The screen and spatial projectors are orthogonal projections of rank two and three, and the TT projection is the orthogonal projection onto the two-dimensional transverse-traceless space

(Adapted basis.) Set $e_0:=n$, $e_3:=s$. By [L1] the space $\mathrm{\Sigma }(n,s)$ is a two-dimensional positive definite subspace of $V$; by [L2] it has an orthonormal basis $(e_1,e_2)$. Then $(e_0,e_1,e_2,e_3)$ is an $\eta$-orthonormal basis with $\eta (e_A,e_B)=\mathrm{diag}(-1,1,1,1{)}_{AB}$, and $\mathrm{span}\{e_1,e_2\}=\mathrm{\Sigma }(n,s)$, $\mathrm{span}\{e_1,e_2,e_3\}=n^{\perp }$.

(Components of the projectors.) In the basis of step 1.1 the components of $n$ and $s$ are $n^A={\delta }_0^A$, $n_A=-{\delta }_A^0$ and $s^A={\delta }_3^A$, $s_A={\delta }_A^3$, so $p^A{}_B={\delta }^A{}_B+n^A{n}_B-s^A{s}_B=\mathrm{diag}(0,1,1,0{)}^A{}_B,h^A{}_B={\delta }^A{}_B+n^A{n}_B=\mathrm{diag}(0,1,1,1{)}^A{}_B\mathrm{.}$ Indeed $n^A{n}_B$ is the endomorphism $u\mapsto n^A\eta (n,u)$, which has the single nonzero entry $-1$ in position $(0,0)$; $s^A{s}_B$ likewise has the single entry $+1$ in position $(3,3)$. As symmetric component matrices, $p_{AB}={\eta }_{AC}{p}^C{}_B=\mathrm{diag}(0,1,1,0{)}_{AB}$ and $h_{AB}=\mathrm{diag}(0,1,1,1{)}_{AB}$.

(Clauses (a) and (b).) From the diagonal forms of step 2.1, squaring gives $p^2=p$ and $h^2=h$; $p$ kills $e_0$ and $e_3$ and fixes $e_1,e_2$, while $h$ kills $e_0$ and fixes $e_1,e_2,e_3$. Hence $\mathrm{im}p=\mathrm{span}\{e_1,e_2\}=\mathrm{\Sigma }(n,s)$ and $\mathrm{im}h=\mathrm{span}\{e_1,e_2,e_3\}=n^{\perp }$, while $\ker p=\mathrm{span}\{e_0,e_3\}=\mathrm{span}\{n,s\}$ and $\ker h=\mathbb{R}{e}_0=\mathbb{R}n$. By [L4] the ranks are the dimensions of the images, $\mathrm{rank}p=2$ and $\mathrm{rank}h=3$. Since $p_{AB}$ and $h_{AB}$ are symmetric matrices, $\eta (pu,v)={\sum }_{A,B}{p}_{AB}{u}^A{v}^B=\eta (u,pv)$ and likewise for $h$; and subtracting the two diagonal forms gives $h^A{}_B-p^A{}_B=\mathrm{diag}(0,0,0,1{)}^A{}_B=s^A{s}_B$. All these are tensor identities, hence hold in every basis by [L3]. This proves (a) and (b).

