Past exam of the lie algebras course of Part C of the mathematics course of the University of Oxford (C2.1)

Original solutions to past examinations of the Lie algebras course of Part C of the mathematics course of the University of Oxford.

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.

Since the general linear Lie algebra ${\mathfrak{gl}}_1(k)$ has zero Lie bracket, identify it with $kz$. A central extension of a Lie algebra is a short exact sequenceof Lie algebra homomorphisms with $\iota (kz)$ contained in the center of a Lie algebra $Z(\tilde{\mathfrak{g}})$. Thus $\pi$ is surjective, its kernel is precisely the indicated line, and that line commutes with all of $\tilde{\mathfrak{g}}$.

Choose a linear map $s:\mathfrak{g}\to \tilde{\mathfrak{g}}$ with $\pi s=\mathrm{id}$. Such a section exists by lifting a basis. Identify $\tilde{\mathfrak{g}}$ with the direct sum of vector spaces $\mathfrak{g}\oplus kz$ by $(x,tz)\mapsto s(x)+t\iota (z)$. The difference $[s(x),s(y)]-s([x,y])$ belongs to the kernel of $\pi$, so there is a unique bilinear map $\alpha :\mathfrak{g}\times \mathfrak{g}\to k$ such thatIt is an alternating bilinear form: $\alpha (x,x)=0$, equivalently $\alpha (x,y)=-\alpha (y,x)$ in characteristic zero. Taking the central component of the Jacobi identity gives the Lie algebra two-cocycle conditionConversely, any alternating bilinear form satisfying this identity defines a Lie algebra byIndeed its Lie bracket is an alternating multilinear map and a bilinear map, and the two components of its Jacobi identity are respectively the Jacobi identity in $\mathfrak{g}$ and the displayed Lie algebra two-cocycle condition. The line $kz$ is central, and projection to $\mathfrak{g}$ gives the required central extension of a Lie algebra.

For later use, replacing $s$ by $s^{\prime }(x)=s(x)+t(x)\iota (z)$, where $t\in {\mathfrak{g}}^{\ast }$ is in the dual space, changes the Lie algebra two-cocycle toThus changes of section add a Lie algebra two-coboundary, up to the conventional sign. A section that is a Lie algebra homomorphism has ${\alpha }^{\prime }=0$ and gives a direct sum of Lie algebras.

Choose a basis $x,y$ of the two-dimensional abelian Lie algebra $\mathfrak{a}$, and lifts with the same names. Every alternating bilinear form is determined by $c=\alpha (x,y)$, and every such form is a Lie algebra two-cocycle because every Lie bracket in $\mathfrak{a}$ is zero. Thus every central extension of a Lie algebra has a basis $x,y,z$ withIf $c=0$, all Lie brackets vanish, so this is the direct sum of Lie algebras $\mathfrak{a}\oplus kz\cong k^3$. If $c\ne 0$, replace the central basis vector by $z^{\prime }=cz$. Then $[x,y]=z^{\prime }$, which is the Heisenberg Lie algebra ${\mathfrak{h}}_3(k)$. This change of basis shows that every nonzero $c$ gives the same Lie algebra isomorphism class.

These two Lie algebras cannot be isomorphic: their derived algebras have dimensions zero and one, respectively, and a Lie algebra isomorphism preserves the derived algebra. Moreover, the Heisenberg Lie algebra cannot be a direct sum of Lie algebras $kz\oplus \mathfrak{a}$, since that direct sum of Lie algebras would have zero Lie bracket. ThereforeHere “up to isomorphism” means the Lie algebra isomorphism class of the middle Lie algebra, allowing a change of the identifications at the ends of the short exact sequence. If instead equivalence of central extensions of a Lie algebra must induce the identity on both $kz$ and $\mathfrak{a}$, the parameter $c$ cannot change: the formula in part (a) has $t([x,y])=0$. That stricter notion has one class for each $c\in k$, rather than two. This distinction explains the convention required by the question.

First put the nonabelian two-dimensional Lie algebra in a convenient form. For any basis $p,q$, let $u=[p,q]=ap+bq\ne 0$. Then $[p,u]=bu$ and $[q,u]=-au$. Since $(a,b)\ne (0,0)$, some linear combination $x$ of $p,q$ has $[x,u]=u$. In particular, $x,u$ are linearly independent. Taking $y=u$ gives a basis withThis is the affine Lie algebra of the line.

