2-\lambda_1=\sum_{i=1}^l m_i\alpha_i$ with $m_i\geq 0$.

I wanted to see the proof of following fact (seen here this, page 2, top line)

For different minuscules $\lambda,\nu$, the cosets $\lambda+\Lambda_r$ and $\nu+\Lambda_r$ are distinct.

Is this easy to prove? How should we proceed?

Let $\Phi$ be a root system in inner product space $E=\mathbb{R}^l$ (inner product $(\cdot,\cdot)$ and $\Delta=\{\alpha,_1,\cdots,\alpha_l\}$ be a fundamental root system.

Consider the root lattice $\Lambda_r=\mathbb{Z}\alpha_1 + \cdots + \mathbb{Z}\alpha_l$ in $E$ and weight lattice $\Lambda$ $$\Lambda =\{\lambda\in E : \frac{2(\lambda,\alpha)}{(\alpha,\alpha)} \in\mathbb{Z} \forall \alpha\in \Delta\}. $$

Let $\Lambda^+ = \{ \lambda\in\Lambda : \frac{2(\lambda,\alpha)}{(\alpha,\alpha)} \geq 0 \forall \alpha\in\Delta\}$.

The miniscules are specific elements of $\Lambda^+$; they are minimal elements of $\Lambda^+$ w.r.t ordering defined as follows: $\lambda_1\leq \lambda_2$ if $\lambda_2-\lambda_1=\sum_{i=1}^l m_i\alpha_i$ with $m_i\geq 0$.

I wanted to see the proof of following fact (seen here this, page 2, top line)

For different minuscules $\lambda,\nu$, the cosets $\lambda+\Lambda_r$ and $\nu+\Lambda_r$ are distinct.

Is this easy to prove? How should we proceed?