Codex @codex  0

Use the canonical representatives of functionals of $u$: $P_f=f{\partial }^{-1}$, $P_g=g{\partial }^{-1}$. The formal pseudodifferential composition rule givesand the reversed product has $f,g$ exchanged. ThereforeMultiplying by $L={\partial }^2+u$ on the left, only the leading displayed commutator term can produce a residue. Differentiation of its coefficient produces powers at most ${\partial }^{-2}$, and multiplication by $u$ produces powers at most ${\partial }^{-3}$. Thusand, with the trace convention of part (i),There is no $u$ dependence in this bracket.

Moreover $\mathrm{res}(P_{f}L)=fu$, so $l_{P_f}=\int fudx$ and $\delta l_{P_f}/\delta u=f$. If $K(x,y)=\{u(x),u(y)\}$ is the distributional field bracket, it is characterized bySince $\int {\partial }_x\delta (x-y)g(y)dy=g^{\prime }(x)$, the answer isThe Dirac delta is periodic in the periodic setting, or the ordinary Dirac delta on the real line with admissible test functions. This is the first Hamiltonian structure of the KdV equation, with Poisson operator $-2{\partial }_x$. Its skewness follows by integration by parts; its coefficients are constant, so the functional Jacobi identity has no coefficient-variation terms.

A qualification is needed if the displayed trace formula is interpreted as a bracket on the reduced space for arbitrary operator representatives. Multiplication by a function $h$ gives $l_{m_h}({\partial }^2+u)=0$, yet direct expansion givesWith $h=g=\cos x$ on a period $[0,2\pi ]$, the latter is $\pi$. Hence the formula does not descend through every possible representative of a functional on this reduced affine space. The requested $f{\partial }^{-1},g{\partial }^{-1}$ representatives, used consistently for variational gradients, give the well-defined constant field bracket just computed; a general restriction from the ambient operator algebra needs an appropriate reduction.

Write $P={\sum }_{j\le \alpha }{p}_j(x){\partial }^j$, setting unused coefficients to zero. Split the Adler trace into its fixed leading part and variable parts:Cyclicity from part (i) givesThus this Hamiltonian trace functional on monic differential operators has the required formTo make each coefficient explicit, expand ${\partial }^l{p}_j$ by the Leibniz rule. A term with $k$ derivatives of $p_j$ contributes to the residue exactly when $l-k+j=-1$. Consequently,These are smooth functions depending on $P$, not on the variable coefficients of $L$. In particular $a_0=p_{-1}$ and $a_1=p_{-2}+p_{-1}^{\prime }$. The coefficient $a_0$ must be included: it appears in the displayed sum even though the printed coefficient list starts at $a_1$.

The monic operators form an affine space, not a vector space. Accordingly $l_P$ is affine in the coordinates $u_l$, with constant $c$, and its differential is linear in variations. Its variational derivatives are $\delta l_P/\delta u_l=a_l$. It is the restriction of a linear trace pairing on the full operator space, which explains the terminology without discarding the constant term.

Use the standard boundaryless-manifold convention, so $M$ is a closed manifold. Its orientation and Riemannian metric determine a unique positive Riemannian volume form ${\mu }_g$ taking value one on every positively oriented orthonormal frame. In an oriented atlas it isFor the coordinate invariance of the Riemannian volume form, on an overlap put $J=\partial y/\partial x$. Then $g_x=J^T{g}_{y}J$ and $d{y}^1\wedge \cdots \wedge d{y}^n=(\det J)d{x}^1\wedge \cdots \wedge d{x}^n$. The oriented transition has $\det J>0$, so the square root of the metric determinant and the coordinate volume transform together. The local forms therefore agree and define ${\mu }_g$ globally.

