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A useful precise form of the transplantation theorem is as follows. Take $N$ congruent copies of a Euclidean tile with labelled flat faces. For each face colour $s$, encode the gluings by a symmetric signed permutation matrix $M_s$: a glued pair gives off-diagonal entries $+1$ in the two corresponding positions; an unglued face gives a diagonal entry $-1$ for a Dirichlet boundary condition, or $+1$ for a Neumann boundary condition. Corresponding face identifications must use the same maps in the tile coordinates. Suppose two assemblies have matrices $M_s$ and $M_s^{\prime }$ and there is an invertible matrix $C$ withThen, for every eigenvalue $\lambda$, $f=(f_1,\dots ,f_N)\mapsto Cf$ is an isomorphism between their Laplacian eigenfunction spaces. The two assemblies are isospectral, including multiplicities.

To see why, pull every tile restriction back to the same Euclidean tile. A boundary value vector $u$ and an outward normal derivative vector $v$ satisfyAt an internal face these are continuity of the function and equality of oppositely directed derivatives. At a diagonal $-1$ they impose $u_i=0$; at a diagonal $+1$ they impose $v_i=0$. The intertwining identities preserve both equations, and the Euclidean Laplacian commutes with constant linear combinations of tile functions. The resulting piecewise function is a weak eigenfunction, with the prescribed boundary conditions, and interior elliptic regularity makes it smooth wherever the assembled manifold is smooth. At polygonal corners one uses the usual weak domain of the Dirichlet Laplacian or Neumann Laplacian, rather than assuming smoothness through a corner. The inverse matrix gives the inverse transplantation.

Here is a completely specified propeller pair. Label the seven triangles $0,\dots ,6$, with central triangle $0$, and colour the three sides $a,b,c$. In the left and right assemblies the internal-edge pairings areEvery pair means reflection of one triangle across the indicated side; the other faces are external boundary. Each assembly has a central triangle and three two-triangle arms. These are the propeller domains. The drawing uses one acute scalene triangle, and each colour always denotes the side opposite the same marked vertex of the starting triangle.

Let $P_s,Q_s$ be the ordinary permutation matrices obtained by adding fixed points to the pairings. The following matrix supplies the Neumann boundary condition transplantation:There is a short way to construct and check it. Start with the support $\{1,2,4\}$ in row $0$, and require the support in row $Q_s(i)$ to be $P_s$ applied to the support in row $i$. Traversing the six edges of the right-hand tree gives exactly the displayed rows; checking the remaining fixed-point relations verifies $N{P}_s=Q_{s}N$ for all three colours. Every row has three ones and every two different rows meet in exactly one column, soThe right-hand side has eigenvalues $9$ once and $2$ six times, proving $N$ is invertible, with $|\det N|=24$. Its incidence pattern is that of the Fano plane, which explains the uniform intersections.

For the Dirichlet boundary condition, put $D=\mathrm{diag}(1,-1,-1,1,-1,1,1)$. In both assemblies this gives opposite signs to adjacent triangles. Thus their signed face matrices are $M_s=-D{P}_{s}D$ and $M_s^{\prime }=-D{Q}_{s}D$: the off-diagonal gluing entries are positive and the unpaired diagonal entries are negative. The matrixthen satisfies $C{M}_s=M_s^{\prime }C$ and is invertible. The transplantation theorem proves equality of the Dirichlet Laplacian spectra; $N$ proves equality of the Neumann Laplacian spectra as well.

Finally vary the three labelled side lengths $(a,b,c)$ of the initial triangle in an open region of strict triangle inequalities, with an acute scalene shape sufficiently close to an equilateral triangle. Reflection constructs both domains continuously from these three independent real parameters. At the equilateral configuration the tiles have disjoint interiors. Choose a sufficiently small neighbourhood of that configuration with every triangle angle ${\alpha }_i$ in $(\pi /4,\pi /2)$. At each central vertex exactly four tile angles meet, forming a sector of angle $4{\alpha }_i<2\pi$, while nonincident pieces stay separated. Thus there is an open three-dimensional parameter region in which both assemblies are embedded planar domains.

