Past exam of the mathematics course of the University of Cambridge / 2008 / iii / Paper 23 (Algebraic Groups)

For a real Lie group $G$, its Lie algebra is the real tangent space $T_{e}G$ at the identity. Each tangent vector extends uniquely to a left-invariant vector field, and the commutator of those fields defines the Lie bracket. For a matrix group this bracket is $[X,Y]=XY-YX$.

For a complex affine algebraic group, the Lie algebra of an affine algebraic group is the complex Zariski tangent space $T_{e}G={\mathrm{Hom}}_{\mathbb{C}}({\mathfrak{m}}_e/{\mathfrak{m}}_e^2,\mathbb{C})$. Its bracket is the commutator of left-invariant derivations. In a faithful matrix realization, one can compute the tangent space using dual numbers: $X$ belongs to it exactly when $I+\epsilon X$ satisfies the group equations modulo ${\epsilon }^2$. This is a complex Lie algebra, whereas compact groups such as $U_n$ and $S{U}_n$ have real Lie algebras.

Differentiate $g^{\ast }g=I$ at the identity to get $X^{\ast }+X=0$. Conversely $e^{tX}$ is unitary for every skew-Hermitian $X$. Differentiating the determinant gives $\det (I+tX)=1+t\mathrm{tr}X+O(t^2)$, while $\det e^{tX}=e^{t\mathrm{tr}X}$ proves sufficiency of the trace condition. Finally, $G{L}_n(\mathbb{C})$ is an open subset of all complex matrices, so its tangent space has no linear constraint. ThereforeTheir dimensions are respectively $n^2,n^2,n^2-1,n^2-1$, with real dimensions for the first and third and complex dimensions for the second and fourth. In every case the Lie bracket is the matrix commutator.

An irreducible complex affine algebraic group is connected. It is a reductive algebraic group when its unipotent radical $R_u(G)$, the largest connected normal unipotent subgroup, is trivial. This is an algebraic condition; in characteristic zero it also implies complete reducibility of finite-dimensional rational representations, as proved in Question 2.

Examples are algebraic tori, $G{L}_n(\mathbb{C})$ and $S{L}_n(\mathbb{C})$, and products of these groups. A diagonal matrix that is unipotent must be the identity, so a torus has no nontrivial unipotent subgroup. For the two matrix groups, let $N$ be a normal unipotent subgroup. The Kolchin theorem supplies a nonzero fixed vector in their standard representation. The subspace fixed by $N$ is invariant under the whole group: for $g\in G$, $n(gv)=g(g^{-1}ng)v=gv$. The standard representation is irreducible, so every vector is fixed by $N$; its faithfulness makes $N$ trivial. A normal unipotent subgroup of a product projects to such a subgroup of each factor, proving the product assertion. By contrast the additive group and a nontrivial upper-unitriangular group are not reductive: their unipotent radicals are the whole group.

For a compact Lie group, take any positive-definite Hermitian inner product $h_0$ on $V$ and average using probability Haar measure:It remains positive definite, because the integrand is positive for a nonzero vector, and translation invariance makes it $G$-invariant. If $W\subset V$ is an invariant subspace, then $W^{\perp }$ is invariant too: $h(gv,w)=h(v,g^{-1}w)=0$ when $v\in W^{\perp }$. Thus $V=W\oplus W^{\perp }$. Induction on dimension proves complete reducibility. This uses the usual continuous finite-dimensional representations of the compact group.

For a connected complex reductive algebraic group, representations are understood to be rational representations. Use a compact real form $K\subset G$ with $\mathrm{Lie}(G)=\mathrm{Lie}(K){\otimes }_{\mathbb{R}}\mathbb{C}$. The structural construction takes a compact real form of the semisimple derived factor and the unit circles in the central torus, then passes through their finite central quotient. This is the compact-form structure of a complex reductive group, rather than an assumption that its given representation is already unitary.