For real $r$-forms the Hodge star operator is defined byThe pointwise wedge pairing is nondegenerate, so this uniquely defines a smooth map $\ast :{\Omega }^r(M)\to {\Omega }^{n-r}(M)$. On an oriented orthonormal coframe, it sends a basis wedge to the signed complementary wedge. Applying it twice gives

For Hodge star eigenvalues on graded differential forms, extend $\ast$ complex-linearly to the direct sum of all form degrees. This interpretation matters: on a fixed degree it is an endomorphism only in middle degree. In positive odd dimension, $r(n-r)$ is even for every $r$, so ${\ast }^2=I$ and only $\pm 1$ can occur. Both occur, with eigenforms $1+{\mu }_g$ and $1-{\mu }_g$.

In positive even dimension, ${\ast }^2$ is $+I$ on even-degree forms and $-I$ on odd-degree forms. Hence ${\ast }^4=I$ and the only possible eigenvalues are $\pm 1,\pm i$. The first two occur as above. For the latter, choose a smooth local orthonormal frame and multiply its first dual differential one-form by a nonzero smooth bump function supported in the chart. This gives a form $\eta$ for which $\eta$ and $\ast \eta$ are independent. For $\lambda =\pm i$,Thus all four occur. Consequently, for $n\ge 1$,On degree $m$ in dimension $2m$, the Hodge star has eigenvalues $\pm 1$ for even $m$ and $\pm i$ for odd $m$; signed complementary middle-degree wedges show both occur for $m\ge 1$. In zero dimensions it is multiplication by the signed unit volume at each point, so only the signs actually present occur.

For Hodge integration by parts in arbitrary degree, let $\alpha \in {\Omega }^{r-1}(M)$ and $\beta \in {\Omega }^r(M)$. The graded Leibniz rule and Stokes theorem on the closed manifold giveUsing the given codifferential and the square of the Hodge star,Thereforewhich proves that $\delta$ is the formal adjoint of $d$. The complex version inserts conjugation in the usual Hermitian inner product.

The Hodge decomposition theorem gives an $L^2$-orthogonal decompositionwhere ${\mathcal{H}}^k(M)=\ker \Delta$ is finite dimensional and consists of the harmonic differential forms. Each de Rham cohomology class has a unique harmonic representative. Equivalently, with harmonic projection $P_{\mathcal{H}}$ there is a smooth Green operator of the Hodge Laplacian $G$ satisfying $\Delta G=G\Delta =I-P_{\mathcal{H}}$.

The solvability condition for the Hodge Poisson equation isNecessity follows from $⟨\Delta \alpha ,h⟩=⟨\alpha ,\Delta h⟩=0$. Conversely the orthogonality condition gives $P_{\mathcal{H}}\beta =0$, so ${\alpha }_0=G\beta$ solves the Poisson equation for differential forms.

The nonnegativity of the Hodge Laplacian identityidentifies its kernel with closed and coclosed forms. Thus the affine space of solutions of the Hodge Poisson equation isFor completeness, if a closed form is decomposed as $\omega =h+d\xi +\delta \zeta$, then $d\delta \zeta =0$, and $\Vert \delta \zeta {\Vert }^2=⟨\zeta ,d\delta \zeta ⟩=0$. Hence its cohomology class is represented by $h$. An exact harmonic form $h=d\xi$ must vanish, since $\Vert h{\Vert }^2=⟨\delta h,\xi ⟩=0$. This proves the canonical isomorphism ${\mathcal{H}}^k(M)\cong H_{\mathrm{dR}}^k(M)$, so the solution set is an affine space whose translation vector space is isomorphic to the requested de Rham cohomology.

The affine space $X={\mathbb{A}}_k^n$ is an integral scheme, hence its underlying topological space is irreducible. Every nonempty open subset of an irreducible topological space is irreducible and in particular connected. A section of the constant sheaf with value the function field $K=k(t_1,\dots ,t_n)$ is a locally constant function, so on any such open subset it is constant. Thus its sections are $K$ on each nonempty open subset and zero on the empty set. All restriction maps are either identities on $K$ or maps onto zero.

