3** $a_j \neq 0$ for all $j$.

By , $a_j^2 \c{j}{i} = -2a_ja_i$ for all $i<j$, hence, $\c{j}{i} \neq 0$ for all $i,j$. Note that $B_2$ is a Hirzebruch surface. Since the diffeomorphism type of a Hirzebruch surface $B_2$ is determined by the parity of $\c{2}{1}$, if $\c{2}{1}$ is even then $B_2 $ is diffeomorphic to ${\mathbb{C}P}^1 \times {\mathbb{C}P}^1$. Hence $B_n$ is diffeomorphic to $B_n'$ with $\c{2}{1}'=0$. Thus we may assume that $\c{2}{1} =0$ for simplicity. But then, by either $a_1$ or $a_2$ is zero, which contradicts to the assumption of Case 3. Therefore we may assume that $\c{2}{1}$ is odd; in fact we may assume that $\c{2}{1}=1$ because the diffeomorphism type of $B_2$ is determined by the parity of $\c{2}{1}$. Since $\c{j}{i} \neq 0$ and $a_j \neq 0$ for all $j$ and $i$, by , $\c{j}{i}a_j = -2a_i$ for all $i<j$. Hence $a_2 = -2 a_1$. Moreover since $\c{3}{1}a_3 = -2a_1$ we have $\c{3}{2}a_3 = -2a_2 = 4a_1$, hence, $\c{3}{2}=-2\c{3}{1}$. We claim that $B_3$ with $$\left(
\begin{array}{ccc}
0 & 1 &c \\
 & 0 & -2c \\
 &  & 0 \\
\end{array}\right)$$ is diffeomorphic to $B_3'$ with $$\left(
\begin{array}{ccc}
0 & 1 & c \\
 & 0 & 0 