Use an oriented geodesic hyperbolic $2s$-gon $D$ with paired edges of equal length. Identify each pair by a length-preserving map reversing the boundary directions, to make an oriented base surface $M_0$ with possibly singular vertices. Assign to the $s$ edge pairs generators $A_1,\dots ,A_s$ of $T$, with inverse labels for reversed crossings. For each $g\in T$, take a copy $D_g$ and glue its positively labelled edge of type $k$ to the corresponding mate in $D_{g{A}_k}$. The generators make the glued surface $M$ connected. Left multiplication $D_g\mapsto D_{hg}$ commutes with these right-labelled gluings and gives the $T$ action with quotient $M_0$.

Existence for an arbitrary finite group can even be arranged with no cone singularities. If $T$ has $d$ generators, take $h\ge \max \{2,d\}$ and a regular hyperbolic polygon with $4h$ edges, all angles $\pi /(2h)$, paired with side word ${\prod }_{i=1}^h{a}_i{b}_i{a}_i^{-1}{b}_i^{-1}$. Its vertices form one cycle of angle $2\pi$, giving a smooth base of genus $h$. Label the $a_i$ by the generators of $T$, padded by identities, and every $b_i$ by the identity. The single vertex word is ${\prod }_i[A_i,1]=1$, so every lifted vertex also has angle $2\pi$. The resulting connected ordinary normal covering map has degree $|T|$ and deck group $T$.

This is the paired-polygon construction of finite surface covers. Interior and open-edge points are regular hyperbolic points; possible singularities occur only at vertices. When the base has cone points, the map is an orbifold covering and may be branched as a map between the underlying topological surfaces. It should not be called an unbranched manifold covering at those branch points.

A vertex cycle of a paired polygon is the collection of vertices identified to one point on $M_0$. Let $r$ be the number of cycles, ${\theta }_j$ the sum of the polygon angles in cycle $j$, and $C_j\in T$ the gluing word recorded on going around it. If $q_j=|C_j|$ is the order of that element, a lift closes after $q_j$ turns. It has angle $q_j{\theta }_j$, and there are $|T|/q_j$ such lifted vertices. The base and lifted cone orders, when they are integer orbifold orders, areAn orbifold homomorphism requires $q_j\mid m_j$. The lifted point is smooth exactly when $q_j{\theta }_j=2\pi$, equivalently $q_j=m_j$ in that case. For arbitrary cone metrics the angle formulas remain valid even when an angle does not correspond to an integer orbifold order.

Counting faces, edges and vertices gives the Euler characteristic of a paired-polygon surface cover:These are ordinary topological Euler characteristics, rather than orbifold Euler characteristics. If the lifted surface is smooth, the equivalent orbifold formula isFor a closed oriented smooth surface, $\chi =2-2g$ defines its genus. These formulas also follow from the Riemann-Hurwitz formula by subtracting the ramification deficits $|T|-|T|/q_j$ over each base vertex.

For $U\le T$, the quotient map $M\to M_1=U\backslash M$ is an ordinary normal covering if and only if $U$ acts freely. All possible point stabilizers in this construction are conjugates of $⟨C_j⟩$, so the concrete necessary and sufficient condition isThis is the freeness criterion for a subgroup of a polygon-cover deck group. Under it the deck group is $U$, the degree is $|U|$, andThere is no requirement $U◃T$ for this map. That different condition characterizes normality of the intermediate cover $M_1\to M_0$. If fixed points are retained and one instead speaks of an orbifold covering, the quotient is regular as an orbifold cover, but ordinary Euler characteristic need not divide by $|U|$; orbifold Euler characteristic does.