The group $K$ is Zariski dense: the Lie algebra of its Zariski closure contains the complex span of $\mathrm{Lie}(K)$, hence all of $\mathrm{Lie}(G)$; connectedness then makes that closure $G$. Average over $K$ as above. The orthogonal complement of a $G$-submodule is $K$-invariant. Its stabilizer is a Zariski-closed subgroup, since preserving a fixed subspace is a system of algebraic matrix conditions. The stabilizer contains dense $K$, so it is all of $G$. This again gives an invariant complement to every submodule and provesIf reductivity is defined to allow finitely many components, first obtain a $G^0$-equivariant projection $P:V\to W$. Average its conjugates over the finite component group. Since $P$ commutes with $G^0$, the average is independent of coset representatives; it is $G$-equivariant and remains the identity on $W$. Its kernel supplies the required complement. Thus the result holds in that convention too.

Let $R=\mathbb{C}[V]=\mathrm{Sym}(V^{\ast })$, with its degree grading, and let $A=R^G$ be the polynomial invariant ring. For a compact group average polynomials over $G$; for a complex reductive group average over its Zariski-dense compact form from part (i). Each degree piece is finite dimensional, so this defines a polynomial and gives the Reynolds operator on a polynomial invariant ringIt preserves degree. For the reductive case compact invariance is full group invariance by Zariski density.

Let $A_{+}$ be the positive-degree invariants and form the ideal $I=R{A}_{+}$. The Hilbert basis theorem makes $R$ a Noetherian ring, so finitely many homogeneous invariants $f_1,\dots ,f_s\in A_{+}$ generate $I$ as an $R$-ideal. Indeed start with a finite ideal-generating set and collect the finitely many invariants occurring in expressions for those generators; then replace them by homogeneous pieces.

We claim that these same $f_i$ generate $A$ as an algebra. For a homogeneous invariant $f$ of degree $d>0$, write $f={\sum }_i{a}_i{f}_i$ with homogeneous coefficients of degree $d-\mathrm{deg}{f}_i$; terms of negative degree are absent. Apply the Reynolds operator on a polynomial invariant ring to get $f={\sum }_i\mathcal{R}(a_i)f_i$. The averaged coefficients are invariant and have smaller degree than $d$. Starting from the constants, induction on degree puts them and then $f$ in $\mathbb{C}[f_1,\dots ,f_s]$. Taking homogeneous pieces covers every invariant. This proves Hilbert finite generation for reductive invariants:Noetherianity alone would not suffice for a subring; the Reynolds module property is the step that makes the induction work.

Here representations mean finite-dimensional rational representations of $S{L}_2(\mathbb{C})$. Put $W={\mathbb{C}}^2$. The irreducibles areThe invariant alternating form on $W$ identifies $W$ with $W^{\ast }$, so these can equally be realized as binary forms of degree $d$. With basis $e_j=x^{d-j}{y}^j$, the standard Lie algebra generators act asThe distinct torus weights and these nonzero lowering and raising coefficients show irreducibility: any nonzero invariant subspace contains a weight vector, and repeated applications of $E,F$ obtain the entire basis.

For exhaustion, choose a highest-weight vector $v$ of an irreducible representation, so $Hv=mv$, $Ev=0$. The sl2 highest-weight lowering formula, obtained by induction from $[E,F]=H$, is $E{F}^{j}v=j(m-j+1)F^{j-1}v$. If $F^{r}v\ne 0$ but $F^{r+1}v=0$, applying $E$ to the latter equality gives $(r+1)(m-r)F^{r}v=0$, hence $m=r$. The vectors $v,Fv,\dots ,F^{r}v$ have distinct weights and span an invariant submodule, so irreducibility makes it the whole representation. After rescaling basis vectors it is $V_r$. Complete reducibility from Question 2 now describes every representation as a finite direct sum of these symmetric powers. Their central element $-I$ acts by $(-1{)}^d$.

The symbolic method for binary-form invariants works as follows. Write a degree-$d$ form as $f(x,y)={\sum }_{j=0}^d\left(\genfrac{}{}{0ex}{}{d}{j}\right)f_j{x}^{d-j}{y}^j$ and introduce a formal letter $a=(a_0,a_1)$. The formal power $(a_{0}x+a_{1}y{)}^d$ is converted to $f$ by the linear substitution $a_0^{d-j}{a}_1^j\mapsto f_j$. This is an umbral linear rule, not a claim that every form is an actual power of a linear form.

