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Sylvester's law of inertia says that any real quadratic form has coordinates in which it is a sum of $p$ positive squares minus $m$ negative squares, with $z$ unused coordinates, and that the numbers $(p,m,z)$ are invariant under every invertible linear change of coordinates. Its rank is $p+m$; its signature of a quadratic form is commonly $p-m$, with $(p,m)$ also called its signature pair.

The symmetric matrix here is $J-I$, where $J=11^T$ and $1=(1,\dots ,1{)}^T$. On the line spanned by $1$, $J-I$ has eigenvalue $n-1$. If ${\sum }_i{x}_i=0$, then $Jx=0$, so that the orthogonal complement has eigenvalue $-1$ with multiplicity $n-1$. Thus the off-diagonal all-ones quadratic form hasFor $n=1$ the form vanishes, so rank and scalar signature are both zero, with inertia $(0,0,1)$. Including this case is necessary because the distinguished eigenvalue then becomes zero.

Suppose $v=(v_1,\dots ,v_r{)}^t$ belongs to the kernel of the displayed matrix $\Lambda$. Define the polynomial $q(z)={\sum }_{k=1}^r{v}_k{z}^k$. Then $q({\lambda }_j)=0$ for all $j$, and $q(0)=0$. These are $r+1$ distinct roots of a polynomial of degree at most $r$. Repeated application of the factor theorem therefore makes $q$ identically zero, so every $v_k=0$. Hence $\Lambda$ is injective and, by finite-dimensional rank-nullity theorem, invertible. Its transpose is invertible with inverse $({\Lambda }^{-1}{)}^t$, proving both kernels are trivial.

For the second matrix, membership of the all-ones vector in the kernel meansGroup the distinct nonzero values as ${\mu }_1,\dots ,{\mu }_r$, with multiplicities $m_1,\dots ,m_r>0$. Zero values contribute nothing to these positive powers. The first $r$ equations areor $\Lambda (\mu {)}^{t}m=0$. The proved transpose-kernel statement implies $m=0$, contradicting its positive entries if $r>0$. Therefore $r=0$ andConversely, all zero values plainly give a zero matrix. This proves the zero power sums force a finite complex multiset to vanish criterion, including repetitions and zero values rather than assuming all eigenvalues distinct.

Set $y=e^{-x/2}v$, the Dirichlet gauge transform for constant drift. Direct differentiation givesThe homogeneous endpoint conditions on $y$ become $v(0)=v(1)=0$. Thus the eigenvalue equation is $v^{\prime \prime }=(\lambda +1/4)v$. To exclude other branches without assuming the eigenvalue real, multiply by $\bar{v}$ and integrate by parts:for a nonzero eigenfunction. Therefore $\lambda +1/4=-k^2$ with $k>0$. The boundary conditions in $v=A\sin (kx)+B\cos (kx)$ give $B=0$ and $k=n\pi$, $n\ge 1$. HenceEquivalently, $L[y]=\lambda y$ is the Sturm-Liouville form $(e^x{y}^{\prime }{)}^{\prime }=\lambda e^{x}y$, with positive weight $e^x$. Its orthogonality relation isIt is the weighted inner product, not the unweighted one, that makes these eigenfunctions orthogonal.

Expand $x{e}^{-x/2}={\sum }_n{b}_n{y}_n$ with the coefficients from part a. Completeness and the nonzero eigenvalues give the Sturm-Liouville eigenfunction expansionThe coefficients are $O(n^{-3})$, so the function series converges uniformly and vanishes at both endpoints. Its first derivative also converges uniformly. Applying the transformed operator in the square-integrable sense gives the part-a expansion of $x$; equivalently $v^{\prime \prime }-v/4=x$ on the interior. Integrating this continuous right-hand side gives the classical twice-differentiable solution there. No nonzero homogeneous solution satisfies both endpoints, so this solution is unique. The mismatch of the forcing's sine series at a single endpoint does not affect the interior equation or these boundary conditions.

