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