Automorphisms of manifold vs automorphisms of Hodge structure

Question

Let $X$ be a compact Kahler manifold. Let $H^k = H^k_{prim} (X)$ be primitive cohomology of $X$ considered as an object in category of of polarized Hodge structures. We have a map \begin{equation*} h_k: Aut (X) \rightarrow Aut(H^k) \end{equation*} Question 1 Does $h_k$ have to be surjective for $k \neq 0, n$?

For boundary cases it obviously fails. You can not act non-trivially on $H^0$ via manifold automorphism. But $-1$ is a automorphism of Hodge structure.

Question 2 Do you know about some analogous statements which are theorems? Could you give source for them?

Answer 1

The homomorphism $h_k: Aut(X) \to Aut(H^k_{prim}(X,\mathbb C))$ is extremely interesting. Let me follow your notation and write $H^k$ for $H^k_{prim}(X,\mathbb C)$. Also, I will content myself to algebraic varieties in the discussion below.

Let me answer your first question. For a large class of varieties, the answer to your question is no. Indeed, the ``general" variety has no non-trivial automorphisms to start with. On the other hand, as you said, there is always the automorphism $-1$ of $H^k$. For example, the general cubic threefold $X$ has no non-trivial automorphisms (so $Aut(X) = \{id_X\}$), whereas $H^3$ is a $10$-dimensional polarized Hodge structure. Thus $h_3$ is not surjective in this case.

As you see, the surjectivity of $h_k$ fails quite often. It does hold in the case of complex tori as pointed out below in the comments.

To answer question 2 partially, let me point out that there is much work done on the kernel of $h_k$. These results (and the methods employed in their proof) might help you find the sort of statement you're looking for. Indeed, if $X$ is a curve of genus at least two, a principally polarized abelian variety, a hypersurface of degree at least three (and dimension at least two), a variety with very ample canonical bundle, or a hyperkahler variety with certain properties, then $h_k$ is injective; see Section 2.5 (especially Prop. 2.15) in http://arxiv.org/abs/1505.02249 .

Answer 2

Surprisingly, I found counterexample even for complex tori. Denote the elliptic curve $\mathbb{C} / < 1, \lambda > $ by $E_{ \lambda }$. Consider complex tori $E_{\lambda_1} \times E_{\lambda_2}$. Moreover suppose that

• $\lambda_1$ and $\lambda_2$ are generic i.e. automorphism group $E_{\lambda_1}$ and $E_{\lambda_2}$ is isomorphic to $\mathbb{Z}/ 2 \mathbb{Z}$.
• $\lambda_1$ and $\lambda_2$ are generic with respect to each other. It means that $\lambda_1 \notin \mathbb{Q} (\lambda_2)$ and $\lambda_2 \notin \mathbb{Q} (\lambda_1)$.

Lemma $Aut(E_{\lambda_1} \times E_{\lambda_2}) = \mathbb{Z}/ 2 \mathbb{Z} \times \mathbb{Z}/ 2 \mathbb{Z}$.

Let me use lemma to prove that for $E_{\lambda_1} \times E_{\lambda_2}$ map $h_2$ is not surjection. Notice that

$$H^2 (E_{\lambda_1} \times E_{\lambda_2}) = \Lambda^2 \big( H^1 (E_{\lambda_1}) \oplus H^1 (E_{\lambda_2}) \big)=$$ $$=\Lambda^2 \big( H^1 (E_{\lambda_1})) \oplus \Lambda^2 \big( H^1 (E_{\lambda_2}) \big) \oplus H^1 (E_{\lambda_1}) \otimes H^1 (E_{\lambda_2}) $$

$\Lambda^2 \big( H^1 (E_{\lambda_1})) \oplus \Lambda^2 \big( H^1 (E_{\lambda_2}))$ has type $\begin{matrix} 0 & 2 & 0 \end{matrix}$. If you pass to primitive cohomology you have to take quotient by Kahler class $\omega \in \Lambda^2 \big( H^1 (E_{\lambda_1})) \oplus \Lambda^2 \big( H^1 (E_{\lambda_2}))$.

Easy to see from Lemma that $Aut(E_{\lambda_1} \times E_{\lambda_2})$ acts on $\Lambda^2 \big( H^1 (E_{\lambda_1})) \oplus \Lambda^2 \big( H^1 (E_{\lambda_2}))$ trivially. But this substructure has $-1$ automorphism which can be continued on hole $H^2$ by acting identity on another summand. So, I contacted automorphism of $H^2$ wich do not come from $Aut(E_{\lambda_1} \times E_{\lambda_2})$.

Proof of lemma.

Each automorphism of complex tori $\mathbb{C}^2 / \Lambda$ can be lifted to automorphism of universal covering $\mathbb{C}^2$, which maps lattice $\Lambda$ to itself. Let prove that in our case this automorphism is given by diagonal matrix. Suppose that $(1, 0) \in \Lambda$ maps to $(x, y) \in \Lambda$ and $y \neq 0$. Then $(\lambda_1, 0) \in \Lambda$ maps to $(\lambda_1 x, \lambda_1 y ) \in \Lambda$.

$$(x, y) \in \Lambda \Rightarrow y \in \mathbb{Q}(\lambda_2)$$ $$(\lambda_1 x, \lambda_1 y ) \in \Lambda \Rightarrow \lambda_1 y \in \mathbb{Q}(\lambda_2)$$

Then $\lambda_1 \in \mathbb{Q}(\lambda_2)$. It contradicts assumption of generic $\lambda_1$ and $\lambda_2$ with respect to each other.