I'm reading about Lie algebras and I don't understand the root space decomposition of $\mathfrak{so}_N(\mathbb{F})$ given in Victor Kac's Lecture 15, Proposition 15.1. (there was also a book which claimed the same thing, but I can't remember it).
In short, it says $\mathfrak{so}_N(\mathbb{F})$ is semisimple for $N\ge 3$. For that, he wants to compute the root space decomposition of $\mathfrak{so}_N(\mathbb{F})$. At the end he states the following
Hence the set of roots is: \begin{align} &N=2n+1:\Delta_{\mathfrak{so}_N(\mathbb{F})}=\{\epsilon_i-\epsilon_j,\epsilon_i,-\epsilon_i,\epsilon_i+\epsilon_j,-\epsilon_i-\epsilon_j\mid i,j\in \{1,\dots,n\},i\neq j\}\\ &N=2n:\Delta_{\mathfrak{so}_N(\mathbb{F})}=\{\epsilon_i-\epsilon_j,\epsilon_i+\epsilon_j,-\epsilon_i-\epsilon_j\mid i,j\in \{1,\dots,n\},i\neq j\} \end{align}
However, we know that for $N=2n+1$, $\dim(\mathfrak{so}_N(\mathbb{F}))=2n^2+n$. On the other hand, we have at least $3n(n-1)+2n=3n^2-n$ roots according to Kac's Lecture which in general is greater than $2n^2+n$. Isn't this a contradiction because each root contributes at least a subspace of dimension $1$?
This question must be really simple to answer but I can't figure it out.
You overcount the $\epsilon_i + \epsilon_j$ and $-\epsilon_i - \epsilon_j$: switching $i$ and $j$ gives the same root. So half those, then the actual number of roots is $2n(n-1)+2n=2n^2$ as it should be.