Past exam of the mathematics course of the University of Cambridge / 2002 / iii / Paper 15 (Geometric Invariant Theory)

A monomial $x^a{y}^b$ is invariant precisely when $a-b$ is divisible by $n$: pulling back along the generator multiplies it by ${\zeta }^{a-b}$. Since distinct monomials are linearly independent, an invariant polynomial is a linear combination of exactly these invariant monomials. If $a\ge b$, write $a-b=kn$ and obtainIf $b\ge a$, the analogous expression uses $y^n$ and $xy$. Thus the polynomial invariant ring is generated bywhich satisfy $uv=w^n$.

To check that this is the only relation, reduce each monomial in $\mathbb{C}[U,V,W]$ using $UV=W^n$ until it has one of the formsTheir images are, respectively, $x^c{y}^c$, $x^{nk+c}{y}^c$ and $x^c{y}^{nk+c}$. These are pairwise distinct monomials in $x,y$, so no nontrivial linear combination of the reduced forms maps to zero. ThereforeThis identifies the determinant-one cyclic quotient of the affine plane with the affine hypersurface $UV=W^n$, and the quotient map is $(x,y)\mapsto (x^n,y^n,xy)$.

Apply the Jacobian criterion to $f=UV-W^n$. Its gradient isOver $\mathbb{C}$, with the printed $n\ge 3$, this vanishes only at $U=V=W=0$. Every other point is smooth. This point is exactly the image of the origin, and its inverse image is only the origin, since $x^n=y^n=0$ forces $x=y=0$. In fact the origin itself is singular: its tangent space has dimension three while the hypersurface has dimension two. Hence the quotient is smooth away from the image of the origin, with that point its unique singularity. Primality is not needed for this calculation; the same argument works for every integer $n\ge 2$.

The multiplicative algebraic group ${\mathbb{G}}_m$ has coordinate ring $\mathbb{C}[t,t^{-1}]$. An algebraic morphism $f:{\mathbb{G}}_m\to {\mathbb{G}}_m$ therefore sends the invertible coordinate on the target to a unit of this Laurent polynomial ring.

Every unit of $\mathbb{C}[t,t^{-1}]$ has the form $c{t}^i$, with $c\in {\mathbb{C}}^{\ast }$ and $i\in \mathbb{Z}$. To prove this, suppose Laurent polynomials $P,Q$ satisfy $PQ=1$. Write their least and greatest active exponents as $a\le b$ and $c\le d$. The least and greatest exponents in their product are $a+c$ and $b+d$, since their endpoint coefficients are nonzero. Both must be zero. Consequentlyso each factor has only one active exponent. This proves the asserted unit description.

Thus $f(t)=c{t}^i$. A group homomorphism sends the identity to the identity, so $f(1)=1$ and $c=1$. Conversely, for every integer $i$, including negative integers and zero, $t\mapsto t^i$ is regular on ${\mathbb{G}}_m$ and satisfies $(st{)}^i=s^i{t}^i$. We obtainThese are exactly the characters of an algebraic torus in the one-dimensional case. The algebraic-morphism requirement is essential to the argument: the map must come from the Laurent coordinate ring.

Consider the subalgebraIt is closed under multiplication because every nonconstant term of its elements is divisible by $x$. Evaluation at $x=0$ defines a homomorphism $A\to \mathbb{C}$ with kernel $I=x\mathbb{C}[x,y]$. Its square is $I^2=x^2\mathbb{C}[x,y]$, soas a complex vector space. In particular, the classes of $x{y}^j$, $j\ge 0$, are linearly independent and span it.

If $A$ were a finitely generated algebra, take a finite list of generators and subtract their scalar images under evaluation. The resulting generators $z_1,\dots ,z_r$ lie in $I$ and still generate $A$ over $\mathbb{C}$. Every element of $I$ is a polynomial in the $z_i$ with zero constant term. Modulo $I^2$, all terms of degree at least two vanish, leaving a linear combination of the classes of $z_1,\dots ,z_r$. This would make $I/I^2$ finite-dimensional, contradicting the displayed infinite basis. Thus this subalgebra is not finitely generated. This is the infinite cotangent space obstructs finite algebra generation argument; it proves failure of finite generation even when the proposed generators are arbitrary polynomials, rather than just monomials.

