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Over an algebraically closed field of characteristic zero, any two Cartan subalgebras of the symplectic Lie algebra are conjugate by the symplectic group. A Cartan subalgebra $\mathfrak{h}$ is an abelian Lie algebra and equals its centralizer of a Lie subalgebra. The symplectic centralizer decomposition therefore forces all its weight spaces in $V$ to be one-dimensional, nonzero, and paired by opposite weights of a representation. Choose a basis vector in each space, normalizing the perfect pairing with its opposite to one. These vectors form a symplectic basis $\mathcal{B}$. In this basis, $\mathfrak{h}$ is contained in the Lie subalgebra of diagonal matrices with entries $(t_1,\dots ,t_n,-t_1,\dots ,-t_n)$. All those diagonal matrices commute with $\mathfrak{h}$, so equality with its centralizer of a Lie subalgebra forces equality with the entire diagonal Lie subalgebra. Repeating for ${\mathfrak{h}}^{\prime }$ gives a second symplectic basis. The linear map sending one symplectic basis to the other belongs to the symplectic group and conjugates the two diagonal Lie subalgebras, proving the claim.

Fix a volume form on a four-dimensional real symplectic vector space. Wedge pairing on its exterior square has signature $(3,3)$. The symplectic group fixes the bivector inverse to the symplectic form. Its orthogonal complement is five-dimensional with signature $(3,2)$ after choosing the appropriate overall sign. The induced representation has kernel $\{\pm I\}$: identity action on the entire exterior square preserves every two-plane and thus every line, forcing a scalar transformation with square one. Its derivative is injective and both group dimensions are ten, so its image is the identity component of $SO(3,2)$.

The identity is symplectic because $I^{T}JI=J$. For two symplectic matrices $M,N$,Thus the set is closed under matrix multiplication. Every member is invertible by part (i). Multiplying $M^{T}JM=J$ on the left by $M^{-T}$ and on the right by $M^{-1}$ givesso it is also closed under inverses. Associativity is inherited from the general linear group. These facts prove that the symplectic group is a subgroup of $\mathrm{GL}(2m,\mathbb{R})$.

Solutions to all three questions of the Lie algebras course of Part C of the mathematics course of the University of Oxford examination, sat on 6 June 2023. Source: original released PDF, bundle pages 307–309; paper pages 1–3. Questions 1 and 2 occupy paper page 2, and Question 3 occupies paper page 3. Module C2.1 and formal paper code A12189W1 identify this Part C paper. The original question numbering is retained.

Throughout, $k$ is an algebraically closed field of characteristic zero. All Lie algebras and vector spaces are finite-dimensional vector spaces except the Lie algebra representation explicitly considered in Question 2(c). We use the operator commutator $[A,B]=AB-BA$ and the standard sl2 triple convention $[H,E]=2E$, $[H,F]=-2F$, $[E,F]=H$. The printed operators in Question 2(c)(i) have a sign inconsistency with that convention; its solution explains the correction. The displayed definition of the symplectic group on paper page 3 also lacks a closing set brace; its intended meaning is the group of invertible symplectic linear maps.

Use the standard structural fact that $\mathfrak{sp}(V)$ is a semisimple Lie algebra over an algebraically closed field of characteristic zero. If $\mathfrak{h}$ is a Cartan subalgebra, part (a)(ii) gives $C_{\mathfrak{sp}(V)}(\mathfrak{h})=\mathfrak{h}$, and $\mathfrak{h}$ is an abelian Lie algebra. Part (c) therefore shows that its weight spaces occur as $n$ pairs of one-dimensional spaces, with no zero weight space.

Choose representatives ${\lambda }_1,\dots ,{\lambda }_n$ of its nonzero weights of a representation. Pick $p_i\ne 0$ in $V_{{\lambda }_i}$ and $q_i\ne 0$ in $V_{-{\lambda }_i}$, scaling $q_i$ so that $\omega (p_i,q_i)=1$. Part (b) shows that all other pairings among different pairs are zero. Each individual weight space is one-dimensional and the bilinear form is alternating, so the ordered basisis a symplectic basis. Every $h\in \mathfrak{h}$ is a diagonal matrix in this basis, with entries ${\lambda }_1(h),\dots ,{\lambda }_n(h),-{\lambda }_1(h),\dots ,-{\lambda }_n(h)$.

Let $D_{\mathcal{B}}$ be the Lie subalgebra consisting of all diagonal matricesEach such matrix lies in the symplectic Lie algebra and commutes with $\mathfrak{h}$. Henceand $\mathfrak{h}=D_{\mathcal{B}}$. In particular, every Cartan subalgebra has dimension $n$; the ${\lambda }_i$ are a basis of ${\mathfrak{h}}^{\ast }$.

