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The printed assumptions omit the orthogonality needed for the claimed local martingales. Merely placing two Brownian motions on the same probability space does not make them independent. Interpret both as Brownian motions relative to a common filtration, and put $C_t=[B,\vartheta {]}_t$. For $X=e^B\cos \vartheta$ and $Y=e^B\sin \vartheta$, the two-variable Itô formula givesThe pure second derivatives cancel because $[B{]}_t=[\vartheta {]}_t=t$, while the mixed partial derivatives are $X_{b\vartheta }=-Y$ and $Y_{b\vartheta }=X$. Thus the mixed quadratic covariation term cannot be omitted.

For a counterexample allowed by the PDF, take $\vartheta =B$. Then $C_t=t$ and $X$ has drift $-Ydt$, while $Y$ has drift $Xdt$. In particular $X_0=1$ makes the latter drift nonzero near zero, so $Y$ is not a local martingale. Nor is $X$: $Y$ cannot be identically zero, so its accumulated continuous drift is not identically zero. Uniqueness of the continuous semimartingale decomposition excludes cancellation by another local martingale term.

The corrected hypothesis is $[B,\vartheta ]\equiv 0$, for example independent Brownian motions in their joint filtration. Conversely, if both were local martingales, uniqueness of the continuous semimartingale decomposition would force $YdC=XdC=0$; since $X^2+Y^2=e^{2B}>0$, this forces $dC=0$. Thus the correction is exactly the condition needed for both local martingales. Under this hypothesis the drift terms vanish and the intended stochastic differential equations are $dX=XdB-Yd\vartheta$ and $dY=YdB+Xd\vartheta$. The continuous integrands are locally bounded, so localization makes these Itô integrals continuous local martingales. Subsequent intended conclusions are proved under this explicitly stated correction.

Use natural units $\hslash =c=1$ and the Minkowski metric $\eta =\mathrm{diag}(1,-1,-1,-1)$ throughout. An Abelian gauge transformation is $A_{\mu }\mapsto A_{\mu }+{\partial }_{\mu }\chi$. For a sufficiently regular $\chi$, commuting the partial derivatives gives $F_{\mu \nu }\mapsto F_{\mu \nu }$, so the Maxwell Lagrangian is invariant pointwise.

Vary the action with respect to the electromagnetic four-potential, using variations of compact support or vanishing on the boundary. The antisymmetry of the electromagnetic field tensor givesAfter integration by parts, $\delta S=\int d^{4}x({\partial }_{\mu }{F}^{\mu \nu })\delta A_{\nu }$. The principle of stationary action therefore gives the source-free Maxwell equations,In the Lorenz gauge, ${\partial }_{\mu }{A}^{\mu }=0$, this becomes $\square A_{\nu }=0$ with $\square ={\partial }_{\mu }{\partial }^{\mu }$. The PDF's name “Lorentz gauge” denotes the usual Lorenz gauge, named after Lorenz. Residual gauge transformations preserve it when $\square \chi =0$; this condition is not itself a complete removal of the gauge freedom.

A Majorana spinor equals its charge conjugation, ${\psi }^c=C{\bar{\psi }}^T=\psi$, with conventional phase choices absorbed into $C$. The two-component worldsheet Majorana fermion therefore has no independent complex conjugate components; in a Majorana representation its components can be real Grassmann variables. The same condition applies to the constant supersymmetry parameter. Grassmann statistics matter in the variation: replacing the spinors by commuting numerical vectors would give the wrong bilinear interchange signs.

The displayed rigid transformation is understood in flat conformal gauge. Take $h_{\mu \nu }={\eta }_{\mu \nu }=\mathrm{diag}(-1,1)$, $\{{\gamma }^{\mu },{\gamma }^{\nu }\}=2{\eta }^{\mu \nu }$, and chooseHere $C={\gamma }^0$ is a valid invariant charge-conjugation form: $C^T=-C$ and $({\gamma }^{\mu }{)}^{T}C=-C{\gamma }^{\mu }$. These concrete matrices make all signs checkable; equivalent Majorana conventions give the same result with their consistently transformed bars.

