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The unbroken group is the stabilizer subgroup of the chosen vacuum expectation value:Its Lie algebra consists of generators $T$ with $T{\varphi }_0=0$. The vacuum manifold is the orbit $G{\varphi }_0\simeq G/H$ under the stated transitivity assumption. Its tangent space is the image of $T\mapsto T{\varphi }_0$, so its dimension is $r=\dim G-\dim H$.

Let $K_{ij}={\partial }_i{\partial }_{j}V({\varphi }_0)$ be the scalar mass matrix, with canonical scalar kinetic terms. Invariance of the scalar potential gives $({\partial }_{i}V)(T_a\varphi {)}_i=0$ for every $\varphi$. Differentiate with respect to ${\varphi }_j$ and evaluate at the minimum, where ${\partial }_{i}V=0$:Thus every independent tangent direction is a zero eigenvector of $K$. These are the Goldstone directions in the scalar mass matrix; for a global symmetry they are $r$ physical massless Goldstone bosons. For a generic nondegenerate minimum transverse to the orbit, they are precisely all the zero eigenvalues. Without that nondegeneracy assumption, additional massless scalars are possible.

For a local gauge group, use the orthogonal unitary-gauge slice near a scalar vacuum. Since ${\varphi }_0^T{T}_a{\varphi }_0=0$ by antisymmetry, the proposed condition is equivalent toIt sets the gauge-orbit components of the scalar fluctuation to zero. This is an admissible local unitary gauge: on a basis of independent broken tangent vectors $v_a=T_a{\varphi }_0$, the variation of these conditions under the broken gauge parameters is $v_a^T{v}_b$. This Gram matrix is invertible on that basis, so the implicit function theorem selects the required gauge transformation near the vacuum. Unbroken transformations remain as a residual $H$ gauge symmetry; no global gauge slice through zeros of the scalar field is being asserted.

Write $\varphi ={\varphi }_0+h$. To quadratic order, the covariant kinetic term containsThe middle term vanishes in this unitary gauge. The gauge-boson mass rank from a real scalar vacuum is therefore determined byThese are squared masses: the physical gauge boson masses are square roots of the matrix's nonzero eigenvalues after diagonalization. For real $c_a$,Its kernel consists exactly of the unbroken generators. With $g\ne 0$, the rank is $r$, so exactly $\dim G-\dim H$ gauge-field combinations become massive. Each removed scalar Goldstone mode supplies the extra longitudinal polarization of a massive gauge boson.

The remaining physical scalar fluctuations lie in the orthogonal complement of the orbit tangent space. They have no massless modes if $K$ is positive definite on that complement. The unconditional absence of massless scalars does not follow from the printed assumptions. For a concrete counterexample, take a real doublet, $G=SO(2)$ andIts entire minimum set is one circle orbit, as required, but its Hessian matrix vanishes at the minimum. Gauging the rotation gives one vector of squared mass $g^2{v}^2$ and removes the angular scalar. The radial scalar remains massless at quadratic order, since its potential starts at fourth order in the radial fluctuation. This is an accidental massless radial mode beyond Goldstone modes. Thus the intended no-massless-scalar conclusion needs nonzero transverse curvature.

In the Standard Model, the Higgs doublet contains four real components and breaks $SU(2{)}_L\times U(1{)}_Y$ to $U(1{)}_{\mathrm{em}}$. Three Goldstone bosons become the longitudinal components of $W^{\pm }$ and $Z$, while the photon remains massless. The remaining radial Higgs boson is massive for the ordinary quartic potential with positive curvature. With $⟨H⟩=(0,v_{\mathrm{EW}}/\sqrt{2}{)}^T$,The colour gauge group is unbroken. This is the Higgs mechanism, with the generic curvature condition realized by the standard Higgs potential.

The displayed matrices describe the orientation-preserving component $SE(2)$ of the full Euclidean group $E(2)=O(2)⋉{\mathbb{R}}^2$. Both have the same Lie algebra, so this naming convention does not affect the calculation. Choose the translation generators $P_1,P_2$ and rotation generator $J$ in the Lie algebra of this Matrix Lie group asTaking matrix commutators givesAll remaining brackets follow from antisymmetry. The corresponding left-invariant vector fields in coordinates $(x,y,\psi )$ areTheir Lie brackets of vector fields obey the same relations. This also fixes which sign of the rotation generator is being used.

