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The Frankl-Wilson theorem. Let $p$ be a prime number, let $L\subseteq {\mathbb{F}}_p$ have $s$ elements, and let $\mathcal{F}\subseteq \mathcal{P}([n])$. Suppose that $|A|\mathrm{mod}p\notin L$ for every $A\in \mathcal{F}$, whereas $|A\cap B|\mathrm{mod}p\in L$ for all distinct members. Thenwith terms beyond $n$ understood as zero.

Work over the finite field ${\mathbb{F}}_p$. For each member define its intersection polynomialReplace every positive power $x_i^d$ by $x_i$ to obtain a multilinear polynomial ${\tilde{f}}_A$ of polynomial degree at most $s$. This Boolean multilinearization preserves its values at every characteristic vector of a set. At the characteristic vector of a set $B$,It is zero if $A\ne B$ and nonzero if $A=B$. Evaluating a proposed linear relation at each $1_B$ therefore establishes linear independence of the ${\tilde{f}}_A$. The vector space of multilinear polynomials of polynomial degree at most $s$ has the basis of monomials ${\prod }_{i\in T}{x}_i$, $|T|\le s$, of size ${\sum }_{j=0}^s\left(\genfrac{}{}{0ex}{}{n}{j}\right)$. This proves the Frankl-Wilson theorem by the polynomial method in combinatorics. The empty product for $s=0$ is $1$, and the argument still applies.

The constant-intersection family bound, directly. Write $\mathcal{F}=\{F_1,\dots ,F_q\}$ and assume $n\ge 1$. The cases $q\le 1$ are immediate. If $\lambda =0$, the nonempty members are pairwise disjoint and there can also be the empty set, givingFor $\lambda >0$, every member has size at least $\lambda$. If every member has size strictly greater than $\lambda$, form the real matrix $X$ whose rows are their characteristic vectors of sets. Its Gram matrix satisfiesFor any nonzero real vector $z$,Thus the rows are linearly independent and $q\le n$. If a member $F_1$ has size $\lambda$, it lies inside every other member. The sets $F_i\setminus F_1$, $i>1$, are nonempty and pairwise disjoint, so $q-1\le n-\lambda$. This also givesThese bounds are sharp: the singletons together with $\varnothing$ attain $n+1$ at $\lambda =0$, while all complements of singletons attain $n$ at $\lambda =n-2>0$ when $n\ge 3$.

The constant t-wise intersection dichotomy. In its substantive form the last argument requires $m\ge t$. The printed statement does not explicitly impose this. Without it the $t$-fold condition can be vacuous and the claimed dichotomy is false: take $t=3$, $m=2$, $n=1$, $A_1=\varnothing$, $A_2=\{1\}$ and $\lambda =1$. There is no $3$-tuple to test, no common $1$-set, and $k=0$, so $m=2>k+t-2=1$. We therefore prove the intended assertion for $m\ge t$ and record this necessary qualification.

If $\lambda =0$, the first alternative always holds with $S=\varnothing$. Suppose $\lambda >0$. Choose $t-2$ indices attaining the minimum intersection size $k$, and writeFor each of the $q=m-t+2\ge 2$ remaining indices put $B_j=K\cap A_j$. The hypothesis gives $|B_j\cap B_{\ell }|=\lambda$ for distinct remaining indices. This reduces the constant t-wise intersection dichotomy to a constant-intersection family bound on the $k$-element set $K$. The two possibilities are treated below.

Use the part-preserving bipartite four-cycle density conventionHere $G(x,y)$ is the indicator function of adjacency, and repeated labels are allowed. This is the homomorphism density convention appropriate to the four-cycle norm. We derive the needed counting and operator norm facts directly.

Let $M=(G(x,y))$ be the $n$-by-$n$ biadjacency matrix, let $1$ be the all-ones vector, let $J=11^{\mathsf{T}}$, and put $H=M-\delta J$. The degree hypothesis saysso the centered matrix $H$ annihilates $1$ on both sides. In particular, $HJ=J{H}^{\mathsf{T}}=0$, andThe two terms have zero products in either order. Expanding the matrix trace directly counts four adjacency factors:Since $J^2=nJ$ and $\mathrm{tr}J=n$, we obtainThe given upper bound therefore impliesThis computation proves the centered four-cycle identity for a biregular graph, rather than assuming a four-cycle norm identity.

