If $G^\vee = SL(n,\mathbb{C})$ and $V = \mathbb{C}^n$, then the Langlands dual of $G^\vee$ is $G = PGL(n,\mathbb{C})$. Denote by $T$ and $T^\vee$ maximal tori in $G$ and $G^\vee$ respectively. The highest weight of $V$ (as a representation of $G^\vee$) is a map
$$\rho^\vee \colon T^\vee \to U(1)$$
which induces a map
$$\rho \colon U(1) \to T.$$
Then $\rho$ can be seen as a clutching function for a holomorphic $G_\mathbb{C}$ bundle $E$ over $\mathbb{C}P^1$. It turns out that there is a moduli space of such bundles $E$, which we will denote by $\mathcal{N}(\rho)$, and we will denote its compactification by $\bar{\mathcal{N}}(\rho)$. In the case of our baby example above, with $G^\vee = SL(n,\mathbb{C})$ and $V = \mathbb{C}^n$ its standard representation, $\mathcal{N}(\rho)$ is actually $\mathbb{C}P^{n-1}$. Part of the geometric correspondence is that the cohomology ring of $\bar{\mathcal{N}}(\rho)$ can be identified with $V$.
This is what I have learned from reading part of a talk by Witten given at Atiyah's 80-th birthday, which can be found for instance at Witten's talk (pdf). I am not sure if the identification above is known or if it is (still) conjectural.
My questions are the following.
- If $G^\vee = SO(2m+1)$ and $V = \mathbb{C}^{2m+1}$ is its standard representation, what would $\bar{\mathcal{N}}(\rho)$ be in that case.
- Similar question if $G^\vee = Sp(m)$ and $V = \mathbb{C}^{2m}$ is its standard representation.
- Similar question if $G^\vee = SO(2m)$ and $V = \mathbb{C}^{2m}$ is its standard representation.
My questions are probably elementary to experts. Thank you!
If I were to guess, I would say that for $G^\vee = SO(2m)$, $\bar{\mathcal{N}}(\rho) = Q^{2m-2} \subset \mathbb{C}P^{2m-1}$, where $Q^{2m-2}$ is a smooth complex quadric. This has the topology of the oriented grassmannian of (oriented) $2$-planes in $\mathbb{R}^{2m}$.
Edit 1: to support my "wild" guess, note that the odd-indexed Betti numbers of $Q^{2m-2}$ are $0$ and all the even-indexed Betti numbers are $1$ except for the middle Betti number $b_{2m-2}$, which is $2$. So the sum of the Betti numbers is thus
$$b_0 + b_2 + \cdots + b_{2m-2} + \cdots + b_{4m-6} + b_{4m-4} = 2(m-1) + 2 = 2m = \operatorname{dim}(\mathbb{C}^{2m}).$$
So at least the dimension of my wild guess is the correct one.
