Abelian variety with CM defined over real numbers

Question

Is there an abelian variety $A/\mathbb R$ of dimension $n$ such that $End_{\mathbb R}(A)\otimes \mathbb Q$ contains a field $K$ of degree $[K:\mathbb Q]=2n$? ($End_{\mathbb R}(A)$ is the ring of $\mathbb R$-endomorphisms of $A$)

Answer 1

No.

Assume for contradiction that such an $A$ exists. First look at the singular cohomology $H^1(A_{\mathbb C}, \mathbb Q)$, which admits an action of $K$ and so is a $K$-vector space. It has dimension $2n$ over $\mathbb Q$ and so is a 1-dimensional $K$-vector space.

Tensoring with $\mathbb C$, we see that $H^1(A_{\mathbb C}, \mathbb C)$, as a vector space with an action of $K$, is a sum of $2n$ eigenspaces of $K$ associated to the $2n$ different embeddings $K \to \mathbb C$.

Now by Hodge theory, $H^1(A_{\mathbb C}, \mathbb C) = H^1(A_{\mathbb C}, \mathcal O_A) + H^0(A_{\mathbb C}, \Omega^1_A)$ with the two summands complex conjugates of each other. So for each eigenvector appearing associated to an embedding appears in $H^1(A_{\mathbb C}, \mathcal O_A)$, the eigenvector associated to the complex conjugate embedding appears in $H^0(A_{\mathbb C}, \Omega^1_A)$, and thus, because the eigenspace is 1-dimensional so there is only one eigenvector up to scaling, no eigenvector associated to the complex conjugate embedding appears in $H^1(A_{\mathbb C}, \mathcal O_A)$.

So $H^1(A_{\mathbb C}, \mathcal O_A)$ is not isomorphic to its complex conjugate as a complex vector space with an action of $K$.

But $H^1(A_{\mathbb C}, \mathcal O_A) = H^1(A_{\mathbb R}, \mathcal O_A) \otimes_{\mathbb R} \mathbb C$ and thus is isomorphic to its complex conjugate. (Here we use that the endomorphisms in $K$ are defined over $\mathbb R$ and thus act on $ H^1(A_{\mathbb R}, \mathcal O_A) $.)

This is a contradiction, so no such $A$ exists.

Answer 2

(Essentially the same argument as the one given by Will Sawin, but perhaps a bit simpler. Further clarification included thanks to comment by Wojowu.)

If $A$ is an abelian variety over a field $k\supset\mathbb{Q}$, then the tangent space $T_0(A)$ at identity is a module over $\mathrm{End}_{k}(A)\otimes\mathbb{Q}$.

Now, if the latter contains a field $K$, then $T_0(A)$ has to have dimension at least 1 over $K$. On the other hand $T_0(A)$ has dimension $\dim(A)$ over $k$. Thus $[K:\mathbb{Q}]\leq\dim(A)$.

Added: Alternate explanation of above.

Consider $A(\mathbb{R})$ as a Lie group with connected component $A(\mathbb{R})_0$. The exponential map $T_0(A)\to A(\mathbb{R})_0$ is the universal covering of a compact torus of real dimension $n=\dim(A)$. It is clear that elements of $\mathrm{End}_{\mathbb{R}}(A)$ lift to this cover; Let $K$ be a subfield of $\mathrm{End}_{\mathbb{R}}(A)$. Note that $\mathcal{O}_K$ is a domain and the covering group (which is $\mathbb{Z}^{n}$) is a module over $\mathcal{O}_K$. Thus the rank of $\mathcal{O}_k$ as a $\mathbb{Z}$ module is at most $n$.