TITLE: Symplectic group and Quaternion Inner product
QUESTION [7 upvotes]: I have a problem understanding a passage from "Naive Lie theory"(by John Stillwell), here is the passage from section $3.9$ ,page $71$:

The idea of treating orthogonal, unitary, and symplectic groups uniformly
  as generalized isometry groups of the spaces $R^n$, $C^n$, and $H^n$ seems to
  be due to Chevalley [1946]. Before the appearance of Chevalley’s book,
  the symplectic group $Sp(n)$ was generally viewed as the group of unitary
  transformations of $C^{2n}$ that preserve the symplectic form:
\begin{equation}
(\alpha_1 \bar\alpha'{_1}-\beta_1 \bar\beta'{_1})+ ...+(\alpha_n \bar\alpha'{_n}-\beta_n \bar\beta'{_n}) 
\end{equation}
where $(α_1,β_1, . . . ,α_n,β_n)$ is the typical element of $C^{2n}$. This element corresponds to the element $(q_1, . . . ,q_n)$ of $H^n$, where:
\begin{equation}
q_i=\begin{bmatrix}
    \alpha_i & -\beta_i  \\
    \bar\beta_i & \bar\alpha_i 
   \end{bmatrix}
\end{equation}
The invariance of the quaternion inner product:
\begin{equation}
q_1 \bar{q}'{_1}+...+q_n \bar{q}'{_n}
\end{equation}
  is therefore equivalent to the invariance of the matrix product:
\begin{equation}
 \left( \begin{array}{ccc}
\alpha_1 & -\beta_1\\
\bar{\beta_1} & \bar{\alpha_1}\end{array} \right)\
 \left( \begin{array}{ccc}\
\bar{\alpha'_1} & \beta'_1\\
-\bar{\beta'_1} & \alpha'_1\end{array} \right)+...+
\left( \begin{array}{ccc}
\alpha_n & -\beta_n\\
\bar{\beta_n} & \bar{\alpha_n}\end{array} \right)\
 \left( \begin{array}{ccc}\
\bar{\alpha'_n} & \beta'_n\\
-\bar{\beta'_n} & \alpha'_n\end{array} \right)
\end{equation}
which turns out to be equivalent to the invariance of the symplectic form.

My problem is that i couldn't see how the invariance of matrix product results in invariance of symplectic form mentioned by the author. I have checked the Chevalley book but i didn't find the specific symplectic form used by Stillwell.
Chevalley and other authors have used a symplectic form that doesn't include complex conjugate. This free complex conjugate symplectic form makes sense because it is the off-diagonal element of matrix product above, .i.e:
\begin{equation}
(\alpha_1 \beta'{_1}-\beta_1 \alpha'{_1})+ ...+(\alpha_n \beta'{_n}-\beta_n \alpha'{_n}) 
\end{equation}
Other authors also use symplectic form without complex conjugate, for example
Modern Geometry
Modern Geometric Structures and Fields
But i don't know how to see the symplectic form defined by Stillwell is remained invariant.
What am i missing?

REPLY [2 votes]: Ok, It seems that my doubt was correct. The from without conjugates is invariant and it is a mistake in book. Probably will be corrected in new editions. It is weird to answer your own question, but since this question doesn't need to be open anymore, I do it :),