For a scalene triangle in this region, the three central vertices are exactly the three reflex boundary vertices: their interior angles are $4{\alpha }_i$, while the other boundary vertices have angles ${\alpha }_i$ or $2{\alpha }_i<\pi$. Hence a planar isometry of the domains must map the central triangle onto the central triangle. Put those two central triangles in the same position. Their unequal side lengths force any such isometry to fix the central triangle pointwise. Yet the first arm across, for example, side $a$ attaches its outer triangle across colour $c$ in $L$, and across colour $b$ in $R$. These occupy different positions, so the identity does not identify the domains. An intrinsic Riemannian isometry between connected planar domains also restricts locally to a rigid Euclidean motion and hence is one globally. The two propellers are nonisometric throughout this open three-parameter family. Scaling is included as one of the parameters; modulo scaling the family has two parameters.

Define the open-string endpoint momentum flux along the open string by $J_m=e{T}^2{X}_m^{\prime }+u{P}_m$. Variation of the Nambu-Goto phase-space action givesThe two last equations are the Virasoro constraints. Integration of the momentum equation givesThus the necessary and sufficient condition for total momentum conservation is equality of the two endpoint fluxes, component by component. Vanishing of both fluxes is a sufficient local boundary condition.

The spatial boundary term in the action variation is $-\int dt[J_m\delta X^m{]}_0^{\pi }$. For free ends the endpoint variations are arbitrary, soIn a gauge with $u=0$ and nonzero $e$ this reduces to the usual Neumann boundary condition $X^{\prime m}=0$. The conclusion is conservation of total momentum, not that total momentum must vanish. The PDF contains the derivative in this conclusion; the TeX transcription drops it.

For endpoints confined to $x^1=0$, the Dirichlet boundary condition fixes $\delta X^1=0$. It therefore imposes no requirement that $J_1$ vanish. Preserving the fixed position requires ${\dot{X}}^1=e{P}^1+u{X}^{\prime 1}=0$ at each endpoint, but that is a different condition. For example, in $u=0$ gauge the endpoint momentum $P^1$ is zero while $e{T}^2{X}^{\prime 1}$ can be nonzero. Consequently ${\mathcal{P}}_1$ is generally not conserved: the normal endpoint forces transfer momentum to the hyperplane. A dynamical D-brane recoils and carries the missing momentum. If the hyperplane is idealized as fixed, its external support absorbs the momentum. The combined string-and-support system conserves momentum; freely varying directions tangent to the hyperplane retain their vanishing-flux condition.

Use the Bromwich inversion formula in the real variable $\omega$, with $p=\gamma +i\omega$:When the relevant Laplace transforms and spatial integral transforms are explicit, every sample of this integrand can be evaluated directly, without a time-stepping approximation to the partial differential equation. Split the initial condition integral at $y=x$: the two pieces of $R_q$ are linear combinations of $e^{\pm qy}$, so truncated spatial exponential transforms suffice. For real data the negative-frequency half is the complex conjugate of the positive-frequency half.

A practical numerical integration is to truncate to $|\omega |\le \Omega$, apply an adaptive quadrature rule, and independently increase $\Omega$ and refine the quadrature. Choose $\gamma >0$ large enough to remain to the right of all singularities, but avoid an unnecessarily large $e^{\gamma t}$. Evaluate hyperbolic function ratios in scaled form; for example,This prevents overflow when $\mathrm{Re}qL$ is large. The boundary contributions at an interior point are damped by $e^{-x\mathrm{Re}q}$ or $e^{-(L-x)\mathrm{Re}q}$, with $\mathrm{Re}q\sim \sqrt{|\omega |/2}$. Near an endpoint more frequencies are needed; at the endpoint itself use the prescribed Dirichlet boundary condition. The initial condition resolvent has a generally only $O(1/\omega )$ tail. Treat its oscillatory tail accurately, subtract a known transform with the same leading term, or evaluate that contribution separately with the heat kernel.

An equally useful check is the eigenfunction expansion. DefineThe causal modal formula isIt follows either from the heat kernel formula or directly from integration by parts against a sine eigenfunction. Exponential-transform formulas evaluate the time integral explicitly when available. The initial condition contribution has a Gaussian function cutoff in $n$ for each positive time. Nonzero endpoint forcing produces a more slowly convergent sine series tail. A boundary lifting $\ell (x,t)=(1-x/L)g_0(t)+(x/L)e^{aL}h(t)$ gives $w=\ell +z$, where $z$ has zero Dirichlet boundary conditions and forcing $-{\ell }_t-a^2\ell$; expanding $z$ gives better convergence and imposes the endpoint values exactly. At very small times the method of images is often more efficient than retaining many modes.