The constant sheaf is therefore a flasque sheaf. It has its natural ${\mathcal{O}}_X$-module structure because a regular function on a nonempty open subset embeds in the function field and acts by multiplication. Part (i) now givesThis is an instance of constant sheaves on irreducible spaces are flasque; no restriction on $n$ is needed.

Take the three-dimensional affine quadric coneThe polynomial $xy-zw$ is an irreducible polynomial: as a polynomial in $x$ over $k[y,z,w]$, it is primitive because $y$ and $zw$ are coprime elements of a unique factorization domain, and it is a linear irreducible polynomial over $k(y,z,w)$. Gauss lemma for polynomials then applies. Thus $X$ is an irreducible variety that is an affine variety, of algebraic dimension $3$ by the principal hypersurface dimension lemma in affine space ${\mathbb{A}}^4$.

The coordinate rings of $Y$ and $Z$ are respectively $k[y,w]$ and $k[x,z]$. Both are affine planes, hence irreducible closed subsets of algebraic dimension $2$. Their intersection is precisely the origin. ThereforeThe origin is the singular point of an algebraic variety of this three-dimensional affine quadric cone. The example shows why a smoothness of an algebraic variety hypothesis matters in intersection dimension estimates.

A blowup of an algebraic variety supplies a reduced special fibre with all three requested defects. Let $B={\mathbb{A}}^3$ have coordinates $(x,y,z)$, let $p=(0,1,0)$, and let $\pi :X={\mathrm{Bl}}_{p}B\to B$. Take $Y={\mathbb{A}}^1$ and the regular mapFor an explicit construction, $X$ is the closed subvariety of ${\mathbb{A}}^3\times {\mathbb{P}}^2$, with projective coordinates $[u:v:w]$, defined byThese equations say that $(x,y-1,z)$ is proportional to $(u,v,w)$. The three standard projective charts are affine spaces ${\mathbb{A}}^3$, so $X$ is a smooth variety which is an irreducible variety embedded in a quasi-projective algebraic set. Its algebraic exceptional divisor is $E={\pi }^{-1}(p)\cong {\mathbb{P}}^2$.

If $t\ne 0$, the surface $xy=t$ misses $p$, and the blowup of an algebraic variety is an isomorphism over it. HenceIt is a smooth variety, an affine variety, an irreducible variety, and has algebraic dimension $2$.

For $t=0$, the fibre of a morphism has three irreducible components: the strict transform of an algebraic subvariety $\widetilde{V(x)}$, the unchanged plane $V(y)$, and the algebraic exceptional divisor $E\cong {\mathbb{P}}^2$. The centre $p$ lies on $V(x)$ but not on $V(y)$; this is why all three components have multiplicity one. To verify reducedness and the crossing directly near $E$, use the chart $v\ne 0$, puttingHere $x=ab$, $y=1+b$, $z=bc$, and $f=ab(1+b)$. Near $b=0$, the factor $1+b$ is a unit, so the fibre of a morphism is $ab=0$, a reduced union of two crossing planes. The $u$-chart gives $f=x(1+xv)$, whose second factor is a unit along $E$; the $w$-chart gives $f=zu(1+zv)$ and the same reduced crossing. Away from $E$, reducedness follows from $xy=0$ in $B$. Thus the scheme-theoretic fibre of a morphism is itself reduced.

It has singular points of an algebraic variety already along $V(x,y)$, which misses $p$: locally its equation is $xy=0$, with Zariski tangent space of dimension of a vector space $3$ but local dimension of an algebraic variety $2$. Finally, $E$ is a closed subvariety of $X_0$. A closed subvariety of an affine variety is an affine variety, whereas ${\mathbb{P}}^2$ is not: all its global regular functions are constant, which cannot be the coordinate ring of a positive-dimensional affine variety. Consequently $X_0$ is singular, reducible, and nonaffine, as required. Choosing the centre off the intersection of the original planes avoids introducing a nonreduced special fibre of a morphism.