Use the natural linearization of a line bundle on ${\mathcal{O}}_X(1)$ induced by the conjugation representation on $M_n\mathbb{C}$. The notation involving the dual does not change this matrix description: the nondegenerate pairing $(A,B)\mapsto \mathrm{tr}(AB)$ identifies the representation with its dual equivariantly. Let $[A]$ be a projective point, so $A\ne 0$.

The Hilbert-Mumford criterion for projective stability says that $[A]$ is not a semistable projective point in geometric invariant theory precisely when some algebraic one-parameter subgroup $\lambda :{\mathbb{G}}_m\to S{L}_n$ sends its affine representative to zero:Equivalently, all active weights for that subgroup are strictly positive. With the convention $\mu ([A],\lambda )=-\min \{\text{active weights}\}$, semistability requires $\mu \ge 0$ for every subgroup, including conjugates of diagonal ones.

Suppose such a zero limit exists. The coefficients $c_j(A)$ inare invariant under conjugation. Continuity along the zero limit gives $c_j(A)=c_j(0)=0$ for every $j$. By the Cayley-Hamilton theorem, $A^n=0$: the matrix is nilpotent.

Conversely, if $A$ is a nonzero nilpotent matrix, choose a basis in which it is strictly upper triangular, for example a Jordan normal form. The change of basis can be chosen in $S{L}_n$: multiply any invertible change-of-basis matrix by a scalar with the required $n$th power, which does not affect its conjugation. In this basis takeThe exponents sum to zero, so this is an algebraic one-parameter subgroup of the special linear group. The $(i,j)$ matrix entry has weightEvery potentially nonzero entry has $i<j$ and therefore positive weight. Thus the conjugated matrix tends to zero. Conjugate this subgroup back to the original basis to obtain the required destabilizing subgroup for $A$.

It follows that the nilpotent cone for conjugation of matrices is exactly the affine null cone, andEach $c_j$ is a homogeneous invariant of degree $j$, so the second description also gives invariant open sets witnessing semistability. Equivalently, a representative has at least one nonzero eigenvalue. Nonzero singular matrices can be semistable; invertibility is not the criterion. For example, a diagonal matrix with diagonal entries $1,0,\dots ,0$ is semistable, while a nonzero strictly upper triangular matrix is unstable.

For a nonzero holomorphic vector bundle $F$ on the curve, its degree of a holomorphic vector bundle and slope of a holomorphic vector bundle areWe obtain a bound independent of the particular holomorphic subbundle by fixing a Hermitian metric on a holomorphic vector bundle $E$ once and for all. Let $D_E$ be its Chern connection, $F_E=D_E^2$ its Chern curvature, and $\omega$ any positive area form on the compact Riemann surface. WriteHere $K$ is a smooth Hermitian endomorphism of $E$. By compactness, its largest eigenvalue is bounded above by some finite constant $C$ on the whole curve.

Let $F\subseteq E$ be a holomorphic subbundle of positive rank $r$, with the induced metric. Use the smooth orthogonal splitting of a vector subbundle $E=F\oplus F^{\perp }$. In this splitting the connection has block formwhere $D_F$ is the projected Chern connection and $B$ is the second fundamental form of a holomorphic subbundle. The latter has type $(1,0)$: the $(0,1)$ part of $D_E$ preserves the holomorphic subbundle. The adjoint $B^{†}$ includes conjugation of the form factor and has type $(0,1)$. Squaring this block matrix givesThis derives the curvature formula for a holomorphic subbundle in the column-vector convention used here. Locally write $B=bdz$. ThenThe degree of a holomorphic vector bundle is the trace-curvature integral, since the trace is the curvature of the determinant line bundle. HenceThe last integrand is a nonnegative real two-form and is independent of the local unitary frame and coordinate. Since $K\le C{I}_E$, the compressed trace is at most $rC$ pointwise. ThereforeThis is the subbundle slope bound from ambient Chern curvature. The right side depends only on $E$ and the initially chosen metrics, and is finite. It proves the requested uniform upper bound for every positive-rank subbundle, without assuming any stability property of $E$ or using a bound that depends on $F$.

