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Call this simultaneous antipodal quotient of two spheres $M$. The simultaneous antipodal map acts freely on $S^2\times S^2$, so $M$ is a connected closed manifold of dimension four. Each antipodal factor has mapping degree $-1$; their product preserves orientation. Hence $M$ is orientable.

The product $S^2\times S^2$ is simply connected, so this double covering space is the universal cover. Consequentlyusing the abelianization of the fundamental group. The Euler characteristic under a finite covering gives $\chi (M)=4/2=2$. Its rational first Betti number is zero, and Poincare duality gives $b_3=b_1=0$ and $b_0=b_4=1$. Thus $b_2=0$ as well.

The universal coefficient theorem for cohomology now gives $H^1(M;\mathbb{Z})=0$ andHere $\mathrm{Hom}(H_2(M;\mathbb{Z}),\mathbb{Z})=0$ because $b_2=0$. Integral Poincare duality further gives $H^3(M;\mathbb{Z})\cong H_1(M;\mathbb{Z})=\mathbb{Z}/2$, and $H^4(M;\mathbb{Z})=\mathbb{Z}$.

Let $a,b,c$ generate these groups in degrees $2,3,4$ respectively. The only potentially nonzero positive-degree cup product is $a^2$. But $2a^2=(2a)a=0$, while $H^4(M;\mathbb{Z})$ has no nonzero torsion subgroup, so $a^2=0$. Every other positive product vanishes by dimension. Thuswith every product of positive-degree elements equal to zero. The degree-three torsion is essential: it would be lost by computing only rational cohomology.

Write $A=S^2\times S^2$ and let $i:A\to X$ be its inclusion after the attachment. The cohomology ring of a product of two spheres iswhere $a,b$ come from the first and second factors and $ab$ is the chosen orientation class. The diagonal pulls both $a$ and $b$ back to the same generator of $H^2(S^2;\mathbb{Z})$.

There is one new three-cell. In the cellular chain complex, its boundary has coordinates $(1,1)$ in the two-dimensional cells. Thus the relevant differential isand the original four-cell still has zero boundary. This gives $H_2(X;\mathbb{Z})=\mathbb{Z}$, $H_3(X;\mathbb{Z})=0$, $H_4(X;\mathbb{Z})=\mathbb{Z}$, and no other positive homology groups. Equivalently, the relative cohomology sequence of $(X,A)$ gives an injective restriction in degree two with image $\mathbb{Z}(a-b)$, and an isomorphism in degree four.

Choose $u\in H^2(X;\mathbb{Z})$ and $v\in H^4(X;\mathbb{Z})$ by $i^{\ast }u=a-b$, $i^{\ast }v=ab$. Naturality of the cup product givesSince restriction is injective in degree four, $u^2=-2v$. All further positive-degree cup products vanish by dimension. ThereforeChanging the sign of $v$ would instead give $u^2=2v$; the displayed sign uses the product orientation $ab$. The factor $2$ is the characteristic feature of a cup square after a diagonal sphere attachment.

The intermediate value theorem states that a real-valued continuous function on a closed interval takes every value between its endpoint values. Explicitly, if $v$ lies between $f(s)$ and $f(t)$, there exists $u\in [s,t]$ with $f(u)=v$. No monotonicity is assumed.

Use the stated Fredholm definition of the essential spectrum, which applies to general bounded operators:For each $z$, part (f) applied to $L-zI$ says that $L-zI$ is Fredholm exactly when $L+K-zI$ is Fredholm. Taking the complement of these equivalent Fredholm conditions givesThis is the general Weyl theorem for bounded compact perturbations. Individual isolated eigenvalues can change; the Fredholm obstruction, and hence this essential spectrum, cannot.

We prove compact perturbation invariance of Fredholm operators using a two-sided inverse modulo compact operators. If $L$ is Fredholm, decomposeThe restriction of $L$ from $(\ker L{)}^{\perp }$ to its closed range is a bounded bijection, so the bounded inverse theorem supplies a bounded inverse there. Extend that inverse by zero on $(\mathrm{ran}L{)}^{\perp }$ to obtain a bounded operator $B$. If $P,Q$ are the orthogonal projections onto $\ker L$ and $(\mathrm{ran}L{)}^{\perp }$, respectively, thenBoth $P$ and $Q$ have finite rank.

