Duals of the spinor representations of $\frak{so}_{2n}$

Question

For the $D_n$-series simple Lie algebra $\frak{so}_{2n}$ a curious phenomenon occurs for the fundamental representations corresponding to the spinor nodes of the Dynkin diagram, which is to say the spinor representations $V_{\pi_{n-1}}$, and $V_{\pi_{n}}$: In the case where $n$ is even both $V_{\pi_{n-1}}$ and $V_{\pi_{n}}$ are self-dual representations, which is to say $$ V_{\pi_{n-1}}^{\vee} \simeq V_{\pi_{n-1}}, ~~~~~ V_{\pi_{n}}^{\vee} \simeq V_{\pi_{n}}. $$ However, in the odd case, the two representations are dual to each other, which is to say $$ V_{\pi_{n-1}}^{\vee} \simeq V_{\pi_{n}}. $$ Is there a conceptual reason why this occurs, why the even and odd case behave differently? Why does the action of the longest element of the Weyl group behave differently in each case.

Answer 1

Here is one explanation, although I am not sure if this is the conceptual explanation you are looking for.

Let $E = \mathbb{R}^n$ with orthonormal basis $\varepsilon_1, \ldots, \varepsilon_l$. You can realize a root system of type $D_n$ as $\Phi = \{ \pm (\varepsilon_i \pm \varepsilon_j) : i \neq j \}$.

The Weyl group $W$ is the group of permutations and sign changes on $\varepsilon_1$, $\ldots$, $\varepsilon_n$ involving only an even number of sign changes. That is, for $\sigma \in W$ you have $\sigma(\varepsilon_i) = c_i \varepsilon_{\pi(i)}$ with $\pi \in Sym_n$, $c_i = \pm 1$ and $c_1c_2 \cdots c_n = 1$.

From this you already see that $-1 \in W$ if and only if $n$ is even.

Also, the weights occurring in the two irreducible spin representations are precisely those of the form $$\lambda_I = \frac{1}{2} (\sum_{i \in I} \varepsilon_i - \sum_{i \not\in I} \varepsilon_i)$$ for a subset $I \subseteq \{1, \ldots, n\}$. Clearly $\lambda_I$ and $\lambda_{J}$ are conjugate under the Weyl group if and only if $|I| = |J| \mod{2}$.

One of the spin representations has highest weight $\lambda = \frac{1}{2}(\sum_{i = 1}^n \varepsilon_i)$ and the other one has highest weight $\mu = \frac{1}{2}(\sum_{i = 1}^{n-1} \varepsilon_i - \varepsilon_n)$. So $\lambda$ and $-\mu$ are conjugate under the Weyl group if and only if $-\mu$ involves an even number of minus signs, equivalently $n$ is odd.

Answer 2

I don't think there is a really satisfying "conceptual" explanation of why the longest element is $-1$ times the Dynkin diagram automorphism for $D_n$ with $n$ odd and is $-1$ times the identity for $D_n$ with $n$ even. But see this answer of Allen Knutson for a "mnemonic": https://mathoverflow.net/a/82168/25028