There is an important qualification to the printed connection prescription. For an arbitrary vector field $Z$ and smooth function $f$, its literal right-hand side obeysAn affine connection instead requires ${\nabla }_{L_i}(fZ)=L_i(f)Z+f{\nabla }_{L_i}Z$. Thus the formula for arbitrary $Z$ defines a connection only when $\lambda =1$. For example, at $\psi =0$, take $i=1$, $f=x$ and $Z=L_1$: the erroneous formula gives $\lambda L_1$, whereas the Leibniz rule requires $L_1$.

The intended one-parameter family is the bracket connection on a Lie group: prescribe the rule for left-invariant $Z$, then extend it by the Leibniz rule. If $[L_i,L_j]=c_{ij}{}^k{L}_k$, this meansExtend linearly over functions in the first slot. This is a well-defined affine connection for every real $\lambda$. The following Ricci tensor calculation applies to this intended family; under the literal arbitrary-$Z$ reading only its $\lambda =1$ member exists.

Use the curvature conventionFor left-invariant vector fields the structure constants of a Lie algebra are constant. The Jacobi identity givesWriting $A=\lambda (\lambda -1)$, the only nonzero curvature actions, apart from antisymmetry in the first two slots, areFor instance, $[[J,P_1],J]=[P_2,J]=P_1$. Contracting the first and output slots gives $\mathrm{Ric}(L_3,L_3)=-2A$, while every other component vanishes. Equivalently, $[[Z,X],Y]={\mathrm{ad}}_Y{\mathrm{ad}}_{X}Z$, soThis is the Killing form. Here ${\mathrm{ad}}_J$ rotates the two-dimensional translation space and vanishes on $J$, so $B(J,J)=-2$; translation generators give all other components zero. ThereforeSince the coframe component dual to $L_3$ is $d\psi$, the same Ricci tensor is $2\lambda (1-\lambda )d\psi \otimes d\psi$ in these coordinates. Reversing the curvature convention reverses this sign.

At both $\lambda =0$ and $\lambda =1$, the full curvature vanishes. At $\lambda =0$, the left-invariant vector fields form a global parallel frame. At $\lambda =1$, every right-invariant vector field is parallel because left- and right-invariant vector fields commute; the connection is flat in that global parallel frame. Flatness does not imply zero torsion: for left-invariant arguments the torsion tensor isThus the two endpoint connections are flat, with opposite nonzero torsion. For comparison, $\lambda =1/2$ is torsion-free but has nonzero curvature, with ${\mathrm{Ric}}_{33}=1/2$. No metric has been specified, so these connections should not automatically be identified with a Levi-Civita connection.

The displayed torus-only Weyl integration formula needs $\phi$ to be a class function, meaning $\phi (xg{x}^{-1})=\phi (g)$. This hypothesis is implicit in the usual formula but is not printed here. We prove the general version for a continuous function, then specialize:Here $dg$ and $dt$ are probability Haar measures; $d\mu$ is the invariant probability measure on the homogeneous space $G/T$; and $W=N_G(T)/T$ is the finite Weyl group.

We use the following standard maximal-torus facts, as permitted: every element of a connected compact Lie group lies in some maximal torus; all maximal tori are conjugate; two elements of a fixed $T$ are conjugate in $G$ exactly when they are in the same $W$-orbit; and the complexified Lie algebra decomposes into ${\mathfrak{t}}_{\mathbb{C}}$ and one-dimensional root spaces. Opposite roots of a root system give real two-dimensional planes in ${\mathfrak{t}}^{\perp }$. At an element $t$ for which no root character has value one, the identity component of its centralizer is $T$.