Set $T=L+K$. Products of a compact operator and a bounded operator are compact: on one side compactness preserves compact images, and on the other the bounded operator maps the unit ball into a bounded ball. Hencewhere $C,D$ are compact.

Here is why these identities force $T$ to be Fredholm. On $\ker T$, $C=-I$, so part (a) and compactness make $\ker T$ finite-dimensional. There is a positive constant $c$ withOtherwise unit vectors $h_n\perp \ker T$ could satisfy $T{h}_n\to 0$. From $h_n=-C{h}_n+BT{h}_n$ and compactness, a subsequence would converge strongly to a unit vector $h\perp \ker T$ with $Th=0$, which is impossible. The lower bound proves closed range: for a convergent sequence $T{h}_n$, first discard the kernel components, then the remaining $h_n$ are Cauchy and their limit maps to the desired range limit.

For finite codimension, take the adjoint operator of $TB=I+D$, giving $B^{\ast }{T}^{\ast }=I+D^{\ast }$. The operator $D^{\ast }$ is compact: the finite-rank approximation proved in part (e) gives finite-rank adjoints converging in operator norm to $D^{\ast }$, and norm limits of compact operators are compact. On $\ker T^{\ast }$, $D^{\ast }=-I$, so this kernel is finite-dimensional. Finally,Because the range is closed, its cokernel is isomorphic to this finite-dimensional orthogonal complement. This verifies all three Fredholm operator conditions for $L+K$.

For the converse, start with $L+K$ and perturb by the compact operator $-K$. Therefore $L$ is Fredholm if and only if $L+K$ is Fredholm, with no self-adjointness assumption.

We first prove the needed Fredholm alternative for a compact operator without assuming self-adjointness. If $C$ is compact, the finite-rank approximation theorem for compact operators on a Hilbert space follows directly by covering the compact closure of $C$'s unit-ball image by finitely many balls of radius $\epsilon$, spanning their centers by a finite-dimensional space $E$, and using the orthogonal projection $P_E$. Then $\Vert C-P_{E}C\Vert \le \epsilon$.

Choose a finite-rank $F$ with $\Vert C-F\Vert <1$, put $E_0=C-F$, and factorThe first factor is invertible by the Neumann series, and $G$ has finite rank. With $V=\mathrm{ran}G$ and $H=V\oplus V^{\perp }$, the second factor has the triangular formIt is injective precisely when the finite-dimensional block $I_V-G{|}_V$ is injective, and onto precisely when that block is onto. These properties are equivalent in finite dimension. This proves that $I-C$ is injective if and only if it is onto; in that case its inverse is bounded.

For $\lambda \ne 0$, apply this result to $C=K/\lambda$. If $\lambda$ lies in the spectrum of a bounded operator $K$, then $K-\lambda I$ is not invertible, hence is not injective. Therefore every nonzero spectral point of $K$ is an eigenvalue. On its eigenspace $E_{\lambda }$, the operator $K$ equals $\lambda I$. Compactness of $K$ would make the closed unit ball of $E_{\lambda }$ compact after dividing by $\lambda$; part (a) forces $\dim E_{\lambda }<\infty$.

It remains to rule out infinitely many distinct eigenvalues away from zero. Suppose $|{\lambda }_n|\ge \epsilon >0$ are distinct, with respective eigenvectors $v_n$. Distinct eigenvalues give linearly independent eigenvectors. Let $V_n=\mathrm{span}(v_1,\dots ,v_n)$ and choose a unit vector $h_n\in V_n\cap V_{n-1}^{\perp }$. Since $V_n$ is invariant and $(K-{\lambda }_{n}I)V_n\subset V_{n-1}$,For $m<n$, $K{h}_m\in V_{n-1}$, and orthogonality gives $\Vert K{h}_n-K{h}_m\Vert \ge |{\lambda }_n|\ge \epsilon$. This contradicts compactness. Thus for each $\epsilon >0$ there are only finitely many distinct spectral points with modulus at least $\epsilon$.

