Open Neumann Classical Mode Expansion

Put $u=\tau +\sigma$, $v=\tau -\sigma$ on the flat target strip [F1], using the light-cone convention [F3]. As in the direct integration of ${\mathrm{\partial }}_u{\mathrm{\partial }}_{v}X=0$ from [F4], every $C^2$ solution is $X=F(u)+G(v)$ for $C^2$ functions $F,G$. Its spatial derivative is $X^{\mathrm{\prime }}=F^{\mathrm{\prime }}(u)-G^{\mathrm{\prime }}(v)$.

At $\sigma =0$, [F5] gives $F^{\mathrm{\prime }}(\tau )=G^{\mathrm{\prime }}(\tau )$ for every $\tau$, so these derivatives are one function $P:\mathbb{R}\to {\mathbb{R}}^{1,D-1}$. At $\sigma =\pi$, [F5] gives $P(\tau +\pi )=P(\tau -\pi )$; replacing $\tau -\pi$ by $s$ proves $P(s+2\pi )=P(s)$. Thus $F^{\mathrm{\prime }}$ and $G^{\mathrm{\prime }}$ are the same $2\pi$-periodic function.

Let $A=(2\pi {)}^{-1}{\int }_{-\pi }^{\pi }P(s)ds$ and define $p=A\mathrm{/}{\alpha }^{\mathrm{\prime }}$. Then $F(u)-{\alpha }^{\mathrm{\prime }}pu$ and $G(v)-{\alpha }^{\mathrm{\prime }}pv$ are $C^2$, $2\pi$-periodic, and have the same derivative; hence they differ only by a constant. Choose constants $x\mathrm{/}2$ so the periodic remainders coincide. Their sum has zero mode $x+2{\alpha }^{\mathrm{\prime }}p\tau$.

Apply [F6] to the common periodic remainder $U$ and write $U(s)={\sum }_{n\mathrm{\ne }0}{a}_n{e}^{-ins}$, after placing its constant Fourier mode into $x$. Define ${\alpha }_n:=n{a}_n\mathrm{/}(i\sqrt{{\alpha }^{\mathrm{\prime }}\mathrm{/}2})$. Then $F(u)=x\mathrm{/}2+{\alpha }^{\mathrm{\prime }}pu+U(u)$ and $G(v)=x\mathrm{/}2+{\alpha }^{\mathrm{\prime }}pv+U(v)$. The periodic terms sum to $2{\sum }_{n\mathrm{\ne }0}{a}_n{e}^{-in\tau }\cos (n\sigma )$. Since $i\sqrt{2{\alpha }^{\mathrm{\prime }}}{\alpha }_n\mathrm{/}n=2a_n$, this is exactly the stated oscillator series. The series is absolutely uniformly convergent by [F6], and real $X$ gives ${\alpha }_{-n}={\alpha }_n^{\ast }$. Each nonconstant cosine has zero spatial average, so the mean is $x+2{\alpha }^{\mathrm{\prime }}p\tau$ and $T{\int }_0^{\pi }\dot{X}d\sigma =2\pi T{\alpha }^{\mathrm{\prime }}p=p$. In conformal gauge the action density is $\mathcal{L}=T({\dot{X}}^2-X^{\mathrm{\prime }2})\mathrm{/}2$, so its derivative with respect to ${\dot{X}}_{\mu }$ is $T{\dot{X}}^{\mu }$, identifying this integral as total target-translation momentum. Its time derivative is $T[X^{\mathrm{\prime }}{]}_0^{\pi }=0$ by the wave equation and Neumann data. The dimensions $[X]=L$ and $[{\alpha }^{\mathrm{\prime }}]=L^2$ then give $[x]=L$, $[p]=L^{-1}$ and dimensionless ${\alpha }_n$. The chiral derivative convergence from [F6] gives $L^2$ convergence on compact time slabs after restriction to the strip.

Differentiating the mode partial sums in [F6] in $L^2$ yields the common chiral derivative $P(u)=\sqrt{{\alpha }^{\mathrm{\prime }}\mathrm{/}2}{\sum }_{m\in \mathbb{Z}}{\alpha }_m{e}^{-imu}$, with ${\alpha }_0=\sqrt{2{\alpha }^{\mathrm{\prime }}}p$. For a component $q=P^{\mu }$, the coefficient of $e^{-imu}$ is $\sqrt{{\alpha }^{\mathrm{\prime }}\mathrm{/}2}{\alpha }_m^{\mu }$. For every finite $N$, orthogonality from [F6] gives $\mathrm{\parallel }q-Q_N{\mathrm{\parallel }}_2^2=\mathrm{\parallel }q{\mathrm{\parallel }}_2^2-\frac{{\alpha }^{\mathrm{\prime }}}{2}{\sum }_{\mathrm{\mid }m\mathrm{\mid }\le N}\mathrm{\mid }{\alpha }_m^{\mu }{\mathrm{\mid }}^2\ge 0$, where $Q_N$ is its symmetric coefficient sum. Summing over target components bounds $({\alpha }^{\mathrm{\prime }}\mathrm{/}2){\sum }_{\mathrm{\mid }m\mathrm{\mid }\le N,\mu }\mathrm{\mid }{\alpha }_m^{\mu }{\mathrm{\mid }}^2$ by $\mathrm{\parallel }P{\mathrm{\parallel }}_{2,E}^2$. Hence the mode sequence is componentwise square-summable.

For fixed $n$, [F7] and step 5.1 make ${\sum }_m{\alpha }_m\cdot {\alpha }_{n-m}$ absolutely convergent. The $L^2$ partial sums $P_N$ satisfy $P_N\cdot P_N\to P\cdot P$ in $L^1$ by [F6] and [F7]. Hence the $n$-th Fourier coefficient of $P\cdot P$ is $({\alpha }^{\mathrm{\prime }}\mathrm{/}2){\sum }_m{\alpha }_m\cdot {\alpha }_{n-m}={\alpha }^{\mathrm{\prime }}{L}_n$. Because $T_{++}=P\cdot P=0$ by [F4], $L_n=0$ for every $n$. The other metric constraint gives the same condition for the identical chiral profile.