The matrix $Q=H{H}^{\mathsf{T}}$ is a positive semidefinite matrix. By the spectral theorem for real symmetric matrices it has nonnegative eigenvalues ${\lambda }_j$, andMoreover, for the Euclidean norm,Indeed $\Vert H^{\mathsf{T}}v{\Vert }_2^2=v^{\mathsf{T}}Qv$, whose maximum on unit vectors is ${\lambda }_{\max }$; the Cauchy-Schwarz inequality characterization $\Vert H{\Vert }_{\mathrm{op}}={\sup }_{\Vert u{\Vert }_2=\Vert v{\Vert }_2=1}|u^{\mathsf{T}}Hv|$ also shows that $H$ and $H^{\mathsf{T}}$ have the same operator norm. Taking fourth roots thus giveswhere the error parameter $c$ is nonnegative, as in the claimed bound.

Finally use the indicator vectors $a=1_A$, $b=1_B$. Their Euclidean norms are $\sqrt{|A|}$ and $\sqrt{|B|}$, so the Cauchy-Schwarz inequality and the operator norm estimate yieldThis is the required discrepancy bound. It also holds when either subset is empty or $\delta =0$. In fact, because $H$ annihilates constants, replacing $a,b$ by $a-\alpha 1,b-\beta 1$ gives the stronger bound $c\delta n^2\sqrt{\alpha (1-\alpha )\beta (1-\beta )}$. This is the spectral discrepancy bound for a biregular graph.

The two displacement vectors $\mathbf{b}-\mathbf{a}$ and $\mathbf{c}-\mathbf{a}$ lie in the required plane. Because the three points are non-collinear, their cross productis nonzero and is a normal vector to that plane. A point with position vector $\mathbf{r}$ lies in the plane exactly whenExpanding the cross product givesThe terms $\mathbf{a}\cdot (\mathbf{a}\times \mathbf{b})$ and $\mathbf{a}\cdot (\mathbf{c}\times \mathbf{a})$ vanish, so the equation of a plane through three points becomesNon-collinearity is important: it ensures this is a genuine plane equation with nonzero normal, rather than the vacuous equality $0=0$.

For the specified coordinates,Their sum is $(3,3,6)$, with Euclidean norm $3\sqrt{6}$. Consequently either orientation of the unit normal is valid:As a check, the scalar triple product is $12$, so the plane equation simplifies to $x+y+2z=4$, which all three supplied points satisfy. The normal direction is $(1,1,2)$.

Let $R$ be an orthogonal matrix describing a Cartesian change of coordinates, with transformed vector components $v_i^{\prime }=R_{ia}{v}_a$ and $w_j^{\prime }=R_{jb}{w}_b$. Their outer product transforms aswhich is the transformation law of a second-order Cartesian tensor. Here “rank two” counts tensor indices, not the rank of a matrix; a nonzero outer product has matrix rank one.

For an isotropic second-rank tensor, invariance means $RT{R}^T=T$ for every proper rotation matrix. The half-turn matrices $\mathrm{diag}(1,-1,-1)$, $\mathrm{diag}(-1,1,-1)$ and $\mathrm{diag}(-1,-1,1)$ force all off-diagonal entries to vanish. Quarter-turns around coordinate axes then force the three diagonal entries to be equal. Conversely $RI{R}^T=I$, so the most general second-rank isotropic tensor isSince $v{w}^T$ has matrix rank at most one, it cannot equal a nonzero scalar multiple of the three-dimensional identity. Thus its isotropy forces $v{w}^T=0$. If $v\ne 0$, a nonzero component $v_i$ makes the entire $i$th row vanish only when $w=0$; the converse is immediate. The required choices are exactlyFor the Levi-Civita symbol, multilinearity and antisymmetry of the determinant givefor every $R\in SO(3)$. This proves the rotation invariance of the Levi-Civita symbol, hence its third-rank isotropy. The convention is invariance under proper rotations: under an orthogonal reflection $\det R=-1$, the sign reverses. Thus it is a pseudotensor if transformations of both orientations are included.

