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Put $B=\partial M$ and let $v$ be the cone vertex. Specify the convention away from the Witt case: first take intersection cohomology to mean the hypercohomology of the intersection complex $I{C}_{\bar{m}}$, normalized in degree zero on the regular stratum. The lower middle perversity at the vertex is $\bar{m}(2m+1)=m-1$. The standard local cone calculation for this convention saysand restriction to $U\simeq B$ is the identity below the cutoff. Here this can also be read directly from the construction $I{C}_{\bar{m}}={\tau }_{\le m-1}R{j}_{\ast }\mathbb{Q}$, with $j:\hat{M}\setminus \{v\}\hookrightarrow \hat{M}$: punctured conical neighbourhoods retract onto $B$, and the truncation discards the degrees above $m-1$.

Define relative intersection cohomology as the hypercohomology of the homotopy fibre of restriction, namely $R\Gamma (\hat{M},\hat{U};IC)=\mathrm{Cone}(R\Gamma (\hat{M};IC)\to R\Gamma (\hat{U};IC))[-1]$. Restricting to the pair $(M,U)$ gives a map to the corresponding ordinary relative cohomology, since $IC{|}_M={\mathbb{Q}}_M$. The induced map of exact triangles gives the requested commutative diagram:Both rows are long exact cohomology sequences of a pair. Commutativity includes the connecting homomorphisms, because they arise from the same map of restriction cones, with a common sign convention. The relative vertical maps are isomorphisms: excision removes the common conical neighbourhood, leaving precisely the manifold pair outside it, where $IC$ is constant. Also the collar retracts onto $B$, so $H^{\ast }(M,U)\cong H^{\ast }(M,B)$.

The only possible nonmanifold stratum is $v$, with link $B$ of dimension $2m$. Since intersection homology of a manifold is ordinary homology, the exact necessary and sufficient condition isThe equivalence with the defining $H_m(B;\mathbb{Q})=0$ follows from the universal coefficient theorem for cohomology over $\mathbb{Q}$. In the Witt case the cone restriction is an isomorphism through degree $m$, since both degree-$m$ groups are zero. Comparing the two exact rows, or applying the Five lemma to each five-term segment, gives $I{H}^i(\hat{M})\cong H^i(M)$ for $i\le m$. For $i>m$, both $I{H}^{i-1}(\hat{U})$ and $I{H}^i(\hat{U})$ vanish, so the top row identifies the absolute group with the relative one. HenceWithout the Witt hypothesis, comparison still gives $I{H}_{\bar{m}}^i(\hat{M})\cong H^i(M)$ for $i<m$. At degree $m$ the upper row readsThe lower row identifies the quotient by $\mathrm{im}\delta$ with the image of the relative-to-absolute map. In higher degrees both cone terms vanish. Thus the full lower-middle Deligne answer isFor clarity, if $I{H}^i$ is instead defined as the linear dual of the lower-middle intersection chains of question 1, it corresponds to the complementary upper middle perversity Deligne sheaf. Its cone group retains $H^m(B)$ as well. The same diagram then givesEquivalently the exceptional group is $H^{m+1}(M,B)/\mathrm{im}(H^m(B)\stackrel{\delta }{\to }{H}^{m+1}(M,B))$. These conventions agree under the stated Witt condition, explaining why that condition removes the ambiguity.

Take the solid torus $M=S^1\times D^2$, whose boundary is the two-dimensional torus $B=T^2$. Its ordinary cohomology is $H^0(M)=H^1(M)=\mathbb{Q}$, zero otherwise; Poincare duality for a manifold with boundary gives $H^2(M,B)=H^3(M,B)=\mathbb{Q}$ and all other relative groups zero. The lower-middle Deligne formula givesThus degrees one and two have different dimensions, so there can be no three-dimensional Poincare duality. In the dual-lower-chain convention the dimensions are $(1,1,0,1)$, which likewise fail duality. Both computations exhibit the obstruction $H^1(T^2)={\mathbb{Q}}^2\ne 0$.

Use rational coefficients and finite singular chains. For an $N$-dimensional stratified pseudomanifold with filtration $X_j$, the lower middle perversity isA singular simplex $\sigma :{\Delta }^i\to X$ is an allowable singular simplex when, for every $k\ge 2$, ${\sigma }^{-1}(X_{N-k})$ lies in the $(i-k+\bar{m}(k))$-skeleton of ${\Delta }^i$; negative-dimensional skeletons are empty. Let $A_i$ be the span of these simplices in $S_i(X;\mathbb{Q})$. The intersection chains areThe condition on $\partial c$ is imposed after combining equal simplices and cancelling coefficients: the faces of an allowable simplex need not themselves be allowable. If $c\in I{S}_i$, then $\partial c\in A_{i-1}$ and $\partial (\partial c)=0$, so $\partial c\in I{S}_{i-1}$. Thus $I{S}_{\ast }$ is a chain subcomplex of $S_{\ast }$, and $I{H}_i(X)=H_i(I{S}_{\ast })$ defines intersection homology.

Write $v$ for the vertex of $cY$. Its codimension is $2n$, so the vertex condition for an allowable $i$-simplex isIn particular all simplices of degree at most $n$ avoid $v$. For $i<n$, both the degree-$i$ chains and the degree-$(i+1)$ chains that can bound them lie in $cY\setminus \{v\}=Y\times (0,1)$. The two chain complexes therefore have the same cycles and boundaries in degree $i$, proving the low-degree isomorphism.

For $i\ge n$, let $z$ be an allowable $i$-cycle. Cone every simplex of $z$ radially to $v$, using the cone on ${\Delta }^i$ as ${\Delta }^{i+1}$, and write the resulting chain as $cz$. At the vertex, coning an old inverse image in the $(i-n-1)$-skeleton gives a subset of the $(i-n)$-skeleton, and the new cone vertex also lies in this skeleton because $i-n\ge 0$. For every other singular stratum, coning the inverse image increases its permitted skeleton dimension by one, exactly as increasing the simplex dimension does. Thus $cz$ is allowable. The usual cone identity is $\partial (cz)=z-c(\partial z)=z$; since $z$ is allowable, the boundary condition for intersection chains holds as well. Every cycle is consequently a boundary. This proves the intersection homology cone formula here:The assumption $\dim Y=2n-1$ entails $n\ge 1$, so no augmented degree-zero exception arises in this coning argument.

For the suspension of a topological space, the two endpoints are collapsed separately, giving two distinct vertices. Use the following standard computational facts for intersection homology: a product with an interval has the same intersection homology as the original stratified space, via the projection and its stratum-preserving homotopy inverse; and an open cover has a Mayer–Vietoris sequence whose overlap map is the signed pair of inclusion maps. Cover the suspension by its two open cones. Their overlap is $Y\times (-\epsilon ,\epsilon )$. Below degree $n$, the overlap map is, under the interval identifications, $a\mapsto (a,-a)$ in $I{H}_i(Y)\oplus I{H}_i(Y)$, hence is injective with cokernel $I{H}_i(Y)$. In degree $n$ both cone groups vanish and the next overlap map, in degree $n-1$, is still injective. Above degree $n$ both adjacent cone groups vanish. Exactness therefore givesThis differs from ordinary reduced suspension homology: below the cutoff the cycles cannot use either vertex, and in the middle degree coning kills them. If $Y$ is a compact oriented Witt space, its Poincare duality pairs $I{H}_i(Y)$ with $I{H}_{2n-1-i}(Y)$. The displayed groups of the suspension are correspondingly paired in complementary degrees $i$ and $2n-i$, while its middle group is zero. The new vertices have even codimension $2n$ and thus impose no additional Witt space condition; the suspension is again Witt. Its middle intersection form has zero rank and, when its dimension is divisible by four, zero signature.

Let $C$ be the smooth algebraic curve. A nonzero regular section of a line bundle has an effective zero divisor, so its degree of a line bundle is nonnegative. Degrees add under tensor products and change sign on taking the dual. We use genuine line subbundles, so their quotients in $E$ are locally free. Both bounds can be proved constructively.

For the upper bound, take a rational frame of $E^{\ast }$ over the function field of $C$. Its two members are rational sections with only finitely many poles. Choose an effective divisor $D$ large enough to bound the pole orders of both rational sections. They become global sections of $E^{\ast }\otimes {\mathcal{O}}_C(D)$ and define a sheaf mapwhich is an isomorphism at the generic point. It is therefore injective as a sheaf map: a section in its kernel would vanish generically in a locally free sheaf, which is a torsion-free sheaf. For any line subbundle $L\subset E$, at least one of its two component maps into ${\mathcal{O}}_C(D)$ is nonzero. Such a map is a nonzero global section of ${\mathcal{O}}_C(D)\otimes L^{-1}$. Its zero divisor on an algebraic curve is effective, soThis upper degree bound for line subbundles on a smooth curve depends only on a choice made for $E$, and applies to every one of its line subbundles.

For the lower-bound question we will actually construct subbundles ${\mathcal{O}}_C(-m{p}_0)$ for every sufficiently large integer $m$, where $p_0$ is a fixed point. First prove the globally generated vector bundle property rather than assuming that negative-degree inclusions automatically are subbundles.

Apply the upper-bound argument to the rank-two vector bundle $F=E^{\ast }\otimes K_C$, where $K_C$ is the canonical bundle. Let $B$ be its upper degree bound. Choose $m$ with $m-1>B$. For any point $p\in C$, a nonzero global section of $F(-m{p}_0+p)$ would give a nonzero mapIts image can be saturated to a saturated line subbundle on a smooth curve $M\subset F$. Locally the local ring of a smooth algebraic curve is a discrete valuation ring: dividing out the common power of a uniformizer makes the image a primitive rank-one summand, and its quotient is free. Thus the saturation is indeed a subbundle. The original map gives a nonzero global section of $M\otimes {\mathcal{O}}_C(-m{p}_0+p)$ and consequently $\mathrm{deg}M\ge m-1$, contradicting the upper bound. Hence $H^0(F(-m{p}_0+p))=0$ for every $p$. By Serre duality,The exact sequenceand its long exact sequence in sheaf cohomology show that evaluation $H^0(E(m{p}_0))\to E(m{p}_0){|}_p$ is onto for every $p$. This proves the global generation of positive twists on a smooth curve with one bound on $m$ that works at every point.

Write $W=H^0(E(m{p}_0))$ and $N=\dim W$. The evaluation map $W\otimes {\mathcal{O}}_C\to E(m{p}_0)$ is surjective with kernel bundle of rank $N-2$. Consider the incidence subsetIt is the total space of this kernel bundle, so $\dim Z=1+(N-2)=N-1$. Its projection into $W$ is closed because $C$ is proper, and its image has dimension at most $N-1<N$. Choose $s$ outside this proper closed subset. It vanishes nowhere. In a local trivialization at any point, one component of $s$ is a unit, so it can be completed to a basis. Therefore ${\mathcal{O}}_C\stackrel{s}{\to }E(m{p}_0)$ is a subbundle with a locally free rank-one quotient. Twisting back givesThis proves the unbounded negative degrees of line subbundles. The nowhere-vanishing construction guarantees a subbundle at every point, including the zeros that might otherwise have occurred in a generic meromorphic inclusion. If the curve has several connected components, apply these constructions on each component and add the degrees; the same upper-bound and no-lower-bound conclusions follow.

Put $A=\mathbb{C}[V]={⨁}_{d\ge 0}{A}_d$ and $R=A^G$. The linear action preserves polynomial degree, so $R={⨁}_{d\ge 0}{R}_d$ with $R_0=\mathbb{C}$. We prove Hilbert finite generation for reductive invariants by combining finite ideal generation in $A$ with a degree-lowering argument in $R$.

Over $\mathbb{C}$, representations of a reductive algebraic group are completely reducible. On each finite-dimensional $A_d$, project its isotypic component for the trivial representation onto itself and its nontrivial summands onto zero. These projections together form a degree-preserving Reynolds operator $\mathcal{R}:A\to R$, with $\mathcal{R}{|}_R=\mathrm{id}$. The crucial additional property isHere is why it holds rather than being an arbitrary projection property. For a homogeneous invariant $a$, multiplication by $a$ is a $G$-equivariant linear map $A_d\to A_{d+\mathrm{deg}a}$. It sends invariant vectors to invariant vectors. On a nontrivial irreducible summand, its composition with projection onto the target's trivial summand is zero, since there is no nonzero equivariant map from a nontrivial irreducible representation to the trivial representation. It therefore commutes with the Reynolds operators. Summing the homogeneous components of $a$ and $h$ proves the displayed module property in general.

Let $R_{+}={⨁}_{d>0}{R}_d$ and form the ideal $I=A{R}_{+}$ in the ambient polynomial ring. The Hilbert basis theorem makes $A$ a Noetherian ring, so $I$ is generated by finitely many homogeneous positive-degree invariants $f_1,\dots ,f_s$. To justify choosing the generators from this particular set, start with any finite ideal generating set; each of its members is a finite sum of multiples of homogeneous members of $R_{+}$, and the union of those finitely many invariants then generates $I$. If $I=0$, there are no positive-degree invariants and the answer is simply $R=\mathbb{C}$.

Otherwise consider a homogeneous invariant $f$ of degree $d>0$. Since $f\in I$, writewith $h_i=0$ when that degree is negative. Taking the degree-$d$ component of any ideal expression gives exactly such an expression. Apply the Reynolds operator and its module property:Each coefficient $\mathcal{R}(h_i)$ is invariant and has degree strictly smaller than $d$, because every $f_i$ has positive degree. Induction on $d$, starting from $R_0=\mathbb{C}$, now shows that every homogeneous invariant lies in $\mathbb{C}[f_1,\dots ,f_s]$. An arbitrary invariant is a finite sum of invariant homogeneous components, soThis proves finite generation as an algebra. No Noetherian property of the polynomial invariant ring was assumed in obtaining the generators.

Using the printed convention that the projective points represent lines of binary quartics, the locus of GIT stable points isThe root at infinity is included. We determine stability by calculating every relevant algebraic one-parameter subgroup weight.

Use the natural action $(g\cdot f)(X,Y)=f(g^{-1}(X,Y))$. Every nontrivial algebraic one-parameter subgroup of $S{L}_2$ is conjugate to ${\lambda }_m(t)=\mathrm{diag}(t^m,t^{-m})$ for some integer $m>0$: its two integral weights of a representation sum to zero, and a basis of weight vectors can be rescaled to have determinant one. WriteIf $r=\min \{j:a_j\ne 0\}$, then $r$ is the multiplicity of a root at the projective point $[1:0]$: in the local coordinate $Y/X$, the polynomial begins with a nonzero multiple of $(Y/X{)}^r$.

The Hilbert-Mumford criterion for projective stability, with $\mu =-\min$(active weights), now givesBeing a GIT stable point requires this number to be strictly positive for every nontrivial subgroup. Conjugating the subgroup lets $[1:0]$ range over the entire projective line. Thus the root-multiplicity criterion for stable binary forms becomes $r_p(f)<2$ at every point $p$. A nonzero degree-four form over $\mathbb{C}$ has four projective roots counted with multiplicity, so this means precisely four distinct roots. This is equivalently nonvanishing of its discriminant.

The boundary cases show why a weak inequality would give the wrong answer. A triple or quadruple root can be moved to $[1:0]$, leaving only strictly positive weights; its affine representative tends to zero under that subgroup, so it is unstable. If all root multiplicities are at most two, all Hilbert-Mumford values are nonnegative and the point is a GIT semistable point. A double root gives a zero value for a suitable subgroup, so never represents a GIT stable point. More concretely, after putting that double root at $[1:0]$ one hasand ${\lambda }_1(t)\cdot f\to a{X}^2{Y}^2$. When there are three distinct roots, this limiting form, representing a GIT semistable point with two distinct roots, is outside the original group orbit, so that group orbit is not closed in the semistable locus. When there are two double roots, a change of coordinates gives $X^2{Y}^2$, which has the positive-dimensional diagonal torus as stabilizer. Neither case represents a GIT stable point.

For a form with four distinct roots, the projective stabilizer is finite: it permutes the four roots, and a Möbius transformation fixing three distinct points is the identity. Thus its image in $PG{L}_2$ injects into the finite permutation group on the roots. The kernel of a group homomorphism for $S{L}_2\to PG{L}_2$ is $\{\pm I\}$, also finite. This agrees with the finite-stabilizer condition in geometric invariant theory. The weight calculation already supplies the required strict stability condition for every conjugate subgroup.

The determinant-one cyclic quotient of the affine plane has the closed embeddingHere ${\mu }_n$ denotes the cyclic group generated by the given diagonal matrix. We establish the presentation of the polynomial invariant ring and verify that its fibres really are the group orbits.

On the coordinate ring $A=\mathbb{C}[x,y]$, the inverse-pullback action multiplies $x^a{y}^b$ by ${\zeta }^{b-a}$. Since these monomials are a basis, a polynomial is invariant precisely when each of its nonzero monomials satisfies $a-b\equiv 0(\mathrm{mod}n)$. If $a\ge b$, such a monomial is $(x^n{)}^{(a-b)/n}(xy{)}^b$; if $b\ge a$, it is $(y^n{)}^{(b-a)/n}(xy{)}^a$. ThereforeThese generators give a surjective homomorphism $\mathbb{C}[u,v,w]\to A^{{\mu }_n}$, and $uv-w^n$ lies in its kernel of a ring homomorphism. To prove there are no further relations, reduce any monomial using $uv=w^n$. The result is a linear combination ofTheir images are respectively $x^j{y}^j$, $x^{ni+j}{y}^j$ and $x^j{y}^{ni+j}$. All have distinct exponent pairs, so they are linearly independent in $A$. Any polynomial in the kernel of a ring homomorphism has zero reduced expression and hence belongs to $(uv-w^n)$. ConsequentlyThis proves the claimed closed subvariety embedding of the affine quotient.

For surjectivity of $\pi$, if $u\ne 0$ choose $x^n=u$ and set $y=w/x$. Then $y^n=w^n/u=v$. If $u=0$, the relation forces $w=0$; choose $x=0$ and any $n$th root of $v$ for $y$. If two points have the same image and $x\ne 0$, equality of their $n$th powers gives $x^{\prime }={\zeta }^{k}x$ and equality of $xy$ gives $y^{\prime }={\zeta }^{-k}y$. If $x=0$, both first coordinates vanish, and the same argument using $y^n$ gives the orbit relation; the case $x=y=0$ has only that one point in its fibre. Thus every fibre is exactly one group orbit.

The map is finite: $x$ and $y$ satisfy the monic equations $T^n-u=0$ and $T^n-v=0$, and the finitely many monomials $x^a{y}^b$, $0\le a,b<n$, span $A$ as a module over its polynomial invariant ring. A finite morphism is closed when it is surjective as here and therefore gives the quotient topology. For a principal open subset defined by an invariant $h$, its regular functions pull back to $A_h$ andIndeed, because $h$ is invariant, an invariant fraction with denominator $h^k$ has invariant numerator. These principal open subsets cover the affine hypersurface, proving the required local equality of invariant regular functions. Together with the orbit fibres, this verifies the geometric quotient, not merely a bijection of underlying sets.