Order the states as sandwich $N$, Hall $H$, and own cooking $C$. The transition matrix, with current state indexing the row, isThe two probabilities $1/4$ in the first row are unconditional alternatives to the probability $1/2$ of returning. Let $a_n=(P^n{)}_{HH}$, so $n$ counts days after the initial Hall lunch. Expanding the determinant gives the characteristic polynomialThe three distinct eigenvalues imply that the sequence has the formusing the specified $0^0=1$. Directly, $a_0=1$, $a_1=1/3$ and $a_2=7/36$. The last two equations give $A=3/14$ and $B=-5/7$; then $C=3/2$. ConsequentlyThe limiting value is also the Hall entry of the stationary distribution $(4/7,3/14,3/14)$. The tiny finite-time correction is retained in the boxed exact probability. Sixty days later means sixty transitions, rather than the sixtieth day including the initial day.

An orthogonal transformation preserves the real inner product. If $W$ is an invariant subspace, its orthogonal complement is invariant too: $\alpha W=W$ in finite dimensions, and for $v\in W^{\perp }$, $w\in W$, we have $⟨\alpha v,w⟩=⟨v,{\alpha }^{-1}w⟩=0$.

Complexify the space and the transformation. There is a complex eigenvalue $\lambda$, and preservation of the Hermitian norm gives $|\lambda |=1$. If $\lambda$ is real, it is $1$ or $-1$, and the real kernel of $\alpha -\lambda I$ contains a nonzero vector. Its span is an invariant line on which $\alpha$ acts as $+1$ or $-1$.

If $\lambda$ is nonreal, write an eigenvector as $u+iv$. Its real and imaginary parts are linearly independent: otherwise a nonzero real vector would have the nonreal eigenvalue. The real plane $W=\mathrm{span}(u,v)$ is invariant, because comparison of real and imaginary parts in $\alpha (u+iv)=\lambda (u+iv)$ gives real combinations of $u,v$. The restriction has determinant $\lambda \bar{\lambda }=1$. In an oriented orthonormal basis of $W$, it is therefore a rotationThere is no invariant real line in this plane: such a line would supply a real eigenvalue, whereas its two eigenvalues are nonreal.

Apply the argument repeatedly to the invariant orthogonal complement. Thus the smallest mutually orthogonal invariant blocks are lines with action $\pm 1$, and planes with a genuine rotation. Planes with angle $0$ or $\pi$ can be decomposed further into lines.

In dimension three a suitably chosen orthonormal basis therefore gives eitherorChanging the orientation of the plane reverses the angle, so this angle range loses no cases. These exhaust rotations, reflections and their possible compositions in three dimensions.

Work in finite dimension over $\mathbb{C}$; the characteristic-zero real version follows by complexification. The Cartan solvability criterion isIts semisimplicity consequence, also often called Cartan's criterion, is $\mathfrak{g}$ is semisimple if and only if its Killing form is nondegenerate. We prove both forms.

The auxiliary results used are as follows. The Lie theorem simultaneously triangularizes every finite-dimensional representation of a complex solvable Lie algebra. The Engel theorem says that a matrix Lie algebra consisting entirely of nilpotent endomorphisms has a common annihilated nonzero vector and a basis making all its matrices strictly upper triangular; in particular it is a nilpotent Lie algebra. Nilpotent means that the lower central series terminates, which implies that the derived series terminates as well. Finally, the Additive Jordan decomposition writes any complex endomorphism uniquely as $x=x_s+x_n$, where $x_s$ is diagonalizable, $x_n$ is nilpotent and they commute. On a generalized eigenspace $V_{\lambda }$, the two parts are $\lambda I$ and $x-\lambda I$, and both are polynomials in $x$. On endomorphisms, $\mathrm{ad}{x}_s$ is diagonalizable, $\mathrm{ad}{x}_n$ is nilpotent, and these are the Jordan parts of $\mathrm{ad}x$: the first assertion follows by splitting into maps between eigenspaces, and the second by expanding repeated commutators with the nilpotent $x_n$.

