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Use the decomposition into Lagrangian subspaces $V=V_{+}\oplus V_{-}$ from part (a). The bilinear form $\Omega$ pairs $V_{+}$ and $V_{-}$ nondegenerately: a vector in either summand annihilating the other already annihilates its own summand and hence all of $V$, so is zero.

Choose a basis $e_1,\dots ,e_n$ of $V_{+}$. The linear mapis an linear isomorphism. Let $f_j$ correspond to the dual basis of the $e_i$. Then $\Omega (e_i,f_j)={\delta }_{ij}$, and the pairings within each summand vanish. Thus

Use the Minkowski metric ${\eta }_{mn}=\mathrm{diag}(-1,1,\dots ,1)$ and units with speed of light one. The contractions are $P^2={\eta }^{mn}{P}_m{P}_n$ and $(T{X}^{\prime }{)}^2=T^2{\eta }_{mn}(X^m{)}^{\prime }(X^n{)}^{\prime }$; $(X^m{)}^{\prime }{P}_m$ and ${\dot{X}}^m{P}_m$ pair a vector with a covector without another metric. The Lagrange multipliers impose the two first-class constraintsThey generate worldsheet diffeomorphisms, so the phase space contains both constrained directions and gauge redundancy. Two first-class constraints remove two canonical pairs, leaving $D-2$ physical degrees of freedom per point.

In Monge gauge, $X^0=t$ and $X^1=\sigma$. Write the transverse canonical variables as $\mathbf{X}=(X^2,\dots ,X^{D-1})$ and $\mathbf{P}=(P_2,\dots ,P_{D-1})$. Solving the first-class constraints givesThe negative root selects positive energy. Substitution into the phase-space action gives the Hamiltonian reductionFor a static segment, $\mathbf{P}=0$, its proper length element is $d\ell =\sqrt{1+|{\mathbf{X}}^{\prime }{|}^2}d\sigma$, and $\mathcal{E}d\sigma =Td\ell$. Thus the string tension is the rest energy per unit proper length. In particular a straight resting segment has $\mathcal{E}=T$. Monge gauge is a local choice on a string embedding map for which $X^1$ is a valid coordinate; it need not cover folded strings or all endpoint configurations.

Reflection in $y=0$ preserves the prescribed translational velocity and axial angular velocity. A force transforms as a polar vector, whereas a torque transforms as an axial vector; consequently $F_y=G_x=G_z=0$. Reflection in $x=0$ reverses both imposed motions but preserves the normal force component. The linearity of Stokes flow then gives $F_z=0$. Thus only $F_x$ and $G_y$ can be nonzero.

There is a sign inconsistency in the printed question. In right-handed coordinates, $\mathbf{\Omega }=\Omega {\mathbf{e}}_y$ gives a rotational velocity $-\Omega a{\mathbf{e}}_x$ at the sphere's lowest point. The printed shear and force formulas instead use the opposite rotational sense. Below, let $\omega$ be the right-handed component along $+{\mathbf{e}}_y$; the paper's displayed formulas are recovered by setting $\omega =-\Omega$.

In the translating frame, the parabolic gap and leading boundary velocities areThe no-slip boundary condition and lubrication theory giveThe gap is stationary in this frame, so ${\nabla }_{\parallel }\cdot \mathbf{q}=0$. The Reynolds lubrication equation is thereforeSince $h_x=x/a$ and ${\nabla }_{\parallel }^{2}h=2/a$,Differentiating the velocity profile gives the leading shear stressesThe first expression becomes the printed formula after $\omega =-\Omega$.

For the logarithmic lubrication resistance of a sphere near a wall, write $\ell =\log (1/ϵ)$. The logarithmic region is $a\sqrt{ϵ}\ll r\ll a$, where $h\sim r^2/(2a)$. ConsequentlyThese follow from $\int r/hdr=a\ell +O(a)$ and $\int r^3/h^{2}dr=2a^2\ell +O(a^2)$, with a fixed small outer cutoff. The general logarithmic radial numerator is $r^{2n-1}$ over $h^n$; the numerator in the printed integration hint appears to have a typographical error.

The force exerted on the fluid through the upper gap boundary includes pressure acting on its slope:The pressure cancels the $x^2/h^2$ term, givingPressure exerts no torque about the centre of a sphere because its traction is radial. To logarithmic order, the shear has lever arm $-a{\mathbf{e}}_z$, soThe other components vanish by parity: $G_x$ involves an integral of the $xy$ term, and $G_z$ has an integrand odd in $y$. The combined hydrodynamic resistance matrix isIts equal off-diagonal coefficients are required by the Lorentz reciprocal theorem for Stokes flow. It is a positive-definite matrix, as required by viscous dissipation.

