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The real vector space of symmetric matrices has dimension $n(n+1)/2$. On its invertible part the derivative of the determinant isAt a determinant-one symmetric matrix, this derivative is nonzero: the symmetric direction $C=B$ gives value $n$. The regular level set theorem therefore yieldsThis applies to every signature component, not only positive-definite matrices.

The special linear congruence action on symmetric matrices preserves symmetry because $(AB{A}^T{)}^T=AB{A}^T$, and preserves determinant because $\det (AB{A}^T)=(\det A{)}^2\det B=1$. The identity acts trivially andThese verify a smooth left Lie group action of the special linear group. By Sylvester's law of inertia, it also preserves the numbers of positive and negative eigenvalues.

For $n=2$, parametrize the symmetric matrices byThus $M$ is exactly the two-sheeted hyperboloidThe two sheets consist respectively of positive-definite and negative-definite matrices. Each is preserved by the special linear congruence action on symmetric matrices. They are individually transitive: on the positive sheet $B=A{A}^T$ with $A=B^{1/2}$ and $\det A=1$; on the negative sheet use $B=-A{A}^T$. The stabilizer of either $I$ or $-I$ is $SO(2)$.

Differentiating the left Lie group action along $\exp (sX)$ gives the fundamental vector fieldFor the three matrices in the question, their coordinate components areA sign convention matters here. With the usual Lie bracket of vector fields $[U,V]f=U(Vf)-V(Uf)$, the fundamental fields of this left action obey $[u_X,u_Y]=-u_{[X,Y]}$. Indeed, for linear coordinate fields $u_X(z)=R_{X}z$, the bracket has coefficient $(R_Y{R}_X-R_X{R}_Y)z$. To obtain a Lie algebra representation rather than an anti-representation, use the infinitesimal left-action sign conventionAn explicit answer is thereforeThese are tangent to the two-sheeted hyperboloid: applying each to $Q=t^2-x^2-y^2$ gives zero. For example,They can be written on either sheet using $t=\pm \sqrt{1+x^2+y^2}$:Here the omitted ${\partial }_t$ component is determined by tangency, and derivatives of $t(x,y)$ must be included when computing brackets in these coordinates.

Direct matrix multiplication givesDirect differentiation of the displayed vector fields gives exactlyFor instance, in ambient $(t,x,y)$ coordinates, $[v_1,v_3]=(-2x,-2t,0)=v_2$. These are the defining relations of the sl2 Lie algebra. The three fields are linearly independent over constant real coefficients: if $a{v}_1+b{v}_2+c{v}_3$ vanishes on a sheet, its ${\partial }_x$ coefficient $(-a+c)y-2bt$ forces $b=0,c=a$, and its ${\partial }_y$ coefficient then equals $-2at$, forcing $a=c=0$. Thus the representation is faithful. Their pointwise span need only have dimension two, consistent with the dimension of each sheet.

For the finite-dimensional Lie algebra representation specified in the PDF, a Lie-invariant bilinear form satisfiesThe dual Lie algebra representation has action $(x\varphi )(w)=-\varphi (xw)$, so the map $T_B:V\to V^{\ast }$ defined by $T_B(v)=B(v,-)$ is an intertwining operator. If $V$ is irreducible and $B\ne 0$, its kernel is zero. Since $V,V^{\ast }$ have equal finite dimension, $T_B$ is an isomorphism, and $B$ is a nondegenerate bilinear form.

For $B_1\ne 0$, the composition $T_{B_1}^{-1}{T}_{B_2}$ is a representation endomorphism. Over an algebraically closed field, the Schur lemma makes it scalar, provingTransposing $B$ gives another Lie-invariant bilinear form. For $B\ne 0$, write $B^T=\mu B$; transposing twice gives ${\mu }^2=1$. In field characteristic different from two, this means $\mu =1$ or $\mu =-1$. Thus the form is a symmetric bilinear form or an alternating bilinear form, respectively. The zero form has both properties. In the second case $B(v,v)=-B(v,v)$ forces $B(v,v)=0$, using the same characteristic assumption.

