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This question is an exact duplicate of:

Recall the definition of the Weyl Chambers: A Weyl Chamber is a region of $V \setminus \bigcup_{\alpha \in \Phi} H_{\alpha}$, where $V$ is underlying Euclidean space, and $H_\alpha$ the hyperplane that is perpendicular to $\alpha$ the elements of root system $\Phi$ in $V$.

I want to show that:

The Weyl chambers are open, convex and connected.

My attempts:

A single hyperplane $H_\alpha$ separates the space $V$ into two half-spaces: positive half-space $H_\alpha^+=\{x \in V \mid (\alpha,x)>0\}$ and negative half-space $H_\alpha^-=\{x \in V \mid (\alpha,x)<0\}$. It is clear that $V \setminus H_\alpha = H_\alpha^+\cup H^-_\alpha.$ As the result, the Weyl chambers can be regarded as components of $$V \setminus \bigcup_{\alpha \in \Phi} H_{\alpha}= \bigcap_{\alpha \in \Phi} \left( V \setminus H_{\alpha} \right)= \bigcap_{\alpha \in \Phi} \left( H_\alpha^+\cup H_\alpha^- \right).$$ By the discussion above, a Weyl chamber is simply a $\textbf{intersection}$ of finite half-spaces. It is the case since one may write $$\left( H_\alpha^+\cup H_\alpha^- \right) \cap \left( H_\beta^+\cup H_\beta^- \right) =\left( H_\alpha^+\cap H_\beta^+ \right) \cup\left( H_\alpha^+\cap H_\beta^- \right) \cup \left( H_\alpha^-\cap H_\beta^+ \right) \cup\left( H_\alpha^-\cap H_\beta^- \right).$$ Therefore openness and convexity of the Weyl chambers are straightforward since half-spaces are open and convex, so are their intersections. Moreover, we would replace $V$ by $\Bbb R^{\dim V}$ since they are isomorphic. Now we can prove that the Weyl chamber is connected by showing that every convex set in $\Bbb R^{\dim V}$ is connected. And I'm sure that proof of this statement existed. So am I done?

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marked as duplicate by André 3000, graydad, Community Jun 1 '15 at 2:29

This question was marked as an exact duplicate of an existing question.

migrated from mathoverflow.net Jun 1 '15 at 1:59

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