First prove the linear trace version: if $L\subseteq \mathfrak{gl}(V)$ andthen $L$ is solvable. Fix $x\in [L,L]$ and let $V=⨁V_{\lambda }$ be its generalized eigenspace decomposition. Define $y$ to act on $V_{\lambda }$ by the scalar $\overline{\lambda }$. We do not assume that $y$, $x_s$ or $x_n$ belongs to $L$.

On $\mathrm{Hom}(V_{\mu },V_{\lambda })$, $\mathrm{ad}x$ has the form $(\lambda -\mu )I+N$ with $N$ nilpotent, whereas $\mathrm{ad}y$ is multiplication by $\overline{\lambda -\mu }$. Choose a polynomial $q$ with $q(d)=\bar{d}$ for every occurring eigenvalue difference $d$, and with all its derivatives of positive order up to the relevant nilpotency index zero at $d$. Such a polynomial exists by Hermite interpolation, or the Chinese remainder theorem for the pairwise coprime powers of $(t-d)$. At $d=0$ its value is zero, so $q(0)=0$. Expanding $q(dI+N)$ now givesBecause $x\in L$ and $[L,L]$ is an ideal, every positive power of $\mathrm{ad}x$ sends $L$ into $[L,L]$. The zero constant term therefore gives $[y,L]\subseteq [L,L]$.

Express $x$ as a sum of commutators ${\sum }_j[a_j,b_j]$ with $a_j,b_j\in L$. Cyclicity of trace and the hypothesis implyBut on $V_{\lambda }$, $xy$ has trace $\dim (V_{\lambda })|\lambda {|}^2$, since the nilpotent part has trace zero. Henceforcing every eigenvalue of $x$ to vanish. Thus every member of the derived algebra $[L,L]$ is a nilpotent endomorphism. By the Engel theorem, $[L,L]$ is nilpotent and therefore solvable; its derived series terminates, and adjoining the initial term $L$ shows that $L$ is solvable.

Conversely, if $L$ is solvable, the Lie theorem makes it upper triangular. Commutators of upper-triangular matrices have zero diagonal, and multiplying such a commutator by an upper-triangular matrix still has zero diagonal. Therefore the trace condition holds. Apply this equivalence to $L=\mathrm{ad}\mathfrak{g}$. Its trace pairing is the Killing form, and solvability of its image is equivalent to solvability of $\mathfrak{g}$: the kernel is its abelian centre, so a terminating derived series in the quotient terminates after at most one more step in $\mathfrak{g}$. This proves the solvability criterion.

For the semisimplicity criterion, let $R$ be the radical of the Killing form. Invariance makes $R$ a Lie algebra ideal. For $x,y\in R$, their adjoint actions on $\mathfrak{g}/R$ vanish, so computing trace in a basis adapted to $R$ gives $B_R(x,y)=B_{\mathfrak{g}}(x,y)=0$. The solvability criterion applied to $R$ shows it is a solvable ideal. Thus a semisimple Lie algebra has $R=0$ and a nondegenerate Killing form.

Conversely, any nonzero solvable ideal has a last nonzero term $A$ in its derived series; this is an abelian ideal of $\mathfrak{g}$. For $x\in A$, $y\in \mathfrak{g}$, the map $\mathrm{ad}x\mathrm{ad}y$ sends $\mathfrak{g}$ into $A$ and vanishes on $A$, so has trace zero. Hence $A\subseteq R$. A nondegenerate Killing form therefore rules out nonzero solvable ideals, exactly the definition of semisimplicity.

For classification, nondegeneracy of the Killing form provides the duality and root-space pairings used to extract a reduced crystallographic root system from a semisimple Lie algebra. Classification then reduces to the finite Dynkin diagrams and reconstruction of the Lie algebra from its root data.

Put $H_{12}=\mathrm{diag}(1,-1,0)$ and $H_{23}=\mathrm{diag}(0,1,-1)$. On the root space spanned by $E_{ij}$, their adjoint eigenvalues are $h_i-h_j$ and $k_i-k_j$; both act as zero on the Cartan subalgebra. Thus the Killing form is the sum of their products over the six roots. The pair of roots $\pm (L_1-L_2)$ contributes $-4$, the pair $\pm (L_2-L_3)$ contributes $-4$, and $\pm (L_1-L_3)$ contributes $2$. HenceThis also agrees with the Killing form of the special linear Lie algebra, $B(A,B)=6\mathrm{tr}(AB)$ for ${\mathfrak{sl}}_3$.