Every alternating bilinear form on this vector space isFor any alternating bilinear form $\alpha$, the expressionis an alternating trilinear form: it is trilinear, and setting any two arguments equal makes the terms cancel. Every alternating trilinear form on a two-dimensional vector space vanishes, since any triple is linearly dependent. Hence every ${\alpha }_c$ satisfies the Lie algebra two-cocycle condition, and there are no others:

For the Lie algebra two-cocycle ${\alpha }_c$ of part (i), the central extension of a Lie algebra has a basis $x,y,z$ and Lie bracketsPut $y^{\prime }=y+cz$. This is still a basis together with $x,z$, andConsequently $⟨x,y^{\prime }⟩$ is a Lie subalgebra isomorphic to the affine Lie algebra of the line. It commutes with $kz$ and has zero intersection with $kz$, so both summands are ideals of a Lie algebra andThis also holds for the stricter equivalence of central extensions of a Lie algebra. Indeed, choose the linear functional $t$ with $t(x)=0$, $t(y)=c$. Since $[x,y]=y$ in $\mathfrak{s}$, the change of section in part (a) gives ${\alpha }_c^{\prime }(x,y)=c-t(y)=0$. Thus every Lie algebra two-cocycle here is a Lie algebra two-coboundary, so $H^2(\mathfrak{s},k)=0$ in Lie algebra cohomology. The explicit map from the split central extension of a Lie algebra sends $x$ to $x$, $y$ to $y+cz$, and $z$ to $z$; it fixes the central line and induces the identity on the quotient Lie algebra.

Let $A$ be a three-dimensional vector space with basis $p,q,z$. Define a bilinear map $\bullet :A\times A\to A$ bywith all other products of basis vectors zero. Equivalently,Every product belongs to $kz$, and every product with an element of $kz$ is zero. Hence both $(u\bullet v)\bullet w$ and $u\bullet (v\bullet w)$ vanish for every $u,v,w\in A$. This proves associativity, so $A$ is an associative algebra. The product is an alternating multilinear map, and each nested product in the Jacobi identity is zero, so the same product is a Lie bracket. It is nonzero because $p\bullet q=z\ne 0$.

There is no multiplicative identity element: any candidate $1_A$ would satisfy $1_A\bullet z=0$, whereas an identity element would require $1_A\bullet z=z$. ThusThis is an associative Lie product; its underlying Lie algebra is the Heisenberg Lie algebra.

For a perfect Lie algebra $\mathfrak{s}$, every element is a sum of Lie brackets. The Adjoint representation satisfies $\mathrm{ad}[x,y]=[\mathrm{ad}x,\mathrm{ad}y]$, and the trace of a commutator is zero. ThereforeSome $\mathrm{ad}x$ must have a nonzero eigenvalue. Otherwise, over the algebraically closed field $k$, every $\mathrm{ad}x$ would be a nilpotent linear map, and the Engel theorem would imply that $\mathfrak{s}$ is a nilpotent Lie algebra. This contradicts perfection: the lower central series of a Lie algebra ${\gamma }_1=\mathfrak{s}$, ${\gamma }_{r+1}=[\mathfrak{s},{\gamma }_r]$ would equal $\mathfrak{s}\ne 0$ at every step instead of terminating.

Choose such an $x$. Since $[x,x]=0$, zero is an eigenvalue of $\mathrm{ad}x$. Its trace is zero and its dimension is three, so the remaining eigenvalues are $\lambda ,-\lambda$ with $\lambda \ne 0$. These three eigenvalues are distinct in characteristic zero, hence $\mathrm{ad}x$ is diagonalizable. Put $h=2x/\lambda$, and choose nonzero eigenvectors $e,f$ withThe zero eigenspace is $kh$, so $h,e,f$ form a basis. Applying the Jacobi identity givesThus $[e,f]=ch$ for some $c\in k$. If $c=0$, all Lie brackets of basis vectors lie in $⟨e,f⟩$, contradicting $[\mathfrak{s},\mathfrak{s}]=\mathfrak{s}$. Therefore $c\ne 0$, and replacing $f$ by $f/c$ gives $[e,f]=h$.

We have constructed an sl2 triple that is a basis. The linear map to the sl2 Lie algebra sendingpreserves all Lie brackets of basis vectors and is bijective. HenceThis is the classification of three-dimensional perfect Lie algebras without appealing to a classification of all simple Lie algebras.

Take the semidirect product of a Lie algebra and a modulewhere $k^2$ is the defining representation of the sl2 Lie algebra. To see directly that this is a Lie algebra, realize $(X,u)$ as the block matrix $\left(\begin{array}{cc}X& u\\ 0& 0\end{array}\right)$. These matrices are closed under the operator commutator, whose Jacobi identity holds in every associative algebra.

