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Let $V$ be a finite-dimensional symplectic vector space over an algebraically closed field of characteristic zero. For $a\in \mathfrak{sp}(V)$, write its Jordan–Chevalley decomposition as $a=S+N$. Applying the symplectic adjoint to this decomposition and using $a^{\ast }=-a$, uniqueness gives $S^{\ast }=-S$ and $N^{\ast }=-N$. Thus both parts belong to the symplectic Lie algebra. The operators $\mathrm{ad}S$ and $\mathrm{ad}N$ commute. The former is diagonalizable on $\mathrm{End}(V)$ in a matrix unit basis for an eigenbasis of $S$, and the latter is a nilpotent linear map, as follows fromThe restrictions to the invariant symplectic Lie algebra retain these properties. Hence these restrictions form the Jordan–Chevalley decomposition of $\mathrm{ad}a$.

If $\mathrm{ad}a$ is diagonalizable, $\mathrm{ad}N=0$. For every $v\in V$, the rank-one map $T_v(w)=\omega (v,w)v$ lies in $\mathfrak{sp}(V)$, since the two terms in $\omega (T_{v}w,u)+\omega (w,T_{v}u)$ cancel by alternation. For $v\ne 0$ its image is $kv$, so commuting with $T_v$ forces $N$ to preserve $kv$. A linear map preserving every line is scalar: compare its action on two linearly independent vectors and on their sum. Since $N$ is a nilpotent linear map, that scalar is zero. Thus $a=S$ is diagonalizable. Conversely, if $a$ is diagonalizable on $V$, its operator commutator action is diagonalizable on $\mathrm{End}(V)$ and therefore on its invariant symplectic Lie algebra. This proves both directions, including $V=0$.

Every finite-dimensional Lie algebra over a field of characteristic zero has a finite-dimensional faithful Lie algebra representation. Thus it embeds as a Lie subalgebra of a matrix algebra, using the matrix commutator as bracket. Faithfulness is crucial for detecting central elements: the Adjoint representation alone kills the center of a Lie algebra.

Over a field of characteristic zero, an alternating polynomial changes sign when two variables are interchanged. It vanishes when two coordinates agree, so each factor $x_i-x_j$ divides it. These distinct linear factors are relatively prime in the polynomial ring, hence their product, the Vandermonde determinant, divides it. The quotient is a symmetric polynomial.

Over a field of characteristic zero, an associative Lie product is a bilinear map that is simultaneously an associative algebra product and a Lie bracket. Such a product has every triple product zero. To prove this, write multiplication by juxtaposition. By associativity and antisymmetry of a Lie bracket,since moving $z$ left through $y$ and then through $x$ introduces two minus signs. Applying antisymmetry of a Lie bracket instead to the two elements $xy$ and $z$ gives $(xy)z=-z(xy)$. Thus $xyz=-xyz$, and $A^3=0$ in characteristic zero.

Conversely, an alternating multilinear map defining a product whose triple products all vanish is associative, because both associations give zero, and satisfies the Jacobi identity, because each nested product gives zero. A nonzero example has basis $p,q,z$, products $pq=z=-qp$, and all other products of basis vectors zero. It is the Heisenberg Lie algebra with its Lie bracket used as the associative algebra product. Its image $A^2=kz$ annihilates $A$ on both sides, and it has no identity element.

Over an algebraically closed field of characteristic zero, every three-dimensional perfect Lie algebra is isomorphic to the sl2 Lie algebra. Indeed the trace of a commutator gives $\mathrm{tr}(\mathrm{ad}x)=0$ for all $x$, because each element is a sum of Lie brackets. If every $\mathrm{ad}x$ were a nilpotent linear map, the Engel theorem would make the Lie algebra nilpotent, contradicting its constant nonzero lower central series of a Lie algebra. Some $\mathrm{ad}x$ therefore has a nonzero eigenvalue $\lambda$. Its other eigenvalues are $0,-\lambda$, since it kills $x$ and has zero trace. They are distinct, so its eigenvectors give a basis $h,e,f$ after scaling $x$ to $h=2x/\lambda$, with $[h,e]=2e$ and $[h,f]=-2f$. The Jacobi identity puts $[e,f]$ in the zero eigenspace $kh$. Its coefficient cannot vanish, since otherwise the derived algebra would lie in $ke\oplus kf$. Rescaling $f$ gives $[e,f]=h$. Thus this basis is an sl2 triple, proving the claimed Lie algebra isomorphism.