The displayed identity is valid for a highest-weight vector $v$, satisfying $Xv=0$ and $Hv=kv$, not for arbitrary $v\in V$. For example, in the defining module with $k=1$, take $v=y$: then $XYv=0$ but the proposed right side for $n=1$ is $v\ne 0$.

For a highest-weight vector, the sl2 triple relations give $H{Y}^{j}v=(k-2j)Y^{j}v$. The operator identitycomes from expanding a commutator with a product. Apply it to $v$, using $Xv=0$. Each term becomes $(k-2(n-1-j))Y^{n-1}v$, so the sum isThis proves the intended sl2 highest-weight lowering formula. The identity valid for an arbitrary vector is instead

On the basis $v,Yv,\dots ,Y^{k}v$ of the irreducible module, $XY$ has eigenvalue $(j+1)(k-j)$ on $Y^{j}v$, for $0\le j\le k$. Summing gives the trace of a raising-lowering product in an irreducible sl2 module:The last equality follows by the formulas for the sums of the first $k$ integers and their squares.

For $G=U(n)$ take the diagonal maximal torusThe spectral theorem for unitary matrices says that every element is conjugate to an element of $T$. A diagonal element with pairwise distinct eigenvalues has centralizer exactly $T$. Its conjugates therefore determine its eigenvalues up to permutation, and the normalizer quotient is $W=N_G(T)/T\cong S_n$. The character lattice of a torus $T$ is ${\mathbb{Z}}^n$, with $e^{\lambda }(z)=z_1^{{\lambda }_1}\cdots z_n^{{\lambda }_n}$. Conjugation on the matrix unit $E_{ij}$ has weight $e_i-e_j$, so these are the roots. Choose $e_i-e_j$ positive for $i<j$; thenDominant integral weights are precisely the integer tuples ${\lambda }_1\ge \cdots \ge {\lambda }_n$. Negative entries are allowed: they correspond to determinant twists, so polynomial representations alone do not exhaust the irreducible representations of $U(n)$.

Normalize Haar measure on $G$ and $T$ to have mass one, and putThe Weyl integration formula for U(n) for a continuous class function isFor a general continuous function, replace $f(t)$ in the torus integral by ${\int }_{G/T}f(gt{g}^{-1})d(gT)$, using the normalized invariant quotient measure.

To prove the formula, consider the conjugation map $q:G/T\times T\to G$, $q(gT,t)=gt{g}^{-1}$. On regular torus elements its fibres have exactly $n!$ points: the possible orderings of the distinct eigenvalues. Equip the group with an invariant inner product on its Lie algebra and split $\mathfrak{u}(n)=\mathfrak{t}\oplus {\mathfrak{t}}^{\perp }$. After translating the derivative back to the identity, its torus directions contribute the identity, while its transverse directions contribute ${\mathrm{Ad}}_{t^{-1}}-I$. Each pair $i<j$ supplies a real two-dimensional off-diagonal plane. On that plane ${\mathrm{Ad}}_t$ is multiplication by $z_i/z_j$, regarded as a plane rotation, so the absolute real determinant of the transverse derivative isThe change-of-variables theorem, divided by the covering multiplicity, now gives the desired weighted integral up to the constant relating the normalized invariant measures. The nonregular locus has measure zero: repeated eigenvalues are the zero-set of the nonzero discriminant, and the torus coincidence sets likewise have measure zero. Thus its omission causes no change to the integral. To determine the normalization, expand $\Delta$ as a sum of $n!$ distinct torus characters. Fourier orthogonality gives ${\int }_T|\Delta {|}^{2}dt=n!$. Applying the formula to $f=1$ fixes the coefficient to $1/n!$. This proves both the class-function and conjugacy-averaged forms.