Using the standard sl2 triple $e,h,f$, its Lie brackets show $[{\mathfrak{sl}}_2,{\mathfrak{sl}}_2]={\mathfrak{sl}}_2$: indeed $h=[e,f]$, $e=[h,e]/2$, and $f=-[h,f]/2$. Also $h=\mathrm{diag}(1,-1)$ is invertible on $k^2$. For any $u\in k^2$, put $v=h^{-1}u$; thenIt follows that the derived algebra contains both ${\mathfrak{sl}}_2\oplus 0$ and $0\oplus k^2$, so $[L,L]=L$: this is a perfect Lie algebra.

However, $0\oplus k^2$ is a nonzero ideal of a Lie algebra with zero Lie bracket. It is therefore a solvable Lie algebra and is contained in the solvable radical of $L$. A semisimple Lie algebra has zero solvable radical, soThis illustrates the perfect semidirect product criterion: the Lie brackets internal to the first summand generate it, while its action generates the second summand.

There is a sign error in the printed assertion under the standard sl2 triple convention. Work first with the operators exactly as printed. Since the $v_i$ form a Hamel basis, their prescribed images extend to linear maps on $V$; each vector is a finite linear combination of basis vectors.

For $i\ge 1$, the operator commutators with $h$ areAt $i=0$, the first identity holds because both sides vanish; the second holds because $hf(v_0)=-2v_1$ and $fh(v_0)=0$. On the other hand, for $i\ge 1$,At $i=0$, both $ef(v_0)=e(v_1)=0$ and $fe(v_0)=0$, so this identity also holds. Thus the actual relations areIn particular, $[e,f]v_1=2v_1\ne -2v_1=h{v}_1$ in characteristic zero; the requested equality cannot hold as printed. Settingcorrects the sign and gives $[H,E]=2E$, $[H,F]=-2F$, $[E,F]=H$. These are the standard sl2 triple relations, so the standard basis of the sl2 Lie algebra acts on $V$ by $E,H,F$, giving a Lie algebra representation.

For completeness, these operators are linearly independent: applying $aE+bH+cF=0$ to $v_0$ gives $c=0$; applying it to $v_1$ then gives $b=0$; applying it to $v_2$ gives $a=0$. Hence the printed operators span a Lie subalgebra isomorphic to the sl2 Lie algebra, even though their printed normalization fails. Replacing the coefficient $i-1$ in the printed definition of $e(v_i)$ by $1-i$ is one equivalent way to repair the statement.

Use the corrected Lie algebra representation $E=-e$, $H=h$, $F=f$ from part (i). A vector subspace is invariant under $E,H,F$ exactly when it is invariant under the printed $e,h,f$, since changing the sign of a linear map does not change its invariant subspaces. DefineThis is a nonzero proper submodule: $H$ preserves each $k{v}_i$, $F$ increases the index, $E(v_1)=0$, and $E(v_i)=-(i-1)v_{i-1}\in W$ for $i\ge 2$.

We must rule out submodules consisting only of mixtures of weight vectors. Let $U\ne 0$ be any submodule, and choose $u={\sum }_{j\in S}{a}_j{v}_j\in U$ with $S$ finite and every $a_j\ne 0$. The eigenvalues $-2j$ of $H$ are distinct. For $i\in S$, the Lagrange interpolation polynomialsatisfies $p_i(H)u=a_i{v}_i$. Since $U$ is invariant under $H$, it is invariant under every polynomial in $H$, so $v_i\in U$.

If $i=0$, repeated applications of $F$ give $F^m{v}_0=m!v_m$, and $U=V$. If $i\ge 1$, repeated applications of $E$ give a nonzero multiple of $v_1$, because the successive coefficients $-(i-1),\dots ,-1$ are nonzero in characteristic zero. Repeated applications of $F$ to $v_1$ then give every $v_j$ with $j\ge 1$, so $W\subseteq U$.

Finally, if such a $U$ contains any vector outside $W$, that vector has a nonzero coefficient of $v_0$; the same polynomial interpolation extracts $v_0$, forcing $U=V$. Hence the complete list of submodules is $0,W,V$. In particular, $W$ is a simple module. The quotient module $V/W$ has basis ${\overline{v}}_0$; $E{\overline{v}}_0=H{\overline{v}}_0=0$ and $F{\overline{v}}_0={\overline{v}}_1=0$. It is the one-dimensional trivial Lie algebra representation.