Every permutation is a product of transpositions, since a cycle can be writtenwith rightmost factors acting first. Define the sign of a permutation to be $(-1{)}^r$ for a product of $r$ transpositions. To prove that it is well defined without assuming parity, let $P_{\sigma }$ be the permutation matrix specified by $P_{\sigma }{e}_i=e_{\sigma (i)}$. Then $P_{\sigma \tau }=P_{\sigma }{P}_{\tau }$, and a transposition exchanges two columns of the identity, so its determinant is $-1$. Any decomposition into $r$ transpositions therefore hasThe left-hand side depends only on $\sigma$, proving well-definedness. Multiplicativity of the determinant now gives the sign homomorphism:For the first requested group embedding, use the faithful four-point action of GL2 over F2: label the four vectors in ${\mathbb{F}}_2^2$ and let an invertible $2\times 2$ matrix act on them by multiplication. Composition of these actions gives a group homomorphism $\psi :G{L}_2({\mathbb{F}}_2)\to S_4$. If a matrix fixes every vector, it fixes both standard basis vectors and is the identity, so the action is faithful. The matrixfixes $(0,0)$ and $(1,1)$ and exchanges $(1,0)$ with $(0,1)$. Its image is a single transposition, hence $\mathrm{sgn}(\psi (B))=-1$. The composite sign map is nontrivial.

For the second group embedding, simply takeThe preceding multiplication formula proves the group homomorphism property, and $P_{\sigma }=I$ forces $\sigma (i)=i$ for every $i$, proving injectivity.

The characteristic polynomial of the given matrix is $(t-1)(t-2{)}^2$. Solving the two eigenvalue equations givesFor example, multiplying the matrix by these three vectors gives respectively $1$, $2$ and $2$ times the original vector. The eigenvectors form a basis: the matrix with these columns has nonzero determinant $-1$. Thus an eigenbasis and its diagonal matrix are

Let $a$ be any constant vector and take $u=\varphi a$. The product rule for divergence gives $\nabla \cdot u=a\cdot \nabla \varphi$. The divergence theorem on a bounded region with piecewise smooth boundary, oriented outward, therefore yieldsSince $a$ is arbitrary, equality of all components provesHere $\varphi$ is continuously differentiable on a neighbourhood of the closed region.

For the side of this right circular cone, the parameter tangents are $x_r=(\cos \theta ,\sin \theta ,\sqrt{3})$ and $x_{\theta }=(-r\sin \theta ,r\cos \theta ,0)$. Their cross product in the outward order isThe sign is outward because the solid right circular cone lies at smaller cylindrical radius for fixed height. Reversing the parameter order reverses the oriented surface element.

To check the integral identity, the closed boundary must include the top Euclidean disk $z=1$, radius $1/\sqrt{3}$, as well as the curved side. For $\varphi =z^2$, horizontal components cancel on integrating $\theta$. The side contribution isOn the top Euclidean disk $\varphi =1$ and $d\mathbf{S}=e_{z}dA$, so its contribution is $(\pi /3)e_z$. The total is $(\pi /6)e_z$. Independently, the cross-section of the solid at height $z$ has area $\pi z^2/3$, and $\nabla \varphi =2z{e}_z$, givingThus the two sides agree. The curved side alone is not a closed surface and does not satisfy this volume identity. The apex has zero area; alternatively one can truncate at height $\epsilon$ and let $\epsilon \downarrow 0$, with the extra boundary contribution vanishing.

Use the Einstein summation convention. Substitute the given basis transformation into the expansion of the vector:By linear independence of the basis vectors, $v_j=R_{ij}{v}_i^{\prime }$. In matrix notation $v=R^T{v}^{\prime }$. A rotation matrix is orthogonal, so multiplication by $R$ givesThe placement of the transpose follows from the specified convention $e_i^{\prime }=R_{ij}{e}_j$. It must not be guessed from a different convention for an active rotation of a vector.