For the low-genus examples takeLet $U_1$ be a nonzero-vector stabilizer and $U_2$ a plane stabilizer. Their actions have seven points, so $|U_1|=|U_2|=24$ and $[T:U_i]=7$. They are almost conjugate subgroups: the numbers of fixed nonzero vectors for a matrix $A$ and its dual action $A^{-T}$ are both $2^{\dim \ker (A-I)}-1$. Thus the two coset permutation characters agree, which is equivalent to Gassmann equivalence. They are not conjugate: a point stabilizer has a common fixed nonzero vector, whereas a plane stabilizer fixes no nonzero vector globally. In particular, since $7\nmid 24$, neither subgroup contains any nonidentity element of a cyclic subgroup of order seven. All cone monodromies of exact order seven therefore act freely on the intermediate covers.

For genus three, start with a hyperbolic triangle whose three angles are $\pi /7$ and double it across one edge. The resulting quadrilateral $D$ has anglesThe side pairings produce three base vertex cycles, each of total angle $2\pi /7$, and an underlying sphere. Choose generators $A,B$ with $|A|=|B|=|AB|=7$; the three vertex words are $A,B,(AB{)}^{-1}$. Hence $s=2$, $r=3$ and all $q_j=7$. The full cover is smooth, andThe hyperbolic polygon area of $D$ is $8\pi /7$, so the seven-sheeted surfaces have area $8\pi$, agreeing with Gauss-Bonnet. Their equal spectra follow from the Sunada theorem. For complete numerical generator data, one possible pair isThese generate all of $T$ and have the required orders. This construction establishes the requested isospectrality; it does not establish nonisometry. In the constant-curvature triangle example a lifted reflection can identify the point and plane covers, as in reflection intertwining of triangle-cover coset actions. Breaking that symmetry with a generic metric gives nonisometric genus-three surfaces, but then the metric need not have constant curvature. Those are distinct assertions.

For genus four, use the cone-torus construction of genus-four Sunada surfaces. Take a regular hyperbolic quadrilateral with four anglesand pair opposite sides in the usual torus pattern $a,b,a^{-1},b^{-1}$. All four vertices belong to one cycle, whose total angle is $2\pi /7$; the underlying base is a torus with one cone point. Choose $A,B$ of order four generating $T$, with commutator $[A,B]$ of order seven. For example,Only the commutator is a cone monodromy, so the order-four side labels do not create cone points of order four. Here $s=2$, $r=1$, $q_1=7$, givingThe quadrilateral has area $12\pi /7$, and each seven-sheeted smooth quotient has area $12\pi$. Again the order-seven cone stabilizer meets neither $U_i$ nontrivially, so the metrics are smooth and the Sunada theorem gives identical spectra. The required group and subgroup orders are therefore 168 and 24 in both constructions; the genus-three side-generator orders are 7 and 7, and the displayed genus-four side-generator orders are 4 and 4, with cone-word order 7 in both cases.

Let $p:N\to M=U\backslash N$ be a finite normal Riemannian covering, so $U$ acts freely by deck isometries. With $x,y$ any lifts of $\bar{x},\bar{y}$, the heat kernel on a finite isometric quotient satisfiesThere is no factor $1/|U|$ in this kernel formula. Simultaneous invariance of $H_N$ under isometries proves independence of the lifts, by reindexing $U$. Each term solves the lifted heat equation. If $F_U$ is a fundamental domain and $f$ a function on $M$, the integral of the right side against $f(\bar{y})$ over $F_U$ equalsbecause the translates $u{F}_U$ tile $N$ up to sets of measure zero. This converges to $f(p(x))$ as $t\downarrow 0$, proving the initial condition. Uniqueness of the heat equation solution on the closed manifold gives the formula. For a noncompact finite cover, use the minimal heat kernels, or compatible Friedrichs heat realizations. The normalized pullback $f\mapsto |U{|}^{-1/2}f\circ p$ is unitary onto the deck-invariant $L^2$ subspace and preserves the Dirichlet energy, so it intertwines the heat semigroups and yields the same kernel formula. No uniqueness of unrestricted noncompact heat-equation solutions is needed.

Set $I_t(g)={\int }_N{H}_N(t,x,gx)dV(x)$ for an isometry $g$. Diagonal integration and the covering degree giveUnlike the kernel formula, the trace formula has an averaging factor: integration of a quotient function over $N$ is $|U|$ times integration over $U\backslash N$.

