some $k<n$.

We may assume that $a_i \neq 0$ for all $i>k$. By , $a_j^2 \c{j}{i} = - 2a_j a_i$ for all $i<j$. Hence, $a_{k+\ell}^2 \c{k+\ell}{k} = -2a_{k+\ell}a_k$ for all $0< \ell \leq n-k$. Since $a_{k+\ell} \neq 0$ and $a_k=0$, $\c{k+\ell}{k}=0$ for all $\ell$. Thus, $\c{k+1}{k} = \c{k+2}{k} = \cdots = \c{n}{k} = 0$. Hence $B_n$ is diffeomorphic to a Bott manifold $B_n'$ with $\Lambda'$ in Corollary \[corollary:columns with zeros\] as the associated matrix.

Let $\{x_1, \ldots, x_n\}$ and $\{y_1, \ldots, y_n \}$ be ordered generator sets of $H^\ast(B_n)$ and $H^\ast(B_n')$ respectively as in . Let $\rho : B_n' \rightarrow B_n$ be the diffeomorphism as indicated in the proof of Corollary \[corollary:columns with zeros\]. Then we can see that $$\begin{aligned}
    \rho^\ast(x_1) &=y_1 \\
    \vdots &\\
    \rho^\ast(x_{k-1}) &= y_{k-1} \\
    \rho^\ast(x_k) &= y_n \\
    \rho^\ast(x_{k+1}) &=y_k \\
    \vdots &\\
    \rho^\ast(x_n) &=y_{n-1}.\end{aligned}$$ Therefore, $\rho^\ast(\alpha) = a_1 y_1 +
\cdots + a_{k-1}y_{k-1} + a_{k+1} y_k + \cdots + a_{n} y_{n-1} +
a_k y_n$. Since $c(\gamma^\alpha \oplus \gamma^{-\alpha})=0$ in $H^\ast(B_n)$, $c(\rho^\ast(\gamma^\alpha \oplus \gamma^{-\alpha}))=0$ in $H^\ast(B_n')$. Since $a_k=0$ from the assumption, we are in **Case 1** for $B_n'$. Therefore $\rho^\ast(\gamma^\alpha \oplus \gamma^{-\alpha})$ is trivial on $B_n'$, and so is $\gamma^\alpha \oplus \gamma^{-\alpha}$ on $B_n$.

**Case 