The two Lie groups have the same local infinitesimal structure, but different global topology. This difference determines which Lie algebra representations integrate to representations of each group.

An element of the SU(2) group is a unitary matrix of determinant one. Orthogonality of its columns and its determinant give the unique formThus its group manifold is the unit three-sphere in ${\mathbb{C}}^2\simeq {\mathbb{R}}^4$: this is SU(2) as the three-sphere. In particular it is compact, connected and simply connected. In terms of the Pauli matrices one may also write $U=a_{0}I-i\mathbf{a}\cdot \mathbf{\sigma }$ with real coefficients satisfying $a_0^2+{\mathbf{a}}^2=1$.

Differentiate $U(t{)}^{†}U(t)=I$ and $\det U(t)=1$ at $U(0)=I$. The tangent space consists of traceless skew-Hermitian matrices:The Lie bracket of a Matrix Lie group is the matrix commutator. With $T_a=-i{\sigma }_a/2$, the Pauli matrix multiplication law givesThis derives the SU(2) Lie algebra as a three-dimensional real Lie algebra. The Hermitian physics generators $J_a={\sigma }_a/2$ instead obey $[J_a,J_b]=i{ϵ}_{abc}{J}_c$; they are $i$ times the skew-Hermitian tangent generators, so these are consistent conventions.

The SO(3) group consists of real orthogonal matrices with determinant one. Differentiating $R(t{)}^{T}R(t)=I$ at the identity givesThe determinant condition gives no additional infinitesimal constraint because a skew-symmetric matrix already has trace zero. Define $L_a\mathbf{v}={\mathbf{e}}_a\times \mathbf{v}$. The cross product identity impliesTherefore the SO(3) Lie algebra has the same structure constants and $T_a\mapsto L_a$ is a Lie algebra isomorphism.

The global relation is the Adjoint double cover from SU(2) to SO(3). For $V=\mathbf{v}\cdot \mathbf{\sigma }$, define $R(U)$ byConjugation preserves the real space of traceless Hermitian matrices and its inner product $\frac{1}{2}\mathrm{tr}(VW)=\mathbf{v}\cdot \mathbf{w}$. Hence $R(U)$ is orthogonal. Continuity and connectedness, together with $R(I)=I$, put it in $SO(3)$. Composition of conjugations makes $R$ a group homomorphism. If $R(U)=I$, then $U$ commutes with every Pauli matrix, hence is scalar; unitarity and determinant one leave precisely $U=\pm I$. The differential sends $T_a$ to $L_a$, so it is an isomorphism. More concretely,induces rotation through angle $\theta$ about $\mathbf{n}$, by the Rodrigues rotation formula. Every three-dimensional rotation has such an axis and angle, proving surjectivity. ConsequentlyThe matrices $U$ and $-U$ are antipodal points on the three-sphere, so the SO(3) group manifold is Real projective space ${\mathbb{RP}}^3$. Equivalently, the closed axis-angle ball $\Vert \theta \mathbf{n}\Vert \le \pi$ has opposite boundary points identified. The fundamental group is ${\pi }_1(SO(3))\simeq {\mathbb{Z}}_2$, whereas ${\pi }_1(SU(2))=0$. A $2\pi$ rotation lifts from $I$ to $-I$; a $4\pi$ rotation returns to $I$. Thus the covering is the universal cover and the groups are not globally isomorphic.