Choose a positive root system ${\Phi }^{+}$. Regard each root $\alpha$ as a character $\alpha :T\to U(1)$ through the Adjoint representation. A convenient globally defined denominator isThe usual Weyl denominator additionally has a unit-modulus factor $e^{\rho }(t)$, with $\rho$ half the sum of positive roots. Where that half-weight is not a global torus character, it may be interpreted on a covering torus; the squared modulus is nevertheless unambiguous. In either convention,

Average an inner product on $\mathfrak{g}$ over the Adjoint representation to make it invariant. It defines a bi-invariant Riemannian metric on $G$ and the quotient metric on $G/T$. Consider the conjugation mapAt $(T,t)$, identify the source tangent directions with $X\in {\mathfrak{t}}^{\perp }$ and $Y\in \mathfrak{t}$, and translate the target tangent vector to the identity by $t^{-1}$. Differentiating givesThe two summands are orthogonal. On a real root plane, $\mathrm{Ad}(t^{-1})$ is a rotation with complex eigenvalues $\alpha (t{)}^{-1},\alpha (t)$. Thus the absolute real Jacobian determinant on this plane is $|1-\alpha (t){|}^2$. The torus direction contributes one. Therefore the conjugation Jacobian for a compact Lie group isThis is also the Jacobian for the normalized measures: the Riemannian submersion has fibers isometric to $T$, so $\mathrm{vol}(G)=\mathrm{vol}(G/T)\mathrm{vol}(T)$; the probability-volume normalization factors cancel.

Let $T_{\mathrm{reg}}$ be the elements for which $J_q\ne 0$. The restriction of $q$ is a local diffeomorphism onto the regular elements of $G$. It is $|W|$-to-one. Indeed, fixing $t_0\in T_{\mathrm{reg}}$, any pair with $xt{x}^{-1}=t_0$ has $xT{x}^{-1}=T$, because conjugation identifies the identity components of the two centralizers. Thus $x\in N_G(T)$, and there is exactly one pair $(xT,x^{-1}{t}_{0}x)$ for each coset in $N_G(T)/T$. This argument does not require the full centralizer of $t_0$ to be connected.

The nonregular locus has measure zero. In $T$ it is a finite union of sets $\alpha (t)=1$ of smaller dimension; their conjugation images are images of smaller-dimensional manifolds under $q$, and hence have zero volume in $G$. The change-of-variables formula for the regular covering therefore givesFor a class function, the inner integral is $\phi (t)$, proving the required formula with its correct hypothesis.

Without conjugation invariance, the torus-only expression can be false. For $G=SU(2)$ with diagonal $T$, take $\phi (g)=|g_{12}{|}^2$. It vanishes on $T$, whereas its group integral is $1/2$: the two squared coordinates of a Haar-distributed unit column have equal means and sum to one. Thus the conjugacy average cannot simply be dropped.

Write the diagonal maximal torus of $U(n)$ as $t=\mathrm{diag}(z_1,\dots ,z_n)$, $|z_i|=1$, and let $\delta =(n-1,n-2,\dots ,0)$. Irreducible complex representations are indexed by dominant integer tuples ${\lambda }_1\ge \cdots \ge {\lambda }_n$, their highest weights. The Weyl character formula in integer-exponent form isThe ratio initially uses distinct $z_i$, and extends to repeated eigenvalues by continuity. Every unitary matrix is conjugate to such a diagonal matrix, so this determines the character on the entire group. The shift $\delta$ differs from the usual Weyl vector $\rho =((n-1)/2,(n-3)/2,\dots ,(1-n)/2)$ by a multiple of $(1,\dots ,1)$; the corresponding determinant factors cancel between numerator and denominator.

We first justify the highest-weight information needed in the proof. Restriction to the torus splits a representation into integer weight spaces. In an irreducible representation choose a lexicographically largest weight $\lambda$ and a nonzero vector $v$ of that weight. A matrix unit $E_{ij}$ with $i<j$ raises the weight by $e_i-e_j$, so it annihilates $v$. The character is invariant under permutations of diagonal coordinates, because permutation matrices conjugate the torus. Hence permutations of $\lambda$ are also weights, and lexical maximality forces ${\lambda }_1\ge \cdots \ge {\lambda }_n$.

Irreducibility implies that $v$ generates the representation under the complexified Lie algebra. Indeed, a Lie-algebra invariant subspace is preserved by exponentials of skew-Hermitian matrices, which generate $U(n)$. Reorder any product of matrix units into lower-triangular units, diagonal units and upper-triangular units, usingInduction on length and on the number of out-of-order pairs gives the required spanning version of the Poincaré-Birkhoff-Witt theorem. Upper-triangular units kill $v$, and diagonal units act by scalars. Every nonempty product of lower-triangular units lowers its weight by a nonzero sum of positive roots. Therefore the top weight space is exactly $\mathbb{C}v$, and its multiplicity is one. This supplies the leading coefficient without assuming a character formula.