Carry out the same construction for ${\mathfrak{h}}^{\prime }$, obtaining a symplectic basis ${\mathcal{B}}^{\prime }=(p_1^{\prime },\dots ,p_n^{\prime },q_1^{\prime },\dots ,q_n^{\prime })$ with ${\mathfrak{h}}^{\prime }=D_{{\mathcal{B}}^{\prime }}$. The unique linear map $g$ sending $p_i$ to $p_i^{\prime }$ and $q_i$ to $q_i^{\prime }$ is invertible and preserves all pairings of basis vectors. By bilinearity, it preserves $\omega$ on all of $V$, so $g$ belongs to the symplectic group $\mathrm{Sp}(V)$. The change of basis givesThus conjugacy of Cartan subalgebras of the symplectic Lie algebra follows by explicitly matching their symplectic bases. If $n=0$, the Lie algebra is zero and the same conclusion holds with $g$ the identity on the zero vector space.

Let $V={\mathbb{R}}^4$ have symplectic basis $(e_1,f_1,e_2,f_2)$. Fix $\mathrm{vol}=e_1\wedge f_1\wedge e_2\wedge f_2$ and define a symmetric bilinear form $Q$ on ${\Lambda }^{2}V$ by $u\wedge v=Q(u,v)\mathrm{vol}$. The pairings between complementary basis bivectors split this six-dimensional space into three hyperbolic planes, so $Q$ has signature $(3,3)$.

The symplectic group preserves volume and the bivector $\Omega =e_1\wedge f_1+e_2\wedge f_2$, obtained from the inverse symplectic form, up to its fixed sign convention. Since $Q(\Omega ,\Omega )=2$, its orthogonal complement $W={\Omega }^{\perp }$ has dimension five and signature $(2,3)$. On $W$ use $-Q$, which has signature $(3,2)$. This constructs the exterior-square double cover of SO0(3,2).

If an element acts trivially on $W$, it also fixes $\Omega$, so its action on the entire exterior square is the identity. Every decomposable bivector then shows that its two-plane is preserved. A line is the intersection of two such planes; hence every line is preserved, and the matrix is scalar. Its exterior-square action forces its scalar square to equal one, giving precisely $\pm I$.

The derivative is injective by the same infinitesimal argument: zero action on ${\Lambda }^{2}V$ forces an infinitesimal scalar, whose induced scalar there is twice its value and therefore zero. The symplectic Lie algebra has dimension $2^2+2(2\cdot 3/2)=10$, from blocks $\left(\begin{array}{cc}A& B\\ C& -A^T\end{array}\right)$ with symmetric $B,C$. The orthogonal Lie algebra in five dimensions also has dimension ten. Thus the image is open. Real symplectic polar decomposition has connected compact factor $U(2)$ and a contractible positive symplectic factor, so $Sp(4,\mathbb{R})$ is connected and its image is exactly the identity component:As in the preceding noncompact cases, this does not identify the quotient with the disconnected full $SO(3,2)$.

Replace $F^n$ by a symplectic vector space $(V,B)$ of dimension $2m$, and elementary row operations by symplectic transvectionsThe cross terms in $B(Tx,Ty)$ cancel, and $B(v,v)=0$, so these maps preserve $B$. Their inverses are $T_{v,-c}$.

They generate $Sp(V)$ by the following symplectic basis elimination. If $B(u,v)\ne 0$, a single transvection with direction $v-u$ and parameter $1/B(u,v)$ sends $u$ to $v$. If $B(u,v)=0$ and $u\ne v$, choose $w$ with both $B(u,w)$ and $B(w,v)$ nonzero: the union of two proper hyperplanes does not exhaust a vector space, even over ${\mathbb{F}}_2$. Two transvections then suffice. Thus one can make a form-preserving linear map fix the first basis vector $e$. Next align its partner $f$, while keeping $e$ fixed. Two partners have difference in $e^{\perp }$, so the preceding single-transvection construction uses a direction in $e^{\perp }$ whenever their mutual pairing is nonzero. If it is zero, the intermediate partner $f+ae$, $a\ne 0$, has nonzero pairing with both. All these transvections fix $e$. Once both $e,f$ are fixed, recurse on their nondegenerate symplectic orthogonal complement. This proves generation over every field.

