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 $A$)



Abelian variety with CM defined over real numbers
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.