Taking $\epsilon =1/j$ shows that the nonzero spectrum is countable; if it is infinite, its distinct elements can be enumerated as a sequence tending to zero. Since the spectrum is closed, zero then also belongs to it. In an infinite-dimensional space zero belongs to the spectrum even if there are only finitely many nonzero spectral points: an invertible compact $K$ would make $I=K^{-1}K$ compact, contradicting part (a). The only possible accumulation point is zero, and every nonzero eigenspace is finite-dimensional. Zero need not be an eigenvalue. The argument uses invariant spans, not an orthogonal eigenbasis, which is unavailable for a general non-self-adjoint compact operator.

Suppose the compact operator $K$ acts on $H$, and $h_n\rightharpoonup h$. The Uniform boundedness principle makes $(h_n)$ bounded, so its images form a relatively compact set. Bounded linearity gives $K{h}_n\rightharpoonup Kh$: for every $g$, $⟨K{h}_n,g⟩=⟨h_n,K^{\ast }g⟩\to ⟨Kh,g⟩$. Every norm-convergent subsequence of $(K{h}_n)$ must therefore converge to $Kh$. If the whole sequence did not converge to $Kh$ in norm, a subsequence staying a fixed positive distance from $Kh$ would have a norm-convergent further subsequence, a contradiction. This proves that compact operators send weak convergence to norm convergence.

Conversely, assume the stated complete continuity property. Its strong limit necessarily equals $Kh$, because bounded linearity still supplies the weak limit. Every bounded sequence in a Hilbert space has a weakly convergent subsequence by weak sequential compactness in a Hilbert space. Applying the assumption to that subsequence gives a norm-convergent subsequence of the images. Thus the image of the unit ball is relatively sequentially compact and hence relatively compact. Complete continuity and compactness are equivalent on a Hilbert space.

If $|z|>\Vert L\Vert$, the Neumann series givesThe series converges in operator norm because $\Vert L/z\Vert <1$. HenceThe same argument applied to small perturbations of an invertible operator shows that the resolvent set is open and the spectrum is closed.

Suppose $L=L^{\ast }$ and $\mathrm{Im}z\ne 0$. The quadratic form $⟨Lh,h⟩$ is real, so the Cauchy-Schwarz inequality impliesThus $L-zI$ is bounded below by $|\mathrm{Im}z|$: it is injective and has closed range. Its adjoint operator $L-\bar{z}I$ has the same lower bound and trivial kernel. Sincethe range is dense as well as closed, and therefore is all of $H$. The inverse has norm at most $|\mathrm{Im}z{|}^{-1}$. For a bounded self-adjoint operator, $\Sigma (L)\subset \mathbb{R}$. The closed-range and adjoint steps are essential: merely excluding nonreal eigenvalues would not exclude all nonreal spectral points.

Work over a complex Hilbert space; for a real space one first complexifies to discuss a complex spectrum. A bounded linear operator $L$ is self-adjoint when $L=L^{\ast }$, where the adjoint operator is defined byEquivalently, $⟨Lh,k⟩=⟨h,Lk⟩$ for every pair of vectors.

The spectrum of a bounded operator isIts complement is the resolvent set. On a Banach space, a bounded bijective operator has a bounded inverse by the bounded inverse theorem, so failure of invertibility is equivalent to failure of bijectivity. An eigenvalue is one possible spectral point, but an injective operator that is not onto can also contribute to the spectrum.

If $H$ is a finite-dimensional Hilbert space, a choice of orthonormal basis identifies its closed unit ball with a closed bounded ball in a finite-dimensional Euclidean space. The Heine-Borel theorem makes it compact in the norm topology.

Conversely, if $H$ is infinite-dimensional, choose a unit vector $e_1$ and inductively choose $e_{n+1}$ orthogonal to the span of $e_1,\dots ,e_n$. The finite span is a proper closed subspace, so its orthogonal complement contains a nonzero vector that can be normalized. The resulting orthonormal sequence lies in the closed unit ball and satisfiesIt has no norm-convergent subsequence. Sequential compactness is equivalent to compactness in a metric space, hence the closed ball is not compact. This proves the norm-compact unit ball criterion: a Hilbert space has a norm-compact closed unit ball exactly when it is finite-dimensional.

For smooth vorticity, parts (b)--(c) and the transport equation give $\omega (t,S_{0,t}(y))={\omega }_{\mathrm{in}}(y)$ and measure preservation. The change of variables formula therefore conserves $\Vert \omega (t){\Vert }_1$ and $\Vert \omega (t){\Vert }_{\infty }$. Parts (d)--(e) give a uniformly bounded velocity and a uniform log-Lipschitz modulus; applying the time-dependent version of part (f) constructs unique global characteristic curves.