Use the action on the points of projective space, whose kernel is the scalar center $Z=\{\lambda I:{\lambda }^2=1\}$. For $m=1$ this is the $S{L}_2$ case. For $m\ge 2$ the action is primitive, although it is generally not two-transitive. The stabilizer of a line $L$ has two orbits on other lines: those orthogonal to $L$ and those not orthogonal to $L$. This follows by extending a hyperbolic pair, or an independent isotropic pair, to a symplectic basis. A block containing $L$ and a second line $M$ contains the entire corresponding orbit of the stabilizer of $L$. The stabilizer of $M$ then supplies a line of the other type: choose a vector orthogonal to one of $L,M$ and not to the other. Nondegeneracy and independence guarantee such a vector, also in dimension four. Hence the block contains both orbits and is the whole set.

The transvections along $L$ form an abelian normal subgroup of its stabilizer; their conjugates generate the group. Thus the Iwasawa simplicity lemma applies as soon as perfectness is established. For $|F|>3$, every transvection lies in an embedded $S{L}_2(F)$ acting on a hyperbolic plane, and 4(c) proves that embedded group perfect.

For the remaining fields, the perfectness of finite symplectic groups has a short direct proof. Over ${\mathbb{F}}_3$ with $m\ge 2$, all parameter-one transvections have the same class $t$ in the abelianization, and $3t=0$. In an isotropic plane with basis $u,v$, the four directions $u,v,u+v,u-v$ give commuting transvections whose product is $1$: their rank-one terms add to zero, sinceover ${\mathbb{F}}_3$. Thus $4t=0$, giving $t=0$. Parameter-two transvections are inverses, so all generators vanish in the abelianization. Over ${\mathbb{F}}_2$ with $m\ge 3$, use the seven nonzero vectors of an isotropic three-space. Their commuting transvections multiply to $1$, since each diagonal coordinate of ${\sum }_{0\ne v}v{v}^{\mathsf{T}}$ occurs four times and each off-diagonal coordinate twice. Their common abelianized class satisfies $2t=7t=0$, and again the group is perfect.

Consequently the projective symplectic group is simple, with the necessary exceptionsThe first two are the linear exceptions; the last is $S{p}_4(2)\cong S_6$, proved in 6(c). The word “projective” matters: the unquotiented symplectic group may have the nontrivial scalar center $\{I,-I\}$.

In weight one, the Hodge numbers are $h^{1,0}=h^{0,1}=r$, and the Hodge filtration is determined by the $r$-plane $H=F^1$. The bilinear orthogonality condition is $\psi (H,H)=0$. An isotropic $r$-plane in a symplectic $2r$-space is a Lagrangian subspace, soThe ordinary Grassmannian has dimension $r^2$. Restriction of the alternating form to an $r$-plane imposes $r(r-1)/2$ independent conditions. Equivalently, near a fixed Lagrangian plane, transverse Lagrangians are graphs of symmetric matrices, with $r(r+1)/2$ independent entries. Therefore

Write ${\mathbb{C}}^{2r}={\mathbb{C}}^r\oplus {\mathbb{C}}^r$ in the block order of the given matrix. On $D$, positivity is $i\psi (v,\bar{v})>0$ for every nonzero $v\in H$. The plane $H$ cannot contain $(x,0)\ne 0$, because that vector has $\psi (v,\bar{v})=0$. Projection onto the second coordinate factor is therefore injective on $H$ and, by equal dimensions, an isomorphism. There is a unique matrix $Z$ such thatFor $v=(Zu,u)$ and $v^{\prime }=(Zw,w)$, direct evaluation givesThus isotropy is exactly $Z=Z^{\mathsf{T}}$. For such $Z$,Hence positivity is precisely positive definiteness of $\mathrm{Im}Z$. Conversely, symmetry and this inequality define a positive Lagrangian plane; it is opposed to its conjugate because $(Z-\bar{Z})u=0$ forces $u=0$. We have explicitly identifiedthe Siegel upper half-space. The choice of the graph $(Zu,u)$, rather than $(u,Zu)$, is what matches the sign of the given symplectic form.

Let an integral element of the symplectic group have block matrixActing on graph vectors givesThe image remains a positive Lagrangian plane, so its second projection is invertible. Thus $CZ+D_0$ is invertible and the induced action isThe symplectic identities give symmetry of this matrix, and preservation of $i\psi (v,\bar{v})$ givesThis is a Hermitian congruence by an invertible matrix, so it proves preservation of the positive domain as well as describing the action. These are coordinates on marked Hodge structures; passing to the integral-group quotient forgets the symplectic marking.

The symplectic group consists of the invertible linear maps preserving the alternating bilinear form:We use row-vector action in this question, which is the convention compatible with its printed upper-triangular flag stabilizer subgroup. Thus matrices preserve a form matrix $J$ by $gJ{g}^T=J$.