Let $A=\sqrt{{\alpha }^{\prime }/2}$ and $B=1/\sqrt{2{\alpha }^{\prime }}$, so ${\alpha }^{\prime }B=A$. Suppressing the common factor $-1/(4\pi {\alpha }^{\prime })$ in the action, the flat kinetic density isThe rigid variations have $\delta X^a=iA\overline{ϵ}{\psi }^a$ and $\delta {\psi }^a=B{\gamma }^{\nu }{\partial }_{\nu }{X}^aϵ$. The supersymmetry variation is even, so its product rule has no extra graded sign. From the Majorana Grassmann bilinear interchange identities,For example the last identity follows by moving the Grassmann-odd parameter through $\psi$, then using $C^T=-C$ and the transpose relations twice. No equation of motion has been used.

The two kinetic variations areThe Clifford algebra turns the first fermionic term plus the bosonic variation into $iA{\partial }_{\nu }{X}_a\overline{ϵ}{\gamma }^{\mu }{\gamma }^{\nu }{\partial }_{\mu }{\psi }^a$. In the second term, commute the partial derivatives and use the same algebra: the antisymmetric gamma product drops out, leaving $iA\square X_a\overline{ϵ}{\psi }^a$. Together they are precisely the rigid worldsheet supersymmetry boundary term:Therefore $\delta S=-(4\pi {\alpha }^{\prime }{)}^{-1}{\int }_{\partial \Sigma }d{\Sigma }_{\mu }{J}^{\mu }$. It vanishes on a closed or periodic worldsheet, for compactly supported changes, or under compatible supersymmetric endpoint conditions. The local total-derivative identity is off shell; the boundary assumptions are needed to call the integrated action invariant.

The flat-gauge qualification is substantive. On an arbitrary curved worldsheet, spinors need a zweibein and spin connection and a constant spinor parameter need not exist. A fully covariant locally supersymmetric action also involves the worldsheet gravitino; the isolated ordinary-derivative expression does not prove rigid invariance for arbitrary $h_{\mu \nu }$. We have proved the intended rigid symmetry of its flat gauge-fixed action, with the fermionic term inside the same integral and overall normalization. If the displayed last term were read as outside the integral, it would not even define an action.

For phenomenology, bosonic string theory has no spacetime fermions and its usual vacuum contains a tachyon. The spinning string has a Ramond sector, whose fermionic zero modes form a spacetime Clifford algebra and give spacetime spinor states, as well as a Neveu–Schwarz sector. In suitable consistent theories the GSO projection removes the tachyonic NS ground state and keeps the appropriate Ramond chirality. The critical dimension of the RNS superstring is 10 rather than 26: $D$ bosons and $D$ real fermions have matter central charge $3D/2$ per chiral sector, while the reparameterization and superconformal ghosts contribute $-26+11=-15$, so the total anomaly cancels at $D=10$.

Suitable GSO-projected superstrings admit spacetime fermions and a tachyon-free spectrum, making them more promising for particle physics than the bosonic string. Rigid worldsheet supersymmetry alone does not establish a tachyon-free spacetime theory or realistic phenomenology. The projection, consistent sectors and compactification in string theory are additional input; spacetime supersymmetry and a realistic four-dimensional spectrum are not automatic consequences of this classical action.

The Jacobian criterion on an affine chart $x_j\ne 0$ says that a point of the projective hypersurface is singular exactly when all derivatives of its dehomogenized equation vanish there. The Euler homogeneous function theorem identity ${\sum }_i{x}_i{f}_{x_i}=df$ then recovers the omitted $j$th derivative on $X$, without dividing by $d$. Thus the correct statement in every characteristic isIf the printed $Z(I)$ denotes a zero locus inside $X$, this proves it as written. If it means the ambient projective zero locus, the assertion is false when the characteristic divides $d$. When the characteristic does not divide $d$, the Euler homogeneous function theorem forces $f=0$ at every common zero of the derivatives, and the intersection with $X$ is redundant.

For a genuine counterexample to the unrestricted ambient interpretation, in characteristic $p$ takeThis is an irreducible homogeneous polynomial: it is primitive and linear in $X_2$ over $k[X_0,X_1]$, with coprime coefficients. At $[1:0:0]$ all partial derivatives vanish, including when $p=2$, but $f=1$. This point is outside $X$. Hence the ambient-locus version needs the displayed intersection or a restriction on the degree.

PutBy the Jacobian criterion, a projective plane curve is singular only at a common projective zero of $F$ and all its partial derivatives. HereTheir simultaneous vanishing forces $X_0=X_1=X_2=0$, which is not a point of projective space. Hence the Fermat quartic $C$ is nonsingular. In particular it is a smooth plane quartic and has geometric genus $(4-1)(4-2)/2=3$.