For representation theory, specify finite-dimensional complex continuous representations. Compactness permits an invariant Hermitian inner product, obtained by averaging against Haar measure, and therefore complete reducibility. Complexifying either real Lie algebra gives the sl2 Lie algebra. Its finite-dimensional irreducibles are indexed by $n\in {\mathbb{Z}}_{\ge 0}$, have highest weight $n$, and have dimension $n+1$. By integration of a Lie-algebra representation, since $SU(2)$ is simply connected, every such Lie algebra representation integrates uniquely. The resulting homogeneous polynomial representation of SU2 isIn the spin angular momentum notation $j=n/2$, its Hermitian $J_3$ eigenvalues are $j,j-1,\dots ,-j$. The central matrix $-I$ acts on the symmetric power by $(-1{)}^n$, so descent of an SU(2) representation to SO(3) occurs exactly when $n$ is even. HenceFor a reducible representation, every summand must satisfy the descent condition. The spin-one-half doublet is a genuine representation of the covering group but does not define a single-valued representation of $SO(3)$; the spin-one triplet does and is its vector representation. The distinction is topological, rather than a difference in their isomorphic Lie algebras.

A two-dimensional Lie algebra representation already provides a counterexample. In the basis $e_1,e_2$ of $V={\mathbb{C}}^2$, defineThe commutator $[\rho (x),\rho (y)]$ is zero, agreeing with $\rho (z)$; the other defining Lie brackets also map to zero. Thus $\rho$ is a Lie algebra homomorphism, although it is not a Faithful Lie algebra representation.

The line $L=\mathbb{C}{e}_1$ is an invariant subspace. Any complementary line has the form $\mathbb{C}(e_2+a{e}_1)$ for some $a\in \mathbb{C}$. Butso no complementary line is an invariant subspace. Equivalently, every invariant line is killed by $\rho (x)$: its restriction to a line is a scalar, and its square is zero, forcing that scalar to be zero. A decomposition into two invariant lines would then force $\rho (x)=0$, a contradiction. This representation is therefore not a semisimple representation. Thus

Use anti-Hermitian representation matrices for the unitary representation, so $R(X{)}^{†}=-R(X)$. Absorb the gauge coupling into the connection, consistently with the transformation law in the PDF. The appropriate gauge covariant derivative isThis has a plus sign because the inhomogeneous part of the connection transformation is $-ϵ{\partial }_{\mu }X$. Its variation isThus gauge covariance of a scalar covariant derivative givesThe cancellation uses that $R$ is a Lie algebra representation, including its commutator identity.

The gauge field strength isIt is the curvature appearing in $[D_{\mu },D_{\nu }]\varphi =R(F_{\mu \nu })\varphi$. The index positions in the PDF are related by the constant Minkowski metric; lowering the index in its transformation law gives the convention used here.

Take signature $(+,-,-,-)$, let $B$ be a real symmetric gauge-invariant internal form, and let $⟨\cdot ,\cdot {⟩}_V$ be the positive Hermitian form preserved by the unitary matter representation. One consistent Lorentz transformation scalar Lagrangian density isHere $g_{\mathrm{YM}}>0$, $m^2\ge 0$, and $\lambda \ge 0$ give a simple stable potential. The coupling is in the kinetic normalization because it was absorbed into $A_{\mu }$; it is not inserted again into $D_{\mu }$. Taking a zero or another gauge-invariant potential is also possible. A positive nondegenerate $B$ gives the usual physical gauge kinetic energy, as for a compact gauge algebra. More generally the invariance calculation below applies to any symmetric invariant bilinear form on a Lie algebra, including the Killing form.

Under the infinitesimal gauge-field transformation law,Invariance and symmetry of $B$ give the cancellation. For matter, both $\varphi$ and $D_{\mu }\varphi$ transform by $R(X)$, so anti-Hermiticity givesfor either pairing used in the Lagrangian. Its potential is a function of the invariant scalar norm. Therefore $\boxed{{\delta }_X\mathcal{L}=0}$, establishing gauge invariance of this Yang-Mills theory coupled to a scalar. Contraction of spacetime indices with the Minkowski metric gives Lorentz invariance.