For a scheme over $\mathbb{R}$, its base change of a morphism of schemes to $\mathbb{C}$ has scheme-theoretic fibre over a point $p$ equal to $\mathrm{Spec}(\kappa (p){\otimes }_{\mathbb{R}}\mathbb{C})$. This is the spectrum of $\kappa (p)[s]/(s^2+1)$. It has two points if $-1$ is a square in the residue field, and otherwise one point with a quadratic field extension. In both cases it is reduced, because the quadratic polynomial has no repeated root in characteristic zero. The distinction applies to generic points as well as closed points.

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.

For a graded Lie algebra over characteristic zero, the associated graded algebra of its universal enveloping algebra under the word-length filtration is its graded symmetric algebra. Ordered homogeneous monomials form a basis, with exponent at most one on each odd Lie basis element. The defining graded commutation relations reorder every word, and the Jacobi identity resolves the triple-reordering ambiguities. In particular the primitive elements of a Hopf algebra of $U(L)$ are precisely $L$: in its associated graded algebra, a primitive polynomial of symmetric length greater than one would have a nonzero mixed term under the diagonal, which is impossible in characteristic zero.

For a Weierstrass equation of an elliptic curve, use the local parameter $t=-x/y$ and put $w=-1/y$. The formal kernel of a minimal Weierstrass equation construction gives $w=t^3+a_1{t}^4+(a_1^2+a_2)t^5+(a_1^3+2a_1{a}_2+a_3)t^6+O(t^7)$. Substitution into the elliptic curve group law yields the displayed initial coefficients and the formal inverse $-t/(1-a_{1}t-a_{3}w(t))$. In characteristic zero, the invariant differential on an elliptic curve becomes $(1+a_{1}t+(a_1^2+a_2)t^2+O(t^3))dt$, giving the formal logarithm $t+a_1{t}^2/2+(a_1^2+a_2)t^3/3+O(t^4)$. The addition coefficients are integral polynomial expressions and hence also apply in positive characteristic of a field.

The Jacobi identity is the compatibility condition on a Lie bracketTogether with an Alternating Lie bracket, it defines a Lie algebra. Unlike associativity, it relates three cyclic bracketings rather than equating two bracketings. Equivalently, using antisymmetry of a Lie bracket,Thus the linear map ${\mathrm{ad}}_x:y\mapsto [x,y]$ is a derivation of a Lie algebra: acting on a bracket is the sum of acting on each argument. This makes infinitesimal symmetries act consistently on each other's products. It also gives $[{\mathrm{ad}}_x,{\mathrm{ad}}_y]={\mathrm{ad}}_{[x,y]}$, the identity that makes the Adjoint representation of a Lie algebra a Lie algebra representation.

For the commutator in an associative algebra, the cancellation is directly visible:Each word occurs once with each sign. This proves the Jacobi identity without a restriction on the characteristic of a field.

For a computed check in the sl2 matrix commutator example,Here the last two terms are individually nonzero over characteristic zero; the identity records their cancellation. In the cross-product Lie algebra, the same cyclic cancellation is the vector triple product identity.

The historical connection is unusually concrete: Lie and Engel's volume I expands nested commutators of differential operators into twelve terms and cancels them cyclically, in Chapter 5, §26, equation (7). See Lie and Engel, §26, in Merker's translation, pp. 105–106.

For the general linear Lie algebra of all $n\times n$ matrices over a field of characteristic zero, write ${\mathrm{ad}}_X=L_X-R_X$ on the matrix space. The traces of $L_X{L}_Y$ and $R_X{R}_Y$ are $n\mathrm{tr}(XY)$, whereas both mixed traces are $\mathrm{tr}X\mathrm{tr}Y$. Expanding gives the displayed Killing form. Its radical is the scalar matrices because $\kappa (X,Y)=0$ for every $Y$ forces $nX-\mathrm{tr}(X)I=0$. Thus the full general linear algebra is not semisimple, even though its traceless summand is semisimple for $n\ge 2$.