Check convergence by increasing both the frequency cutoff and quadrature resolution, and compare with a separately truncated causal modal or image-kernel evaluation. Causality in finite-time Laplace contour inversion means a leftward contour deformation must respect both resolvent operator poles and growth of the transformed data. In particular, the transform of data extended by zero after $T$ can grow like $e^{-pT}$ in the left half-plane: for $t<T$ one cannot close that contour indiscriminately and discard its large arc. The modal forcing integral with upper limit $t$ avoids this causality error.

Put $a=\alpha /2$ and apply the Dirichlet gauge transform for constant drift $w(x,t)=e^{ax}u(x,t)$. Taking derivatives directly removes the advection term and gives the damped heat equation $w_t=w_{xx}-a^{2}w$. Thus $w_0(y)=e^{ay}{u}_0(y)$, while the two Dirichlet boundary conditions become $w(0,t)=g_0(t)$ and $w(L,t)=e^{aL}h(t)$.

The Dirichlet heat kernel on an interval and its damped version areThe method of images gives an alternative, often better at short times:To determine the boundary signs, multiply the equation for $w$ by the backward heat kernel $K_a(x,y,t-s)$ and use integration by parts in $y$. Since the kernel vanishes at $y=0,L$, the surviving boundary expression is $K_{a,y}(x,0,t-s)g_0(s)-K_{a,y}(x,L,t-s)e^{aL}h(s)$. Consequently an integral representation containing only the given data isThis is a Dirichlet boundary-forcing heat-kernel formula. For $0<x<L$ the short-time Gaussian decay makes the forcing integral well defined. The endpoint values are interior limits of the complete formula: evaluating a termwise sine series at an endpoint before taking the time integral loses the nonzero boundary values. At the initial corners a continuous classical solution requires $u_0(0)=g_0(0)$ and $u_0(L)=h(0)$; otherwise the same formula describes the solution away from those corners.

Here is a second integral representation, useful when the integral transforms are explicit. Extend $g_0,h$ by zero after $T$; the values of the extension cannot affect a solution at $t<T$. Write their Laplace transforms as $G(p),H(p)$ and set $q=\sqrt{p+a^2}$ using the principal square root. The transformed Dirichlet Green function for $-{\partial }_x^2+q^2$ isIt vanishes at both endpoints and its first derivative in $x$ jumps by $-1$, which verifies the sign of its unit source. The resolvent kernel for Dirichlet advection-diffusion on an interval gives the transformed solutionThe Bromwich inversion formula therefore yieldsThe square-root notation creates no genuine branch singularity here: all three kernels are even in $q$. Their only spatial resolvent operator poles are $p=-a^2-q_n^2$. This transformed formula and the causal heat kernel formula represent the same solution in the usual smooth-data class.

Let $\Omega$ be a bounded piecewise smooth domain, $f=u{|}_{\partial \Omega }$ and $q={\partial }_{n}u$, with $n$ the outward normal vector. Integrating the divergence form from part (i) and using the divergence theorem gives two global relations for a linear boundary value problem:These conjugate global relations for the modified Helmholtz equation are identities between the prescribed Dirichlet boundary condition and its unknown normal derivative; neither is a pointwise boundary equation. Both hold for every nonzero boundary spectral parameter under the usual trace regularity assumptions.

If $u$ is real, its two boundary traces $f,q$ are real. Since ${\tilde{v}}_{\lambda }=\overline{v_{\overline{\lambda }}}$ and the outward normal vector is real,Thus apply complex conjugation to the first global relation for a linear boundary value problem, and replace its parameter by $\overline{\lambda }$, to get the second. For complex $u$, conjugation would also replace its traces by their conjugates, so that argument requires the stated reality assumption. There is also an algebraic redundancy in this particular parametrization: ${\tilde{v}}_{\lambda }=v_{1/\lambda }$, hence $\tilde{\mathcal{G}}(\lambda )=\mathcal{G}(1/\lambda )$ even for complex data. This does not change the requested conjugation identity.

Label the sides by $L,R,B,T$, and parametrize the two vertical sides by $s=y\in [-1,1]$, the horizontal sides by $s=x\in [-1,1]$. Their outward normal derivatives are $q_L=-u_x(-1,s)$, $q_R=u_x(1,s)$, $q_B=-u_y(s,-1)$ and $q_T=u_y(s,1)$. Let $f_j$ denote the prescribed side values. Assume compatible, sufficiently regular boundary traces; the square's corners have zero arclength measure and do not require separate normal values.

