If $F$ has a particular form, then the wave equation $\square u = F(u,u')$ has a global solution for sufficiently small $C_0^\infty$ Cauchy data. Here $u'=(\partial_tu,\partial_1u,\dots,\partial_nu)$. Klainerman's null condition states that for $n=3$ the quadratic part of $F$ must be a bilinear form $Q(u',u')$ satisfying $$ Q(\xi,\xi)=0 \qquad\text{whenever}\qquad \xi_0^2=\xi_1^2+\xi_2^2+\xi_3^2. \tag{1} $$ The only such forms are linear combinations of the following seven null forms: $$ Q_0(\xi,\eta)=\xi_0\eta_0-(\xi_1\eta_1+\xi_2\eta_2+\xi_3\eta_3) $$ and $$ Q_{ab}(\xi,\eta)=\xi_a\eta_b-\eta_a\xi_b, \qquad 0 \leq a < b \leq 3. $$ The following facts are clear to me:
- The set of all bilinear forms on $\mathbb{R}^{1+3}$ may be identified with a 16-dimensional vector space.
- Those forms satisfying the null condition form a subspace.
- Both $Q_0$ and $Q_{ab}$ satisfy the null condition.
- The list $Q_0,Q_{01},\dots,Q_{23}$ is linearly independent.
But how does (1) imply that the subspace has dimension seven?
Scratched this out over the weekend. Looking at components, write $$ Q = \begin{bmatrix} a_{00} & a_{01} & a_{02} & a_{03} \\ a_{10} & a_{11} & a_{12} & a_{13} \\ a_{20} & a_{21} & a_{22} & a_{23} \\ a_{30} & a_{31} & a_{32} & a_{33} \end{bmatrix} $$
Reduction. Assuming $\xi\neq0$ implies $\xi_0\neq0$ by (1). Dividing through by $|\xi_0|$ yields the equivalent condition $$ Q(\xi,\xi)=0 \qquad\text{whenever}\qquad 1=\xi_1^2+\xi_2^2+\xi_3^2. \tag{2} $$ Proceed investigating six cases by zeroing out components of $\xi$.
Case 1. Let $\xi_2=\xi_3=0$, which implies $\xi_1=\pm1$. Imposing $Q(\xi,\xi)=0$, we see that $$ a_{00}+a_{11}\pm(a_{01}+a_{10})=0. $$ Adding and subtracting gives the two relations \begin{align} a_{00}+a_{11} &= 0 \\ a_{01}+a_{10} &= 0. \end{align}
Cases 2 and 3. By symmetry from case 1, setting $\xi_1=\xi_3=0$ and $\xi_1=\xi_2=0$ produces \begin{align} a_{00}+a_{22} &= 0 \\ a_{00}+a_{33} &= 0 \\ a_{02}+a_{20} &= 0 \\ a_{03}+a_{30} &= 0. \end{align}
Case 4. Let $\xi_3=0$ and $\xi_1=\xi_2=\pm1/\sqrt{2}$. Imposing $Q(\xi,\xi)=0$ and applying the previous relations involving indices 1 and 2, we see that $$ a_{21}+a_{12}=0. $$
Cases 5 and 6. From case 4 and symmetry \begin{align} a_{31}+a_{13} &= 0 \\ a_{32}+a_{23} &= 0. \end{align}
Therefore any bilinear form satisfying (1) must have the form $$ Q = \begin{bmatrix} a_{00} & a_{01} & a_{02} & a_{03} \\ -a_{01} & -a_{00} & a_{12} & a_{13} \\ -a_{02} & -a_{12} & -a_{00} & a_{23} \\ -a_{03} & -a_{13} & -a_{23} & -a_{00} \end{bmatrix}. $$ This can be written as a linear combination of the seven null forms. Since each null form satisfies (1) in general, so does every $Q$ of the above form.