In the rotational convention of the printed force formula, set $\Omega =-\omega$ and measure the scalar couple along $-{\mathbf{e}}_y$, so $G=-G_y$. ThenFor sedimentation under the buoyancy-adjusted weight $F$, a uniform sphere is torque-free. Thus $\omega a=U/4$, and the force balance givesThe rotational scalar in the paper's displayed-force convention is $\Omega =-U/(4a)$. The sphere rolls in the right-handed $+{\mathbf{e}}_y$ sense but still slips: its lowest point has laboratory velocity $U-a\omega =3U/4$. The signs and coefficients of the right-handed resistance agree with the rigid-wall terms of Bertin et al., equations (4.4)–(4.5), after reversing the forces and torques there from fluid-on-sphere to sphere-on-fluid.

Suppose $y$ in the capped simplex has a coordinate strictly between zero and one. Since ${\sum }_i{y}_i=k$ is an integer, it cannot have exactly one such coordinate: the other coordinates would contribute an integer sum. Hence there are distinct $i,j$ with $0<y_i,y_j<1$.

Choose $0<\epsilon <\min \{y_i,1-y_i,y_j,1-y_j\}$ and set $y^{\pm }=y\pm \epsilon (e_i-e_j)$, where $e_i,e_j$ are standard basis vectors. Both perturbed vectors satisfy the coordinate bounds and have the same coordinate sum. They are distinct and $y=(y^{+}+y^{-})/2$, so $y$ is not an extreme point.

Conversely, every zero-one vector in this convex polytope is an extreme point. If it were a nontrivial convex combination of two feasible vectors, each coordinate equal to zero would force both corresponding coordinates to be zero, and each coordinate equal to one would force both to be one. Thus the two vectors would equal the original one. The extreme points are therefore exactly the indicators of $k$-element subsets.

The required linear program isThere are $2n$ inequality constraints and one equality constraint, all independent of $x$.

The feasible set is the capped simplex, a nonempty compact set. A linear function attains its maximum at an extreme point; the previous part identifies these as the indicators of $k$-element subsets. At such a vector, the objective is the sum of the selected coordinates. Maximizing selects the $k$ largest coordinates, including when some are negative or tied. Equivalently, $f_k$ is the support function called the sum of the largest components.

For a proper three-graph colouring $c$, associate the colors with the three roots of unity $d(i)=\exp (2\pi i(c(i)-1)/3)$, viewed as unit vectors in ${\mathbb{R}}^2$. Define $U_{ij}=2⟨d(i),d(j)⟩$ using the real inner product.

This is twice a Gram matrix, so it is a positive semidefinite matrix. Its diagonal is $2$. On every edge, the endpoint colors differ, so their vectors make angle $2\pi /3$ or $4\pi /3$ and $U_{ij}=2\cos (2\pi /3)=-1$. Thus $t=3$ and this $U$ are feasible in the second semidefinite program, provingThis is the regular simplex construction of a strict vector coloring.

Let $w=(1,2,1,0,0{)}^T$. Direct multiplication gives $Hw=(0,0,0,2,2{)}^T$, so $w^{T}Hw=0$.

Suppose $H=P+N$, where $P$ is a positive semidefinite matrix and $N$ is a symmetric nonnegative matrix. Since $w\ge 0$, both $w^{T}Pw$ and $w^{T}Nw$ are nonnegative. Their sum is zero, so both vanish. In particular,Every coefficient $w_i{w}_j$ in this sum is strictly positive and every $N_{ij}$ is nonnegative, forcing $N_{ij}=0$ for $1\le i,j\le 3$.

Now apply the same argument to all five cyclic shifts of $w$. The Horn copositive matrix is cyclically invariant, so every shifted vector also has zero quadratic form. It follows that the entries of $N$ vanish on every cyclic block of three consecutive indices. Every pair of indices on a five-cycle lies in such a block, hence $N=0$.

This would imply $P=H$. But the zero quadratic form of a positive semidefinite matrix would then force $Hw=0$, contradicting $Hw=(0,0,0,2,2{)}^T$. Therefore $H$ lies outside the positive-semidefinite-plus-nonnegative cone. By the previous equivalence, its quartic form is a nonnegative polynomial that is not a sum of squares polynomial.

Let $f_1,f_2\in V^{\ast }$ satisfy $f_1(v)f_2(v)=0$ for every vector $v$. If $f_1=0$ there is nothing to prove. Otherwise choose $u$ with $f_1(u)\ne 0$. For every $v\in \ker f_1$ and every real number $t$,so the hypothesis gives $f_2(u+tv)=0$. Taking two values of $t$ shows that $f_2(u)=f_2(v)=0$. Sincewe obtain $f_2=0$. Thus one of the two linear functionals is zero.

Sylvester's law of inertia says that, in a suitable basis, every real quadratic form isThe rank is $r=p+m$, and the signature is $\sigma =p-m$; both are independent of the chosen basis.