For sine collocation of square modified Helmholtz global relations, use Legendre polynomials for the known Dirichlet boundary condition and a Fourier sine series for the unknown normal derivative:The known coefficients are $d_{j\ell }=(2\ell +1){\int }_{-1}^1{f}_j{P}_{\ell }ds/2$. The sine functions have ${\int }_{-1}^1{\varphi }_m{\varphi }_{n}ds={\delta }_{mn}$ and form a complete basis in $L^2(-1,1)$. Choosing them for the normal derivative does not impose zero flux at a corner: the expansion is an $L^2$ representation, and endpoint values are not determined by it. If preserving corner values of the approximated Dirichlet trace is necessary, subtract its endpoint-interpolating line before polynomial approximation, and add that line back. All subsequent known-data integrals can alternatively be evaluated with the exact $f_j$.

Define the entire function basis transformsAt $\tau =\pm i{p}_m$ the quotient is evaluated by its removable limit or by the defining integral. For example, ${\mathsf{P}}_0=2\sinh \tau /\tau$, and polynomial expansion of $P_{\ell }$ expresses every ${\mathsf{P}}_{\ell }$ in derivatives of ${\mathsf{P}}_0$. PutFor compactness write $A=A(\lambda )$ and $B=B(\lambda )$. Substituting these expansions into the first global relation for a linear boundary value problem givesThe signs are fixed by the outward normal vectors, rather than by a choice of traversal direction. The second approximate global relation for a linear boundary value problem replaces $A$ by $-A$, leaving $B$ unchanged:Here $\simeq 0$ records omitted boundary-expansion tails. In a finite spectral method the selected equations are set equal to zero to solve for the $4N$ unknown real coefficients $c_{jm}$. For real data the two families obey the same complex conjugation relation as their exact counterparts.

A particularly simple constructive choice uses paired collocation points for a global relation. For $n=1,\dots ,N$, putFour suitable sets, one for each side, areThe braces in each set range over all $n$. For large $p_n/k$, compute $r_n^{-1}=k/({\omega }_n+p_n)$ to avoid cancellation. The corresponding $(A,B)$ pairs are, respectively,Each pair has the same exponentially growing normal direction and opposite tangential frequencies. Take the phase-weighted difference of the two collocated global relations for a linear boundary value problem. For example, the bottom pair gives the adjoint testThe other three phase-weighted differences giveThus the four spectral sets supply four real sine-tested equations per mode. We retain these $4N$ equations; requiring every uncombined complex equation as well would instead form an overdetermined finite approximation. For real traces each retained equation is the appropriate real combination of its conjugate pair.

The key advantage is that $v_{Bn},v_{Tn}$ vanish on the vertical sides, while $v_{Ln},v_{Rn}$ vanish on the horizontal sides. Orthogonality then eliminates all off-mode unknowns. Define the known right-hand sidesThe retained global relations for a linear boundary value problem areFor example, with $f_{jn}={\int }_{-1}^1{f}_j^{(M)}(s){\varphi }_n(s)ds$, the bottom and top right-hand sides are explicitlyFor completeness the left and right right-hand sides areAll these integrals use known polynomial approximations or the exact Dirichlet boundary condition.

Scale each row by $e^{-{\omega }_n}$. Every block becomesThere is just one off-diagonal entry in each row. This proves diagonal dominance of paired square global-relation collocation. Therefore the assembled $4N\times 4N$ system is strictly diagonally dominant, with row margin $1-{\delta }_n>1-e^{-\pi }$. This proof applies to the explicitly paired and scaled collocation system, rather than assuming arbitrary raw spectral rows are diagonally dominant. Each block is a positive-definite matrix and have spectral condition number of a positive-definite matrixFor instance, setting ${\rho }_{jn}=e^{-{\omega }_n}{R}_{jn}$ gives the concise answerwith the identical formula for the opposite vertical sides. If the known-data integrals use exact $f_j$, these are exactly the first $N$ sine coefficients of the true normal traces: no discarded unknown mode contributes to a retained row. With a finite Legendre polynomial approximation, the error is solely that of the known-data approximation and the final normal-trace truncation. These square modified Helmholtz Dirichlet-to-Neumann coefficients determine the unknown boundary normal derivative; Green's third identity or the separated-variable construction can then recover the interior solution.