We can now derive the character formula from integration, rather than merely quote it. Use the basic highest-weight theorem: a finite-dimensional irreducible representation has a unique highest weight $\lambda$ with a one-dimensional highest-weight space; all its other weights are $\lambda$ minus a nonnegative sum of positive roots. Its character is Weyl invariant. Also, character orthogonality for compact groups gives ${\int }_G|{\chi }_{\lambda }{|}^2=1$. This orthogonality follows by averaging the representation on $\mathrm{End}(V_{\lambda })$: the integral projects onto intertwining operators, whose dimension is one by the Schur lemma.

Let $\delta =(n-1,n-2,\dots ,0)$ and, for a strictly decreasing integer tuple $\eta$, put $A_{\eta }(z)=\det (z_i^{{\eta }_j})$. The finite Laurent polynomial $\Delta {\chi }_{\lambda }$ is alternating. Every alternating Laurent polynomial is a linear combination of the $A_{\eta }$: collect its monomials by permutation orbits, and observe that an orbit with repeated exponents has zero coefficient by antisymmetry. Different $A_{\eta }$ are orthogonal on $T$ and each has squared norm $n!$.

The coefficient of $z^{\lambda +\delta }$ in $\Delta {\chi }_{\lambda }$ is exactly one. Indeed, a contribution from the term $z^{w\delta }$ would require a weight $\lambda +\delta -w\delta$ in $V_{\lambda }$. For $w\ne 1$, the vector $\delta -w\delta$ is a nonzero nonnegative combination of the positive simple roots, so that weight lies above $\lambda$ and is impossible. The identity permutation supplies the highest-weight coefficient one. ConsequentlyApply the integration formula to $|{\chi }_{\lambda }{|}^2$. Torus orthogonality givesAll the other coefficients vanish, provingThe quotient extends across repeated eigenvalues because its product with the denominator is the already defined continuous character. Equivalently, alternant divisibility makes it a Laurent polynomial invariant under permutations; multiply by a determinant power if negative exponents must first be removed.

Finally, $\delta -\rho =\frac{n-1}{2}(1,\dots ,1)$ is fixed by the Weyl group, so its common factor cancels in the quotient. Since $A_{\rho }=e^{\rho }{\prod }_{\alpha >0}(1-e^{-\alpha })$, the determinant expression is exactlyUsing $\delta$ in the derivation avoids treating a possibly half-integral $\rho$ as an actual character of the torus. At the identity, taking the leading Vandermonde term of the two determinants gives the accompanying dimension formula

There are two necessary qualifications to the literal formulation. The whole group $G$ is itself a closed normal subgroup, so the claim about containment in the centre must concern proper closed normal subgroups. Also, the homomorphism conclusion needs a nontrivial target $H$: for example, $SU(2)\to \{1\}$ has surjective differential onto the zero Lie algebra, but its group kernel is all of $SU(2)$ and no quotient by a finite subgroup is a one-point group. We prove the precise assertions, including these exceptional cases.

Write $\mathfrak{g}=\mathrm{Lie}(G)$. The centre $\mathfrak{z}(\mathfrak{g})$ is invariant under the Adjoint representation, so irreducibility makes it either zero or all of $\mathfrak{g}$. In the latter case $\mathfrak{g}$ is Abelian. For connected $G$, the adjoint action is then trivial: the group is generated by exponentials, and ${\mathrm{Ad}}_{\exp X}=e^{\mathrm{ad}X}=I$. A nonzero trivial representation is irreducible only in dimension one. A connected compact one-dimensional Lie group is a circle, so $G\cong U(1)$.

For every other $G$, the Lie algebra of its centre is zero. The centre is a closed compact Lie subgroup; a zero-dimensional compact Lie group is finite. Let $N$ be a closed normal subgroup. The closed-subgroup theorem identifies $\mathfrak{n}=\mathrm{Lie}(N)$ as an adjoint-invariant subspace of $\mathfrak{g}$, hence $\mathfrak{n}=0$ or $\mathfrak{n}=\mathfrak{g}$. If the latter holds, $N$ contains an identity neighbourhood and is open, so connectedness forces $N=G$. If $N$ is proper, it therefore has zero Lie algebra and is finite. For any $a\in N$, the continuous map $g\mapsto ga{g}^{-1}$ takes connected $G$ into the finite set $N$, so it is constant and equal to $a$. ThusProper closed subgroups of the circle are finite as well, and of course central.