For completeness, every dominant integer tuple occurs. Put $d_j={\lambda }_j-{\lambda }_{j+1}\ge 0$ and form the representationIts lexicographically highest weight is $\lambda$, with one-dimensional highest space, represented by the tensor products of $e_1\wedge \cdots \wedge e_j$. A negative determinant exponent is a legitimate one-dimensional unitary-group character. Averaging an inner product over normalized Haar measure gives invariant orthogonal complements, so the tensor representation splits into irreducible summands. One of them has highest weight $\lambda$. Uniqueness will also follow from the formula and Schur orthogonality relations.

Define the alternant $A_{\kappa }(z)=\det (z_i^{{\kappa }_j})$ for a strictly decreasing integer tuple $\kappa$. The denominatoris the Vandermonde determinant, and its modulus is the denominator modulus in the Weyl integration formula. The Laurent polynomial ${\chi }_{\lambda }{A}_{\delta }$ is alternating. Any alternating Laurent polynomial decomposes as a finite summonomials with repeated exponents have zero coefficient, and the remaining permutation orbits are indexed uniquely by strictly decreasing tuples.

The lexicographically highest monomial in the product is $z^{\lambda +\delta }$ with coefficient one. Indeed, the highest monomial of the character is $z^{\lambda }$ with coefficient one, and the unique lexical maximum among permutations of $\delta$ is $\delta$ itself. It follows that $c_{\lambda +\delta }=1$.

Fourier orthogonality on the torus, whose probability measure is ${\prod }_{i}d{\theta }_i/(2\pi )$, gives alternant orthogonality on a unitary torus:Expand both determinants: an integral survives exactly when the two exponent tuples are identical up to permutation. Strict decreasing order then forces $\kappa =\eta$, and its $n!$ matching terms each contribute one.

Finally, Schur orthogonality relations give ${\int }_{U(n)}|{\chi }_{\lambda }{|}^2=1$. Applying the allowed Weyl integration formula to this class function givesSince the coefficient $c_{\lambda +\delta }$ already equals one, every other coefficient vanishes. Thus ${\chi }_{\lambda }{A}_{\delta }=A_{\lambda +\delta }$, proving the formula. If two irreducible representations had the same highest weight, their characters would now coincide; the Schur orthogonality relations imply that they are isomorphic. Apparent singularities at repeated diagonal eigenvalues are removable because the quotient equals the actual finite Laurent-polynomial character. No dimension formula has been assumed.

The Kronecker delta is the identity map on the defining representation. Its transformation is $A^{\alpha }{}_{\gamma }{\delta }^{\gamma }{}_{\delta }(A^{-1}{)}^{\delta }{}_{\beta }={\delta }^{\alpha }{}_{\beta }$. For the alternating Levi-Civita symbol, the determinant identity givesThe lower-index alternating tensor transforms with three inverse matrices and acquires $\det (A^{-1})=1$ instead. Thus ${\delta }^{\alpha }{}_{\beta }$, ${ϵ}^{\alpha \beta \gamma }$ and ${ϵ}_{\alpha \beta \gamma }$ are invariant tensors of SU(3). These identities use unimodularity; unitarity identifies the dual representation with the complex-conjugate defining representation.

For a mixed SU(3) tensor representation $T^a{}_b$, the matrix trace is invariant. HencegivesThe trace is the singlet. The traceless matrix transforms by conjugation, so it is the adjoint representation of SU(3). It is irreducible: an invariant complex subspace is invariant under commutators with the complexified algebra ${\mathfrak{sl}}_3$, and would be an ideal in that simple Lie algebra. Only the zero and full traceless-matrix spaces are possible.