Work with real-valued smooth functions on the bounded region enclosed by $S$, with the oriented surface element pointing outward. Set $\psi =\varphi -{\varphi }_1$. It is a harmonic function in $V$ and has zero Dirichlet boundary conditions on $S$. Applying Green's first identity to $\psi$ with itself givesBoth terms on the right vanish by the boundary condition and harmonicity. The integrand is nonnegative and continuous, so $\nabla \psi =0$ throughout $V$. Consequently $\psi$ is constant on each connected component; every bounded component meets the prescribed boundary, where that constant is zero. ThereforeThis is the energy proof of Uniqueness of the Dirichlet problem. Connectedness of $V$ is not essential if the boundary values are prescribed on every component. Boundedness, or suitable decay and integrability at infinity, is needed for the boundary/energy argument; the finite enclosed-volume interpretation is used here.

Take a smooth variation $u_{\epsilon }=u+\epsilon \varphi$ with $\varphi =0$ on the boundary. Assume enough differentiability for the following operations, for example a smooth bounded open set, $u\in C^2$, and a twice continuously differentiable integrand. Differentiation under the integral givesBy integration by parts, the boundary term vanishes and the first variation isStationarity for every such variation, including every compactly supported smooth one, and the fundamental lemma of the calculus of variations give the Euler-Lagrange equationFor the quadratic integrand, $L_u=2k^{2}u$, $L_{u_x}=2u_x$, and $L_{u_y}=2u_y$. Thuswith the prescribed Dirichlet boundary condition. For real $k$, a stationary function is also the unique minimizer, if it exists: for any nonzero admissible $\varphi$, the linear term vanishes and $I[u+\varphi ]-I[u]={\int }_{\mathcal{D}}(|\nabla \varphi {|}^2+k^2{\varphi }^2)>0$. This last positivity uses the zero boundary values; it also holds at $k=0$.

The coefficients in the spherical Laplacian are independent of $\varphi$, so $[\Delta ,{\partial }_{\varphi }]=0$. Also ${\partial }_{\varphi }$ is tangent to both boundary spheres; differentiating the zero boundary trace gives another zero trace. Therefore $w={\partial }_{\varphi }u$ satisfies $\Delta w={\partial }_{\varphi }f$ and the same homogeneous Dirichlet boundary conditions.

In Cartesian coordinates ${\partial }_{\varphi }=x_1{\partial }_2-x_2{\partial }_1$, a globally smooth rotation field even where angular coordinates are singular. Permuting coordinate axes gives $R_{12}$, $R_{23}$ and $R_{31}$. For each, the rotational commutator for the Laplacian follows directly from $\Delta (x_i{\partial }_{j}u)=x_i{\partial }_j\Delta u+2{\partial }_i{\partial }_{j}u$; antisymmetrizing cancels the extra terms. ThusIf $f$ is radial, all these right-hand sides vanish. Uniqueness from the energy estimate makes every $R_{ij}u$ zero. At each point of a sphere the rotation fields span its tangent plane, so all tangential derivatives of $u$ vanish. Each sphere is connected, giving $\boxed{u(x)=U(|x|)}$. This proves radiality without assuming it as a separation-of-variables ansatz.

Set $\hslash =1$. Work on the strip $0\le \sigma \le \pi$ in conformal gauge, with worldsheet signature $(-,+)$ and target signature $(-,+,\dots ,+)$. In light-cone gauge in string theory, the independent fields are the $d=D-2=24$ transverse coordinates. Put the Neumann boundary condition at $\sigma =0$ and the Dirichlet boundary condition at $\sigma =\pi$; reversing the endpoints exchanges cosine and sine modes without changing the spectrum. The fixed endpoint is $X^i(\tau ,\pi )=y^i$.

Varying the transverse Polyakov action gives $({\partial }_{\tau }^2-{\partial }_{\sigma }^2)X^i=0$ and the spatial boundary contribution $-(2\pi {\alpha }^{\prime }{)}^{-1}\int d\tau [X^{\prime i}\delta X^i{]}_0^{\pi }$. At the free endpoint this vanishes precisely when $X^{\prime i}=0$; at the fixed endpoint $\delta X^i=0$. Separation of variables then gives $\cos (r\sigma )$ with $\cos (r\pi )=0$, hence $r=n+\frac{1}{2}$, $n=0,1,\dots$. These are Neumann-Dirichlet open-string boundary conditions. No dynamical transverse worldsheet zero mode survives: a constant solution must equal the prescribed $y^i$, and a linear-in-$\sigma$ solution violates the free-end condition.