Now let $d\Phi$ be surjective. The submersion theorem makes $\Phi (G)$ contain an identity neighbourhood in $H$, so it is an open subgroup. Since $H$ is connected, $\Phi (G)=H$. Its group kernel $F$ is closed and normal. If $H$ is nontrivial, $F$ is proper and the preceding argument gives a finite central group kernel; this also holds for $G\cong U(1)$, whose proper closed subgroups are finite. The Lie-group first isomorphism theorem givesIf $H$ is trivial, the correct statement is instead $H=G/G$.

For the Spin groups, use the real Clifford algebra with $v^2=-\Vert v{\Vert }^2$. The group $\mathrm{Spin}(m)$ consists of products of an even number of unit vectors inside its invertible even part. Conjugation on the embedded ${\mathbb{R}}^m$ defines the standard double covering $\mathrm{Spin}(m)\to SO(m)$ with group kernel $\{\pm 1\}$. The Clifford relations show that a unit-vector reflection, with the usual twisted conjugation, is a hyperplane reflection; the even products give orientation-preserving transformations. The Cartan–Dieudonné theorem states that every orthogonal transformation is a product of reflections, proving surjectivity. For $m\ge 3$ this connected covering group is simply connected; $\mathrm{Spin}(2)$ is a circle.

Here is an explicit exterior-square construction of the requested six-dimensional map. Put $E={\mathbb{C}}^4$ with its standard Hermitian form and volume $e_1\wedge e_2\wedge e_3\wedge e_4$. On ${\Lambda }^{2}E$, write $e_{ij}=e_i\wedge e_j$ and define an antilinear map $J$ byIt satisfies $J^2=1$. Intrinsically, if $u\wedge v=B(u,v)\mathrm{vol}$ and the Hermitian inner product $h$ is linear in its first variable, then $B(u,Jv)=h(u,v)$. Because $SU(4)$ preserves both $B$ and $h$, it commutes with $J$. This is the real structure of the SU4 exterior square.

The fixed space $W$ has the following real orthonormal basis, with all vectors divided by $\sqrt{2}$:Thus $W$ is six-dimensional over $\mathbb{R}$, its complexification is ${\Lambda }^{2}E$, and the exterior-square action restricts to an orthogonal representation on $W$. Its determinant is one because $SU(4)$ is connected. We obtain $R:SU(4)\to SO(W)\cong SO(6)$.

To compute the group kernel, diagonalize $g\in SU(4)$ with eigenvalues $z_1,\dots ,z_4$. If $R(g)=I$, its complexified exterior-square action is the identity, hence $z_i{z}_j=1$ for every $i<j$. Comparing pairs shows that all $z_i$ are equal, and their common value has square one. A unitary matrix with that single eigenvalue is scalar, so $\ker R=\{\pm I_4\}$. Conversely, those two scalar matrices plainly act trivially on the exterior square.

The differential of $R$ is injective because its group kernel is finite. Both Lie algebras have dimension $15$, so the differential is an isomorphism. Its image is consequently open in connected $SO(6)$, hence all of $SO(6)$. ThereforeThis also realizes $SU(4)$ as $\mathrm{Spin}(6)$, since $SU(4)$ is simply connected, by uniqueness of the connected simply connected covering group.

For $v=(v_1,v_2,v_3)$ define the skew-symmetric matrixThen $\hat{v}x=v\times x$. This is a linear isomorphism ${\mathbb{R}}^3\to \mathfrak{so}(3)$, with inverse $\varphi (\xi )=({\xi }_{32},{\xi }_{13},{\xi }_{21})$. Since rotations preserve the cross product,Thus $\varphi$ intertwines the Adjoint representation with the standard representation. The vector triple-product identity gives, for every $x$,Consequently $[\hat{v},\hat{w}]=\widehat{v\times w}$ and the chosen normalization satisfiesNo undetermined scale remains in this explicit matrix convention.