For the Adjoint representation, define the Lie algebra structure constants byInternal superscripts here are basis labels, as in the paper; spacetime indices alone are raised with the Minkowski metric. The adjoint covariant derivative and curvature have componentsThese formulas involve no extra factor of $i$, because the generators use the Lie-algebra convention rather than Hermitian physics generators.

To make the Killing-form sign explicit, write $B=s\kappa$ and $⟨\varphi ,\psi {⟩}_V=s{\kappa }^{ab}({\varphi }^a{)}^{\ast }{\psi }^b$. The literal contraction requested in the paper corresponds to $s=1$. For a compact real form of a complex semisimple gauge algebra, its actual positive inner product has $s=-1$, since the Killing form is negative definite. With a complex adjoint scalar, the fully expanded Killing-form Lagrangian for an adjoint scalar isAll repeated internal labels are summed; $s^2=1$ explains the absence of $s$ from the quartic term. Invariance is equivalently the component identity $f^{ca}{}_d{\kappa }^{db}+f^{cb}{}_d{\kappa }^{ad}=0$. For a real adjoint scalar, omit complex conjugation and insert $1/2$ in the scalar kinetic and quadratic mass terms; the same invariance proof applies.

The PDF's identification of the Killing form with an inner product thus requires a sign convention. It also requires nondegeneracy if it is to give ordinary kinetic terms: on an Abelian gauge factor the Killing form vanishes, so an additional invariant metric is needed there. These qualifications do not alter the covariance identities or the formal gauge invariance of the displayed Killing-form contractions.

A finite-dimensional Lie algebra representation is a homomorphism $\rho :L\to \mathfrak{gl}(V)$. It is irreducible when $V$ has no invariant subspaces other than $0$ and $V$.

The algebra ${\mathfrak{sl}}_2$ has basis $e,f,h$ withFor every $n\ge 0$, its $(n+1)$-dimensional irreducible module $V(n)$ has basis $v_0,\dots ,v_n$ and actionwith out-of-range vectors zero. Any nonzero invariant subspace contains a weight vector; repeated application of $e$ reaches $v_0$, and repeated application of $f$ then generates the whole module, proving irreducibility. The adjoint module of ${\mathfrak{sl}}_2$ is $V(2)$, so every ideal is an invariant subspace and ${\mathfrak{sl}}_2$ is simple.

For a root $\alpha$, nondegeneracy of the Killing pairing between $L_{\alpha }$ and $L_{-\alpha }$ allows choices $e_{\alpha },f_{\alpha }$ withAfter rescaling, $(e_{\alpha },f_{\alpha },h_{\alpha })$ obey the ${\mathfrak{sl}}_2$ relations, giving a copy of ${\mathfrak{sl}}_2$ in $U_{\alpha }$.

Restrict the adjoint representation of $L$ to this copy. Finite-dimensional ${\mathfrak{sl}}_2$ theory shows that the $\alpha$-string through zero has one nontrivial summand $V(2)$, with weight spaces $L_{-\alpha }$, $\mathbb{C}{h}_{\alpha }$, and $L_{\alpha }$. Any additional vector in $L_{\alpha }$ would generate another weight-two summand and another independent zero-weight coroot, contradicting nondegeneracy of the root--coroot pairing on $H$. Hence

For $\beta \notin \mathbb{Z}\alpha$, the spaceis stable under $e_{\alpha },f_{\alpha },h_{\alpha }$. Adjacent raising and lowering maps are nonzero until the endpoints, and every root space is one-dimensional, so $W$ is a simple ${\mathfrak{sl}}_2$-module of highest weight $p+q$. Its lowest and highest $h_{\alpha }$-weights giveand therefore $\beta (h_{\alpha })=q-p$. Since $[L_{\alpha },L_{-\alpha }]=\mathbb{C}{h}_{\alpha }$, every $x$ in this bracket line satisfies

Write $D=\mathrm{ad}y$. The generalized eigenspace decomposition of the linear map $D$ isfor any sufficiently large $N$. Because $D$ is a derivation, the generalized-eigenspace bracket lemma givesConsequently $L_{0,y}$ is a Lie subalgebra.