// Target: lie-theory.bigb

For simply connected $Y$, rational Hurewicz homomorphism identifies ${\pi }_{\ast }(\Omega Y)\otimes \mathbb{Q}$ with the primitive elements of a Hopf algebra of $H_{\ast }(\Omega Y;\mathbb{Q})$ and induces an isomorphism $U({\pi }_{\ast }(\Omega Y)\otimes \mathbb{Q})\cong H_{\ast }(\Omega Y;\mathbb{Q})$ of graded Hopf algebras. The bracket on homotopy is the Samelson product. The underlying algebra theorem says that a connected graded cocommutative Hopf algebra over characteristic zero is the universal enveloping algebra of its primitive Lie algebra. These statements let a free tensor algebra of loop-space homology detect a free graded Lie algebra of homotopy.

Newton's identities let two different summaries of the same polynomial roots talk to each other. Elementary symmetric polynomials record products of distinct root positions; power sums record sums of individual powers, $p_r={\sum }_{i=1}^n{x}_i^r$. Either can be easier to obtain in a particular problem. The identities turn one into the other, often avoiding any need to find the individual polynomial roots.

Write $e_0=1$ and $e_m=0$ for $m>n$. For every positive integer $m$, the identity isIt is a polynomial identity over the integers, so it remains true on substitution in any commutative ring. To solve it for $e_m$ by division, however, $m$ must have a multiplicative inverse. Over a field of characteristic zero, the first $n$ power sums recover every elementary symmetric polynomial. The same recovery works when the positive characteristic of a field is greater than $n$; small characteristic needs care.^[ref]

The first three relations are $e_1=p_1$, $2e_2=e_1{p}_1-p_2$, and $3e_3=e_2{p}_1-e_1{p}_2+p_3$. When division by two and three is permitted, this becomesFor the three roots $1,2,3$, their power sums are $p_1=6$, $p_2=14$, and $p_3=36$. The formulas give $e_2=(36-14)/2=11$ and $e_3=(216-252+72)/6=6$, recovering the polynomial $z^3-6z^2+11z-6$ from aggregate data.

The beauty is that the summaries do not depend on the order of the polynomial roots. In symmetric polynomials, one can work with quantities that survive every relabeling; in linear algebra, traces of matrix powers give the same kind of aggregate information about eigenvalues. The identities convert those summaries into coefficients of a characteristic polynomial.

A short derivation packages the identities into a formal power series. Let $E(t)={\prod }_i(1+x_{i}t)={\sum }_{m\ge 0}{e}_m{t}^m$. Since its constant coefficient is one, it has an inverse as a formal power series, and its formal derivative satisfiesMultiply by $E(t)$ and compare the coefficient of $t^{m-1}$. This gives Newton's identities without a convergence assumption or division by $m$. The division issue arises only when using them to recover $e_m$. The logarithmic-derivative approach is also developed in Newton's Identities (January 2019 notes) by Milan Mossé (§4, pp. 6–8).

The alternate name Newton-Girard identities recalls Albert Girard's earlier work. A modern historical account credits formulas through sixth powers to Albert Girard in 1629 and discusses their later appearance in Isaac Newton's Arithmetica universalis.^[ref] This citation refers to that account; the original historical editions have not been inspected here.

A number field is a finite field extension $K/\mathbb{Q}$ of the rational numbers. Its degree of a field extension $[K:\mathbb{Q}]$ is its dimension as a vector space over $\mathbb{Q}$. Every element is an algebraic number; the primitive element theorem also says that $K=\mathbb{Q}(\theta )$ for some single algebraic number $\theta$.

For example,The basis $1,\sqrt{2}$ makes arithmetic explicit: use $(\sqrt{2}{)}^2=2$ to reduce products. Every nonzero element has its inverse in the same field:The denominator cannot vanish unless $a=b=0$, since $\sqrt{2}$ is not a rational number. For instance, $(3+\sqrt{2}{)}^{-1}=(3-\sqrt{2})/7$ belongs to the number field, although it is not an algebraic integer.

A degree-$n$ number field has exactly $n$ distinct field embeddings into the complex numbers fixing $\mathbb{Q}$, because every finite field extension in characteristic zero is a separable field extension. For $\mathbb{Q}(\sqrt{2})$ these are the two real mapsBoth preserve addition and multiplication. Keeping both values, rather than just the positive choice of $\sqrt{2}$, is what makes the field norm and the Minkowski embedding of a number field useful.

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.

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.

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.

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 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.