The first law of thermodynamics isso, regarding $S$ as a function of $(T,V)$,Equality of mixed partial derivatives gives the thermodynamic derivation of blackbody radiation pressureWriting $a={\pi }^2/(15{\hslash }^3{c}^3)$, we have $E=aV{T}^4$, and henceEquivalently, $d(P/T)/dT=a{T}^2$, soThe condition $P=o(T)$ as $T\to 0$ forces $C=0$. Therefore the radiation pressure is

Fordirect differentiation givesandThus ${\theta }_t=D{\theta }_{xx}$.

Its total mass is found from the Gaussian integral:The unit-mass solution converging to the Dirac delta function as $t\downarrow 0$ is therefore the heat kernel

For the second initial condition, the heat-kernel solution is the convolutionWith $u=(x-y)/(2\sqrt{Dt})$ and the definition of the error function, this becomesThis is the heat equation with interval-indicator initial data for $a=1$.

At $t=0$ the graph is the rectangle of height one on $[-1,1]$. For every $t>0$ it is smooth, positive and even, with its maximum at $x=0$. As time increases the graph broadens and its maximum falls, while its total area remains $2$; pointwise it tends to zero as $t\to \infty$.

Let $ϵ$ be the group parameter. The infinitesimal generator of a Lie point symmetryhas flow equationsTherefore it generates the scaling group

Here ${\xi }^t=t$, ${\xi }^x=x$, and $\eta =0$. The total-derivative formulas for the second prolongation of a Lie point symmetry giveandThusFor the wave equation expression $F=u_{tt}-u_{xx}$,It therefore vanishes whenever $F=0$, proving that $V$ generates the simultaneous spacetime scaling symmetry of the wave equation.

The independent invariant of the scaling orbits is the similarity variablewhile $u$ itself is invariant. Seek a group-invariant solution $u(t,x)=f(z)$. Direct partial differentiation givesThe wave equation reduces toHence, on any interval avoiding $z=\pm 1$,

In components,because the Levi-Civita symbol is antisymmetric in $j,k$ while equality of mixed partial derivatives makes ${\partial }_j{\partial }_k\varphi$ symmetric. Hence

A gauge transformation isBecause partial derivatives commute, the added contribution to $F_{\mu \nu }$ isso the electromagnetic field tensor is gauge invariant.

DefineSince ${\partial }_0=c^{-1}{\partial }_t$ and $A_0=-\Phi /c$,

The identityfollows directly from $F=dA$. Its spatial and mixed components areFor the other two equations define the four-currentThen ${\partial }_{\nu }{F}^{0\nu }={\mu }_0{j}^0$ givesand ${\partial }_{\nu }{F}^{i\nu }={\mu }_0{j}^i$ givesThis is the Covariant Maxwell equation with the minus-plus-plus-plus metric.

Usingone findsThis is the electromagnetic energy density.

For a null vector, the trace term in $T^{\mu \nu }{k}_{\mu }{k}_{\nu }$ vanishes. PutAntisymmetry gives $q\cdot k=0$. A vector orthogonal to a null vector has nonnegative Minkowski norm, soExplicitly, in a frame with $k^{\mu }=\kappa (1,1,0,0)$,Hence the null energy condition for the electromagnetic field is strict whenever the contraction is nonzero:

Differentiability at $x\in {\mathbb{R}}^2$ means that there is a linear map $Df{|}_x:{\mathbb{R}}^2\to \mathbb{R}$ such thatThe partial derivatives are $D_{i}f(x)=Df{|}_x(e_i)$ when the derivative exists; independently, they may be defined by the corresponding one-variable difference quotients.

Suppose the partial derivatives exist near $x=(x_1,x_2)$ and are continuous at $x$. Apply the one-dimensional mean value theorem along the two coordinate segments from $x$ to $x+h$. It gives intermediate points ${\xi }_h,{\eta }_h\to x$ such thatContinuity of the partials makes the remainder after subtractingequal to $o(\Vert h\Vert )$. Hence $f$ is differentiable and this is its derivative.

For the given function, writing $r=\sqrt{x^2+y^2}$, at $r>0$ we haveAt the origin both partial derivatives are zero, since $f(h,0)/h$ and $f(0,h)/h$ tend to zero. Neither partial is continuous there: along its corresponding coordinate axis the cosine term oscillates without a limit. Neverthelessso $f$ is differentiable at the origin with derivative zero. This illustrates that existence of partial derivatives does not imply their continuity.