Take the compact Riemann surface $X={\mathbb{P}}^1$ and the holomorphic vector bundle $E={\mathcal{O}}_X\oplus {\mathcal{O}}_X$. For every positive integer $m$, use homogeneous coordinates $[Z_0:Z_1]$ and the mapThe two sections $Z_0^m,Z_1^m$ never vanish together on ${\mathbb{P}}^1$. Thus this map has rank one at every point; locally one of its two components is invertible in suitable line-bundle frames, so its image is a genuine holomorphic subbundle and its quotient is a locally free sheaf. More explicitly, there is an exact sequenceExactness follows fibrewise from the absence of a common zero and hence also locally as a sequence of vector bundles. The image $F_m$ is isomorphic to ${\mathcal{O}}_X(-m)$ and has rank one, soThus there need not be a lower bound for subbundle slopes, even for a fixed trivial rank-two bundle. Merely multiplying a section of one fixed subline by a function with zeros would not suffice: that could give only a subsheaf with a torsion quotient. The nowhere-simultaneously-zero pair above avoids that problem and realizes unbounded negative degrees of line subbundles.

Let $\rho :{\pi }_1(X)\to U(V)$ be the unitary representation. The associated unitary flat holomorphic vector bundle iswhere deck transformations act on the first factor and $\rho$ acts on the second. Local lifts to the universal cover give holomorphic orthonormal frames with locally constant unitary transition matrices. The induced Hermitian metric is therefore well defined, and the induced flat unitary connection is its Chern connection.

Let $s$ be a global holomorphic section. In one of these parallel orthonormal frames write $s=(f_1,\dots ,f_r)$, where every $f_j$ is holomorphic. Its squared norm $u=|s{|}^2={\sum }_j|f_j{|}^2$ is a globally defined smooth function because the transition matrices are unitary. In a local complex coordinate,Choose any smooth conformal metric on the curve, locally $\kappa (z)|dz{|}^2$. Its Laplacian satisfiesOn the compact curve without boundary, Stokes theorem gives ${\int }_X\Delta ud\mathrm{vol}=0$. A continuous nonnegative function with zero integral vanishes everywhere, so each $f_j^{\prime }=0$. Thus the section is locally constant in every parallel frame and is globally parallel. This proves that holomorphic sections of a unitary flat bundle are parallel.

A parallel section is determined by its value $v$ at a chosen base point. Following it around a loop returns the value $\rho (\gamma )v$, so a global parallel section exists exactly when $\rho (\gamma )v=v$ for every $\gamma \in {\pi }_1(X)$. Conversely, such a fixed vector gives the constant equivariant section on $\tilde{X}$, which descends holomorphically. Evaluation and this construction are inverse linear maps, giving the canonical identificationUnitarity is what makes the local squared norms glue and the integrated nonnegative identity available; no assumption that the universal cover is compact is made.

The flat unitary connection on $E_V$ is its Chern connection, since it preserves the induced Hermitian metric and its $(0,1)$ part is the Dolbeault operator. Its curvature is zero. The trace-curvature formula for the degree of a holomorphic vector bundle therefore givesFor any positive-rank proper holomorphic subbundle $F\subset E_V$, equip $F$ with the induced metric and use the block-connection calculation of Question 4(a). Since the ambient curvature vanishes, its Chern curvature iswhere $B=bdz$ is the second fundamental form of a holomorphic subbundle. ConsequentlyThe term being integrated is nonnegative, so its sign gives the inequality for every such subbundle, not just for parallel subbundles. By the definition of a semistable holomorphic vector bundle,If semistability is formulated using coherent subsheaves instead, saturate a subsheaf on the smooth curve. The saturation has the same rank, at least as large a degree, and is a subbundle because a torsion-free sheaf on a smooth curve is locally free. The inequality for the saturated subbundle therefore also bounds the original subsheaf. Finally, equality for a subbundle forces $B=0$, so it is parallel; invariant subspaces of $V$ correspond to these degree-zero parallel subbundles. Semistability is the conclusion required here, rather than stability for every unitary representation.