For the cubic tensor, define its symmetric and antisymmetric first-pair parts byand let $S^{abc}=T^{(abc)}$ be its completely symmetric part. This has $\left(\genfrac{}{}{0ex}{}{5}{3}\right)=10$ independent components and transforms as the symmetric cubic representation of SU(3). Its irreducibility can be seen in the homogeneous cubic-polynomial realization: the simultaneous highest-weight conditions $z_1{\partial }_{z_2}P=0$ and $z_2{\partial }_{z_3}P=0$ leave only $P\propto z_1^3$. A finite-dimensional unitary representation is completely reducible, and each irreducible summand has a highest-weight vector, so this single highest-weight line implies a single irreducible summand. Lowering generates all ten monomials.

Remove this fully symmetric part from $T_{+}$, putting $M^{abc}=T_{+}^{abc}-S^{abc}$. DefineIts trace vanishes because $M$ is symmetric in its first two indices. Invariance of $ϵ$ makes $O$ transform as one upper and one lower index. The inverse map isIndeed contraction with ${ϵ}_{ebc}$ gives $(2O^a{}_e+O^a{}_e-{\delta }^a{}_e\mathrm{tr}O)/3=O^a{}_e$. The inverse has the required first-pair symmetry and zero completely symmetric part. Thus this eight-dimensional remainder is another adjoint representation of SU(3).

For the antisymmetric part, defineNow decompose $N^c{}_d=U^c{}_d+{\delta }^c{}_d\mathrm{tr}N/3$, with $U$ traceless. The first term supplies a second independent octet; the trace supplies a singlet proportional to ${ϵ}^{abc}$. With $q={ϵ}_{abc}{T}^{abc}/6=\mathrm{tr}N/3$, the explicit cubic SU3 tensor projections reconstruct the entire tensor:All projections commute with the SU(3) action because they use permutations and invariant tensors. Their inverse maps and the dimension count $10+8+8+1=27$ establishThe two octets are equivalent as group representations but distinct subspaces in this decomposition; one comes from the first-pair symmetric sector and the other from the first-pair antisymmetric sector. General permutations can mix these equivalent copies.

In the quark model, the up quark, down quark and strange quark form the defining triplet of approximate flavor symmetry. Three-quark baryons therefore have precisely these possible flavor representations. The fully symmetric decuplet corresponds to the observed spin-$3/2$ baryon decuplet: the four Delta baryons, three ${\Sigma }^{\ast }$ states, two ${\Xi }^{\ast }$ states and the Omega baryon. Their strangeness rows have flavor hypercharge and isospin $(Y,I)=(1,3/2),(0,1),(-1,1/2),(-2,0)$; $uuu$ and $sss$ occupy the extreme ${\Delta }^{++}$ and ${\Omega }^{-}$ weights.

The observed spin-$1/2$ baryon octet contains the nucleons, three Sigma baryons, the Lambda baryon and two Xi baryons. In its $Y=0$ row, ${\Sigma }^0$ and $\Lambda$ share the same third-component/hypercharge weight but have different total isospin. Approximate flavor symmetry explains multiplet organization, not exact equality of all their masses: the strange-quark mass and electromagnetic interactions break it.

The two octet copies in the flavor tensor do not require two identical ground-state octets. A three-quark colour singlet is antisymmetric in color; for a symmetric ground-state orbital wavefunction, the Pauli exclusion principle requires a symmetric combined spin-flavor state. The mixed flavor octet tensors combine with the mixed spin-$1/2$ tensors to supply that symmetry. The symmetric flavor decuplet combines with symmetric spin $3/2$. Together they form the symmetric spin-flavour SU6 representation, with $10\times 4+8\times 2=56$ spin-flavor states. A completely antisymmetric flavor singlet cannot appear in this spatial ground state because three spin-$1/2$ indices have no fully antisymmetric spin state: ${\bigwedge }^3{\mathbb{C}}^2=0$. Three-quark flavour singlet configurations instead require a suitable excited spatial/spin symmetry. Flavor SU(3) here is distinct from the exact color gauge symmetry; treating them as the same group action would misidentify the physical multiplets.

For a Cartan-Weyl basis, the Jacobi identity givesThus the bracket has joint Cartan weight $\alpha +\beta$. If $\beta =-\alpha$, it belongs to the zero-weight space, which is the Cartan subalgebra, and so is a linear combination ${\tilde{\alpha }}_i{H}_i$. If $\alpha +\beta$ is a nonzero root, its root space is one-dimensional and the bracket is proportional to $E_{\alpha +\beta }$. If that nonzero sum is not a root, the bracket vanishes. This proves the asserted forms without assuming every pair of roots has a root sum.

