Harmonic-gauge plane-wave amplitudes split uniquely into a pure plane-wave gauge part and a transverse-traceless representative, with a two-dimensional space of polarizations

(Existence of the adapted frame.) A vector $n$ with $k(n)\mathrm{\ne }0$ and the required properties exists: choose any future unit timelike $v$; if $k(v)\mathrm{\ne }0$ take $n:=v$; if $k(v)=0$, nondegeneracy of $\eta$ and $k\mathrm{\ne }0$ give a vector $w$ with $k(w)\mathrm{\ne }0$, and for small $t\mathrm{\ne }0$ the vector $v+tw$ is future timelike and has $k(v+tw)=tk(w)\mathrm{\ne }0$; rescaling it to unit length gives $n$ by [L1]. Put $\omega :=-k(n)\mathrm{\ne }0$ and $s:={\eta }^{-1}(k)\mathrm{/}\omega -n$, where ${\eta }^{-1}(k)$ is the vector with components ${\eta }^{ab}{k}_b$. Then $\eta (n,s)=k(n)\mathrm{/}\omega -\eta (n,n)=(-\omega )\mathrm{/}\omega +1=0$ and $\eta (s,s)={\eta }^{-1}(k,k)\mathrm{/}{\omega }^2-2k(n)\mathrm{/}\omega +\eta (n,n)=0-2(-\omega )\mathrm{/}\omega -1=1$, so $s$ is a unit spacelike vector orthogonal to $n$; moreover $k=\omega (\eta (n,\cdot )+\eta (s,\cdot ))$ by construction. By [L5] the pair extends to an $\eta$-orthonormal basis with $e_0=n$, $e_3=s$; in that basis the frame components of the covector are $k_A=k(e_A)=\omega (\eta (n,e_A)+\eta (s,e_A))=\omega (-1,0,0,1)$ and the raised components are $k^A={\eta }^{AB}{k}_B=\omega (1,0,0,1)$. This proves (a).

(The pure gauge amplitudes.) Let $H_{ab}=k_a{Z}_b+k_b{Z}_a$. Its trace is $\mathrm{tr}H=2k^a{Z}_a=:2{\kappa }_Z$, so ${\bar{H}}_{ab}=k_a{Z}_b+k_b{Z}_a-{\eta }_{ab}{\kappa }_Z$; hence $k^a{\bar{H}}_{ab}=k^a{k}_a{Z}_b+k^a{k}_b{Z}_a-k_b{\kappa }_Z=k_b{\kappa }_Z-k_b{\kappa }_Z=0$ because $k^a{k}_a=0$ and $k^a{k}_b{Z}_a=k_b{\kappa }_Z$. Thus $G\subseteq V_0$. If $H=0$, then testing in the frame of step 1.1 on $e_0$ gives $0=H(e_0,\cdot )=-\omega Z_{\bullet }+Z_0{k}_{\bullet }$; evaluating on $e_0$ gives $0=-2\omega Z_0$, so $Z_0=0$, and substituting back gives $\omega Z_b=0$ for every $b$, so $Z=0$. Hence $Z\mapsto (k_a{Z}_b+k_b{Z}_a)$ is injective, and its image $G$ is isomorphic to ${\mathbb{R}}^4$, so $\dim G=4$ by [L3].

(The TT amplitudes.) If $X\in T$, then $\mathrm{tr}X=0$, so $\bar{X}=X$, and $X(e_0,\cdot )=X(e_3,\cdot )=0$; in frame components $k^a{\bar{X}}_{ab}=\omega (X_{0b}+X_{3b})=0$ for every $b$. Hence $T\subseteq V_0$. By [L2], $\dim T=2$. If $X\in T\cap G$ with $X=k_a{Z}_b+k_b{Z}_a$, then testing on $e_0$ as in step 2.1 gives $Z_0=0$ and then $Z=0$, so $X=0$: $T\cap G=\{0\}$.