The quotient theorem for Cartesian tensors here says: if contracting $T_{ij}$ with every vector $v_j$ gives the components $w_i=T_{ij}{v}_j$ of a vector in every right-handed orthonormal basis, then $T$ obeys the second-order tensor transformation law. Conversely that transformation law guarantees the contraction is a vector. One may equivalently test that $u_i{T}_{ij}{v}_j$ is a scalar for every pair of vectors $u,v$.

By part (a), $v^{\prime }=Rv$ and $w^{\prime }=Rw$. The contraction property in the new basis saysSince this holds for every test vector, $T^{\prime }R=RT$. Using $R^{-1}=R^T$ givesFor the converse, $T^{\prime }{v}^{\prime }=RT{R}^{T}Rv=RTv=Rw$, exactly the vector law. For the equivalent scalar formulation, invariance of $u^{T}Tv$ for every $u,v$ gives $u^T{R}^T{T}^{\prime }Rv=u^{T}Tv$, forcing the same matrix identity.

The quantifier “every” matters: an arbitrary nonzero matrix can annihilate one fixed vector, so a single successful contraction cannot establish tensoriality. The stated bases test proper rotations; transformation under orthogonal reflections would require an additional hypothesis if that were wanted. “Second-order” counts tensor indices, rather than the matrix rank of a particular component matrix.

The successive differences of the constant entries give a simple LDL decomposition:For example, entry $(i,j)$ of $LD{L}^T$ is the sum of the first $\min (i,j)$ entries of $D$, directly verifying every entry of $A$. Because $L$ is an invertible matrix, $x^{T}Ax=(L^{T}x{)}^{T}D(L^{T}x)$ is strictly positive for every nonzero vector exactly when every diagonal entry of $D$ is positive. ConsequentlyAt $\lambda =14$ it is a positive semidefinite matrix and singular; for $\lambda <14$ its quadratic form takes both positive and negative values.

When $\lambda =30$, taking positive square roots of $D$ yields the Cholesky decomposition

A real positive-definite kernel is a symmetric function $k:\mathcal{X}\times \mathcal{X}\to \mathbb{R}$ such that, for every finite collection $x_1,\dots ,x_n$ and every $a\in {\mathbb{R}}^n$,Thus every kernel matrix is a positive semidefinite matrix; “positive definite” here does not require strict positivity. A Reproducing kernel Hilbert space is a Hilbert space $\mathcal{H}$ of functions on $\mathcal{X}$ in which every point evaluation is a continuous linear functional. Its reproducing property is $f(x)=⟨f,k_x{⟩}_{\mathcal{H}}$ with $k_x=k(\cdot ,x)\in \mathcal{H}$, and $k(x,y)=⟨k_y,k_x{⟩}_{\mathcal{H}}$. For a sum of positive-definite kernels, the quadratic form of its kernel matrix is a sum of nonnegative quadratic forms. Symmetry also survives summation. Consequently $\boxed{k=\sum _j{k}_j\text{ is a positive-definite kernel}.}$

For the representer theorem, put $V=\mathrm{span}\{k_{x_1},\dots ,k_{x_n}\}$. This is a closed finite-dimensional vector space. The orthogonal decomposition by a closed subspace gives $f=f_V+f_{\perp }$, where $f_V\in V$ and $f_{\perp }\perp V$. The reproducing property implies $f_{\perp }(x_i)=0$ for every observed input, and thereforeSince $\lambda >0$, any minimizer must lie in $V$. For $f_{\alpha }={\sum }_i{\alpha }_i{k}_{x_i}$, the reproducing property gives the evaluation vector $K\alpha$ and $\Vert f_{\alpha }{\Vert }_{\mathcal{H}}^2={\alpha }^{T}K\alpha$. Hence $Q_1(f_{\alpha })=M(\alpha )$. If $\hat{f}$ minimizes $Q_1$, its representation yields a minimizing $\hat{\alpha }$. Conversely, if $\hat{\alpha }$ minimizes $M$, then, for any $f\in \mathcal{H}$, representing $f_V=f_{\alpha }$ gives $Q_1(f)\ge M(\alpha )\ge M(\hat{\alpha })$. ThusThis is an equivalence of minimizers, not an existence or uniqueness assertion: the loss function is arbitrary. A singular kernel matrix also allows nonunique coefficients, since differences in $\ker K$ represent the zero function. Strict convexity of the loss was not used.

