Let $X$ be a smooth projective curve of genus $\geq 1$ over $\mathbb{C}$, $H^\cdot=H^\cdot(X)$, and $K$ be the kernel of cup product $\cup: H^1\otimes H^1\rightarrow H^2$. Consider the extension of Hodge structures
\begin{equation}
0\longrightarrow K\longrightarrow H^1\otimes H^1\stackrel{\cup}{\longrightarrow} H^2\cong\mathbb{Z}(-1)\longrightarrow 0.
\end{equation}
This of course splits over $\mathbb{R}$. Does it split over $\mathbb{Q}$ as well? Equivalently, is there a Hodge class $\xi$ in $H^1\otimes H^1$ such that 
\begin{equation}
\int\limits\_X\cup(\xi)\neq 0 ~?
\end{equation}


I tried taking $\xi$ to be the $H^1\otimes H^1$ Kunneth component of the class of the diagonal embedding $\Delta(X)$ of $X$ in $X^2$, but I don't seem to be able to show
$$\int\limits\_{\Delta(X)}\xi\neq 0.$$



Does this extension of Hodge structures split over $\mathbb{Q}$?
Venkataramana gave an Hodge-theory argument. There is also a proof by intersection theory that your class $\xi$ gives a splitting.


Observe that the other two Kunneth components of the diagonal are a horizontal and vertical fiber. So $\xi$ is the diagonal minus a horizontal fiber and minus a vertical fiber. Now you're integrating that over the diagonal, which is the same as intersecting with the diagonal. $$\Delta(X) \cdot \Delta(X) =2-2g$$ The intersection of the diagonal with a horizontal or vertical line is $1$, from $1$ transverse point. So all in all $$\Delta(X) \cdot \xi = -2g$$$g \geq 1$, so that is nonzero.


