Consider a holomorphic vector bundle $\pi:E\rightarrow X$ of complex rank $m$ over a Kaehler manifold $X$. Can we find a Thom form $\Theta$ of $E$ such that as a form on the complex manifold $E$, it is a $(m,m)$-form, i.e. a form of pure bi-degree ? Maybe we do not need $X$ to be Kaehler, but just complex. How about the Mathai-Quillen Thom form?
It seems that for the Euler class of $E$, by using the Chern-Weil theory, we can get a representatives of pure bidegree. By taking a hermitian metric on $E$ and the corresponding hermitian connection $\nabla$, which is of $(1,0)$-type, the cuvature matrix $\Omega$ is a matrix of $(1,1)$-form on $X$. The Euler class can be represented by $P_m(\Omega)$, where $P_m$ is the degree $m$ symmetric polynomial. And by the explicit expression we see that $P_m(\Omega)$ is a $(m,m)$-form on $X$.