Closed Classical Mode Expansion

Put $u=\tau +\sigma$ and $v=\tau -\sigma$, using the light-cone convention [F3]. The free wave equation in [F4] is ${\mathrm{\partial }}_u{\mathrm{\partial }}_{v}X=0$. Therefore ${\mathrm{\partial }}_{u}X$ is independent of $v$; integrate it in $u$ to obtain $X(\tau ,\sigma )=F(u)+G(v)$ for $C^2$ functions $F,G:\mathbb{R}\to {\mathbb{R}}^{1,D-1}$.

Closed-string periodicity gives $F(u+2\pi )-F(u)=G(v)-G(v-2\pi )$ for every $u,v$. The left side depends only on $u$ and the right side only on $v$, so both equal one constant vector $C$. Hence each chiral function has increment $C$ over $2\pi$; subtracting its linear part of slope $C\mathrm{/}(2\pi )$ leaves a $C^2$, $2\pi$-periodic function. Set $p=C\mathrm{/}(\pi {\alpha }^{\mathrm{\prime }})$ and choose the additive constants so both chiral zero modes contain $x\mathrm{/}2$.

Apply [F5] to the two periodic remainders. Writing their Fourier series in the phases $e^{-inu}$ and $e^{-inv}$, respectively, and defining ${\tilde{\alpha }}_n$ and ${\alpha }_n$ by multiplying the corresponding coefficients by $n\mathrm{/}(i\sqrt{{\alpha }^{\mathrm{\prime }}\mathrm{/}2})$, gives exactly the displayed expansion, with ${\alpha }_0={\tilde{\alpha }}_0=\sqrt{{\alpha }^{\mathrm{\prime }}\mathrm{/}2}p$. The series are absolutely and uniformly convergent by [F5]. Since the remainders are real componentwise, their Fourier coefficients at opposite indices are conjugates, giving the stated reality conditions. The periodic remainders have zero average, so the spatial average is $x+{\alpha }^{\mathrm{\prime }}p\tau$ and $T{\int }_0^{2\pi }\dot{X}d\sigma =2\pi T{\alpha }^{\mathrm{\prime }}p=p$. Directly, the conformal action density is $\mathcal{L}=T({\dot{X}}^2-X^{\mathrm{\prime }2})\mathrm{/}2$, whose derivative with respect to ${\dot{X}}_{\mu }$ is $T{\dot{X}}^{\mu }$, identifying this integral as total target-translation momentum; its time derivative is $T[X^{\mathrm{\prime }}{]}_0^{2\pi }=0$ by the wave equation and periodicity. Since $[X]=L$ and $[{\alpha }^{\mathrm{\prime }}]=L^2$, the expansion gives $[x]=L$, $[p]=L^{-1}$ and dimensionless oscillator coefficients. Chiral $L^2$ convergence from [F5] also gives spacetime $L^2$ convergence on compact time slabs by periodicity and integration in time.

Differentiating the partial sums in step 3.1 gives $Y(v):={\mathrm{\partial }}_{-}X=\sqrt{{\alpha }^{\mathrm{\prime }}\mathrm{/}2}{\sum }_{m\in \mathbb{Z}}{\alpha }_m{e}^{-imv}$ in the $L^2$ limit, and analogously ${\mathrm{\partial }}_{+}X=\sqrt{{\alpha }^{\mathrm{\prime }}\mathrm{/}2}{\sum }_m{\tilde{\alpha }}_m{e}^{-imu}$. For a component $q=Y^{\mu }$, the Fourier coefficient of $e^{-imv}$ is $\sqrt{{\alpha }^{\mathrm{\prime }}\mathrm{/}2}{\alpha }_m^{\mu }$. For each finite $N$, the orthogonality in [F5] 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 the finite 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 $\mathrm{\parallel }Y{\mathrm{\parallel }}_{2,E}^2$. The same argument for ${\mathrm{\partial }}_{+}X$ bounds $\tilde{\alpha }$, so both mode sequences are componentwise square-summable.

For fixed $n$, [F6] and step 4.1 make ${\sum }_m{\alpha }_m\cdot {\alpha }_{n-m}$ absolutely convergent. If $Y_N$ is the $N$-th Fourier partial sum of $Y$, then $Y_N\to Y$ in componentwise $L^2$ by [F5], and [F6] gives $Y_N\cdot Y_N\to Y\cdot Y$ in $L^1$. Each finite product has $n$-th Fourier coefficient $({\alpha }^{\mathrm{\prime }}\mathrm{/}2){\sum }_{\mathrm{\mid }m\mathrm{\mid },\mathrm{\mid }n-m\mathrm{\mid }\le N}{\alpha }_m\cdot {\alpha }_{n-m}$; absolute convergence lets $N\to \mathrm{\infty }$. Because $T_{--}=Y\cdot Y=0$ by [F4], its every Fourier coefficient vanishes, so $L_n=0$. The same argument with ${\mathrm{\partial }}_{+}X$ gives ${\tilde{L}}_n=0$.