The set $I(K)=\{x\in L:[x,K]\subseteq K\}$ is the normalizer of a Lie subalgebra $K$. Certainly $L_{0,y}\subseteq I(L_{0,y})$. Conversely, if $x\in I(L_{0,y})$, then $y\in L_{0,y}$ givesOn the direct sum of the nonzero generalized eigenspaces, $D$ is invertible. Hence the nonzero-eigenvalue component of $x$ vanishes, and

Now let $K$ be a Lie subalgebra containing $L_{0,y}$. Since $y\in K$, the subspace $K$ is $D$-invariant. The generalized zero eigenspace of the induced map on $L/K$ is the image of $L_{0,y}$, hence is zero. If $x\in I(K)$, then $Dx=[y,x]\in K$, so $x+K$ lies in that zero eigenspace. Thus $x\in K$ and

A nilpotent Lie algebra is one whose lower central serieseventually reaches zero. Suppose $L$ is nilpotent and $K⊊L$. Choose the least $r\ge 2$ for which ${\gamma }_r(L)\subseteq K$. Then ${\gamma }_{r-1}(L)⊈K$, and anysatisfies $[x,K]\subseteq [L,{\gamma }_{r-1}(L)]={\gamma }_r(L)\subseteq K$. Therefore $x\in I(K)\setminus K$, proving the normalizer condition for a nilpotent Lie algebra

It remains to prove the converse needed here. The Engel lemma states that if a finite-dimensional Lie algebra of linear maps consists of nilpotent maps, then the maps have a common nonzero vector in their kernels. To prove it, induct on the dimension of the algebra. For a maximal proper subalgebra $H$, induction applied to the action of $H$ on $L/H$ produces $x\notin H$ with $[H,x]\subseteq H$. Thus $H$ is an ideal of codimension one. Induction also gives a nonzero common kernelThe ideal property makes $W$ invariant under $L$; a nilpotent representative of a basis of $L/H$ has a nonzero kernel on $W$, yielding a vector killed by all of $L$.

Apply the lemma to the Adjoint representation. It produces a nonzero element of the center of a Lie algebra. Induction on $\dim L$, followed by passage to the quotient by this center, proves Engel theorem: if every $\mathrm{ad}y$ is nilpotent, then $L$ is nilpotent. The hypothesis $L_{0,y}=L$ says exactly that every $\mathrm{ad}y$ is nilpotent, so

For a smooth Lie-group representation $D:G\to GL(V)$, define its derived representationDifferentiation makes $dD$ linear. Applying $D$ to the group commutator curveand taking the mixed derivative at $(0,0)$ givesThus $dD:\mathfrak{g}\to \mathfrak{gl}(V)$ is a Lie algebra representation.

For a finite-dimensional Lie algebra representation $\varphi :\mathfrak{g}\to \mathfrak{gl}(V)$, the Trace form of a Lie algebra representation isThe Killing form is the trace form of the Adjoint representation of a Lie algebra:A bilinear form $B$ is $\mathfrak{g}$-invariant whenequivalently $B([x,y],z)=B(x,[y,z])$. For a trace form this follows from cyclicity of trace:

Bilinearity and alternatingness ofare immediate. For three elements, the $\mathfrak{g}$ component of the Jacobi sum vanishes by the Jacobi identity in $\mathfrak{g}$. In the $V$ component, the coefficient of a vector such as $u$ isbecause the action is a Lie algebra representation; the other terms cancel cyclically in the same way. Hence the bracket satisfies Jacobi and defines the semidirect product of a Lie algebra and a module $\mathfrak{g}⋉V$.