Suppose now $U\le T$, with $T$ any larger group of isometries of the closed manifold $N$. Change variables $x=hy$ and use simultaneous heat-kernel invariance to obtainThus $I_t$ is a class function, and the Sunada orbital heat-trace formula can be grouped by conjugacy classes $C$ of $T$:Only classes meeting the finite subgroup $U$ contribute, so this class sum is finite even if the larger isometry group $T$ is infinite.

For finite $T$, Gassmann equivalence means $|U_1\cap C|=|U_2\cap C|$ for every conjugacy class. Summing these counts gives $|U_1|=|U_2|$, so the displayed formula gives identical heat traces. Their spectra, with multiplicities, agree: as $t\to \infty$, the constant limit identifies the multiplicity of zero; after subtracting it, the slowest-decaying exponential identifies the least positive eigenvalue and its multiplicity. Repeating determines every subsequent eigenvalue. This proves the Sunada theorem in the stated covering situation.

There is an important convention when the universal cover $N$ has an infinite deck group $T$. For compact quotient manifolds, the subgroups must have finite index; the appropriate generalized Gassmann condition is equality of the characters of the finite coset permutation representations, not equality of possibly infinite cardinalities of conjugacy-class intersections. Let $K$ be the intersection of the two subgroup cores. Each core is the kernel of the corresponding finite coset action, so $K$ is normal and of finite index in $T$, and $K\subseteq U_1\cap U_2$. The manifold $X=K\backslash N$ is then a compact finite normal cover of $M_0$. The finite group $G=T/K$ acts on $X$, and $H_i=U_i/K$ have equal coset permutation characters in $G$. For a finite group, the fixed-coset count isEquality at the identity gives equal subgroup orders, and then equality for all $g$ gives almost-conjugacy. Applying the proved finite theorem to $H_i\backslash X=U_i\backslash N$ establishes isospectrality also in this finite-index interpretation. It avoids taking the divergent heat trace of a noncompact universal cover.

To realize a finitely presented group $T=⟨a_1,\dots ,a_r\mid R_1,\dots ,R_s⟩$, begin with the closed smooth four-manifold $B={\#}_{j=1}^r(S^1\times S^3)$, whose fundamental group is free on the $a_j$. Represent the relator words by disjoint smoothly embedded circles; general position permits this in dimension four, including separating self-intersections of representatives. Their oriented normal rank-three bundles are trivial, so perform framed surgeries replacing $S^1\times D^3$ by $D^2\times S^2$. Removing the circles' tubular neighborhoods leaves the fundamental group unchanged, by codimension-three general position. The Seifert-van Kampen theorem shows that each replacement kills exactly the normal closure of its relator. The resulting closed smooth $M_0$ has ${\pi }_1(M_0)\cong T$. This gives a concrete closed-manifold realization of a finitely presented fundamental group. Equip it with any smooth Riemannian metric and lift that metric to the universal and intermediate covers; all covering maps become local isometries.

To ensure nonisometry, one must choose nonconjugate subgroups. If $U_1=U_2$, or they are conjugate, the quotients are isometric for every lifted metric, so Gassmann equivalence alone cannot imply nonisometry. For nonconjugate subgroups, choose the base metric to have no nonidentity local isometry between open sets, using the Sunada local isometry lemma in dimension at least two. If $F:M_1\to M_2$ were an isometry, a local inverse of the first covering followed by $F$ and the second covering would be a local isometry of $M_0$, hence the identity. Thus the covering maps satisfy $p_{2}F=p_1$ everywhere. Classification of connected covers then forces $U_1,U_2$ to be conjugate in ${\pi }_1(M_0)=T$, a contradiction. Therefore nonconjugate almost-conjugate subgroups and a locally rigid base metric give isospectral nonisometric quotients.