The module extension does not split. If $V=W\oplus U$ for a submodule $U$, then $U$ would be a one-dimensional trivial Lie algebra representation, since it maps isomorphically to $V/W$. But $\ker H=k{v}_0$ and $F{v}_0=v_1\ne 0$, so no such $U$ exists. ThusHere semisimple module means a direct sum of modules that are simple modules, equivalently a module in which every submodule has a complementary submodule. This is the nonsplit zero-weight Verma extension for sl2: rescaling to $w_i=i!v_i$ gives $F{w}_i=w_{i+1}$ and $E{w}_i=i(1-i)w_{i-1}$. Its infinite-dimensional vector space is outside the hypotheses of the Weyl complete reducibility theorem.

The normalizer of a Lie subalgebra and the centralizer of a Lie subalgebra areThe second is always contained in the first. Conversely, let $x\in N_{\mathfrak{g}}(\mathfrak{a})$ and $a\in \mathfrak{a}$. Then $[a,x]\in \mathfrak{a}$, and, because $\mathfrak{a}$ is an abelian Lie algebra,By the definition of a toral subalgebra, $\mathrm{ad}a$ is diagonalizable. A diagonalizable linear map $T$ satisfies $\ker T^2=\ker T$: in an eigenbasis, both kernels consist of exactly the eigenvectors whose eigenvalue is zero. Thus $[a,x]=0$. This holds for every $a\in \mathfrak{a}$, so $x\in C_{\mathfrak{g}}(\mathfrak{a})$ andThis is the normalizer-centralizer equality for a toral abelian subalgebra; it does not require $\mathfrak{g}$ to be a semisimple Lie algebra.

Use the general definition of a Cartan subalgebra: a nilpotent Lie algebra that is a Lie subalgebra and equals its normalizer of a Lie subalgebra. In a semisimple Lie algebra, the stated structural fact says every Cartan subalgebra $\mathfrak{h}$ is a toral subalgebra and an abelian Lie algebra. Part (i) therefore givesConversely, suppose $\mathfrak{h}$ is a toral subalgebra and an abelian Lie algebra with $C_{\mathfrak{g}}(\mathfrak{h})=\mathfrak{h}$. An abelian Lie algebra is a nilpotent Lie algebra, because its lower central series of a Lie algebra vanishes at the second term. Part (i) gives $N_{\mathfrak{g}}(\mathfrak{h})=C_{\mathfrak{g}}(\mathfrak{h})=\mathfrak{h}$. Thus $\mathfrak{h}$ satisfies the defining conditions of a Cartan subalgebra, proving

We use the supplied weight decomposition $V={⨁}_{\lambda \in \Psi (\mathfrak{a})}{V}_{\lambda }$. There is a small distinction to justify: the definition of a toral subalgebra uses the Adjoint representation, while these weight spaces use the defining representation. For the symplectic Lie algebra, adjoint semisimplicity in the symplectic Lie algebra implies that each $a\in \mathfrak{a}$ is diagonalizable on $V$ as well. Here is a direct verification.

Define the symplectic adjoint $a^{\ast }$ by $\omega (av,w)=\omega (v,a^{\ast }w)$. The symplectic Lie algebra condition is $a^{\ast }=-a$. Write the Jordan–Chevalley decomposition in $\mathrm{End}(V)$ as $a=S+N$, with $S$ diagonalizable, $N$ a nilpotent linear map, and $SN=NS$. Uniqueness of the Jordan–Chevalley decomposition, applied to $a^{\ast }=-a$, gives $S^{\ast }=-S$ and $N^{\ast }=-N$. Thus $S,N\in \mathfrak{sp}(V)$. On $\mathrm{End}(V)$, $\mathrm{ad}S$ is diagonalizable by using a matrix unit basis in an eigenbasis of $S$, and $\mathrm{ad}N$ is a nilpotent linear map, sincevanishes for large $m$. These operators commute and preserve $\mathfrak{sp}(V)$, so their restrictions are the Jordan–Chevalley decomposition of $\mathrm{ad}a$. Because $\mathrm{ad}a$ is diagonalizable, its nilpotent part is zero: $\mathrm{ad}N=0$ on $\mathfrak{sp}(V)$.