With the stated normalization, put $H_{\alpha }=2{\alpha }_i{H}_i/{\alpha }^2$. The commutators becomeThe adjoint relations make $J_3=H_{\alpha }/2$, $J_{+}=E_{\alpha }$, $J_{-}=E_{-\alpha }$ the usual SU(2) Lie algebra generators. Hence every finite-dimensional SU(2) irreducible component has $J_3$ eigenvalues $m=-j,-j+1,\dots ,j$, with $2j$ an integer. Therefore $H_{\alpha }$ has integer eigenvalues $2m$, including in the finite-dimensional Adjoint representation. Algebraically this is the sl2 subalgebra associated with a root.

If $[H_{\alpha },X_{\lambda }]=\lambda X_{\lambda }$, another use of the Jacobi identity givesThe bracket is therefore zero or lies in the eigenvalue-$\lambda +2$ space. Within one irreducible SU(2) ladder it is proportional to the next chosen vector $X_{\lambda +2}$. If the full eigenspace has multiplicity, the weight statement is the general one; proportionality to one named vector presupposes the ladder choice. Similarly $E_{-\alpha }$ lowers this eigenvalue by two, and the ladder weights run from $-n$ to $n$ in steps of two.

For a root vector $E_{\beta }$, its $H_{\alpha }$ weight isThe adjoint SU(2) ladder therefore provesFor nonparallel $\alpha ,\beta$, raising and lowering $E_{\beta }$ move its joint Cartan weight through the root string $\beta +k\alpha$. The finite irreducible ladder has weights symmetric about zero. Starting at $m_{\alpha \beta }$, reaching its partner $-m_{\alpha \beta }$ requires $k=-m_{\alpha \beta }$. The corresponding nonzero vector lies in the root space of $\beta -m_{\alpha \beta }\alpha$, proving the Weyl reflection property:There is no zero joint root in this string when $\alpha ,\beta$ are nonparallel. Its one-dimensional root spaces prevent two irreducible ladders with the same weight parity from overlapping, so the ladder argument applies to the entire string. For $\beta =\pm \alpha$, reflection simply interchanges the two opposite roots, which already exist by the adjoint relation.

For angle and length restrictions, set $p=2\alpha \cdot \beta /{\alpha }^2$ and $q=2\alpha \cdot \beta /{\beta }^2$. These Cartan integers have the same sign when nonzero and obeyFor nonparallel, nonorthogonal roots, $pq$ is an integer strictly between zero and four. The possibilities are consequently:

Orthogonal roots have $p=q=0$ and $\theta =\pi /2$. Parallel roots have $pq=4$; the integer pairs would allow only length multiples $1,2,1/2$. The stated reducedness condition excludes the latter two, so parallel roots are equal or opposite, with angles zero or $\pi$ and equal lengths. ThusNot every root system realizes every listed angle.

The orthogonal pair alone gives no ratio through $q/p$. To finish the length restriction for a simple Lie algebra, use irreducibility of its root system. Let $U$ span the Weyl group orbit of $\alpha$. Both $U$ and $U^{\perp }$ are invariant under the orthogonal root reflections. If a root had nonzero projections in both, reflection of its projection in $U$ would have a nonzero $U^{\perp }$ component, violating invariance. Hence every root is entirely in one of these subspaces. A proper nonzero $U$ would split the root system into two orthogonal parts, contradicting irreducibility. This proves that the Weyl orbit spans an irreducible root space.

In particular, some orbit root $\gamma$ is nonorthogonal to any chosen $\beta$, and ${\gamma }^2={\alpha }^2$. Applying the previous integer-pair restriction to $\gamma ,\beta$ therefore also handles orthogonal $\alpha ,\beta$:There can be at most two different root lengths: three lengths would have ratios two and three relative to the shortest, but then the longest-to-middle squared-length ratio would be $3/2$, which is forbidden. A reducible collection of orthogonal rank-one factors could have independently chosen scales; the simple-algebra assumption excludes that exception.

