Hodge loci

• Hodge loci
• Review of algebraic groups

Two main goals:

1. cover content on Hodge loci from Voisin,
2. review a bit on algebraic groups (probably basics).

Hodge loci#

Recall the setup: $\pi \colon X \to B$ a smooth projective morphism between varieties over $\mathbb{C}$. We have a local system $$ H^i_{\mathbb{C}} = R^i \pi_\ast \mathbb{C}, $$ and the holomorphic vector bundle $\mathcal{H}$ is obtained by tensoring with $\mathcal{O}_ B$. We also have Griffiths transversality.

For a point $0 \in B$, one can pick a contractible open neighborhood $U \ni 0$ and (canonically) trivialize the local system. This allows us to extend cohomology classes uniquely to sections on $U$. Given $\alpha \in H^{2k}(X_0, \mathbb{Q})$, we define $U_{\alpha}$ to be the set of $t$ such that $\alpha$ stays Hodge, equivalently, such that $\tilde{\alpha}_ t \in F^k H^{2k}_ \mathbb{Q}$.

The Hodge loci are the union of the $U_{\alpha}$ which are not equal to $U$.

Lemma 1. Each $U_{\alpha}$ is an analytic subset of $U$, which can be defined locally by most at most $h^{k-1, k+1}$ many equations.

Due to the nature of the filtration, this number is the rank of $F^{k-1} / F^k$.

This result has the following cool corollary: For $t \in U_{\alpha}$, consider the Zariski tangent space at $t$. This is $$ T_{U_{\alpha}, t} = \lbrace v \in T_{u, t}: \alpha f(v) = 0 \text{ for all } f \in I_{U_{\alpha}} \rbrace. $$ We have a map $$ {}^t \overline{\nabla} \colon F^k \mathcal{H}_ t / F^{k+1} / \mathcal{H}_ t \to F^{k-1} \mathcal{H}_ t / F^k \mathcal{H}_ t \otimes \Omega_{U, t}. $$

Lemma 2. The tangent space $T_{U_{\alpha}, t}$ is the kernel of ${}^t \overline{\nabla}(\alpha)$.

Corollary 3. The analytic subset $U_{\alpha}$ is smooth of codimension $h^{k-1, k+1}$, in case the map ${}^t \overline{\nabla}(\alpha)$ is surjective.

Review of algebraic groups#

Definition 4. An algebraic group $G$ is unipotent if for any representation $r \colon G \to \GL(V)$, one of the following equivalent conditions holds:

• $r(G)$ lies in the unipotent upper triangular matrices relative to some basis.
• $r(G)$ fixes a nonzero vector.

Proposition 5. Unipotence is preserved by quotients and extensions.

If a property of groups is preserved by quotients and extensions, then there is a unique maximal normal subgroup with this property. Otherwise, if $H, N$ both have the property, then we can write $$ 1 \to N \to HN \to H / (H \cap N) \to 1. $$ And this would imply $HN$ also has the property and is normal.

Therefore there is a unique maximal normal connected unipotent subgroup $R_u(G)$, which we call the unipotent radical of $G$.

Definition 6. $G$ is reductive if $R_u(G_{\bar{k}}) = e$.

By design, we have that $G / R_u(G)$ is reductive—the product of two subgroups, one normal and unipotent, the other merely unipotent, is unipotent.

In addition to unipotence, solvability is preserved by quotient and extension. Thus, there is also a solvable radical.

Definition 7. $G$ is semisimple if $R(G)$ (the solvable radical) is trivial.

Maybe say $k = \bar{k}$ from now on.

Definition 8. A subgroup $P$ of $G$ is parabolic if one of the following equivalent conditions holds:

• $G / P$ is complete.
• $G / P$ is projective. [quasiprojective is automatic]
• $P$ contains a Borel (maximal solvable) subgroup.

The interesting direction is the Borel fixed point theorem: Suppose that $G / P$ is complete. Then any Borel $B$ acts on the left (the quotient is on the right), and this therefore has a fixed point, which corresponds to a conjugate of $B$ contained in $P$.

Proposition 9. If $G$ is defined over $\mathbb{Q}$, then there is a maximal reductive subgroup $L$, such that $$ G = L \cdot R_u(G) = R_u(G) L = L \ltimes R_u(G). $$

This is called the Levi decomposition, and $L$ is the Levi subgroup.

For $\mathbb{Q}$, all maximal $\mathbb{Q}$-split tori are conjugate by a point of $G(\mathbb{Q})$ (true for any field). This gives rise to the rank of a group over $\mathbb{Q}$, which is the rank of a maximal $\mathbb{Q}$-split torus.

Definition 10. We say that $G$ is anisotropic over $\mathbb{Q}$ if $r_{\mathbb{Q}}(G) = 0$.

For the rest of the talk, assume that $G$ is a connected reductive linear algebraic group such that its center is anisotropic over $\mathbb{Q}$. This is true if $G$ is semisimple, because then there are no normal solvable groups (thus, no tori in the center).

Let $S$ be a maximal $\mathbb{Q}$-split torus on $G$. Then the adjoint action of $S$ on the Lie algebra $\mathfrak{g}$ gives a decomposition $$ \mathfrak{g} = \mathfrak{g}_ 0 \oplus \sum_{\alpha \in \Phi(G, S)} g_{\alpha}, $$ where $$ g_{\alpha} = {X \in \mathfrak{g}: \mathrm{Ad}(s) X = \alpha(s) X \text{ for all } s \in S }. $$ Fixing an ordering on $\Phi(G, S)$, we have the set $\Phi^+(G, S)$ of positive roots. Define $\mathfrak{n}$ to be the Lie algebra spanned by the positive root spaces, and let $N$ be the corresponding subgroup. This Lie algebra is nilpotent because only finitely many $g_{\alpha}$ are nonzero. Thus, and since there is a well-defined exponential map (?), $N$ is unipotent.

Let $Z(S)$ denote the centralizer of $S$. Then $N Z(S)$ is a group because $Z(S)$ normalizes $N$, and in fact this is a minimal parabolic subgroup. In fact, any minimal parabolic subgroup arises in this fashion relative to some choice of positive roots.

Let $\Delta$ denote the set of simple roots. For $I \subset \Delta$, let $S_I$ be the identity component $$ \lbrace g \in S, \alpha(g) = 1 \text{ for all } \alpha \in I \rbrace. $$ We have $P_I = N Z(S_I)$, for $S_I$ as above. Note that $S_{\emptyset} = S$. All contain $S$, since $Z(S_I) \supset Z(S)$.