Let $W = \mathbb{F}_{2^{m + 1}} \oplus \mathbb{F}_{2^{m + 1}}$ be a $2(m+1)$-dimensional vector space over $\mathbb{Z}_2$ equipped with a symplectic form $ \langle \cdot , \cdot \rangle : W \times W \rightarrow \mathbb{Z}_2$ defined by the field trace as follows: for any elements $\vec{w} = (\alpha; \beta), \vec{w}' = (\alpha'; \beta') \in W$, \begin{align} \langle \vec{w}, \vec{w}' \rangle = Tr _{\mathbb{F}_{2^{m + 1}}/\mathbb{Z}_2}(\alpha \beta' - \alpha' \beta). \end{align} Then, $W$ possesses a symplectic basis $\mathcal{B} = \{ \vec{e}_1, \ldots, \vec{e}_{m+1}, \vec{f}_1,\ldots, \vec{f}_{m+1} \}$ where $\{\vec{e}_1, \ldots, \vec{e}_{m+1} \}$ and $\{\vec{f}_1,\ldots, \vec{f}_{m+1} \}$ span the first and the second factor, respectively, such that \begin{align} \langle \vec{w}, \vec{w}' \rangle = \sum_{i = 1}^{m+1}(a_ib'_i - a'_ib_i), \end{align} where $\vec{w} = \sum_{i = 1}^{m+1}(a_i\vec{e}_i + b_i\vec{f}_i)$ and $\vec{w}' = \sum_{i = 1}^{m+1}(a'_i\vec{e}_i + b'_i\vec{f}_i)$.

Next, set a quadratic form \begin{align} q(\vec{w}) = \sum_{i = 1}^{m + 1}a_i b_i + (a_1 + b_1). \end{align} Then \begin{align} \langle \vec{w}, \vec{w}' \rangle = q(\vec{w}) + q(\vec{w}')+q(\vec{w} + \vec{w}'). \end{align} From this setup, I know that $|\{ \vec{w} \in W : q(\vec{w}) = 1 \}| = 2^m(2^{m + 1} + 1)$ and $| \{ (0, \lambda) \in W : \lambda \neq 0 \text{ and } q((0, \lambda)) = 1 \} | = 2^m$.

For each $\alpha \in \mathbb{F}_{2^{m + 1}}$, let $A_\alpha = \{ (\lambda, \alpha \lambda) \in W : \lambda \neq 0 \text{ and } q((\lambda, \alpha \lambda)) = 1 \}$.

How do we show that for any $\alpha_1, \alpha_2 \in \mathbb{F}_{2^{m + 1}}$, the cardinalities of $A_{\alpha_1}$ and of $A_{\alpha_2}$ are the same, i.e. $|A_{\alpha_1}| = |A_{\alpha_2}|$? In particular, they should equal $2^m$.


Based on the way you have your bilinear and quadratic forms set up, I think you can define $q((x,y)) = \mathrm{Tr}(xy)$ for $(x,y) \in W$ (where $\mathrm{Tr}: \mathbb{F}_{2^{m+1}} \to \mathbb{F}_{2}$ is the trace function). (This isn't quite right; it defines a hyperbolic quadric while yours defines an elliptic...)

Given this, $|A_{\alpha}| = |\lambda \in \mathbb{F}_{2^{m+1}} : \mathrm{Tr}(\alpha\lambda^{2}) = 1\rbrace|$. Since squaring gives a bijection on $\mathbb{F}_{2^{m+1}}$, this is equal to $|\lambda \in \mathbb{F}_{2^{m+1}} : \mathrm{Tr}(\alpha\lambda) = 1\rbrace|$. Regardless of the choice of $\alpha \neq 0$, $\{\alpha\lambda : \lambda \in \mathbb{F}_{2^{m+1}} \} = \mathbb{F}_{2^{m+1}}$, so this set will always have the same size.

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  • $\begingroup$ Not sure about this representation of $q$ actually, the $(a_{1}+b_{1})$ in your formula maybe throws it off... $\endgroup$ – Morgan Rodgers Oct 21 '17 at 14:32
  • $\begingroup$ Are the function $q$ you defined and $q$ I have defined the same? and what about $A_\alpha$? $\endgroup$ – NongAm Oct 21 '17 at 15:05
  • $\begingroup$ @NongAm If the function $q$ I have defined is the same as the one you have defined, then $A_{\alpha}$ is the same. That's something you need to check. $\endgroup$ – Morgan Rodgers Oct 21 '17 at 15:42
  • $\begingroup$ @NongAm Since you are counting over all $\lambda$, every field element gets represented once; $\beta \lambda = \alpha (\alpha^{-1}\beta\lambda) = \alpha\lambda^{\prime}$, where $\lambda^{\prime} = \alpha^{-1}\beta\lambda$. $\endgroup$ – Morgan Rodgers Oct 21 '17 at 16:11
  • $\begingroup$ You may need to adjust something though, I think my function $q$ is different than yours, disagreeing on points with $a_{1} \neq b_{1}$; this may be just a few cases to consider separately. $\endgroup$ – Morgan Rodgers Oct 21 '17 at 16:14

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