Now use the stated minimum-norm description of a sum of reproducing-kernel Hilbert spaces. Let $\hat{f}={\sum }_j{\hat{f}}_j$. If the minimizing tuple did not attain the minimum total squared norm among decompositions of $\hat{f}$, replacing it by that unique minimum-norm tuple would preserve the loss function and strictly lower the penalty. This contradicts optimality. ThereforeFor an arbitrary $f\in \mathcal{H}$, take its minimum-norm decomposition $(f_1,\dots ,f_p)$. Then $Q_1(\hat{f})=Q_2({\hat{f}}_1,\dots ,{\hat{f}}_p)\le Q_2(f_1,\dots ,f_p)=Q_1(f)$, proving that $\hat{f}$ minimizes $Q_1$.

Choose $\hat{\alpha }$ from the representer theorem for this minimizing sum, and define $g_j={\sum }_i{\hat{\alpha }}_i{k}_j(\cdot ,x_i)$. Write $K_j$ for the component kernel matrices. ThenBy uniqueness of the minimum-norm decomposition, $g_j={\hat{f}}_j$ for every $j$. This proves the common representer coefficients for a sum of kernels conclusion:The same coefficient vector works simultaneously. Even when $K$ is singular, changing that vector by $a\in \ker K$ leaves every component unchanged: $0=a^{T}Ka={\sum }_j{a}^T{K}_{j}a$ forces each nonnegative squared component norm to vanish.

Put $\hat{\Sigma }=X^{T}X/n$. The Debiased Lasso correction isFor the Nodewise Lasso, let $r_j=X_j-X_{-j}{\hat{\gamma }}^{(j)}$. Its Karush-Kuhn-Tucker conditions areso $z_{jk}=\mathrm{sgn}({\hat{\gamma }}_k^{(j)})$ at a nonzero coefficient and $|z_{jk}|\le 1$ at a zero coefficient. Multiplying by ${\hat{\gamma }}^{(j)}$ gives the nodewise-Lasso residual identityWe require nonzero design columns, a necessary qualification missing from the unrestricted printed formulation. Because ${\lambda }_j>0$, ${\hat{\tau }}_j^2=0$ would force both $r_j=0$ and ${\hat{\gamma }}^{(j)}=0$, and hence $X_j=0$. Thus nonzero columns suffice for ${\hat{\tau }}_j^2>0$ even when $p>n$ or the full Gram matrix is singular. A zero column instead gives ${\hat{\tau }}_j^2=0$ and its coefficient is unidentifiable from $Y$; the proposed normalization and confidence interval do not exist there.

For each $j$, define $c_j\in {\mathbb{R}}^p$ by $(c_j{)}_j=1$ and $(c_j{)}_{-j}=-{\hat{\gamma }}^{(j)}$. Set row $j$ of $\hat{\Theta }$ to $c_j^T/{\hat{\tau }}_j^2$. Since $X{c}_j=r_j$,This row construction is the nodewise normalization used by van de Geer and coauthors. The matrix $\hat{\Theta }$ need not be symmetric or be an actual inverse.

Substitute the linear model into the correction. The exact decomposition isIn the fixed-design normal linear model, or with $\epsilon \mid X\sim N_n(0,{\sigma }^{2}I)$ and nodewise estimates and tuning chosen from $X$ alone, a linear transformation of a multivariate normal distribution gives $W\mid X\sim N_p(0,{\sigma }^2\hat{\Omega })$. Marginal normality of $\epsilon$ without this conditional assumption would not suffice for a response-dependent random design. No independence between $W$ and $\Delta$ is asserted. The diagonal cancellation and the Holder inequality give the debiased-Lasso remainder bound

For a coordinate variance, ${\hat{\Theta }}_j{X}^T=r_j^T/{\hat{\tau }}_j^2$, whenceWriting $z_{1-\alpha /2}$ for the corresponding standard normal quantile, an approximate confidence interval isEquivalently its half-width is $z_{1-\alpha /2}\sigma \Vert X_j-X_{-j}{\hat{\gamma }}^{(j)}{\Vert }_2/(n{\hat{\tau }}_j^2)$. We assume $0<\alpha <1$ and $\sigma >0$. For asymptotic coverage, the standardized remainder ${\Delta }_j/(\sigma \sqrt{{\hat{\Omega }}_{jj}})$ must have convergence in probability to zero and the coordinate variance must be nondegenerate. Exact conditional normality of $W$ alone does not give finite-sample exact coverage for $\hat{b}$.