For a coefficient polynomial of degree $r$, polarize it into $r$ copies of the form, represented by letters $a,b,\dots$. Each letter must occur with degree $d$. In characteristic zero, polarization and evaluation on equal forms recover the original coefficient polynomial. The basic invariant expressions in the letters are bracketsThey are invariant because $\det g=1$. Invariant tensors on copies of a two-dimensional space are spanned by pairwise alternating contractions; equivalently, the invariant polynomials in the letters are generated by these brackets. One can see the tensor statement from the two-row Young-symmetrizer description of the determinant powers in tensor space. Their basic relations include $(ab)=-(ba)$ and $(ab)(cd)-(ac)(bd)+(ad)(bc)=0$. Multiply brackets, ensure the required degree in each letter, symmetrize letter copies as needed, and apply the umbral substitutions. This gives all coefficient invariants, and also works for joint invariants of several forms.

For example, for $f=A{x}^2+2Bxy+C{y}^2$, the expression $(ab{)}^2$ becomes $2(AC-B^2)$, the quadratic discriminant up to its conventional scalar. Contractions leaving variable indices unpaired give covariants. A useful systematic construction is the transvectant of binary forms:For forms of degrees $m,n$, this is an equivariant form of degree $m+n-2r$. The corresponding tensor decomposition isIts character identity follows by multiplying the weight sums; complete reducibility then gives the decomposition, and alternating contractions supply the displayed maps onto its summands.

A vector in a rational representation is an unstable vector in a rational representation when $0\in \overline{Gv}$, with Zariski closure. For $G=S{L}_n(\mathbb{C})$, the Hilbert-Mumford criterion for the affine null cone isEvery algebraic one-parameter subgroup is conjugate to $\mathrm{diag}(t^{a_1},\dots ,t^{a_n})$ with integer $a_j$ summing to zero. This follows by decomposing the standard representation into characters of the multiplicative group, then imposing determinant one. In its weight decomposition, a limit of zero means that only strictly positive weights occur.

The reverse implication in the criterion is immediate: the subgroup lies in $G$ and supplies points of the orbit tending to zero. For the harder implication, the complex orbit is locally closed and Zariski open in its closure. A nonempty Zariski-open subset of a complex irreducible variety is dense also in the usual topology. Thus choose $g_{j}v\to 0$ in the ordinary vector-space topology. Singular value decomposition, with determinant phases adjusted, givesUse an $S{U}_n$-invariant Hermitian inner product on $V$. By compactness, after a subsequence ${\ell }_j\to \ell$. Put $w=\ell v$, and decompose it into the weights of the diagonal torus, $w={\sum }_{\chi }{w}_{\chi }$. These weight spaces are orthogonal because the unitary diagonal torus preserves the inner product. HenceFor every $\chi$ with $w_{\chi }\ne 0$, its final norm is bounded away from zero for large $j$, forcing $\chi (b_j)\to -\infty$. There are only finitely many such weights, so choose a single real vector $b$ with sum zero and $\chi (b)<0$ for all of them. The inequalities are strict with integer coefficients. A nearby rational vector still satisfies them; multiply its negative by a common denominator to obtain integers $a_r$ of sum zero with $\chi (a)>0$ throughout this support. Then $\mu (t)=\mathrm{diag}(t^{a_r})$ takes $w$ to zero. Conjugating gives $\lambda (t)={\ell }^{-1}\mu (t)\ell$ taking $v$ to zero, completing the proof. The zero vector itself satisfies the criterion trivially.

For degree-six binary forms, a nontrivial algebraic one-parameter subgroup of $S{L}_2$ is, after conjugation and choosing its direction, $\mathrm{diag}(t^a,t^{-a})$ with $a>0$. On $x^{6-j}{y}^j$ the weight is $a(6-2j)$. Strictly positive weights permit only $j=0,1,2$, so the form is divisible by $x^4$. Conversely every such form tends to zero under that subgroup. Changing basis replaces $x$ by any linear factor. The root-multiplicity criterion for unstable binary forms therefore givesA triple root alone is insufficient: $x^3{y}^3$ has weight zero and is not unstable. Nor does merely having a repeated root suffice.

An affine algebraic group is a linearly reductive algebraic group if every finite-dimensional rational representation is completely reducible; equivalently the functor of invariants is exact. It is a geometrically reductive algebraic group if, whenever $0\ne v\in V^G$, there is a homogeneous invariant polynomial $F\in k[V{]}^G$ of positive degree with $F(v)\ne 0$. Linear reductivity permits degree one; geometric reductivity allows higher degree.