(Solving for the TT representative.) Let $X\in V_0$ and let $Z$ be any vector. Define $X_{ab}^{\mathrm{\prime }}:=X_{ab}-k_a{Z}_b-k_b{Z}_a$. In frame components with $k_0=-\omega$, $k_3=\omega$, the four equations $X_{0b}^{\mathrm{\prime }}=0$ read $X_{0b}+\omega Z_b-k_b{Z}_0=0$, that is $b=0:X_{00}+2\omega Z_0=0;b=1:X_{01}+\omega Z_1=0;b=2:X_{02}+\omega Z_2=0;b=3:X_{03}+\omega Z_3-\omega Z_0=0,$ whose unique solution is the displayed $Z$ (the matrix of the system is triangular with diagonal entries $2\omega ,\omega ,\omega ,\omega$, all nonzero). Set $X^{\mathrm{\prime }}$ accordingly; then $X^{\mathrm{\prime }}\in V_0$ because $V_0$ is a subspace containing $X$ and $G$ (steps 2.1, 3.1). We show $X^{\mathrm{\prime }}\in T$. First, the harmonic-gauge condition $k^a{\bar{X}}_{ab}^{\mathrm{\prime }}=0$ in frame components reads $\omega ({\bar{X}}_{0b}^{\mathrm{\prime }}+{\bar{X}}_{3b}^{\mathrm{\prime }})=0$. For $b=1,2$ this is $X_{31}^{\mathrm{\prime }}+X_{01}^{\mathrm{\prime }}=0$ and $X_{32}^{\mathrm{\prime }}+X_{02}^{\mathrm{\prime }}=0$, because ${\eta }_{31}={\eta }_{32}=0$; since $X_{01}^{\mathrm{\prime }}=X_{02}^{\mathrm{\prime }}=0$ by construction, $X_{31}^{\mathrm{\prime }}=X_{32}^{\mathrm{\prime }}=0$. For $b=0$ it gives ${\bar{X}}_{00}^{\mathrm{\prime }}+{\bar{X}}_{30}^{\mathrm{\prime }}=0$; here ${\bar{X}}_{30}^{\mathrm{\prime }}=X_{30}^{\mathrm{\prime }}-\frac{1}{2}{\eta }_{30}\mathrm{tr}{X}^{\mathrm{\prime }}=X_{03}^{\mathrm{\prime }}=0$ and ${\bar{X}}_{00}^{\mathrm{\prime }}=X_{00}^{\mathrm{\prime }}-\frac{1}{2}{\eta }_{00}\mathrm{tr}{X}^{\mathrm{\prime }}=\frac{1}{2}\mathrm{tr}{X}^{\mathrm{\prime }}$, so $\mathrm{tr}{X}^{\mathrm{\prime }}=0$. For $b=3$ it gives ${\bar{X}}_{03}^{\mathrm{\prime }}+{\bar{X}}_{33}^{\mathrm{\prime }}=0$ with ${\bar{X}}_{03}^{\mathrm{\prime }}=X_{03}^{\mathrm{\prime }}=0$ and ${\bar{X}}_{33}^{\mathrm{\prime }}=X_{33}^{\mathrm{\prime }}-\frac{1}{2}\mathrm{tr}{X}^{\mathrm{\prime }}=X_{33}^{\mathrm{\prime }}$, so $X_{33}^{\mathrm{\prime }}=0$. Substituting $\mathrm{tr}{X}^{\mathrm{\prime }}=X_{11}^{\mathrm{\prime }}+X_{22}^{\mathrm{\prime }}$ (which holds because $X_{00}^{\mathrm{\prime }}=X_{33}^{\mathrm{\prime }}=0$) gives $X_{11}^{\mathrm{\prime }}+X_{22}^{\mathrm{\prime }}=0$. Therefore $X^{\mathrm{\prime }}(e_0,\cdot )=0$, $X^{\mathrm{\prime }}(e_3,\cdot )=0$ and $\mathrm{tr}{X}^{\mathrm{\prime }}=0$: $X^{\mathrm{\prime }}\in T$.

(Uniqueness and the projection.) If $X^{\mathrm{\prime }\mathrm{\prime }},X^{\mathrm{\prime }}$ are two TT representatives of the same $X$, their difference lies in $T\cap G=\{0\}$ by step 3.1 and the definition of $G$; so the representative is unique. Writing $\pi (X):=X^{\mathrm{\prime }}$, the assignment is the composition of the linear maps $X\mapsto (X,Z(X))$ and $(X,Z)\mapsto X-kZ-kZ$ with $Z$ the linear solution of step 4.1, hence linear; and $\pi (X)=X$ for $X\in T$ because then $Z=0$ solves $X_{0b}^{\mathrm{\prime }}=X_{0b}=0$ (uniqueness of the solution). Consequently every $X\in V_0$ is $X=\pi (X)+(X-\pi (X))$ with $\pi (X)\in T$ and $X-\pi (X)\in G$, so $V_0=T+G$; combined with $T\cap G=\{0\}$ this is an internal direct sum, and by [L3] $\dim V_0=\dim T+\dim G=6$. The map $V_0\mathrm{/}G\to T$, $X+G\mapsto \pi (X)$, is a linear isomorphism, and by [L2] the map $T\to {\mathbb{R}}^2$, $X\mapsto (X_{11},X_{12})$, is a linear isomorphism. This proves (b) and (c).

(Differential form and the gauge transformation.) For $h_{ab}=X_{ab}\cos (k\cdot x)$ we have ${\bar{h}}_{ab}={\bar{X}}_{ab}\cos (k\cdot x)$ and, by [L4], ${\mathrm{\partial }}^a{\bar{h}}_{ab}=k^a{\bar{X}}_{ab}(-\sin (k\cdot x))$; this vanishes identically exactly when $k^a{\bar{X}}_{ab}=0$ for every $b$, that is, exactly when $X\in V_0$. With $Z$ as in step 4.1 and ${\xi }_b=Z_b\sin (k\cdot x)$, the product rule and [L4] give ${\mathrm{\partial }}_a{\xi }_b=k_a{Z}_b\cos (k\cdot x)$, hence ${\mathrm{\partial }}_a{\xi }_b+{\mathrm{\partial }}_b{\xi }_a=(k_a{Z}_b+k_b{Z}_a)\cos (k\cdot x)$ and $h_{ab}^{\mathrm{\prime }}=(X_{ab}-k_a{Z}_b-k_b{Z}_a)\cos (k\cdot x)=X_{ab}^{\mathrm{\prime }}\cos (k\cdot x)$ with $X^{\mathrm{\prime }}\in T\subseteq V_0$, so $h^{\mathrm{\prime }}$ is in harmonic gauge. This proves (d).

(Conclusion.) Steps 1.1, 2.1, 3.1, 4.1, 5.1 and 6.1 prove the four clauses; the arguments use finite linear algebra in an adapted orthonormal frame and the coordinate calculus of the flat model only. No field equation and no physical assumption enters, and no choice beyond the inherited countable choice of the cited finite-dimensional suppliers is used.