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edited Jul 29 at 11:21

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Here's a naive approach which I think works in the simply-laced cases, unless I've made a silly mistake somewhere. The restriction to simply-laced cases allows me to placeplay fast and loose with the root-coroot pairings vs. the inner product on roots, and also assume that all roots have length 2.

Assume $\{\alpha, \beta\} \neq \{\gamma, \delta\}$, but that $\mathrm{ht}(\alpha)=\mathrm{ht}(\gamma)$. Since $\alpha+\beta-\gamma=\delta$ is a root, then $$ (\alpha+\beta-\gamma, \alpha+\beta-\gamma)=(\delta,\delta)=2. $$

Expanding this out, using all roots have the same length, we get $$ (\alpha,\beta)-(\alpha, \gamma)-(\beta,\gamma)=-2, $$ or that $(\alpha,\beta)-(\alpha,\gamma)=(\beta,\gamma)-2$. Since $\alpha+\beta \not \in \Phi^+$, necessarily $(\alpha, \beta) \geq 0$. Similarly, since $\alpha \neq \gamma$ and $\mathrm{ht}(\alpha)=\mathrm{ht}(\gamma)$, $\alpha-\gamma \not \in \Phi \sqcup \{0\}$ so that $(\alpha, \gamma) \leq 0$.

In total, then, $(\alpha, \beta)-(\alpha, \gamma) \geq 0$, so that $(\beta, \gamma)-2 \geq 0 \implies (\beta, \gamma) \geq 2$; since we are in the simply-laced case, necessarily $(\beta, \gamma)=2$ and thus $\beta =\gamma$, contradicting $\{\alpha, \beta\} \neq \{\gamma, \delta\}$.

Here's a naive approach which I think works in the simply-laced cases, unless I've made a silly mistake somewhere. The restriction to simply-laced cases allows me to place fast and loose with the root-coroot pairings vs. the inner product on roots, and also assume that all roots have length 2.

Assume $\{\alpha, \beta\} \neq \{\gamma, \delta\}$, but that $\mathrm{ht}(\alpha)=\mathrm{ht}(\gamma)$. Since $\alpha+\beta-\gamma=\delta$ is a root, then $$ (\alpha+\beta-\gamma, \alpha+\beta-\gamma)=(\delta,\delta)=2. $$

Expanding this out, using all roots have the same length, we get $$ (\alpha,\beta)-(\alpha, \gamma)-(\beta,\gamma)=-2, $$ or that $(\alpha,\beta)-(\alpha,\gamma)=(\beta,\gamma)-2$. Since $\alpha+\beta \not \in \Phi^+$, necessarily $(\alpha, \beta) \geq 0$. Similarly, since $\alpha \neq \gamma$ and $\mathrm{ht}(\alpha)=\mathrm{ht}(\gamma)$, $\alpha-\gamma \not \in \Phi \sqcup \{0\}$ so that $(\alpha, \gamma) \leq 0$.

In total, then, $(\alpha, \beta)-(\alpha, \gamma) \geq 0$, so that $(\beta, \gamma)-2 \geq 0 \implies (\beta, \gamma) \geq 2$; since we are in the simply-laced case, necessarily $(\beta, \gamma)=2$ and thus $\beta =\gamma$, contradicting $\{\alpha, \beta\} \neq \{\gamma, \delta\}$.

Here's a naive approach which I think works in the simply-laced cases, unless I've made a silly mistake somewhere. The restriction to simply-laced cases allows me to play fast and loose with the root-coroot pairings vs. the inner product on roots, and also assume that all roots have length 2.

Assume $\{\alpha, \beta\} \neq \{\gamma, \delta\}$, but that $\mathrm{ht}(\alpha)=\mathrm{ht}(\gamma)$. Since $\alpha+\beta-\gamma=\delta$ is a root, then $$ (\alpha+\beta-\gamma, \alpha+\beta-\gamma)=(\delta,\delta)=2. $$

Expanding this out, using all roots have the same length, we get $$ (\alpha,\beta)-(\alpha, \gamma)-(\beta,\gamma)=-2, $$ or that $(\alpha,\beta)-(\alpha,\gamma)=(\beta,\gamma)-2$. Since $\alpha+\beta \not \in \Phi^+$, necessarily $(\alpha, \beta) \geq 0$. Similarly, since $\alpha \neq \gamma$ and $\mathrm{ht}(\alpha)=\mathrm{ht}(\gamma)$, $\alpha-\gamma \not \in \Phi \sqcup \{0\}$ so that $(\alpha, \gamma) \leq 0$.

In total, then, $(\alpha, \beta)-(\alpha, \gamma) \geq 0$, so that $(\beta, \gamma)-2 \geq 0 \implies (\beta, \gamma) \geq 2$; since we are in the simply-laced case, necessarily $(\beta, \gamma)=2$ and thus $\beta =\gamma$, contradicting $\{\alpha, \beta\} \neq \{\gamma, \delta\}$.

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created Jul 29 at 10:59

550
3
9

Here's a naive approach which I think works in the simply-laced cases, unless I've made a silly mistake somewhere. The restriction to simply-laced cases allows me to place fast and loose with the root-coroot pairings vs. the inner product on roots, and also assume that all roots have length 2.

Assume $\{\alpha, \beta\} \neq \{\gamma, \delta\}$, but that $\mathrm{ht}(\alpha)=\mathrm{ht}(\gamma)$. Since $\alpha+\beta-\gamma=\delta$ is a root, then $$ (\alpha+\beta-\gamma, \alpha+\beta-\gamma)=(\delta,\delta)=2. $$

Expanding this out, using all roots have the same length, we get $$ (\alpha,\beta)-(\alpha, \gamma)-(\beta,\gamma)=-2, $$ or that $(\alpha,\beta)-(\alpha,\gamma)=(\beta,\gamma)-2$. Since $\alpha+\beta \not \in \Phi^+$, necessarily $(\alpha, \beta) \geq 0$. Similarly, since $\alpha \neq \gamma$ and $\mathrm{ht}(\alpha)=\mathrm{ht}(\gamma)$, $\alpha-\gamma \not \in \Phi \sqcup \{0\}$ so that $(\alpha, \gamma) \leq 0$.

In total, then, $(\alpha, \beta)-(\alpha, \gamma) \geq 0$, so that $(\beta, \gamma)-2 \geq 0 \implies (\beta, \gamma) \geq 2$; since we are in the simply-laced case, necessarily $(\beta, \gamma)=2$ and thus $\beta =\gamma$, contradicting $\{\alpha, \beta\} \neq \{\gamma, \delta\}$.