Root system for Lie algebras, Does $t_{\alpha+\beta}=t_\alpha+t_\beta$? For a root system, where $\alpha,\beta \in \Delta$ and $\alpha+\beta\in \Delta$
Does $t_{\alpha+\beta}=t_\alpha+t_\beta$?
Where $t_\alpha$ I know is denoted in different ways depending on author, one author calls these root vectors. I.e. for $\alpha\in\Delta$, $t_\alpha$ spans $\mathfrak{h}$, $\quad t_\alpha \in [\frak{g_\alpha,g_{-\alpha}}]$
I tried to manipulate things using $h_\alpha = \frac{2}{\langle \alpha,\alpha\rangle} t_\alpha,\quad \alpha\in \Delta\quad [e_\alpha,e_{-\alpha}]=h_\alpha$, expanding with the Jacobi identity, $[e_\alpha,e_\beta]=N_{\alpha,\beta} e_{\alpha+\beta}$.
I am starting to think it is false, than
Additional notation help $e_\alpha$ is an eigenvector common to all elements of the Cartan subalgebra $\mathfrak{h}$

You don't really give a definition for $t_\alpha$, and I have not seen this notation before. But I assume that $h_\alpha$ and $e_\alpha$ have their standard meaning, and take the relation $t_\alpha = \frac{\langle \alpha, \alpha \rangle}{2}h_\alpha$ you give as definition, where I further assume that $\langle \cdot ,\cdot\rangle$ is an invariant scalar product on the root system.
Then the one thing to note is that for such an invariant scalar product, one has an identification of the coroot $\check\alpha(\cdot)$ with $\displaystyle\frac{2 \langle\alpha,\cdot \rangle}{\langle\alpha,\alpha\rangle}$. Whereas on the Lie algebra level, one has
$$\check\alpha(\gamma) = \gamma(h_\alpha)$$
for all roots $\gamma$, or in other words, $\check\alpha$ is the evaluation at $h_\alpha$.
Putting that together, one gets that the evaluating any root $\gamma$ (or actually any $l \in \mathfrak{h}^*$) at $t_\alpha$ is just $\langle \alpha, \gamma\rangle$ (or $\langle \alpha, l\rangle$). Since the scalar product is bilinear, the assertion follows. (And $t_\alpha \leftrightarrow \alpha$ gives an identification of the $t_\alpha$ with the original root system, like $h_\alpha \leftrightarrow \check\alpha$ describes the dual root system.)