Identify $SU(2)$ with the unit quaternions usingThe relations $jz=\overline{z}j$ and $j^2=-1$ verify multiplication, and every matrix in $SU(2)$ has this form. If $v$ is imaginary, then $\overline{v}=-v$. Quaternionic conjugation reverses product order and $q^{-1}=\overline{q}$, soThus conjugation preserves $\mathrm{Im}\mathbb{H}\cong {\mathbb{R}}^3$. Multiplicativity of the quaternion norm shows that this action preserves the inner product. The unit quaternions form the connected three-sphere, so its determinant, equal to one at the identity, is always one. We obtain a homomorphism $C:SU(2)\to SO(3)$.

Its group kernel consists of unit quaternions commuting with every imaginary quaternion. Commuting with both $i$ and $j$ forces a quaternion to be real, so the group kernel is exactly $\{\pm 1\}$. To prove surjectivity explicitly, let $u$ be a unit imaginary quaternion and put $q=\cos (\theta /2)+u\sin (\theta /2)$. For imaginary $u,v$, multiplication satisfies $uv=-u\cdot v+u\times v$. Expanding $qv{q}^{-1}$ therefore givesThis is the Rodrigues rotation formula about axis $u$. Every element of $SO(3)$ has such an axis-angle description: an odd-dimensional orthogonal matrix of determinant one has an eigenvector of eigenvalue one, and its action on the perpendicular plane is a plane rotation. Hence $C$ is onto, andThis is the quaternionic form of the Adjoint double cover from SU(2) to SO(3). Its differential sends an imaginary quaternion $u$ to $2\hat{u}$, because $[u,v]=2u\times v$, consistent with the bracket normalization used above.

Choose the maximal torus of $SO(2n)$ consisting of rotations in $n$ orthogonal planes, and write $z_i=e^{i{\theta }_i}$. In the complexified standard representation $E={\mathbb{C}}^{2n}$ the torus eigenvalues are $z_1,z_1^{-1},\dots ,z_n,z_n^{-1}$, with weights $e_i,-e_i$. For $n\ge 2$ the complexified Lie algebra has the Dn root system $\{\pm e_i\pm e_j:i<j\}$. The root-space decomposition theorem splits this Lie algebra into its Cartan subalgebra and one-dimensional root spaces. The Cartan subalgebra contributes the zero weight with multiplicity $n$. Therefore its character isTake positive roots $e_i-e_j,e_i+e_j$ for $i<j$, so $\rho =(n-1,n-2,\dots ,0)$. For $n>2$ the root $\theta =e_1+e_2$ is a highest weight: its root vector is killed by every positive root vector, since $\theta +\alpha$ is never a root for positive $\alpha$. To verify irreducibility without relying on an unstated simplicity theorem, apply complete reducibility and compare dimensions. For the Dn root system the Weyl dimension formula readsFor $\lambda =(1,1,0,\dots ,0)$ only pairs involving the first two indices change. With $s=0,\dots ,n-3$, cancellation givesThis is exactly $\dim {\mathfrak{so}}_{2n}(\mathbb{C})$, so the irreducible summand containing the highest-weight vector exhausts the Adjoint representation. Thus for $n>2$ the complexified Adjoint representation is irreducible, with highest weight $e_1+e_2$.

For $n=2$, the roots split into the two independent systems $\{\pm (e_1+e_2)\}$ and $\{\pm (e_1-e_2)\}$. The Adjoint representation splits into two irreducibles of dimension three, with respective highest weights $(1,1)$ and $(1,-1)$. Equivalently these are the self-dual and anti-self-dual exterior two-forms in four dimensions. Their characters areThey add to the displayed adjoint character. This agrees with the Chiral decomposition of the complexified so4 Lie algebra.