Count ordered symplectic bases. There are $q^{2m}-1$ choices for the first nonzero vector $e_1$. Nondegeneracy makes the equation $(e_1,f_1)=1$ a nonzero linear-functional equation, with $q^{2m-1}$ solutions. Their span is a nondegenerate plane; its orthogonal complement is symplectic of dimension $2m-2$. Repeating there givesEach symplectic basis is the image of a fixed one under exactly one form-preserving map, justifying the count as a group order. If $q=p^a$, every factor $q^{2i}-1$ is prime to $p$, so the exact $p$-part is $q^{m^2}$.

Taking diagonal blocks is a homomorphism $P\to L$. Its kernel is $Q$, and block-diagonal inclusion is a section. For any $g\in P$, remove its diagonal element $\ell \in L$ by $g{\ell }^{-1}\in Q$. Also $Q\cap L=1$. ThusUsing $|{\mathrm{GL}}_k(q)|=q^{k(k-1)/2}{\prod }_{i=1}^k(q^i-1)$ and part (a), the product simplifies toThe exponent simplification is $2k(m-k)+k(k+1)/2+k(k-1)/2+(m-k{)}^2=m^2$.

For an independent orbit-stabilizer theorem count, choose an ordered independent isotropic tuple $f_1,\dots ,f_k$. After $r$ choices, their span has $q^r$ elements and its orthogonal complement has dimension $2m-r$. The next choice therefore has $q^{2m-r}-q^r$ possibilities. The number of tuples isEvery totally isotropic $k$-space has $|{\mathrm{GL}}_k(q)|$ ordered bases, so the number of such spaces isEach tuple extends to a symplectic basis by successively choosing paired partners and taking orthogonal complements. Consequently the symplectic group is transitive on these spaces. The stabilizer subgroup of $W$ also preserves $W^{\perp }$, and so is exactly $P$. Dividing $|{\mathrm{Sp}}_{2m}(q)|$ by $N_k$ reproduces the boxed answer. Dividing by $T_k$ instead would count the pointwise stabilizer subgroup of the ordered tuple, a different subgroup.

WriteThe identity belongs to $G$. If $A,B\in G$, thenTaking determinants in $A^T\Omega A=\Omega$ shows that $A$ is invertible, and multiplying that identity by $A^{-T}$ and $A^{-1}$ gives $(A^{-1}{)}^T\Omega A^{-1}=\Omega$. Thus $G$ is a group.

Let ${\mathrm{Skew}}_{2n}(\mathbb{R})$ denote the vector space of skew-symmetric $2n\times 2n$ matrices and defineThen $G=F^{-1}(\Omega )$. At $A\in G$,Given any skew-symmetric matrix $S$, set $H=AB$ with $B=-\frac{1}{2}\Omega S$. Since $A^T\Omega A=\Omega$, a direct calculation givesThe derivative is therefore surjective at every point of $F^{-1}(\Omega )$. The regular level set theorem proves that the symplectic group is an embedded submanifold of $M_{2n}(\mathbb{R})$.

Positive complex Lagrangian subspaces for the integral symplectic form $\left(\begin{array}{cc}0& -I\\ I& 0\end{array}\right)$ are graphs $H_Z=\{(Zu,u):u\in {\mathbb{C}}^r\}$. Isotropy is the symmetry of $Z$, and $iQ(v,\bar{v})=2u^{\mathsf{T}}(\mathrm{Im}Z)\bar{u}$ gives positivity. For $g=\left(\begin{array}{cc}A& B\\ C& D\end{array}\right)$ in the symplectic group, the action is $Z\mapsto (AZ+B)(CZ+D{)}^{-1}$.

Vertices are Lagrangian subspaces of a $2m$-dimensional symplectic vector space, joined when they intersect in codimension one. A pair with intersection dimension $m-r$ admits a symplectic basis in which $U=⟨e_1,\dots ,e_m⟩$ and $W=⟨f_1,\dots ,f_r,e_{r+1},\dots ,e_m⟩$. Replacing $e_i$ successively by $f_i$ gives an isotropic path of length $r$, while one edge can increase intersection dimension with a fixed target by at most one. Equal-distance ordered pairs have the same adapted form, so the symplectic group acts distance-transitively.

A linear map $A:(V,{\omega }_V)\to (W,{\omega }_W)$ is symplectic when it pulls back the target symplectic form to the source form. Nondegenerate bilinear form makes such a map injective. In equal finite dimensions it is an isomorphism. The automorphisms form the symplectic group and preserve the symplectic orientation on the ambient space and on every symplectic subspace.

A symplectic vector bundle has a smooth fiberwise symplectic form, making each fiber a symplectic vector space. Its structure group reduces to the symplectic group. A symplectic normal bundle is an important example.