To conclude that $N=0$ without an assumption on the center of a Lie algebra, observe that each map $T_v(w)=\omega (v,w)v$ lies in $\mathfrak{sp}(V)$. If $v\ne 0$, its image is $kv$, because $\omega$ is a nondegenerate bilinear form. Since $N$ commutes with $T_v$, it preserves $kv$. Hence $Nv\in kv$ for every $v$, forcing $N$ to be a scalar linear map: compare its action on two linearly independent vectors and their sum. A scalar nilpotent linear map is zero. For $V=0$ the assertion is immediate. This proves the needed diagonalizability. Commuting diagonalizable linear maps admit simultaneous diagonalization, giving the displayed weight decomposition.

Now let $v\in V_{\lambda }$ and $w\in V_{\mu }$. For every $a\in \mathfrak{a}$, preservation of the alternating bilinear form givesIf $\lambda +\mu \ne 0$ as a linear functional, choose $a$ on which it is nonzero; then $\omega (v,w)=0$. ThereforeFor $v\ne 0$ in $V_{\lambda }$, the fact that $\omega$ is a nondegenerate bilinear form supplies a $w\in V$ with $\omega (v,w)\ne 0$. Decomposing $w$ into weight spaces, only its $V_{-\lambda }$ component can contribute. Thus $V_{-\lambda }\ne 0$, and $\Psi (\mathfrak{a})=-\Psi (\mathfrak{a})$. The same argument on both sides shows that $\omega :V_{\lambda }\times V_{-\lambda }\to k$ is a perfect pairing, so $\dim V_{\lambda }=\dim V_{-\lambda }$. On $V_0$, the restriction of $\omega$ is a nondegenerate bilinear form that is an alternating bilinear form, so $\dim V_0$ is even.

Choose one representative ${\lambda }_i$ from each pair of nonzero weights of a representation, for $1\le i\le r$. HereThe linear map $\mathfrak{a}\to k^r$ sending $a$ to $({\lambda }_1(a),\dots ,{\lambda }_r(a))$ is injective: an element in its kernel acts by zero on every nonzero weight space, and also on $V_0$, so it is the zero element of $\mathfrak{a}\subseteq \mathrm{End}(V)$. Taking dimensions yieldsThe final inequality follows because each pair contributes at least two to $\dim V=2n$. Equivalently, the nonzero weights of a representation span the dual space ${\mathfrak{a}}^{\ast }$; they need not be linearly independent at this stage.

Choose representatives ${\lambda }_1,\dots ,{\lambda }_r$ as in part (b). We use the symplectic centralizer decompositionTo prove it, a linear map commuting with every $a\in \mathfrak{a}$ must preserve every weight space: if $v\in V_{\lambda }$, then $a(Tv)=T(av)=\lambda (a)Tv$. Conversely, preserving the weight spaces makes a linear map commute with the scalar action of every $a$ on each space.

On $V_0$, the symplectic Lie algebra condition says exactly that the restriction belongs to $\mathfrak{sp}(V_0)$. On $V_{{\lambda }_i}\oplus V_{-{\lambda }_i}$, use the perfect pairing from part (b) to identify $V_{-{\lambda }_i}$ with the dual space $V_{{\lambda }_i}^{\ast }$. The action $A_i$ on $V_{{\lambda }_i}$ can be arbitrary, but its action on $V_{-{\lambda }_i}$ must be $-A_i^{\ast }$, where $A_i^{\ast }$ is the dual map. Indeed,determines $B_i=-A_i^{\ast }$ uniquely. There are no additional conditions between different pairs because $\omega$ vanishes between different pairs. Every choice of these blocks therefore gives an element of the centralizer of a Lie subalgebra. The block correspondence preserves Lie brackets, proving the claimed direct sum of Lie algebras.

The general linear Lie algebra ${\mathfrak{gl}}_m(k)$ is an abelian Lie algebra exactly when $m\le 1$: for $m\ge 2$, the matrix units satisfy $[E_{11},E_{12}]=E_{12}\ne 0$. A nonzero symplectic vector space has a symplectic basis and contains a two-dimensional symplectic subspace. The symplectic Lie algebra acting on that plane, extended by zero on its symplectic orthogonal complement, contains the matrices $H=\mathrm{diag}(1,-1)$ and $E=E_{12}$ with $[H,E]=2E\ne 0$. Thus $\mathfrak{sp}(V_0)$ is an abelian Lie algebra exactly when $V_0=0$.

Since different summands in a direct sum of Lie algebras commute, we obtain the precise criterionEquivalently, there are $2n$ distinct nonzero weights of a representation, each with a one-dimensional weight space. The condition concerns multiplicities in the defining representation; it does not assert that $\dim \mathfrak{a}=n$ for every such $\mathfrak{a}$.

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.