Over the algebraically closed field, an algebraic torus is $T\cong ({\mathbb{G}}_m{)}^r$. Its coordinate ring is the Laurent polynomial ring. A rational action writes the coaction of a vector as a finite sum $\delta (v)={\sum }_{\chi }{v}_{\chi }\otimes t^{\chi }$. The coaction identity and linear independence of Laurent monomials give $\delta (v_{\chi })=v_{\chi }\otimes t^{\chi }$ and $v={\sum }_{\chi }{v}_{\chi }$. ThusEach weight space has scalar action, so any vector-space complement within each weight space is invariant. This proves complete reducibility, and hence tori are linearly reductive in every characteristic. The argument uses characters themselves, not their Lie algebra differentials, which could coincide modulo $p$.

We now prove geometric reductivity of $G=S{L}_2(k)$ by determinant separation for positive-characteristic SL2. Fix a rational $G$-module $V$ and a nonzero fixed vector $v$. For the diagonal torus, choose a weight-zero functional $l\in V^{\ast }$ with $l(v)=1$, using the torus decomposition just proved. Define matrix coefficientsThese are regular functions on $G$, and $\Phi (hw)(g)=\Phi (w)(h^{-1}g)$ makes $\Phi$ equivariant for left translations. Moreover $\Phi (w)(gt)=\Phi (w)(g)$ because $l$ is torus-invariant. Thus $\Phi$ maps into $k[G{]}^T$ for the right torus action, and $\Phi (v)=1$.

Write a matrix's two columns as $a=(a_0,a_1)$ and $b=(b_0,b_1)$, with $\Delta =a_0{b}_1-a_1{b}_0=1$. Right torus multiplication scales these columns by $t$ and $t^{-1}$. Every right-torus-invariant regular function has a weight-zero polynomial lift, hence a sum of polynomials of bidegrees $(j,j)$ in $(a,b)$. To see that a weight-zero lift exists, take any lift and retain its zero-weight component; the ideal $(\Delta -1)$ is torus-stable and the quotient map preserves weights. Multiply each piece by ${\Delta }^{n-j}$ to put the entire sum in bidegree $(n,n)$ without changing its restriction to $G$.

Let $P_n$ be the space of degree-$n$ polynomials in one column. Restriction gives an injective equivariant map $P_n\otimes P_n\to k[G{]}^T$. Indeed if a bidegree-$(n,n)$ polynomial vanishes on determinant-one matrices, scale the first column of any invertible matrix to make its determinant one. Homogeneity then makes the polynomial vanish on every invertible matrix, a dense open subset, so it is zero. Denote the image by $R_n$. These spaces increase because multiplication by $\Delta$ realizes $R_{n-1}\subset R_n$, and their union is $k[G{]}^T$. The finite-dimensional image $\Phi (V)$ therefore lies in one $R_n$, and in any sufficiently larger filtration piece.

Choose $q=p^s$ with $q-1$ at least that bound, and set $n=q-1$. The identity $(1+z{)}^q=1+z^q$ in characteristic $p$ givesThe invariant tensortherefore defines an invertible equivariant linear map $D:P_n^{\ast }\to P_n$: its matrix is anti-diagonal with all entries nonzero. Interpret any tensor $B\in P_n\otimes P_n$ as a map $P_n^{\ast }\to P_n$ and send it to $B{D}^{-1}\in \mathrm{End}(P_n)$. Equivariance of $D$ makes this identification intertwine the diagonal tensor action with conjugation on endomorphisms. Crucially, $D$ restricts to the constant function $1$ on $G$, and is sent to the identity endomorphism.

Under the resulting equivariant linear map $V\to \mathrm{End}(P_n)$, the vector $v$ is sent to $I$. Since $\dim P_n=n+1=q$, the endomorphism determinant is a homogeneous polynomial of positive degree $q$, invariant under conjugation. Pull it back to $V$. The resulting polynomial satisfiesThis constructs the required invariant for every fixed nonzero vector and proves $S{L}_2(k)$ is geometrically reductive, including characteristic two. No averaging by a group order or division by $q$ is used; those operations would fail in the field's characteristic.