Let's consider $\mathfrak{so}(5)$ as the Lie algebra of $\mathrm{SO}(5)$, where the symmetric bilinear form is $x_1y_5+\cdots +y_1x_5$. Then the maximal torus is given by $$\left(\begin{array}{cccccc} a &&&&\\ &b &&&\\ &&1 && \\ &&& b^{-1} &\\ &&&&a^{-1}\ \end{array}\right)$$.
The Lie algebra is the same, with $1$ replaced by $0$ and inverses with minus signs. The representations of this torus are the weight space, $a^{m}b^{n}$ for some $m,n$. But this isn't the weight space for the Lie algebra, as we need to include $(1/2, 1/2)$ to get the spin representations.
What am I missing about the relation of weights to characters of maximal tori? Does this only hold for simply connected groups? And if so, is there an easy way to discover that $SO(5)$ is not simply connected without constructing the Spin groups?
Sorry to resurrect such an old post...
The matrix you wrote is not in $\text{SO}(5)$, as it not an orthogonal matrix.
Only for simply connected Lie groups can a representation of the Lie algebra be lifted to a representation of the Lie group. $\text{SO}(5)$ is not simply connected. So not every representation of its Lie algebra $(\text{B}_2)$ can be lifted to a representation of $\text{SO}(5)$. On the other hand, $\text{Sp}(4)$ is simply connected, and its Lie algebra $(\text{C}_2)$ is the same as $\text{B}_2$. So $\text{Sp}(4)$ is a universal of $\text{SO}(5)$. The natural $4$-dimensional representation of $\text{Sp}(4)$ is the spin representation. The representation $\wedge^2 4$ is of dimension $6$ and contains the trivial representation, namely the line defined by the element that corresponds to the invariant symplectic form in $4$. The quotient representation is $5$-dimensional and is the defining representation for $\text{SO}(5)$.