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

1. Dec 30, 2011

### naele

1. The problem statement, all variables and given/known data
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
$$s_i\delta=\delta-\alpha_i$$

2. Relevant equations

3. The attempt at a solution
It's clear that
$$s_i\delta=\delta - \sum_{\alpha\neq \alpha_i} \frac{\alpha\cdot\alpha_i}{\alpha_i^2}\alpha_i - \alpha_i$$

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.

2. Dec 31, 2011

### fzero

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.

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.

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook