Past exam of the mathematics course of the University of Cambridge / 2006 / iii / Paper 22 (Smooth Four-manifolds)

Choose a complex projective line in each Complex projective plane summand. Make the connected sum at small disks on these lines, so the lines themselves join through the neck to form a smoothly embedded sphere $S\cong S^2$. Its class is the sum $h_1+h_2$ of the two line classes. The intersection form is diagonal with entries one, henceThe oriented normal bundle of $S$ is a rank-two real bundle with Euler number two, because its Euler number equals the self-intersection number. Oriented plane bundles over $S^2$ are classified by that integer, so this normal bundle is isomorphic to $T{S}^2$. The boundary of a closed tubular neighborhood $N$ is consequently the unit tangent bundle of $S^2$.

A unit tangent pair $(x,v)$ determines the oriented orthonormal frame $(x,v,x\times v)$ in ${\mathbb{R}}^3$. Thus $UT{S}^2\cong SO(3)\cong {\mathbb{RP}}^3$, the latter identification coming from the double covering by unit quaternions. Therefore $\partial N$ is the required smoothly embedded ${\mathbb{RP}}^3$. Its complement is disconnected: the nonempty interior of $N$ and the nonempty exterior $X\setminus N$ are disjoint open subsets of $X\setminus \partial N$ that exhaust it. The neighborhood can be chosen small enough that its exterior is nonempty. This is the construction of separating projective three-space in a positive connected sum.

In the cohomology basis $h_1,\dots ,h_n$ supplied by the projective lines in the summands, the intersection form isFor each permutation $\sigma \in S_n$, the linear map $h_i\mapsto h_{\sigma (i)}$ preserves this pairing. Distinct permutations induce distinct maps, and their compositions agree with permutation composition. Hence the permutation matrices give a subgroup $G\cong S_n$ of the isometry group of $Q_X$.

To realize these isometries smoothly, describe $X$ as a punctured $S^4$ with $n$ punctured copies of ${\mathbb{CP}}^2$ attached to its boundary spheres. An orientation-preserving ambient isotopy of $S^4$ can interchange any two chosen disjoint attachment balls while carrying their parametrizations along with them: move their centers along disjoint arcs and extend the motion to small balls. There is room to make these arcs disjoint in dimension four. Extend the endpoint diffeomorphism across the two exchanged punctured copies using the chosen identifications of those identical oriented summands; use the identity on the remaining copies. Collar coordinates make the maps fit smoothly along the gluing spheres. This exchanges the two line classes without changing their signs. Since transpositions generate $S_n$, composing these diffeomorphisms realizes each element of $G$. Pullback permutes the cohomology basis by the inverse permutation, which still realizes the same subgroup and all its elements. These are permutation diffeomorphisms of identical connected summands.

Let $f:S^2\times S^2\to {\mathbb{CP}}^2\#{\mathbb{CP}}^2$ have mapping degree $d$. Naturality of the cup product and evaluation on the fundamental class implyThe two factor classes give the source matrix $H=\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$, while the target matrix is $I_2$. If $P$ is the integral matrix for the pullback in these bases, thenTaking determinants gives $-(\det P{)}^2=d^2$, whose left side is nonpositive and whose right side is nonnegative. Thus $d=0$. There is no map of nonzero degree, including negative degree. This degree constraint from intersection forms is a cohomological argument and does not assume that $f$ is smooth.

Use the convention $\nabla =d+A$ with anti-Hermitian connection one-forms. For a unitary complex line bundle, $A$ is locally an $i\mathbb{R}$-valued one-form and $F_{\nabla }=dA$, since the Lie algebra is abelian. Under a change of unitary frame, the vector-bundle curvature transforms by conjugation, which is trivial for $U(1)$. Hence $F_{\nabla }$ is a globally defined imaginary-valued two-form, and $i{F}_{\nabla }$ is an ordinary real two-form. Equivalently, $\mathrm{End}(L)$ has its canonical scalar trivialization, so no nontrivial coefficient bundle remains.

Locally $d{F}_{\nabla }=d^{2}A=0$, and these equalities patch globally. If another unitary connection is ${\nabla }^{\prime }=\nabla +i\beta$ for a global real one-form $\beta$, thenThus the real vector-bundle curvature is closed and its de Rham cohomology class is independent of the unitary connection.

With the First Chern class normalization $c_1(L{)}_{\mathbb{R}}=[i{F}_{\nabla }/(2\pi )]$, the transverse section also identifies this class as $\mathrm{PD}[\Sigma ]$. Indeed, the complex orientation of each normal disk makes the section wind positively once around a transverse zero. The Euler class of the underlying oriented plane bundle counts these zeros with their induced orientation, and equals $c_1(L)$. This records the role of the section and orientation hypotheses; connection independence itself did not require choosing a section.

Start with any unitary connection ${\nabla }_0$. The Hodge decomposition theorem gives a unique real harmonic differential form $\eta$ representing $[i{F}_{{\nabla }_0}]$. Since they have the same de Rham cohomology class, $\eta -i{F}_{{\nabla }_0}=d\beta$ for a smooth real one-form $\beta$. Set $\nabla ={\nabla }_0-i\beta$. The vector-bundle curvature difference formula givesThis proves existence of a harmonic-curvature unitary line connection for any metric and any complex line bundle.

Two such unitary connections have the same vector-bundle curvature because harmonic representatives are unique. Their difference is therefore $i\gamma$ with $d\gamma =0$. When $b_1(X)=0$, write $\gamma =d\phi$ globally. The bundle gauge transformation $u=e^{i\phi }$ changes $\nabla$ to $u^{-1}\nabla u=\nabla +id\phi$. Thus the harmonic-curvature connection is unique up to gauge. On a general base, closed forms modulo the integral-period forms coming from circle-valued gauge transformations leave the torus $H^1(X;\mathbb{R})/(2\pi H^1(X;\mathbb{Z}))$ of possible gauge classes; the hypothesis removes this freedom.

The reverse-oriented Complex projective plane has intersection form $(-1)$, so $b^{+}=0$, while $b_1=0$. The Hodge star splits harmonic two-forms into self-dual and anti-self-dual subspaces, and their dimensions are the positive and negative indices of the intersection form. Therefore every harmonic two-form is an anti-self-dual two-form for every chosen metric.

By part (b), every complex line bundle has a unitary connection with harmonic real vector-bundle curvature, so this connection is an ASD connection. Conversely, an ASD connection on a complex line bundle has $dF=0$ and $d\ast F=-dF=0$, so its vector-bundle curvature is harmonic. The uniqueness in part (b) therefore applies to all ASD connections on that fixed complex line bundle. Each complex line bundle has exactly one gauge orbit of ASD connections.

Let $a,b\in H^2(S^2\times S^2;\mathbb{Z})$ be the two factor classes, normalized by $a^2=b^2=0$ and $⟨ab,[X]⟩=1$. Choose the complex line bundle $L$ with $c_1(L)=a+b$, for example the tensor product of pullbacks of degree-one complex line bundles from the two factors. ThenIf $L$ had an ASD connection, the real closed form $\alpha =iF/(2\pi )$ would represent this class and satisfy $\ast \alpha =-\alpha$. Hencea contradiction. This complex line bundle admits no ASD connection for any metric of the given product orientation. For the reverse product orientation, take $c_1(L)=a-b$ instead, which then has square $+2$. This is the positive-square obstruction to ASD line connections.

The suggested rank-two construction gives the same obstruction. The split bundle $E=L\oplus L^{\ast }$ has trivial determinant andAn ASD connection on $L$ would induce an ASD connection on $E$, but its instanton number would be $\Vert F_E{\Vert }_2^2/(8{\pi }^2)\ge 0$, contradicting $c_2(E)[X]=-2$.

Write the flat connection as $d+A_{x}dx+A_{y}dy$ in a global unitary frame. Solve the matrix ordinary differential equationSmooth dependence on parameters gives a smooth solution on the entire open square. Since $A_x$ is skew-Hermitian and trace free, differentiating $u^{\ast }u$ and $\det u$ shows that $u(x,y)\in SU(2)$. Under the bundle gauge transformation $u$, the transformed connection hasIts vector-bundle curvature is still zero. The remaining vector-bundle curvature equation is ${\partial }_x{A}_y^{\prime }=0$, so $A_y^{\prime }=B(y)$ is independent of $x$. Now solve $v^{\prime }(y)=-B(y)v(y)$ with $v(0)=I$. Again $v\in SU(2)$; this second bundle gauge transformation, independent of $x$, leaves $A_x^{\prime }=0$ and makes $A_y^{\prime \prime }=0$. Thus the original connection is gauge equivalent to the trivial connection. Both transformations exist on the whole square, not just in a small coordinate neighborhood.

For a connected base with a chosen frame at a base point, flat connections modulo the based unitary gauge group correspond to flat holonomy representations ${\pi }_1(X,x)\to G$. Removing the frame, or allowing the full unitary bundle gauge group, also divides by conjugation in $G$. On a prescribed underlying bundle, retain only those representations whose associated flat bundle has that bundle type.

The correspondence can be seen directly. Zero vector-bundle curvature makes parallel transport invariant under homotopies of paths with fixed endpoints, giving a group homomorphism on loops. Conversely, a representation $\rho$ gives the flat bundle $(\tilde{X}\times G)/{\pi }_1(X)$, with the deck action on the first factor and $\rho$ on the second, and the product horizontal distribution descends. If two framed connections have identical holonomies, compare their parallel transports along a path from the base point to each point. The comparison is path independent and defines a bundle gauge transformation equal to the identity at the base point. These constructions are inverse. This is framed flat connections and holonomy.

The punctured torus retracts onto a wedge of two circles, so its fundamental group is the free group $F_2$. A representation into $SU(2)$ is specified freely by the two generator images $(A,B)$, with no relation. Every resulting principal $SU(2)$ bundle is trivial, since a bundle with connected structure group over a graph is trivial. Thus no representation is excluded by the fixed trivial-bundle condition. Holonomy depends continuously on a connection, and the flat-bundle construction gives locally continuous choices of representatives as the two matrices vary; local choices of paths in $SU(2)$ give this continuity without requiring a global logarithm. ConsequentlyThe last identification is the description of $SU(2)$ by unit quaternions. The unbased quotient used next is simultaneous conjugation, not independent conjugation of the two generator images.

By part (b), the moduli space is $M(T)=SU(2{)}^2/SU(2)$ under simultaneous conjugation. Regard $SU(2)$ as the unit quaternions and writewith $u,v\in {\mathbb{R}}^3$. Conjugation rotates both imaginary vectors by the same element of $SO(3)$. Their lengths and inner product determine an orbit, even when one vector vanishes or the vectors are collinear: an isometry between the two spanned planes can be extended to an orientation-preserving rotation of ${\mathbb{R}}^3$. Thus complete invariants areEvery allowed triple is realized by a pair of vectors with the prescribed lengths and dot product. The induced continuous bijection from the compact quotient to this closed subset of ${\mathbb{R}}^3$ is a homeomorphism.

Let $U$ be the subset where $A\ne \pm 1$, equivalently $-1<a<1$. It is open and dense because the excluded first holonomies can be approximated by noncentral ones. Rotate $u$ to the positive $i$ axis, so $A=a+\sqrt{1-a^2}i$. Its remaining stabilizer rotates the $j,k$ plane. If $B=b+ti+yj+zk$, the orbit is determined by $(b,t)$, andwith the remaining radius $\sqrt{1-b^2-t^2}$. These coordinates give the corrected resultContinuity of the inverse follows by taking $y=\sqrt{1-b^2-t^2}$ and $z=0$ as a representative. The complement consists of $A=1$ and $A=-1$, separately. In either case conjugation classes of $B$ are parametrized by its real part $b\in [-1,1]$. Therefore $M(T)\setminus U$ is two disjoint closed intervals, as requested.

The printed $S^2$ factor is incorrect: the factor must be a closed disk. The stabilizer acts by conjugation, fixing the circle of quaternions $b+ti$; it is not the free Hopf circle action on $S^3$. To rule out a different choice of dense open set with the printed topology, we can also identify the whole space. Replace $c$ by $d=ab+c$. The invariant region becomesThese are the Gram matrices of the three unit vectors $1,A,B$ in ${\mathbb{R}}^4$. Conversely, a positive-semidefinite matrix of this form can be realized by such vectors, hence by the original quaternion data. The set $\mathcal{C}$ is compact and convex and contains the identity matrix as an interior point in its three coordinates. Radial parametrization of a convex body therefore identifies it with a closed three-ball. ThusThis also agrees with Theorem 6.5 of the primary character-variety paper.

An open subset homeomorphic to the boundaryless three-manifold $S^2\times (0,1)$ cannot contain a boundary point of this closed ball. Otherwise a local chart, followed by inclusion into ${\mathbb{R}}^3$, would be a continuous injection from an open subset of ${\mathbb{R}}^3$ whose image contains a ball-boundary point while staying inside the closed ball, contradicting invariance of domain. Its complement would therefore contain the entire boundary sphere. But a sphere cannot embed in two disjoint intervals: its connected image would lie in one interval and be a nondegenerate compact interval, whose interior points disconnect it, unlike $S^2$. This proves that the literal printed pair of claims is impossible, and supplies the complete corrected SU2 character variety of the punctured torus description.

Use anti-Hermitian $SU(2)$ vector-bundle curvature and the fundamental matrix trace, sois an integer on a closed oriented four-manifold. The supplied formula's phrase “extending $W$” is a typographical error in the original PDF: the connection extends the boundary connection $\nabla$. An extension exists, for example on the trivial bundle over $B^4$, by extending a boundary connection through a collar and cutting it off farther inside.

For well-definedness, glue two choices $W_0$ and $-W_1$ along their identified boundary bundles. Connections may be made product connections on a collar while keeping their boundary values, and the transgression identity below shows that this does not alter the relevant integral. The glued bundle over the resulting closed four-manifold has an integer Second Chern number. Therefore the two extension integrals differ by an integer, proving the value of $\mathrm{CS}$ is independent of all choices in $\mathbb{R}/\mathbb{Z}$. In particular, different extensions on one fixed bundle with the same boundary value have zero difference by transgression and Stokes theorem.

To compute the derivative, extend the boundary variation $a$ to an adjoint-valued one-form $\tilde{a}$ on $W$ and set ${\tilde{\nabla }}_t=\tilde{\nabla }+t\tilde{a}$. The vector-bundle curvature derivative is $\dot{F}=d_{\tilde{\nabla }}\tilde{a}$. Invariance of the trace and the Bianchi identity giveThe second equality follows from the covariant product rule; its other term contains $d_{\tilde{\nabla }}F=0$. Integrating and applying Stokes theorem, with the given boundary orientation, yieldsThis derivative means the derivative of any local real lift of the circle-valued functional; changing that lift by an integer changes no derivative.

If a bundle gauge transformation $u$ extends to $\tilde{u}$ on $W$, use ${\tilde{u}}^{-1}\tilde{\nabla }\tilde{u}$ as the extension of the transformed boundary connection. Vector-bundle curvature is conjugated, so its trace square is unchanged pointwise. $\mathrm{CS}(u\cdot \nabla )=\mathrm{CS}(\nabla )$. This proof even preserves the chosen extension's real integral, before reducing modulo integers. The local Chern-Simons three-form transgresses the same characteristic form and is the boundary expression used in the compactness argument below.

Let $k=e(E)=c_2(E)[X]=1$, where the underlying oriented real rank-four bundle has its complex orientation. The charge-one Uhlenbeck-Donaldson compactness for charge-one ASD connections assertion is that any sequence of ASD connections $A_n$ has a subsequence with one of these alternatives: modulo bundle gauge transformations, it converges smoothly on all of $X$ to an ASD connection of charge one; or it converges smoothly on compact subsets of $X\setminus \{x\}$ to a connection that extends smoothly on a charge-zero bundle and is flat, withBundle identifications are made on the complement of the bubble point in the second alternative. In general the flat limit can have nontrivial holonomy; trivial charge does not imply trivial connection on an arbitrary base. The closure of the charge-one moduli space in the ideal-instanton topology adds only a subset of $M_0\times X$, and is compact. Here $M_0$ denotes flat gauge classes on charge-zero bundles.

The local analytic inputs are the following, stated so that the compactness conclusion is not itself assumed. A Uhlenbeck small-energy Coulomb gauge on a four-ball exists whenever ${\int }_B|F_A{|}^2$ is below a universal local threshold; it obeys $d^{\ast }a=0$, the usual normal boundary condition, and $\Vert a{\Vert }_{W^{1,2}}\le C\Vert F_A{\Vert }_2$ after rescaling. For ASD connections, the local small-energy regularity estimates bound every vector-bundle curvature derivative on a smaller ball:Together with elliptic regularity in the Coulomb gauge, these give bounds on all derivatives of local connection matrices on smaller balls. Smooth fixed metrics permit these estimates on sufficiently small coordinate balls. Finally, the Uhlenbeck removable singularity theorem for ASD connections states that a smooth finite-energy ASD connection on a punctured four-ball extends smoothly after a gauge change and bundle extension. Rellich compactness and elliptic bootstrapping are used in passing from bounded local representatives to smooth subsequential limits. These are local estimates and an extension theorem, not the global bubbling theorem being proved.

For the proof, put the inner product $⟨U,V⟩=-\mathrm{Tr}(UV)$ on the anti-Hermitian Lie algebra. Since $\ast F_A=-F_A$,Thus every $A_n$ has energy $8{\pi }^2$. Pass to a weakly convergent subsequence of the nonnegative curvature-energy measures. Let $S$ be the points at which the limit measure has an atom at least the small-energy threshold. There are finitely many, because the total mass is fixed.

At every point outside $S$, a sufficiently small ball has limit mass less than the threshold; choose its boundary to have zero limit mass. The corresponding energies for all large $n$ are then below the threshold. The local Coulomb gauges and the stated estimates yield smooth subsequential convergence on smaller balls. A countable cover and diagonal extraction give these limits everywhere outside $S$. On overlaps, the transition gauges satisfy the first-order equation relating the two connection matrices. Their derivatives are bounded by the already controlled matrices, and the compactness of $SU(2)$ controls their zeroth-order values. Extracting limits of these transitions patches the local limits into a smooth bundle and ASD connection $A_{\infty }$ on $X\setminus S$. On a fixed compact subset, close transition cocycles identify the original and limiting bundles by smooth near-identity isomorphisms; one can construct these by embedding the bundles into one trivial Hermitian bundle and using the polar decomposition of the nearby orthogonal projections. This yields actual bundle identifications for the convergent connections, not merely separate local limits.

The limiting vector-bundle curvature has finite energy by lower semicontinuity. Apply the removable-singularity input at each point of $S$. We obtain a smooth extended bundle $E_{\infty }$ and ASD connection on all of $X$. Strong convergence away from $S$ shows that the remaining measure is a nonnegative sum of atoms:Discard any artificial zero-defect points if they occur in a preliminary choice of exceptional set.

Here is the topological argument that makes the defects integral, rather than merely small positive real numbers. Around one point $p$, choose a ball $B$ whose boundary contains no exceptional point. Both the original and the extended limiting bundles are trivial over $B$, but their gauges identifying the boundary can differ by a map $S^3\to SU(2)$. The Second Chern number of the clutching bundle obtained by gluing two copies of $B$ is an integer. Using the Chern-Simons three-form on the boundary givesThe $o(1)$ term is the difference of the boundary Chern-Simons integrals in the convergent gauges, hence tends to zero by smooth boundary convergence. The integer is the clutching contribution. The left side tends to ${\beta }_p/(8{\pi }^2)$, so the closedness of $\mathbb{Z}$ gives ${\beta }_p=8{\pi }^2{m}_p$ for an integer $m_p\ge 1$.

Taking total masses now givesThere are only two possibilities. If the sum is zero, there is no concentration; the same local gauges cover all of $X$, giving smooth gauge convergence on the original bundle. If the sum is nonzero, it consists of one integer one, and $k(E_{\infty })=0$. The limiting vector-bundle curvature therefore vanishes, and all energy is the single atom $8{\pi }^2{\delta }_x$. This proves the alternatives.

Finally, flat gauge classes on a compact base form a compact space: choose finitely many generators of its fundamental group, identify flat classes with the closed representation subset of a finite product of $SU(2)$ subject to its relations, and quotient by the compact conjugation action. All flat $SU(2)$ bundles here have zero second Chern class and the corresponding charge-zero topological type. Thus the ideal stratum $M_0\times X$ is compact. Combining its subsequential compactness with the two alternatives proves compactness of the closure in the ideal-instanton topology. No gluing theorem asserting that every formal ideal point is realized is needed for this compactness statement.

For the adjoint real bundle $\mathrm{ad}E$, the ASD deformation complex iswhere $d_A^{+}=P_{+}{d}_A$ and $P_{+}=(1+\ast )/2$. On an adjoint section $\xi$, the vector-bundle curvature identity gives $d_A^2\xi =[F_A,\xi ]$. Since $P_{+}$ acts only on the two-form part and $F_A^{+}=0$,This proves it is a complex. Its three cohomology spaces are the infinitesimal stabilizers $H_A^0$, infinitesimal deformations modulo gauge $H_A^1$, and obstructions $H_A^2$.

The associated gauge-fixed elliptic differential operator isThe ASD deformation index formula isHere the Atiyah-Singer index theorem is used for the formula, with $p_1(\mathrm{ad}E)=-4c_2(E)$ and $e(E)=c_2(E)[X]$. For the simply connected negative-definite base in this question, $b_1=b^{+}=0$, so $\mathrm{ind}{D}_A=8e(E)-3$. The alternating Euler characteristic of the three-term complex is the negative of this operator index; stating which index is being used avoids a sign ambiguity.

By definition $H_A^0=\ker d_A$ consists of parallel adjoint sections. Evaluation at one point identifies it with the subspace of $\mathfrak{su}(2)$ fixed by the full holonomy group. For holonomy equal to a maximal circle, write its elements as $\mathrm{diag}(\lambda ,{\lambda }^{-1})$. The fixed Lie algebra isThe off-diagonal real plane rotates with weight two and has no nonzero vector fixed by the whole circle. A parallel section is uniquely determined by its value, and every invariant value gives a globally well-defined parallel section by parallel transport. Therefore ${\dim }_{\mathbb{R}}{H}_A^0=1$. The full-circle holonomy hypothesis is essential: central holonomy would fix all of $\mathfrak{su}(2)$.

The Fredholm Kuranishi reduction has the following form. Let $F$ be a smooth map between real Hilbert spaces with $F(0)=0$ whose derivative $L=DF(0)$ is a Fredholm operator. Split the source as $K\oplus V$, where $K=\ker L$, and the target as $R\oplus C$, where $R=\mathrm{im}L$ and $C$ represents the finite-dimensional cokernel. The restriction $L:V\to R$ is a bounded isomorphism. The implicit function theorem applied to $P_{R}F(k+v)=0$ solves it uniquely near zero as $v=v(k)$, with $v(0)=Dv(0)=0$. Thusis a smooth finite-dimensional obstruction map, and the local zero set of $F$ is the graph over ${\kappa }^{-1}(0)$. If a compact group acts preserving the problem, average inner products to choose invariant splittings; uniqueness in the implicit function theorem makes the construction equivariant.

Apply this to the vector-bundle curvature equation near $A$. Complete connections in $L_{\ell }^2$ and the unitary bundle gauge group in $L_{\ell +1}^2$ with integer $\ell \ge 3$, so the required multiplication and gauge operations are smooth. The Coulomb slice for unitary connections imposes $d_A^{\ast }a=0$. Its local existence follows from the gauge-fixing equation: the derivative in the gauge direction is $d_A^{\ast }{d}_A$, invertible on the orthogonal complement of its parallel kernel by the elliptic estimate. The local slice theorem then says that its remaining identifications are precisely by $\mathrm{Stab}(A)$.

On this slice the equation isIts kernel is the harmonic representative space $H_A^1$, and its cokernel is $H_A^2$. The hypothesis $H_A^2=0$ makes the obstruction target zero. Therefore the solutions near $A$ are an equivariant smooth graph over $H_A^1$, and the gauge-orbit neighborhood is a neighborhood of zero in $H_A^1/\mathrm{Stab}(A)$.

It remains to compute this representation, not just its dimension. A reducible SU2 connection with circle holonomy preserves $E=L\oplus L^{-1}$. In this splitting,The diagonal part is a trivial real connection. Its degree-one deformation space is the space of ordinary harmonic one-forms. To check this directly, if $d^{\ast }a=0$ and $d^{+}a=0$, then $da$ is an exact anti-self-dual two-form. Stokes theorem givesso $da=0$ and $a$ is harmonic. This neutral deformation space vanishes because $b_1(X)=0$. Hence all of $H_A^1$ is in the off-diagonal part, where the complex structure of $L^2$ commutes with both operators; it is a complex vector space.

Using part (a), part (b), and $H_A^2=0$,Parameterize the stabilizer by $u_{\lambda }=\mathrm{diag}({\lambda }^{-1},\lambda )$, with $\lambda \in S^1$. Under our convention $A^u=u^{-1}Au+u^{-1}du$, its action on an off-diagonal perturbation isConsequently the equivariant graph identifies a neighborhood of $[A]$ with a neighborhood of the origin inThis proves the exact scalar weight requested, with a stabilizer parametrization consistent with the chosen gauge convention. The ineffective kernel is $\{\pm 1\}$. The orbit space is also the cone on ${\mathbb{CP}}^{d-1}$, because the weight-two circle has the same nonzero orbits as the ordinary scalar circle. This is the local cone at an unobstructed reducible SU2 instanton.