In the book Multivariate Public Key Cryptography, the author describes the polynomials in the cryptographic system Unbalanced Oil and Vinegar in the following way:
Define $V=\{1,\dots,v\}$ and $O=\{v+1,\dots,n\}$. The central map is $\mathcal{F}=(f^{(1)},\dots, f^{(o)})$ where the polynomials $f^{(i)}$ are of the form:
$$f^{(i)} = \sum_{j,k\in V} \alpha^{(i)}_{j,k}x_jx_k + \sum_{j\in V,k\in O}\beta^{(i)}_{j,k}x_jx_k + \sum_{j\in V\cup O}\gamma^{(i)}_j x_j + \delta^{(i)}\quad (i=1,\dots,o).$$
The author then makes the following claim:
The homogeneous part of the polynomials $f^{(i)}$ can be rewritten as $$\hat{f}^{(i)}(\mathbf{x}) = \mathbf{x}^T\cdot F^{(i)}\cdot \mathbf{x}$$ with $$F^{(i)} = \begin{pmatrix}\star_{v\times v} & \star_{v\times o}\\ \star_{o\times v} & 0_{o\times o}\end{pmatrix}$$ where $\star$ means non-zero elements.
My question is then: Is the last part of the matrix full of zeros because there is no term in the $f^{(i)}$ of the form $\sum_{j,k\in O}\text{[something]}$?
Yes, that's right. In fact we may write $f^{(i)}(\mathbf{x})=\mathbf{x}^T F^{(i)}\mathbf{x}+(\mathbf{y}^{(i)})^T\mathbf{x}+\delta^{(i)}$ where
$$ F^{(i)}=\begin{bmatrix} A^{(i)} & B^{(i)} \\ 0 & 0 \end{bmatrix} $$
and $A^{(i)}=[\alpha_{jk}^{(i)}]$, $B^{(i)}=[\beta_{jk}^{(i)}]$, $\mathbf{y}^{(i)}=[\gamma_j^{(i)}]$. It's possible to write the corner blocks $B$ and $0$ differently and get the same result, for example with $\frac{1}{2}B$ and $\frac{1}{2}B^T$ (resp.). If the corner blocks are $C$ and $D$ (resp.), the only constraint is that $C+D^T=B$ to get the same quadratic form.
To understand why this is all true, set $|V|=2,|O|=2$ and write everything out explicitly.