(The TT projection.) Let $A\in \mathcal{S}$ have components $A_{AB}$ in the basis of step 1.1. By step 2.1, $({\mathrm{\Pi }}_{n,s}A{)}_{AB}=p_A{}^C{p}_B{}^D{A}_{CD}-\frac{1}{2}{p}_{AB}{p}^{CD}{A}_{CD},$ and since $p^{CD}=p_{CD}=\mathrm{diag}(0,1,1,0{)}_{CD}$, the only nonzero components of ${\mathrm{\Pi }}_{n,s}A$ are $({\mathrm{\Pi }}_{n,s}A{)}_{11}=\frac{1}{2}(A_{11}-A_{22}),({\mathrm{\Pi }}_{n,s}A{)}_{22}=-\frac{1}{2}(A_{11}-A_{22}),({\mathrm{\Pi }}_{n,s}A{)}_{12}=({\mathrm{\Pi }}_{n,s}A{)}_{21}=A_{12}\mathrm{.}$ Thus ${\mathrm{\Pi }}_{n,s}A$ is supported on $\mathrm{\Sigma }(n,s)$, is symmetric there, and has vanishing screen trace: it is TT. Applying the same display to $A^{\mathrm{\prime }}={\mathrm{\Pi }}_{n,s}A$, whose components satisfy $A_{11}^{\mathrm{\prime }}=-A_{22}^{\mathrm{\prime }}$ and $A_{12}^{\mathrm{\prime }}=A_{21}^{\mathrm{\prime }}$, leaves them unchanged, so ${\mathrm{\Pi }}_{n,s}^2={\mathrm{\Pi }}_{n,s}$; and if $A$ is TT then $A_{11}=-A_{22}$, $A_{12}=A_{21}$ and all other components vanish, so ${\mathrm{\Pi }}_{n,s}A=A$. Conversely ${\mathrm{\Pi }}_{n,s}A=A$ forces the displayed form, which is TT. The displayed parametrization shows that $T_{n,s}$ consists exactly of the symmetric matrices with $A_{11}=-A_{22}$, $A_{12}=A_{21}$ and all other components zero; these form a two-dimensional space. This proves (c).

(Orthogonality and amplitudes.) For spatial symmetric $A,B$ define $⟨A,B⟩=h^{ac}{h}^{bd}{A}_{ab}{B}_{cd}$; in the adapted basis this is ${\sum }_{i,j=1}^3{A}_{ij}{B}_{ij}$. It is bilinear and symmetric, and $⟨A,A⟩=A_{11}^2+A_{22}^2+A_{33}^2+2A_{12}^2+2A_{13}^2+2A_{23}^2$, which is nonnegative and vanishes exactly when all spatial entries vanish; because $A(n,\cdot )=0$, this is exactly $A=0$. Thus it is a positive definite inner product. For a TT tensor $B$ only $B_{11}=-B_{22}$ and $B_{12}=B_{21}$ can be nonzero. By step 3.2, the screen block of $A-{\mathrm{\Pi }}_{n,s}A$ is $\frac{1}{2}(A_{11}+A_{22})\mathrm{diag}(1,1)$ and its off-diagonal screen entries vanish; its remaining spatial entries pair with zero entries of $B$. Therefore $⟨A-{\mathrm{\Pi }}_{n,s}A,B⟩=\frac{1}{2}(A_{11}+A_{22})(B_{11}+B_{22})=0$. Since ${\mathrm{\Pi }}_{n,s}A\in T_{n,s}$ by step 3.2, [L2] identifies the restriction of ${\mathrm{\Pi }}_{n,s}$ to the spatial tensors as the orthogonal projection onto $T_{n,s}$. Finally, a TT tensor is $B=B_{11}(e^1\otimes e^1-e^2\otimes e^2)+B_{12}(e^1\otimes e^2+e^2\otimes e^1)$, its amplitudes are $(A_{+},A_{\times })=(B_{11},B_{12})$, and the map is linear and bijective onto ${\mathbb{R}}^2$ by the parametrization of step 3.2. If $(e_1^{\mathrm{\prime }},e_2^{\mathrm{\prime }})$ is the rotated screen basis of The wave screen of an observer and propagation direction, and the transverse-traceless projection of a symmetric two-tensor clause (c), then bilinearity gives $A_{+}^{\mathrm{\prime }}=B(e_1^{\mathrm{\prime }},e_1^{\mathrm{\prime }})=B_{11}({\cos }^2\theta -{\sin }^2\theta )+2B_{12}\sin \theta \cos \theta =B_{11}\cos 2\theta +B_{12}\sin 2\theta ,$ $A_{\times }^{\mathrm{\prime }}=B(e_1^{\mathrm{\prime }},e_2^{\mathrm{\prime }})=-B_{11}\sin 2\theta +B_{12}\cos 2\theta ,$ using the addition formulas of [L5] at $2\theta =\theta +\theta$. This proves (d).

(Conclusion.) Steps 3.1, 3.2 and 4.1 prove clauses (a)-(d); all computations were performed in one adapted orthonormal basis and are identities between tensors, so they hold in every basis by [L3]. No choice principle, field equation or physical assumption is used.