For a rough Yudovich characteristic flow, state the needed temporal hypothesis explicitly:The usual weak Euler solution class provides these conditions and the initial time trace. On $[0,T]$, put $A(t)=\Vert \omega (t){\Vert }_1+\Vert \omega (t){\Vert }_{\infty }$. The velocity bounds are $|u(t,x)|\le CA(t)$ and $|u(t,x)-u(t,y)|\le CA(t)\mu (|x-y|)$, with $A$ integrable on this interval.

For completeness, mollify $u$ spatially. The resulting smooth velocities have trajectories and obey the same speed and modulus bounds, up to one common constant. Their speeds have the integrable majorant $CA(t)$, giving uniform boundedness and equicontinuity of trajectories. Arzela-Ascoli theorem supplies a uniformly convergent subsequence. The modulus estimate controls $u(t,X_{\epsilon })-u(t,X)$, while the spatial mollification error tends to zero with an integrable majorant. Thus the limit solvesReplace $Ct$ by $C{\int }_0^{t}A(s)ds$ in the Osgood uniqueness criterion proof. It gives uniqueness and prevents finite-time escape. The characteristics are globally uniquely defined for each given Euler solution in the usual Yudovich class. This constructs the flow for a fixed solution; it is not by itself the entire uniqueness proof for the nonlinear Euler equation.

Merely saying that both spatial norms are finite separately at each time does not state the local temporal bound or measurability used in this argument. The printed formulation must be interpreted in the usual solution class, or supplemented with these temporal conditions. Also, an initial datum here is a function of $x\in {\mathbb{R}}^2$; the extra time variable in the displayed initial-data space is extraneous.

Literally, the printed hypothesis with positive $C$ is impossible for pairs at distance less than one, even when $U$ is constant. Use the corrected log-Lipschitz modulus $\mu$ from part (e). Local existence for the continuous finite-dimensional vector field follows from the Peano existence theorem. Boundedness gives $|X(t)-X(s)|\le \Vert U{\Vert }_{\infty }|t-s|$. Thus a trajectory cannot escape to infinity in finite time; at a finite endpoint it has a limit, and local existence there extends it. Solutions exist globally.

For uniqueness, let $X,Y$ have the same initial point and set $\delta (t)=|X(t)-Y(t)|$. This absolutely continuous function satisfies, almost everywhere,This is the Osgood uniqueness criterion, rather than the usual linear Gronwall inequality, because the velocity need not be Lipschitz continuous. To see the argument directly, put $w_{\epsilon }=\delta +\epsilon$. As long as $w_{\epsilon }\le 1$, monotonicity of $\mu$ givesFor $z_{\epsilon }=1-\log w_{\epsilon }$, this becomes $z_{\epsilon }^{\prime }\ge -C{z}_{\epsilon }$. Integration givesFor any fixed finite interval, this upper bound tends uniformly to zero as $\epsilon \downarrow 0$. Taking $\epsilon$ small prevents a first exit through $w_{\epsilon }=1$, so the bound remains valid throughout that interval. Thus $\delta =0$. Backward time has the same proof. Every initial point has exactly one global trajectory.

The decisive fact is ${\int }_0^{1}dr/\mu (r)=\infty$. For different nearby initial points, the same comparison givesuntil the separation reaches one. This quantitative continuous dependence makes the flow a homeomorphism, although it need not be a smooth diffeomorphism.

For $0<r<1$, $r\log r<0$. The printed upper bound is negative whenever $\Omega \ne 0$, so it cannot bound a nonnegative difference. The correct small-distance modulus is $r\log (1/r)$, and the radius in the hint must also be positive.

Write $h=|x_1-x_2|\le e^{-1}$, $x_3=(x_1+x_2)/2$, and $M=h\log (1/h)$. Then $h\le M\le e^{-1}$. Denote the planar vorticity velocity kernel by $K$. It satisfies $|K(z)|\le C/|z|$ and $|DK(z)|\le C/|z{|}^2$. Split the integral at radii $M$ and $1$ about $x_3$.