Write $X^i-y^i={\sum }_{r>0}{q}_r^i(\tau )\cos (r\sigma )$. The orthogonality relation ${\int }_0^{\pi }\cos (r\sigma )\cos (s\sigma )d\sigma =(\pi /2){\delta }_{rs}$ reduces the action to independent harmonic oscillators:The canonical commutation relations $[q_r^i,P_s^j]=i{\delta }^{ij}{\delta }_{rs}$ determine normalized annihilation operators at $\tau =0$Thus $q_r^i=\sqrt{2{\alpha }^{\prime }/r}(a_r^i{e}^{-ir\tau }+a_r^{i†}{e}^{ir\tau })$. These formulas derive the quantization from the action rather than import integer-moded open-string rules.

For the usual phased string oscillators, set ${\alpha }_r^i=-i\sqrt{r}{a}_r^i$ and ${\alpha }_{-r}^i=i\sqrt{r}{a}_r^{i†}$ for $r>0$. Then the half-integer open-string oscillator expansion and algebra areThe corresponding field momentum density is ${\Pi }_i={\dot{X}}^i/(2\pi {\alpha }^{\prime })$. Completeness of the mixed-boundary eigenfunctions gives $[X^i(\sigma ),{\Pi }_j({\sigma }^{\prime })]=i{\delta }_j^i{\delta }_{\mathrm{ND}}(\sigma ,{\sigma }^{\prime })$, where ${\delta }_{\mathrm{ND}}=(2/\pi ){\sum }_{r>0}\cos (r\sigma )\cos (r{\sigma }^{\prime })$. This is a distribution identity on the mixed-boundary function space, not an unrestricted value at a fixed endpoint with a Dirichlet boundary condition.

Classically, the transverse Virasoro algebra zero-mode generator isThere is no transverse momentum term. If the two light-cone directions are common directions with Neumann boundary conditions, the full zero-mode constraint adds ${\alpha }^{\prime }{p}_{\parallel }^2=-2{\alpha }^{\prime }{p}^{+}{p}^{-}$: $L_{0,\mathrm{cl}}={\alpha }^{\prime }{p}_{\parallel }^2+L_{0,\mathrm{cl}}^{\perp }$. This assumption about the longitudinal directions is needed to interpret oscillator levels as target-space masses.

Quantizing the symmetrically ordered transverse generator gives $L_0^{\perp }=N+E_0$, where the string level operator is $N={\sum }_{i,r>0}r{a}_r^{i†}{a}_r^i$. Each harmonic oscillator contributes $r/2$ to the vacuum energy. Use zeta function regularization and the Riemann zeta function with the actual half-integer spectrum:Consequently the Neumann-Dirichlet string zero-point energy isIn the common convention $L_0={\alpha }^{\prime }{p}_{\parallel }^2+N-a$, the normal-ordering constant of a string is $a=-\frac{1}{2}$. This positive shift differs from the $-1$ vacuum energy of 24 integer-moded transverse bosons. The difference between those two vacuum energies is $3/2$. To distinguish zero-mode conventions, the ND twist conformal weight is $1/16$ per transverse boson. With 24 bosons, the plane matter generator is $L_0^{\mathrm{plane}}={\alpha }^{\prime }{p}_{\parallel }^2+N+3/2$; its physical open-string condition $L_0^{\mathrm{plane}}-1=0$ is exactly the strip/light-cone constraint ${\alpha }^{\prime }{p}_{\parallel }^2+N+1/2=0$ used here. The transverse plane and strip constants differ by the central charge shift $c_{\perp }/24=24/24=1$. A common exponential frequency cutoff independently gives ${\sum }_{r>0}r{e}^{-\epsilon r}={\epsilon }^{-2}+1/24+O({\epsilon }^2)$, confirming the finite part and avoiding invalid termwise manipulation of divergent sums.

Let $a_r^i|0⟩=0$. The lowest levels of a fully transverse ND bosonic string areThe third level has two $r=1/2$ excitations; there is no $r=1$ oscillator. Its indices are symmetric because the creation operators commute. With $L_0=0$ and the common longitudinal momentum convention, the rest energies and masses obeyAt fixed positive $p^{+}$ the corresponding light-cone energies are $p^{-}=(N+1/2)/(2{\alpha }^{\prime }{p}^{+})$; the table lists excitation levels, not a spectrum that remains discrete if longitudinal momentum is varied continuously.

The surviving transverse rotations form $SO(24)$. The ground state is a scalar, the next level its vector, and the third level the symmetric square of the vector. Separating its trace givesThe trace state is proportional to ${\sum }_i(a_{1/2}^{i†}{)}^2|0⟩$; subtracting this trace gives the 299-dimensional symmetric traceless square.