One sufficient random-design regime is the Gaussian identity design for debiased Lasso: take independent rows $X_i\sim N_p(0,I_p)$, independently of $\epsilon$, fixed $\sigma >0$, $p=p_n\to \infty$, $s=s_n\ge 1$, andChoose $\lambda =C\sigma \sqrt{\log p/n}$ and all ${\lambda }_j=D\sqrt{\log p/n}$ for sufficiently large fixed $C,D$. Here is why this regime has the requested remainder bound. Uniformly over columns, their squared norms divided by $n$ tend to one, while all off-diagonal sample cross-correlations have magnitude at most a constant times $\sqrt{\log p/n}$ with probability tending to one. Taking $D$ larger makes the zero vector satisfy all nodewise Karush-Kuhn-Tucker conditions strictly, so ${\hat{\gamma }}^{(j)}=0$ for every $j$ on that event, and ${\hat{\tau }}_j^2=\Vert X_j{\Vert }_2^2/n\ge 1/2$ eventually. This also gives ${\hat{\Omega }}_{jj}=1/{\hat{\tau }}_j^2\to 1$ uniformly.

In this regime, the entrywise entrywise concentration of a Gaussian Gram matrix and Lasso cone condition argument in Gaussian identity design for debiased Lasso gives a Compatibility condition for the Lasso bounded away from zero with probability tending to one. Together with the noise-score event and the Basic inequality for the Lasso, this yields, for a fixed $C_1$,For example, the cone inequality and compatibility bound give $\Vert \hat{\beta }-{\beta }^0{\Vert }_1\le 12\lambda s/{\varphi }^2$ on their joint event. Therefore, with a fixed $A\ge 2D{C}_1$,The stated sparsity condition makes this remainder vanish and supplies the preceding confidence-interval requirement. Both dimensions may grow: $p=n^2$ and $s=\lfloor n^{1/4}\rfloor$ satisfy these assumptions. The concentration and Gaussian-design compatibility results are sufficient standard conditions here; the question does not require their proofs. The displayed rate is a high-probability fixed-constant bound, rather than only a claim of boundedness in probability.

The commutator with the scalar metric term vanishes. Reorder the third gamma matrix using the Clifford algebra twice:Subtract the last term and multiply by $1/2$ to obtainThis is the infinitesimal vector transformation law implemented by the Spinor representation of the Lorentz group. The first contraction is ${\eta }^{\nu \rho }$ as printed in the PDF; the TeX conversion's ${\eta }^{\rho \rho }$ is erroneous and would fail, for example, when $\mu =0,\nu =1,\rho =2$: the actual commutator is zero.

Insert ${\Omega }_{jk}=-{ϵ}_{jkl}{\varphi }^l$ and the result for $S^{jk}$ into the Spinor representation of the Lorentz group. The contraction of two Levi-Civita symbols ${\sum }_{j,k}{ϵ}_{jkl}{ϵ}^{jkm}=2{\delta }_l{}^m$ givesThe matrix exponential of this block diagonal matrix is the stated $O$, so ${\psi }^{\prime }(x)=O\psi ({\Lambda }^{-1}x)$. The plus sign in its exponent follows from both minus signs above; it is tied to the rotation-parameter convention in the PDF.

For a rotation through $2\pi$ about $z$, the Pauli matrix has eigenvalues $\pm 1$, hence $e^{i\pi {\sigma }^3}=-I_2$. The spacetime rotation is the identity, whileThis is the spinor sign under a full spatial rotation. A vector returns to itself already at $2\pi$. The comparison concerns the exponentiated Lorentz group representations: a Lie algebra by itself has no distinguished element called a $2\pi$ rotation. The sign records the nontrivial lift of this rotation loop to the Spin group.

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.