Work on a closed compact smooth Riemannian manifold $M$ without boundary, as in the usual discrete spectral setting here. Use the paper's $\Delta ={\mathrm{tr}}_g{\mathrm{Hess}}_g$, and put $P=-\Delta$. The Riemannian heat kernel $H(t,x,y)$ is the smooth kernel, for $t>0$, representing the solution ofThus $({\partial }_t-{\Delta }_x)H=0$ and its distributional initial value in $x$ is the Dirac delta distribution at $y$. The heat trace isThe heat invariants are the coefficients $a_r(M)$ in the short-time heat trace expansion specified below.

To derive its spectral form, take an orthonormal eigenbasis $P{\varphi }_j={\lambda }_j{\varphi }_j$, with $0={\lambda }_0\le {\lambda }_1\le \cdots$, repeating eigenvalues according to multiplicity. For the initial datum ${\varphi }_j$, the function $e^{-t{\lambda }_j}{\varphi }_j$ solves the heat equation. Uniqueness identifies it with the heat-kernel solution. Expanding arbitrary initial data in this basis therefore gives the spectral expansion of the Riemannian heat kernel:Standard elliptic estimates and eigenvalue growth imply smooth convergence for $t$ bounded away from zero: polynomial bounds on derivatives of eigenfunctions are dominated by the exponential factors. On the diagonal all summands are nonnegative, so monotone convergence also justifies integration without needing an interchange of conditionally convergent terms. Since each eigenfunction has unit $L^2$ norm,Equivalently, if the paper's Laplacian eigenvalues are written ${\nu }_j=-{\lambda }_j$, the formula is ${\sum }_j{e}^{t{\nu }_j}$; a growing exponential is not the heat semigroup.

Here are the needed heat kernel expansion facts, stated without proof. On a closed $d$-dimensional smooth Riemannian manifold,with the diagonal remainder uniform in $x$ after any finite truncation. The coefficients are smooth local geometric quantities; $u_1=R/6$. Compactness therefore allows integration term by term:These integrated coefficients are the heat invariants. The uniform leading remainder in particular gives $Z_M(t)=(4\pi t{)}^{-d/2}(\mathrm{Vol}(M)+O(t))$.

Isospectral manifolds have identical heat traces by the derived spectral formula. If their dimensions were $d<e$, multiplying the common trace by $t^{d/2}$ would give a finite positive limit on the $d$-dimensional manifold but divergence on the $e$-dimensional one. Thus the dimensions are equal, and equality of the leading coefficient gives equal Riemannian volume. Explicitly,The absence of a boundary fixes this particular integer-power expansion. With a boundary one must first specify boundary conditions and include the appropriate half-integer boundary terms; the leading dimension-and-volume conclusion remains valid in the usual compact smooth setting.

For a smooth function $f$ and the Levi-Civita connection $\nabla$, the Riemannian Hessian is the covariant two-tensorThe connection rules show that this is linear over smooth functions in both vector-field arguments, so it is a tensor. Torsion-freeness gives its symmetry:Its metric trace defines the paper's Laplace-Beltrami operator, $\Delta ={\mathrm{div}}_g{\mathrm{grad}}_g$; its eigenvalues on a closed manifold are nonpositive. We use $P=-\Delta$ when listing nonnegative eigenvalues or writing $e^{-tP}$.

At $p$, choose an orthonormal basis $e_1,\dots ,e_d$ of the tangent space and unit-speed geodesics ${\gamma }_i(t)={\exp }_p(t{e}_i)$. Differentiating twice along a curve givesThe covariant acceleration vanishes on a geodesic. Taking the trace at $t=0$ proves the geodesic trace formula for the Laplace-Beltrami operator:The last equality is Taylor's formula along the geodesics, and expresses the Laplacian entirely through nearby function values.