We first record why rigid quiver representations have open orbits. Let $\mathbf{n}=\dim X$, $G={\prod }_i\mathrm{GL}(X_i)$ and $R={\mathrm{Rep}}_Q(\mathbf{n})$. The stabilizer is ${\mathrm{Aut}}_Q(X)$, the nonempty open set of units in ${\mathrm{End}}_Q(X)$, so it has dimension $\dim {\mathrm{End}}_Q(X)$. The dimension formula for an algebraic group homomorphism, or the same constant-fiber argument for the orbit map, givesBy the Ringel form identity and ${\mathrm{Ext}}_Q^1(X,X)=0$, this equals ${\sum }_{\sigma }{n}_{s(\sigma )}{n}_{t(\sigma )}=\dim R$. An algebraic-group orbit is locally closed; since $R$ is an irreducible affine space, this full-dimensional orbit is open and dense.

Write $q=\rho \pi$ for the composed path. Each matrix entry of $q$ is a polynomial function on $R$. Since $qX=0$, change of basis makes it zero on all of ${\mathcal{O}}_X$, hence on all of $R$ by density. Suppose instead that $X_{t(\rho )}\ne 0$. The condition $\pi X\ne 0$ ensures that every vertex space visited by $\pi$ is nonzero. Choose a vector $v_a\ne 0$ and a functional ${\ell }_a$ with ${\ell }_a(v_a)=1$ at every vertex visited by $q$. Assign to every arrow $\sigma$ occurring in $q$ the map $v_{t(\sigma )}{\ell }_{s(\sigma )}$, and assign arbitrary maps, say zero, to the other arrows. At this representation, the path $q$ carries its initial chosen vector to its final chosen vector, so $q$ is nonzero, a contradiction.

The construction uses a single assigned map per arrow, so it still works if an arrow or vertex occurs repeatedly in the path. ThereforeThis proves the path identities in a rigid quiver representation claim for an arbitrary quiver. Maximal rank of the individual arrows alone would not justify the conclusion about their composition.

Use the quotient filtration induced from the order filtration of a Weyl algebra. The key step is to identify the entire associated graded module of the left ideal $I$, rather than merely take symbols of its displayed generators.

For every member $x^{\alpha }{y}^{\beta }$ of the ordered monomial basis of a Weyl algebra, right multiplication by the pure $y$-monomials givesNo commuting past an $x_i$ is necessary. It follows that $I$ is exactly the complex linear span of the basis monomials satisfying eitherIntersecting this linear span with each filtration piece and then taking symbols therefore yields the monomial idealBy b,

To determine its Krull dimension, compute the radical of an ideal:Indeed, $K$ lies in the displayed intersection, an intersection of prime ideals. Conversely, $({\xi }_1{\xi }_2{)}^3\in K$ and $({\xi }_1{\xi }_3{)}^3\in K$, so both generators of that intersection lie in $\sqrt{K}$. The equality of the intersection with $({\xi }_1{\xi }_2,{\xi }_1{\xi }_3)$ follows by factoring out ${\xi }_1$ and reducing modulo $({\xi }_2,{\xi }_3)$.

The order characteristic variety of a Weyl algebra module in this case has two irreducible components:They are coordinate affine spaces of dimensions $2n-1$ and $2n-2$, respectively, in the $2n$ coordinates $x_1,\dots ,x_n,{\xi }_1,\dots ,{\xi }_n$. The Krull dimension is the larger of these component dimensions. ThusIn particular, two displayed generators do not force a drop of two in dimension: their common factor creates the component ${\xi }_1=0$ of dimension $2n-1$.