If $q=f_1{f}_2$, its polar form has image contained in $\mathrm{span}\{f_1,f_2\}$, so $r\le 2$. If $r=2$, the two functionals must be independent and, after taking their sum and difference, $q$ is a difference of two squares. Hence $\sigma =0$. If $r=1$, then $|\sigma |=1$. In every case,Conversely, this inequality leaves onlyup to a linear change of coordinates. These factor respectively as $0$, $(\pm x_1)x_1$, and $(x_1-x_2)(x_1+x_2)$. Therefore

Finally suppose that $q$ takes both positive and negative values. Sylvester's law supplies normalized basis vectorswith $q(e_i)=1$, $q(f_j)=-1$, and the $z_l$ spanning the radical. The $p+m+k$ vectorsare linearly independent and all satisfy $q(v)=0$. They form the required basis of isotropic vectors.

Choose an orthonormal eigenbasis $e_i$ of $G$ and set $u_i=e_i$, $v_i=A{e}_i/\sqrt{{\lambda }_i}$. Part (a) shows that the $v_i$ form an orthonormal eigenbasis of $H$. If $U$ and $V$ have these vectors as columns, then they are unitary andThis is the singular value decomposition.

Herewith eigenvalues $4,1$ and corresponding normalized vectors $(1,-1{)}^T/\sqrt{2}$ and $(1,1{)}^T/\sqrt{2}$. A compatible choice isDirect multiplication gives $V^{†}AU=D$.

Expanding with the Levi-Civita symbol and the product rule givesThe Stokes theorem stateswhere the boundary orientation follows the right-hand rule about $n$.

Apply it to $F=m\times v$. On $S$, $m=n$ and $v\cdot n=0$. Substitution of the vector identity and contraction with $n$ leavesThe boundary integrand satisfies $(n\times v)\cdot dx=v\cdot (dx\times n)$, proving the formula. The vector $dx\times n$ is the outward co-normal in the tangent plane of $S$.

For the hemisphere, $n=e_r$ and $v=r\sin \theta e_{\theta }$. The projected divergence is $2\cos \theta$, so the left side isOn the positively oriented equator, $dx=R{e}_{\varphi }d\varphi$, hence $dx\times n=R{e}_{\theta }d\varphi$ and the boundary integral is also $2\pi R^2$.

Momentum conservation makes the sum of the two photon momenta parallel to $u$, so all three vectors are coplanar. Put the photons on opposite sides of the incident direction. Transverse and longitudinal momentum and energy conservation giveEliminating $E_1,E_2$ yieldsUsing the half-angle identities on the left gives exactly

For $\psi (x)=C{e}^{-\alpha x^2}$,The stationary Schrodinger equation becomesHence, choosing the positive root and a positive normalization constant,The state is even, so $⟨x⟩=⟨p⟩=0$. The Gaussian integrals giveand therefore $\Delta x\Delta p=\hslash /2$. These values are collected in gaussian eigenstate for a quadratic potential.

Finally, equality in the derivation of the uncertainty relation requires the centred vectors to be linearly dependent with a purely imaginary proportionality constant. Thus for some $s>0$,In position space this sayswhose normalizable solutions areThus every saturating state is Gaussian up to translation, a plane-wave factor, and an overall phase, as in the equality case of the Heisenberg uncertainty relation.

Dot the Ampère-Maxwell equation with $E$, dot Faraday's law with $B/{\mu }_0$, and use the stated vector identity. The result is the local balance lawHere $S$ is the Poynting vector. Integration over $V$ gives the Poynting theoremThe field energy decreases through outward electromagnetic energy flux and through work done on charges.

Choose bases of $V$ and $W$, of dimensions $v$ and $w$. A linear map is uniquely determined by the arbitrary images of the $v$ basis vectors, each with $w$ coordinates, so

Put $a=\dim A$ and $b=\dim B$, and extend bases of $A,B$ to bases of $V,W$. The condition $\varphi (A)\subseteq B$ forces the $(w-b)a$ matrix entries from $A$ to a complement of $B$ to vanish and imposes no other restriction. Hence

For the last part let $r=\dim (S\cap T)=s+t-v$. ChooseA map in $Y$ sends $S\cap T$ into itself, $S_0$ into $S$, and $T_0$ into $T$, with these choices independent. Therefore

The minimum-cost flow problem chooses, for each directed edge $(i,j)\in E$, a flow $x_{ij}$ and solvessubject toFeasibility requires ${\sum }_i{b}_i=0$.

Set $y_{ij}=x_{ij}-M_{ij}$. Thenand its balance vector isThe objective becomes $\sum C_{ij}{y}_{ij}$ plus the constant $\sum C_{ij}{M}_{ij}$. Translation by $M$ is a bijection between feasible flows and preserves their ordering by cost, so the transformed zero-lower-bound problem is equivalent.