For stabilization, let $X_1,X_2$ be the assumed isospectral nonisometric four-dimensional tori, and let $k=n-4>0$. Take the same small flat torus $F_{\epsilon }={\mathbb{R}}^k/\epsilon {\mathbb{Z}}^k$ as an extra factor. On a Riemannian product,If ${\varphi }_j$ and ${\psi }_{\ell }$ are eigenfunctions with eigenvalues ${\lambda }_j$ and ${\mu }_{\ell }$, then ${\varphi }_j{\psi }_{\ell }$ has eigenvalue ${\lambda }_j+{\mu }_{\ell }$. Products of orthonormal eigenbases are complete in the product $L^2$ space, so this lists all eigenvalues with multiplicities, including coincidences between different sums. Consequently the two products are isospectral.

One must also prove that multiplying by a common factor has not accidentally made the two manifolds isometric. ChooseThe injectivity radii are positive because the factors are compact. A closed product geodesic of length less than $\delta$ projects to a closed geodesic of length less than $\delta$ in $X_i$. Such a projected geodesic must be constant: otherwise its initial vector over one period would be a nonzero vector of norm below the injectivity radius mapping back to its starting point under the exponential map. Thus all sufficiently short closed geodesics lie in the $F_{\epsilon }$ directions. Conversely the $k$ coordinate circles, of length $\epsilon$, pass through every point and span those directions.

It follows that the smooth distribution tangent to the small flat-torus factor is characterized intrinsically by the span of tangent vectors to closed geodesics of length below $\delta$. A hypothetical product isometry would preserve this smooth distribution and its orthogonal complement. The latter has leaves $X_i\times \{q\}$; the isometry and its inverse take whole leaves to whole leaves, thereby giving an isometry $X_1\cong X_2$, a contradiction. This proves isospectral stabilization by a small flat torus:This argument works even without assuming that the original four-dimensional torus metrics are flat.

The central problem of spectral geometry is how much of a Riemannian metric is determined by the eigenvalues, with multiplicities, of its Laplace-Beltrami operator. For a membrane with fixed boundary, the squared vibration frequencies are proportional to the Dirichlet eigenvalues. Kac's 1966 paper brought this inverse problem into focus: the full sequence of frequencies might encode a domain's shape, but recovering a few geometric quantities is much weaker than recovering an isometry class.

The first major line of development extracts geometric information from spectral sums. On a smooth closed $d$-dimensional manifold, the heat trace has the short-time expansionwhere ${\lambda }_j$ are the nonnegative eigenvalues of $P=-\Delta$ and $R$ is scalar curvature. The coefficients are heat invariants, formed by integrating local curvature expressions. They recover dimension, Riemannian volume and total scalar curvature. For smooth planar domains with the Dirichlet boundary condition,Thus area and boundary length are audible. The constant term also contains topological information through Gauss-Bonnet theorem, with corner corrections when the boundary is polygonal. The Weyl law is the corresponding leading eigenvalue-counting result. Neither the leading term nor the whole list of local heat coefficients automatically reconstructs global geometry.

In the 1970s, the wave trace introduced a complementary link with global closed geodesics. The distributionis spectral, while its singularities are governed by periodic geodesic motion. The 1975 Duistermaat-Guillemin work made this relationship precise. Its trace formula places singularities at geodesic lengths and computes their coefficients under suitable clean or nondegenerate hypotheses. When a coefficient is nonzero, the corresponding length is audible. Possible cancellation must be addressed before identifying the entire length spectrum with the singular support. This approach led to spectral rigidity questions, rather than just the recovery of integral curvature quantities.

The second main line constructs counterexamples to metric determination. A precedent was the 1964 flat-torus example: different sixteen-dimensional Euclidean lattices could give the same Laplacian eigenvalues. Subsequent lattice constructions lowered the dimension; Conway and Sloane's 1992 work exhibited four-dimensional Euclidean lattices with matching lattice theta series. For a flat torus ${\mathbb{R}}^d/\Lambda$, the spectrum is the multiset $4{\pi }^2|\xi {|}^2$ for $\xi \in {\Lambda }^{\ast }$, so equality of lattice theta series of the dual lattices gives isospectrality even when the Euclidean lattices are not related by an orthogonal transformation.