Since $-\theta T_1=i\theta {\sigma }_1/2$, the same disentangling identity applies in every Lie algebra representation. The rightmost $E_{+}$ kills $|\Lambda ⟩$, the leftmost $E_{-}$ kills $⟨\Lambda |$, and the middle factor acts by its highest weight. Therefore

For a finite-dimensional Lie algebra representation $\rho$ on $V$, the Trace form of a Lie algebra representation isWrite again $A=\rho (a)$, $B=\rho (b)$, and $C=\rho (c)=[A,B]$. The operator $C$ commutes with both $A$ and $B$. Direct use of the cyclic property of the trace givesIn the last line, cyclicity and $AC=CA$ turn $\mathrm{tr}(ABC)$ into $\mathrm{tr}(BAC)$. Thus the nonzero vector $c\in \mathfrak{g}$ is orthogonal to the basis $a,b,c$, and hence to all of $\mathfrak{g}$. The bilinear form is therefore degenerate.

Use the Polynomial representation of the Heisenberg Lie algebra on the infinite-dimensional polynomial ring $\mathbb{C}[x]$:The product rule gives $[d/dx,x]=1$, so this is a Lie algebra representation. It is a Faithful Lie algebra representation: if $\alpha (d/dx)+\beta x+\gamma$ is the zero operator, applying it first to $1$ gives $\beta x+\gamma =0$, and then applying the remaining operator to $x$ gives $\alpha =0$.

To prove irreducibility, let $W$ be a nonzero invariant subspace and choose a nonzero polynomial in $W$ of least degree. If its degree were positive, repeated differentiation would produce a nonzero element of smaller degree, so $W$ contains a nonzero constant. Invariance under multiplication by $x$ then puts every monomial $x^n$ in $W$, and hence $W=\mathbb{C}[x]$.

The Adjoint representation of a Lie algebra isIt is linear. The Jacobi identity givesso it is a Lie algebra representation.

For a nontrivial finite-dimensional anti-Hermitian Lie algebra representation of a compact Lie algebra that is simple, its Trace form of a Lie algebra representation is a positive multiple of the Killing form. Define $T$ by $H(X,Y)=\kappa (TX,Y)$. Invariance makes $T$ commute with the Adjoint representation of a Lie algebra. Symmetry makes $T$ self-adjoint for $-\kappa$, so its eigenspaces are ideals. Simplicity implies $T=\mu I$. The kernel of the representation is an ideal and hence vanishes; therefore $H(X,X)=-\mathrm{Tr}(d(X{)}^{†}d(X))<0$ for nonzero $X$, proving $\mu >0$.

// Target: lie-theory.bigb

For a Lie algebra representation with generators referred to an orthonormal invariant form, the sum ${\sum }_a{t}_a{t}_a$ commutes with every generator. Indeed, the commutator is proportional to a contraction of antisymmetric structure constants of a Lie algebra with the symmetric product $t_a{t}_b+t_b{t}_a$. The Schur lemma makes it scalar in an irreducible complex representation; that scalar is its quadratic Casimir eigenvalue. With trace normalization $\mathrm{tr}(t_a{t}_b)=C(R){\delta }_{ab}$, taking the trace gives $C(R)\dim \mathfrak{g}=C_2(R)\dim R$.

For a subrepresentation $W\subseteq V$, the quotient vector space $V/W$ inherits the action. For a Lie algebra representation, the formula is $x(v+W)=xv+W$, well-defined because $W$ is invariant.

For a Lie algebra representation of $\mathfrak{g}$ on $V$, the semidirect product $\mathfrak{g}⋉V$ is the vector space $\mathfrak{g}\oplus V$ with bracketThe subspace $V$ is an abelian ideal of a Lie algebra.

A Lie algebra representation is simultaneously upper triangular precisely when it preserves a complete flag, with $\dim V_j=j$. A basis adapted to the invariant flag makes every representing matrix upper triangular. For a complex finite-dimensional representation of a solvable Lie algebra, the Lie theorem provides a common eigenvector; iteration on the quotient representations constructs the flag.