Choose a generic linear functional to define positive roots. The simple roots are those positive roots which cannot be written as sums of two positive roots; they form a basis in which all root coefficients are integers of one sign. Distinct simple roots have nonpositive inner product: if their inner product were positive, their root string would contain their difference, and whichever difference is positive would decompose one of them. In rank two, irreducibility excludes the orthogonal case. The allowed simple-root angles are therefore $120^{\circ },135^{\circ },150^{\circ }$, giving the three irreducible diagrams in the classification of rank-two root systems: A2 root system, B2 root system and G2 root system. Reflections in the two simple roots generate their respective six, eight and twelve roots.

Here are explicit simple-root choices and positive roots; adding their negatives gives each complete diagram.

The nonnegative coefficients in each positive-root list verify the chosen simple roots explicitly. The orthogonal four-root system $A_1\times A_1$ is rank two but reducible, so it is not a diagram for the simple algebra assumed here.

Define the Cartan matrix in the column-coroot convention:Using the ordered roots above givesThe alternative row-coroot convention transposes these matrices. Stating the convention and root ordering avoids a spurious disagreement in the unequal-length cases.

Define the raw Killing form byFor a finite-dimensional real or complex Lie algebra in characteristic zero, it is nondegenerate exactly when the algebra is a semisimple Lie algebra. For a real compact semisimple algebra it is negative definite. Consequently the positive compact Killing metric is $g=-K$: for example on $\mathfrak{su}(2)$ with $[e_i,e_j]={\epsilon }_{ijk}{e}_k$, $K_{ij}=-2{\delta }_{ij}$ and $-K$ gives a Riemannian metric on $SU(2)$. An overall minus sign in the term “Killing metric” is common here; raw trace $K$ itself is negative definite in this example.

For a Lorentzian metric use $\mathfrak{sl}(2,\mathbb{R})$ with $[h,e]=2e$, $[h,f]=-2f$, $[e,f]=h$. Direct traces givewhose metric signature is $(2,1)$. For a non-Abelian degenerate example, the three-dimensional Heisenberg Lie algebra has $[x,y]=z$ central. Every product of two adjoint maps is zero, so $K=0$.

For any adjoint-invariant nondegenerate metric on a Lie group, the Koszul formula on left-invariant fields simplifies toTake $R(X,Y)Z={\nabla }_X{\nabla }_{Y}Z-{\nabla }_Y{\nabla }_{X}Z-{\nabla }_{[X,Y]}Z$. The Jacobi identity givesTaking the trace of $Z\mapsto R(Z,X)Y=-\mathrm{ad}Y\mathrm{ad}X(Z)/4$ yieldsThus the metric $g=K$ is a Killing-form Einstein metric with Einstein constant $-1/4$. With $g=-K$ the same connection has constant $+1/4$. This proves the Einstein manifold assertion for every semisimple group, including products when the unscaled Killing form is used on all factors.

There is also a different, flat metric-preserving connection. Declare the global left-invariant frame parallel:Its curvature is zero because applying two derivatives to coefficient functions and subtracting their commutator gives zero. Its metric derivative vanishes since $g_{ab}$ is constant in this frame. Its torsion tensor on invariant fields isFor each fixed $c$, antisymmetry gives $T_a{}^b{}_c=C_c{}^b{}_a$, so this is the matrix of $\mathrm{ad}{e}_c$. The torsion contractions of a left-parallel Lie-group connection therefore areThe zero trace follows because a semisimple algebra is perfect: every element is a sum of brackets, and the adjoint of a bracket is a matrix commutator with zero trace. If the metric convention is $g=-K$, the second contraction is $-g_{cd}$. These flat translation connections on a Lie group are not the torsion-free Levi-Civita connection; nonzero torsion permits flatness alongside the curved Killing metric.

Here the Cartan subalgebra of a finite-dimensional complex semisimple Lie algebra is a maximal toral subalgebra: it is abelian, its adjoint operators are simultaneously diagonalizable, and its centralizer is itself. These standard properties give the root-space decompositionThe roots of a root system $\alpha$ are the nonzero linear functionals in ${\mathfrak{h}}^{\ast }$ for which ${\mathfrak{g}}_{\alpha }\ne 0$; $\Phi$ is their finite set. Each root space is one-dimensional. The zero weight space is the centralizer of $\mathfrak{h}$, hence exactly $\mathfrak{h}$. Thus the displayed sum is a direct sum of simultaneous eigenspaces of the Adjoint representation. It is the Cartan root-space decomposition meant here.