A systematic geometric construction arrived with the Sunada theorem in 1985. For a common Riemannian cover with finite isometry group $T$, almost conjugate subgroups have the same number of invariant vectors in each Laplacian eigenspace. Their freely acting quotients are therefore isospectral manifolds. This reduces an analytic equality to a finite-group representation identity. It produces many examples on surfaces and in higher dimensions, but nonconjugacy of the subgroups must be supplemented by an argument excluding extra quotient isometries. Transplantation theorem provides an explicit version: piecewise eigenfunctions are recombined by a fixed invertible linear transformation that respects the gluing and boundary conditions.

The planar problem itself was answered negatively by Gordon, Webb and Wolpert in 1992: noncongruent simply connected polygonal plane domains have equal Dirichlet spectra. This is stronger than counterexamples among abstract manifolds or flat tori, since these are genuine two-dimensional planar membranes. It does not settle every restricted version, such as asking for two domains within a specified smooth or convex class.

By the 1990s, isospectrality was also known to coexist with continuous metric variation and changes in local geometry. For example, a 1997 construction gives continuous isospectral families on products of spheres and tori whose members need not be locally isometric; in some examples the maximum scalar curvature changes. Conversely, rigidity and compactness results show that an isospectral class is not unrestricted. The 1988 surface compactness theorem makes fixed-spectrum sets of closed orientable surface metrics compact modulo diffeomorphisms. Compactness does not mean there is only one metric, or even finitely many metrics. By 2005, the subject therefore combined local heat invariants, global geodesic information, group-theoretic constructions, explicit transplantation and geometric rigidity. The spectrum determines substantial geometry, while generally failing to determine the full metric.

Work with a finite-dimensional complex semisimple Lie algebra. The Dynkin diagram is obtained from its intrinsic root system, with several structure theorems entering the construction.

First choose a Cartan subalgebra $\mathfrak{h}$. The semisimple structure theory gives the root-space decompositionThe nonzero roots span ${\mathfrak{h}}^{\ast }$, root spaces have dimension one, and every root has its negative. The rank is $r=\dim \mathfrak{h}$.

Next use the Killing form. Its nondegeneracy on $\mathfrak{h}$ and its positive-definite restriction to the real coroot span identify the real root span $E=\mathbb{R}\Phi$ with a Euclidean space. Explicitly, if $t_{\lambda }={\mathcal{B}}^{-1}(\lambda )$, put $(\lambda ,\mu )=B(t_{\lambda },t_{\mu })$. The sl2 subalgebra associated with a root supplies its normalized coroot $H_{\alpha }$ and the integral pairingThe root-string theorem makes the reflection $s_{\alpha }(\beta )=\beta -\beta (H_{\alpha })\alpha$ permute $\Phi$. Together with the absence of multiples other than $\pm \alpha$, this shows that $\Phi$ is a finite reduced root system and a crystallographic root system. These properties come from semisimple Lie algebra theory, not from the diagram's definition.

Choose a vector in $E$ that is not perpendicular to any root. Define the positive roots by positive inner product with this vector. The simple roots are those positive roots that cannot be expressed as sums of two positive roots. Root-system theory shows that they form a basis $\Delta =\{{\alpha }_1,\dots ,{\alpha }_r\}$, and each root is an integer combination of them with coefficients all of one sign. Distinct simple roots have nonpositive inner product.

Use the standard Cartan matrix conventionThere is one diagram vertex for each simple root. For $i\ne j$, the integral off-diagonal entries are nonpositive andConnect the two vertices with respectively zero, one, two or three edges. The corresponding angles are $90^{\circ },120^{\circ },135^{\circ },150^{\circ }$. A multiple edge carries an arrow toward the shorter root. For joined vertices, the length ratio is determined by $A_{ij}/A_{ji}=({\alpha }_i,{\alpha }_i)/({\alpha }_j,{\alpha }_j)$; hence the diagram records both angle and relative length.