The final assertion is false. DefineAgain $|g|\le r^2$, so $g$ is differentiable at the origin and it is smooth elsewhere. But along the $x$-axis,is unbounded near zero, and similarly $D_{2}g(0,y)$ is unbounded. Thus both partial derivatives can be unbounded in every neighbourhood of a point even when the function is differentiable everywhere.

Write $H=\dot{a}/a$. The nonzero Christoffel symbols for the flat FLRW metric areThe mixed Einstein tensor isFor a mixed tensor,For $\mu =t$, direct substitution givesFor each spatial $\mu$, homogeneity makes the partial derivatives vanish and the remaining diagonal connection terms cancel pairwise. This verifies the flat FLRW Einstein-tensor divergence directly.

In vacuum,The $tt$ equation isChoosing the positive root gives the expanding de Sitter scale factor in flat slicingand the spatial equation is then automatically satisfied.

A rational map $X⇢Y$ between irreducible projective varieties is an equivalence class of morphisms from nonempty open subsets of $X$ to $Y$, with representatives identified when they agree on a nonempty open subset. It is regular at $p$ if some representative is defined on a neighborhood of $p$.

Forthe coordinates vanish together exactly at $(0:1:0)$ and $(1:0:0)$. Near the first point, the paths $z=0$ and $x=0$ have limiting images $(1:0:0)$ and $(0:0:1)$; near the second, the paths $z=0$ and $y=0$ have limiting images $(1:0:0)$ and $(0:1:0)$. No continuous extension exists there. Elsewhere the coordinates do not vanish together, so the map is regular.

DefineWhere all coordinates are nonzero,and similarly $\varphi \psi$ is the identity. Thus $\varphi$ is birational and is an isomorphism on ${\mathbb{P}}^2\setminus Z(xyz)$.

For irreducibility of $P=x^2{z}^4-x^3{y}^3+z^6$, dehomogenize at $z=1$. Over $k(x)$ this is, up to a unit,The rational function on the right is not a cube, since its valuation at either root of $x^2+1$ is one. The cubic is therefore irreducible over $k(x)$; Gauss's lemma gives irreducibility in $k[x,y]$. Since $z$ does not divide $P$, homogenization preserves irreducibility.

Writing $(u:v:w)=(xy:xz:z^2)$ transforms the equation intoIts partial derivatives are $-3u^2$, $2vw$, and $v^2+3w^2$ up to common signs. Simultaneous vanishing forces $u=v=w=0$, impossible projectively, so $C$ is nonsingular. The inverse $\psi$ restricts on dense open subsets, proving that $V$ and $C$ are birational.

Differentiability at $p$ means $f(p+h)=f(p)+Lh+o(|h|)$ for a linear map $L=Df(p)$; partial derivatives are its values on coordinate vectors. Integrating the partial derivatives along the two coordinate segments and using continuity at $p$ proves the stated sufficient condition. Here $f(x,y)=g(x)+g(y)$ with $g(t)=t^2\sin (1/t)$ and $g(0)=0$. Since $g\prime (0)=0$, $f$ is differentiable everywhere. But $g\prime (t)=2t\sin (1/t)-\cos (1/t)$ does not approach $0$ at zero, so $f$ is $C^1$ exactly where $xy\ne 0$.

In the affine chart $z=1$, the equation is $y^2=x^2(x+1)$, irreducible because $x+1$ is not a square in $k(x)$. The partial derivatives show the unique singular point is $P=[0:0:1]$; the point at infinity is smooth.

A line $y=tx$ through $P$ meets the curve again atThus $t\mapsto [t^2-1:t(t^2-1):1]$ is birational, with inverse $t=y/x$ away from $P$. The incidence closure has one point over every smooth $Q$ and two points over $P$, corresponding to $t=\pm 1$. This is the normalization of a nodal curve and separates the two branches.

Dominated by projective space does not imply being projective space. A smooth quadric surface $Y\simeq {\mathbb{P}}^1\times {\mathbb{P}}^1$ is rational and hence dominated by ${\mathbb{P}}^2$, but is not any ${\mathbb{P}}^k$; for example its Picard rank is $2$, not $1$.

Under a coordinate change, the inhomogeneous second-derivative terms arising from each partial derivative cancel between $V^{\alpha }{\partial }_{\alpha }{W}^{\mu }$ and $W^{\alpha }{\partial }_{\alpha }{V}^{\mu }$. The remainder transforms with one contravariant Jacobian, so the commutator is a vector field.