On $|y-x_3|\ge 1$, every point of the segment joining $x_1$ and $x_2$ remains at least $|y-x_3|-h/2\ge (1-h/2)|y-x_3|$ from $y$. The mean value theorem bounds the kernel difference by $Ch/|y-x_3{|}^2$, giving a contribution at most $Ch\Vert \Omega {\Vert }_1$.

On $M\le |y-x_3|\le 1$, the same estimate holds because $h/2\le M/2$. Polar integration givesHere $\log (1/M)\le \log (1/h)$ because $M\ge h$.

On $|y-x_3|<M$, use the sum of the two kernel magnitudes instead of their derivatives. This disk lies inside a disk of radius $M+h/2\le 3M/2$ about either $x_i$. Hence its contribution is bounded by $CM\Vert \Omega {\Vert }_{\infty }$. Combining the three regions yieldsAt coincident points the difference is zero. At larger distances use the boundedness from part (d). A convenient global log-Lipschitz modulus isgiving $|U[\Omega ](x)-U[\Omega ](y)|\le C(\Vert \Omega {\Vert }_1+\Vert \Omega {\Vert }_{\infty })\mu (|x-y|)$. This is positive, continuous and nondecreasing. It also proves continuity of $U[\Omega ]$.

The planar vorticity velocity kernel has magnitude $(2\pi |z|{)}^{-1}$. Split its defining integral into $|x-y|<R$ and $|x-y|\ge R$ for any $R>0$. The singular part is absolutely integrable in two dimensions:In particular $R=1$ provesWhen both norms are nonzero, optimization in $R$ also gives $\Vert U[\Omega ]{\Vert }_{\infty }\le \sqrt{2/\pi }\Vert \Omega {\Vert }_1^{1/2}\Vert \Omega {\Vert }_{\infty }^{1/2}$. If either norm is zero, $\Omega =0$ almost everywhere. Absolute convergence supplies a well-defined velocity at every $x$, with these uniform bounds.

Let $S_{s,t}(y)$ be the position at time $t$ of the characteristic curve satisfying $X(s)=y$ and $\dot{X}=u(t,X)$. Smooth uniqueness implies $S_{t,s}=S_{s,t}^{-1}$. Differentiating in the initial point gives the variational equationThe Liouville formula for a fundamental matrix, or the derivative formula for a determinant, therefore givesBecause $u=(-{\varphi }_{x_2},{\varphi }_{x_1})$, the mixed second derivatives cancel and $\mathrm{div}u=0$. The initial Jacobian determinant is one, soThus this Hamiltonian flow preserves planar Lebesgue measure. This calculation assumes smoothness as in the part; a rough-flow conclusion later must be justified by approximation rather than by differentiating a merely continuous velocity.

Take canonical coordinates $q=x_1$ and $p=x_2$. The Hamiltonian system isIn the standard Hamilton's equations $\dot{q}={\partial }_{p}H$ and $\dot{p}=-{\partial }_{q}H$, its Hamiltonian isAn arbitrary function of $t$ can be added without affecting the characteristic curves. The Hamiltonian generally depends on time through $\omega$ and need not be conserved along a trajectory.

Use the conventional Laplace operator $\Delta ={\partial }_{x_1}^2+{\partial }_{x_2}^2$. The fundamental solution of the Laplace equation in two dimensions is $G(x)=(2\pi {)}^{-1}\log |x|$, with $\Delta G={\delta }_0$. To check its sign, integrate the outward normal derivative around a circle: ${\int }_{\partial B_{\epsilon }}{\partial }_n\log |x|ds=2\pi$. Applying integration by parts against a smooth compactly supported test function and shrinking the circle gives the distributional identity.

The printed stream function has $\varphi =-G\ast \omega$. For each multi-index $\beta$ with $|\beta |\le 2$, transfer derivatives to the compactly supported $C^2$ function:The logarithmic singularity is locally integrable. On a bounded region in $x$, the convolution uses one fixed bounded region in the integration variable; splitting off a small disk around the singularity proves continuity of these derivatives. Thus $\varphi \in C^2$, and convolution of the distributional identity givesThe asserted plus sign is false for the printed kernel and the conventional Laplacian. Any nonzero smooth compactly supported $\omega$ is a counterexample. To obtain $\Delta \varphi =\omega$, replace the kernel's minus sign by a plus sign, or explicitly adopt the negative Laplace operator. Below, retain the printed kernel and velocity convention. With that convention, $\mathrm{curl}u=\Delta \varphi =-\omega$, so $\omega$ is the negative of conventional scalar vorticity. This sign does not change the transport or flow arguments.