For example, ${\mathfrak{sl}}_3$ has simple roots $L_1-L_2$ and $L_2-L_3$, giving two equal-length vertices joined by one edge: type $A_2$. For ${\mathfrak{sp}}_4$, take ${\alpha }_1=L_1-L_2$ and ${\alpha }_2=2L_2$. Thenso the two vertices have a double edge directed toward ${\alpha }_1$, the shorter root: type $C_2$.

Finally, orthogonal irreducible components of the root system correspond to the simple ideals of a Lie algebra in the semisimple Lie algebra, so its diagram is the disjoint union of their connected diagrams. Conjugacy of Cartan subalgebras and the Weyl group action on choices of positive roots make the resulting diagram independent of these choices up to isomorphism. Thus the vertices, edge multiplicities, arrows and connected components encode the simple-root geometry and the simple-ideal decomposition.

Work over $\mathbb{C}$ with finite-dimensional Lie algebras and Lie algebra representation spaces. The adjoint form of Engel theorem isHere a nilpotent Lie algebra has terminating lower central series, ${\gamma }_1\mathfrak{g}=\mathfrak{g}$, ${\gamma }_{j+1}\mathfrak{g}=[\mathfrak{g},{\gamma }_j\mathfrak{g}]$. We prove the stronger linear statement: if $V\ne 0$, $L\subseteq \mathfrak{gl}(V)$ and every member of $L$ is a nilpotent endomorphism, then there is a nonzero $v\in V$ killed by all of $L$, and $L$ can be simultaneously represented by strictly upper triangular matrices.

For the proof of Engel theorem by induction and normalizers, first establish a useful nilpotence fact. If $x^m=0$, left and right multiplication by $x$ on $\mathrm{End}(V)$ commute. Expanding their difference givesFor $N=2m-1$, every summand contains a power of $x$ at least $m$, so $(\mathrm{ad}x{)}^{2m-1}=0$. Restrictions and induced quotient maps remain nilpotent.

Now prove the common-kernel statement, the Engel lemma, by induction on $\dim L$. The zero algebra is immediate. Choose a maximal proper Lie subalgebra $M$. Its action on $L/M$ by commutators consists of nilpotent maps by the preceding fact. Applying the lower-dimensional induction hypothesis to its image gives a nonzero coset $y+M$ with $[M,y]\subseteq M$. HenceThe normalizer of a Lie subalgebra is a Lie subalgebra by the Jacobi identity. Maximality implies $N_L(M)=L$, so $M$ is an ideal of a Lie algebra. Moreover $L/M$ has no proper nonzero Lie subalgebra, and every one-dimensional subspace is a Lie subalgebra. Thus $\dim L/M=1$.

Induction also provides a nonzero common annihilatorIt is stable under $L$: for $m\in M$, $x\in L$, $v\in W$,because $[m,x]\in M$. Choose $x$ representing a basis of $L/M$. The restriction of the nilpotent endomorphism $x$ to the nonzero space $W$ has nonzero kernel. A vector in this kernel is killed by both $M$ and $x$, hence by all of $L$. This finishes the induction. When an action is not faithful, induction is applied to its image, whose dimension is no larger than $\dim M$; no faithfulness assumption is hidden.

Starting with the common-kernel line, apply the same statement to each successive quotient Lie algebra representation space. It constructs a complete flag of invariant subspacesAn adapted basis represents every member of $L$ by a strictly upper triangular matrix. Every product of $d$ such matrices is zero, so every iterated commutator with $d$ entries is zero as well.

Finally, if every $\mathrm{ad}x$ on $\mathfrak{g}$ is nilpotent, apply the linear statement to the Adjoint representation. With $d=\dim \mathfrak{g}$, every product of $d$ adjoint operators vanishes, so ${\gamma }_{d+1}\mathfrak{g}=0$. Conversely, if ${\gamma }_{c+1}\mathfrak{g}=0$, then $(\mathrm{ad}x{)}^{c}y=0$ for all $x,y$, proving the other implication. This proves Engel's theorem and its simultaneous-triangularization form. The hypothesis concerns every element of the algebra, not merely a chosen set of generators.