The bracket respects weight addition: $[{\mathfrak{g}}_{\alpha },{\mathfrak{g}}_{\beta }]\subseteq {\mathfrak{g}}_{\alpha +\beta }$, where ${\mathfrak{g}}_0=\mathfrak{h}$ and a missing weight means the zero space. This follows by applying the Jacobi identity with $H\in \mathfrak{h}$ to the bracket of two root vectors. It explains how the vector-space decomposition encodes the Lie algebra structure.

For the embedded Lie group, the tangent space $T_{e}G$ consists of velocities ${\gamma }^{\prime }(0)$ of smooth curves in $G$ satisfying $\gamma (0)=e$. The ambient coordinates identify these velocities with vectors in ${\mathbb{R}}^n$, but the tangent space itself is intrinsic. Let $c_g(a)=ga{g}^{-1}$. Since this conjugation map fixes $e$, its derivative defines the Adjoint representation of a Lie group:The inverse is ${\mathrm{Ad}}_{g^{-1}}$, and $c_{gh}=c_g\circ c_h$ gives ${\mathrm{Ad}}_{gh}={\mathrm{Ad}}_g{\mathrm{Ad}}_h$. Thus this is a smooth group homomorphism into the linear automorphisms of $T_{e}G$.

Differentiate that representation at the identity. The tangent space of $GL(T_{e}G)$ at its identity is $\mathrm{End}(T_{e}G)$, so $\mathrm{ad}=(d\mathrm{Ad}{)}_e$ is a linear map from $T_{e}G$ into those endomorphisms. The Lie bracket isEquivalently, if $\eta (0)=e$ and ${\eta }^{\prime }(0)=Y$, this is the mixed derivative at $(0,0)$ of $\gamma (s)\eta (t)\gamma (s{)}^{-1}$ in ambient coordinates. For a matrix Lie group, conjugation is ordinary matrix conjugation and this formula reduces to $[X,Y]=XY-YX$.

The checks required are that the derivative is independent of the representing curves, that it is bilinear and alternating, and that it satisfies the Jacobi identity. One must also check that each ${\mathrm{Ad}}_g$ preserves this bracket. Equivalently identify $X$ with its left-invariant vector field $X^L(g)=(d{L}_g{)}_{e}X$; the bracket above agrees with the Lie bracket of vector fields at $e$. These checks make $T_{e}G$ a Lie algebra and upgrade the adjoint maps to its Lie algebra automorphisms.

Consider the orientation-preserving affine group of the real line, realized asThe coordinates $(\log a,b)$ identify it smoothly with ${\mathbb{R}}^2$, so this Lie group is connected. Choose the Lie algebra element $E$ and the group element $D$:Direct matrix multiplication gives ${\mathrm{Ad}}_{D}E=DE{D}^{-1}=2E$.

If a bi-invariant Riemannian metric existed, every conjugation map $x\mapsto Dx{D}^{-1}$ would be an isometry fixing the identity. Its differential there, the Adjoint representation of a Lie group, would preserve the positive-definite inner product at the identity. Thusa contradiction since $E\ne 0$. Therefore this connected Lie group admits no bi-invariant Riemannian metric. This is the adjoint dilation obstruction to a bi-invariant Riemannian metric, an elementary alternative permitted by the question. The positive definite Riemannian metric interpretation matters; no claim about indefinite metrics is needed.

For a connected smooth Riemannian manifold $(M,g)$ without boundary, let $I(M)$ be the group of all global diffeomorphisms $f$ with $f^{\ast }g=g$, endowed with its compact-open topology. Equivalently these maps preserve the intrinsic distance; the Myers-Steenrod theorem supplies smoothness of distance-preserving bijections and makes $I(M)$ a finite-dimensional Lie group acting smoothly. The group is metric-dependent: a topological manifold can carry a very symmetric metric or one with no continuous symmetries. Connectedness matters here. With infinitely many identical disconnected components, permutations of those components can produce a group which is not a finite-dimensional Lie group.