Representations of $\mathrm{SO}_n$ versus representations of $\mathrm{Spin}_n$ $\DeclareMathOperator\SO{SO}\DeclareMathOperator\Spin{Spin}$Since $\Spin_n$ is a compact simply connected simple Lie group, its irreducible representations are equivalent to the irreducible representations of its Lie algebra $\mathfrak{so}_n$. $\Spin_n$ is the universal cover of $\SO_n$, another compact simple Lie group. But $\SO_n$ is not simply connected so it has "fewer irreducible representations". In this answer
to Shared representation of $SU(2)$ and $SO(3)$, it is discussed which representations of $\Spin_3 = \operatorname{SU}_2$ "descend" to representations of $\SO_3$. In short, it is those "labelled by even powers of the fundamental representation" of $\mathfrak{sl}_2$.
Question: How does this look for higher $\SO_n$? What are the integral positive weights of $\mathfrak{so}_n$ that give well-defined representations of $\SO_n$?
Guess Based on some basic calculations, and by looking at the question Non-faithful irreducible representations of simple Lie groups, I would guess that the "descendable" representations are those with weight
$$
\lambda = \sum_{i=1}^l a_i \varpi_i, ~~ \textrm{ satisfies } a_l \in 2\mathbb{N}_0,
$$
when $l = \operatorname{rank}(\mathfrak{so}_n)$.
 A: Your answer is correct for $n$ odd. For $n$ even you instead need the sum of the coefficients of the last two weights $a_{l-1} + a_{l}$ to be even. Either way you are effectively asking for the non-trivial element of the centre of $\mathrm{Spin}(n)$ to act trivially. Since it acts as $-1$ on the spin representation (or on each half spin representation for $n$ even), intuitively we need an "even number of them".
For example let $n$ be odd and $\Delta$ the spin representation of $\mathrm{Spin}(n)$. $\Delta$ does not have a representation of $SO(n)$ but the tensor product $\Delta \otimes \Delta$ does. As do its subrepresentations $\bigwedge^2 \Delta$, $S^2\Delta$.
A: This is maybe easier to see if you think about weights in terms of the usual Cartan of $\mathfrak{so}_{n}$.  Choose a basis $e_i$ for $\mathbb{C}^{n}$ and the form so that the dual basis is $e_i^*=e_{n+1-i}$.  The Cartan is then diagonal matrices so that $a_{n+1-i}=-a_i$, i.e $\mathbb{C}^m$ where $m=\lfloor \frac{n}{2}\rfloor$ and the dual Cartan can be identified with this space as well by the usual inner product.  Under this identification, the weight lattice of $\mathrm{SO}_{n}$ is just $\mathbb{Z}^{m}$, so the interesting question is how the weight space of $\mathrm{Spin}_n$ is different.  I believe it's always the lattice $\Lambda$ of vectors with all entries from $\mathbb{Z}$ or all from $\mathbb{Z}+\frac{1}{2}$: most of the simple coroots are of the form $\epsilon_i-\epsilon_{i+1}$, and thus say that consecutive entries have integer difference, and the last simple root is either $2\epsilon_{m}$ or $\epsilon_{m}+\epsilon_{m-1}$, both of which are satisfied if and only if the element is in $\Lambda$.
