Showing that half-sum of positive roots is the sum of fundamental weights

[naele]

Homework Statement

Let L be a simple compact Lie group, and \Delta_+ is the set of positive roots. I have previously shown that if \alpha\in\Delta_+ and \alpha_i is a simple root, then s_i\alpha\in \Delta_+ where s_i is the Weyl reflection associated with \alpha_i.

Now, let \delta = \frac{1}{2}\sum_{\alpha\in\Delta_+}\alpha. I want to show that
<br /> s_i\delta=\delta-\alpha_i<br />

Homework Equations

The Attempt at a Solution

It's clear that
<br /> s_i\delta=\delta - \sum_{\alpha\neq \alpha_i} \frac{\alpha\cdot\alpha_i}{\alpha_i^2}\alpha_i - \alpha_i<br />

But I have no idea how to show that \alpha\cdot\alpha_i=0\quad \forall\alpha\neq\alpha_i. I cannot make appeal to the fact that delta might be a sum of fundamental weights because that's what I need to show later on.

 

[fzero]

Homework Statement

Let L be a simple compact Lie group, and \Delta_+ is the set of positive roots. I have previously shown that if \alpha\in\Delta_+ and \alpha_i is a simple root, then s_i\alpha\in \Delta_+ where s_i is the Weyl reflection associated with \alpha_i.

To be precise, s_i\alpha\in \Delta_+ for \alpha\neq \alpha_i. That is, s_i reflects \alpha_i \rightarrow -\alpha_i but permutes the \alpha\neq \alpha_i into one another.

Now, let \delta = \frac{1}{2}\sum_{\alpha\in\Delta_+}\alpha. I want to show that
<br /> s_i\delta=\delta-\alpha_i<br />

Homework Equations

The Attempt at a Solution

It's clear that
<br /> s_i\delta=\delta - \sum_{\alpha\neq \alpha_i} \frac{\alpha\cdot\alpha_i}{\alpha_i^2}\alpha_i - \alpha_i<br />

But I have no idea how to show that \alpha\cdot\alpha_i=0\quad \forall\alpha\neq\alpha_i. I cannot make appeal to the fact that delta might be a sum of fundamental weights because that's what I need to show later on.

In light of the comments above, it's more straightforward to note that

s_i \delta = \frac{1}{2}\sum_{\alpha\in\Delta_+}s_i\alpha =\frac{1}{2} \left( \sum_{\alpha\neq \alpha_i} \alpha - \alpha_i \right).

Restoring \delta in an obvious way gives the required result.