Symplectic group and Quaternion Inner product

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:

$$(\alpha_1 \bar\alpha'{_1}-\beta_1 \bar\beta'{_1})+ ...+(\alpha_n \bar\alpha'{_n}-\beta_n \bar\beta'{_n})$$

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:

$$q_i=\begin{bmatrix} \alpha_i & -\beta_i \\ \bar\beta_i & \bar\alpha_i \end{bmatrix}$$

The invariance of the quaternion inner product:

$$q_1 \bar{q}'{_1}+...+q_n \bar{q}'{_n}$$ is therefore equivalent to the invariance of the matrix product:

$$\left( \begin{array}{ccc} \alpha_1 & -\beta_1\\ \bar{\beta_1} & \bar{\alpha_1}\end{array} \right)\ \left( \begin{array}{ccc}\ \bar{\alpha'_1} & \beta'_1\\ -\bar{\beta'_1} & \alpha'_1\end{array} \right)+...+ \left( \begin{array}{ccc} \alpha_n & -\beta_n\\ \bar{\beta_n} & \bar{\alpha_n}\end{array} \right)\ \left( \begin{array}{ccc}\ \bar{\alpha'_n} & \beta'_n\\ -\bar{\beta'_n} & \alpha'_n\end{array} \right)$$

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:

$$(\alpha_1 \beta'{_1}-\beta_1 \alpha'{_1})+ ...+(\alpha_n \beta'{_n}-\beta_n \alpha'{_n})$$

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?

• Notice that: $$q_1 \bar{q}'{_1}= \left( \begin{array}{ccc} \alpha_1 & -\beta_1\\ \bar{\beta_1} & \bar{\alpha_1}\end{array} \right)\ \overline{\left( \begin{array}{ccc}\ \alpha'_1 & -\beta'_1\\ \bar{\beta'_1} & \bar{\alpha'_1}\end{array} \right)}=\left( \begin{array}{ccc} \alpha_1 & -\beta_1\\ \bar{\beta_1} & \bar{\alpha_1}\end{array} \right)\ \left( \begin{array}{ccc}\ \bar{\alpha}'_1 & -\bar{\beta}'_1\\ \beta'_1 & \alpha'_1\end{array} \right)$$ however I do not know if I understand correctly your mean by $\bar{A}$. – Sepideh Bakhoda Aug 2 '15 at 8:25
• I think $\bar{A}$ means quaternion conjugate(not just complex conjugate), because it's a quaternion inner product , notice that if we had a one dimensional space, i.e. $H$, then $q_1 \bar{q_1}$ should give the quaternion norm(which is$\alpha \bar{\alpha}+\beta \bar{\beta}$) on diagonal elements and zero on off-diagonal elements, the $\bar$ over a complex number is complex conjugate and a $\bar$ over a quaternion is quaternion conjugate. – user258300 Aug 2 '15 at 16:41
• The $\bar{q}$ is defined on page 58 of "Naive Lie Theory" in the same way that i used it here, – user258300 Aug 2 '15 at 16:49