For a finite-dimensional memoryless quantum channel $\Lambda$, the Holevo-Schumacher-Westmoreland theorem identifies its product-state classical capacity with the optimized output Holevo quantity:With product codewords and a collective measurement on the outputs, every rate below this value is achievable with vanishing error; no larger rate can be reliable under that product-input restriction. For a fixed classical-quantum output alphabet the corresponding optimized entropy difference gives its classical coding capacity. If entangled inputs across channel uses are also allowed, the general unassisted capacity is the regularized value $C(\Lambda )={\lim }_{n\to \infty }{n}^{-1}{\chi }^{\ast }({\Lambda }^{\otimes n})$. The requested calculation below concerns the single-use optimized product-input expression, so no unproved additivity assumption is needed.

For a Killing vector field $\xi$, set $K^{\mu \nu }={\nabla }^{\mu }{\xi }^{\nu }=-K^{\nu \mu }$. Choose surface orientation conventions so that the Komar charge, up to overall normalization, isThis is the Komar current from trace-reversed stress-energy construction. The equality is the spacetime Stokes theorem. With $[{\nabla }_{\mu },{\nabla }_{\nu }]V^{\rho }=R^{\rho }{}_{\sigma \mu \nu }{V}^{\sigma }$, the Killing equation and ${\nabla }_{\nu }{\xi }^{\nu }=0$ give $J_{\xi }^{\mu }=R^{\mu }{}_{\nu }{\xi }^{\nu }$. In four dimensions, at zero cosmological constant, the Einstein field equations give the currentReversing orientation changes the common overall sign, not conservation. The contracted Bianchi identity givesAn isometry preserves scalar curvature, and the second term contracts symmetric and antisymmetric tensors. Equivalently stress-energy conservation, the Killing equation, and ${\mathcal{L}}_{\xi }T=0$ cancel the matter expression; its trace derivative needs this last observation. Hence $\boxed{{\nabla }_{\mu }{J}_{\xi }^{\mu }=0}$. Nonzero cosmological constant adds $\Lambda {\xi }^{\mu }$ to the geometric current, also divergence-free.

The qualification concerns gravitational energy. Local matter energy exists: an observer $u$ measures $T_{\mu \nu }{u}^{\mu }{u}^{\nu }$. What is absent is a generally covariant local gravitational stress-energy density with the required universal conservation properties. The Equivalence principle allows the connection to vanish at a point in freely falling coordinates; standard gravitational energy expressions depend on coordinates. Curvature does remain, but supplies no unique gravitational energy tensor of the required kind. A nonstationary geometry also lacks a preferred time-translation Killing vector field. For a trial timelike $t$, stress-energy conservation gives ${\nabla }_{\mu }(T^{\mu \nu }{t}_{\nu })=T^{\mu \nu }{\nabla }_{(\mu }{t}_{\nu )}$, not generally zero.

An internal charge instead has a current ${\nabla }_{\mu }{j}^{\mu }=0$ independently of spacetime time translations. Its hypersurface flux is conserved with suitable boundary conditions. For electric charge, the Gauss law expresses it as a surface flux. Observer-dependent charge density does not prevent a well-defined total charge.

For an asymptotically flat spacetime, take asymptotically Cartesian slice coordinates with ${\gamma }_{ij}={\delta }_{ij}+h_{ij}$ and appropriate falloff. In $c=1$ units, the ADM energy isIt is an asymptotic gravitational energy. The dominant energy condition requires $-T^{\mu }{}_{\nu }{u}^{\nu }$ to be future-directed nonspacelike or zero for every future timelike $u$, in particular $T_{\mu \nu }{u}^{\mu }{u}^{\nu }\ge 0$. Together with the constraint equations, completeness and appropriate asymptotic/boundary hypotheses, it yields $E_{\mathrm{ADM}}\ge |{\mathbf{P}}_{\mathrm{ADM}}|\ge 0$. The energy condition alone is not the complete positive-energy theorem.