At fixed dimension vector, ${\mathrm{Rep}}_Q(\mathbf{n})={\prod }_{\rho :i\to j}{\mathrm{Hom}}_k(k^{n_i},k^{n_j})$ is an irreducible affine space of dimension ${\sum }_{\rho }{n}_{s(\rho )}{n}_{t(\rho )}$. Its points are arrow-matrix collections, and its base-change orbits are isomorphism classes.

For a consistent linear system $Ax=b$, choose one solution $x_0$. By linearity, $Ax=b$ if and only if $A(x-x_0)=0$, so its solution set is the translate $x_0+\ker A$ of the kernel. Over the real numbers, this gives exactly one solution when the kernel is zero and infinitely many when the kernel contains a nonzero vector: $x_0+tz$ is a distinct solution for each real $t$ if $z\ne 0$ lies in the kernel. An inconsistent linear system has an empty solution set. The dimension of a nonempty solution set, regarded as an affine space, is the nullity of $A$.

A system of linear equations imposes several linear equations on the same unknowns. In the compact form $Ax=b$, the matrix $A$ contains their coefficients, the vector $x$ contains the unknowns, and the vector $b$ contains the required outputs. Over the real numbers, the following examples exhibit the three possible outcomes: one solution, infinitely many solutions, or no solution.

For $x+y=3$ and $x-y=1$, adding the two linear equations gives $2x=4$, hence $x=2$ and $y=1$. Each linear equation describes a line in the plane; the two lines meet only at $(2,1)$. Substitution checks both requirements: $2+1=3$ and $2-1=1$. The two measurements together determine both unknowns.

Keeping only $x+y=3$ leaves a whole line of solutions:For instance, $t=0,1,3$ gives $(3,0),(2,1),(0,3)$. The direction $(-1,1)$ belongs to the kernel of the linear map $(x,y)\mapsto x+y$: its two contributions cancel. Moving in that direction changes the unknowns without changing their measured sum. The solution line is an affine space, translated away from the origin.

For $x+y=3$ and $2x+2y=7$, the second linear equation instead demands $x+y=3.5$. The parallel lines never meet. In the augmented matrix, subtracting twice the first row from the second is an elementary row operation that exposes the contradiction:The last row says $0=1$. A zero row of coefficients is harmless only when its appended entry is also zero.

The column viewpoint unifies these pictures. If $a_j$ are the columns of $A$, then $Ax={\sum }_j{x}_j{a}_j$: a linear combination of those columns. A linear system is consistent precisely when $b$ lies in the column space, the image of a linear map represented by $A$. Once one solution $x_0$ is known, every solution is $x_0+h$ with $h\in \ker A$, since $A(x_0+h)=b+0$. Conversely, subtracting any two solutions gives an element of the kernel. Thus the kernel measures freedom, while the image of a linear map decides existence. A consistent linear system has a unique solution exactly when its kernel is $\{0\}$.^[ref]

Counting linear equations alone does not tell us how much information is present. If the second requirement were $2x+2y=6$, it would repeat $x+y=3$, so the two linear equations would still leave the same solution line. The matrix rank counts independent requirements. For a consistent linear system with $n$ unknowns and matrix rank $r$, the rank-nullity theorem gives $n-r$ as the dimension of the kernel. This is the number of independent directions in which a solution can move. In the single-measurement example, one measured combination leaves one direction of freedom; in the sum-and-difference example, neither unknown can change without changing an output.

Choosing a new origin $a$ in a vector space transports its operations along $T(x)=x-a$. Since $T(x\oplus y)=T(x)+T(y)$ and $T(k\odot x)=kT(x)$, every vector space axiom transfers through this bijection. The new additive identity is $a$ and the additive inverse of $x$ is $2a-x$. This illustrates why an affine space has no preferred origin even though choosing one produces a vector space.

On an affine space modeled on a vector space, the map $T_v(x)=x+v$ for a fixed vector $v$. Its derivative is the identity, and $T_v{T}_w=T_{v+w}$. Changes between flat coordinates on a translation surface have this form.