For a direct character comparison, put $s(z)={\sum }_i(z_i+z_i^{-1})$. The eigenvalue formula for an exterior square givesThis is precisely ${\chi }_{\mathrm{ad}}$. More intrinsically, let $B$ be the nondegenerate symmetric bilinear form preserved by $SO(2n)$ and defineThis endomorphism is skew with respect to $B$, so $\Psi$ maps ${\Lambda }^{2}E$ to $\mathfrak{so}(E,B)$. In an orthonormal basis the bivectors $e_i\wedge e_j$ map to a basis of skew skew-symmetric matrices, proving that it is an isomorphism. Since $B$ is invariant,Thus the exterior square and the complexified Adjoint representation are naturally isomorphic, explaining the character equality without a weight calculation.

The symmetric square is not irreducible. The inverse metric tensor $\Omega$ is a nonzero invariant vector, and contraction with $B$ splits off its trivial line:The contraction sends $\Omega$ to $2n$, so this is indeed a direct sum. A highest-weight vector $f_1\otimes f_1$, where $f_1$ has torus weight $e_1$, lies in the traceless part because $B(f_1,f_1)=0$, and has highest weight $2e_1$. The same dimension formula gives, for $n\ge 2$,This equals $\dim {\mathrm{Sym}}^{2}E-1=n(2n+1)-1$. Complete reducibility therefore proves that the traceless part is irreducible, including the dimension-nine case for $SO(4)$. In particular,If the circle case $n=1$ is included, the adjoint character is $1$ and the symmetric square instead has three one-dimensional weights $2,0,-2$.

Take the nonnegative scalar Laplace-Beltrami operator $\Delta =-\mathrm{div}\nabla$ on a closed compact Riemannian manifold, and include every eigenvalue with its multiplicity. The spectrum contains substantial global geometric information, but generally does not determine the Riemannian metric up to isometry. Specifying the operator and its domain matters: the spectrum on functions is different information from the spectra of the Hodge Laplacian on differential forms, and a boundary requires fixed boundary conditions.

The Weyl law makes dimension and volume audible. If $N(\lambda )$ counts eigenvalues at most $\lambda$, thenwhere ${\omega }_d$ is the volume of the Euclidean unit ball. Thus the growth exponent recovers $d$, and its leading coefficient recovers $\mathrm{Vol}(M)$. The multiplicity of zero gives the number of connected components: a function in the kernel has $\int |\nabla f{|}^2=0$ and hence is constant on each component.

More detailed invariants come from the heat trace, which is determined directly by the spectrum:The first heat invariants for a closed manifold areHere $R$ is scalar curvature, $\mathrm{Ric}$ the Ricci curvature, and $\mathrm{Rm}$ the Riemann curvature tensor. A divergence term in the local coefficient integrates to zero on a closed manifold. The expansion supplies further integrated curvature expressions, rather than a pointwise curvature map.

On a closed surface, $R=2K$, $|\mathrm{Ric}{|}^2=2K^2$ and $|\mathrm{Rm}{|}^2=4K^2$, so the same coefficients becomeThe Gauss-Bonnet theorem gives $\int KdA=2\pi \chi (M)$; hence the constant term of the heat trace is $\chi (M)/6$. In particular the Euler characteristic, and for a connected orientable surface its genus, are spectrally determined. There is even spectral determination of hyperbolic curvature on a surface: if this spectrum agrees with that of a curvature-$-1$ surface, the coefficients give $\int K=-A$ and $\int K^2=A$, whenceThus $K=-1$ everywhere. This determines local curvature, while leaving room for globally nonisometric hyperbolic surfaces.

Geodesic information is also accessible in important settings. For closed hyperbolic surfaces, the Selberg trace formula makes the scalar spectrum equivalent to the unmarked length spectrum with its appropriate multiplicities. This does not supply a marked length spectrum: it does not identify which group element or intersecting curve has a given length. For general manifolds, one must not simply assert this equivalence; the singularities of the wave trace can be affected by degeneracies and cancellations.

The historic development separates these positive results from the inverse problem. Weyl's 1911 eigenvalue asymptotics established the leading geometric information. The Minakshisundaram-Pleijel heat expansion of 1949 brought local curvature into the short-time analysis. In 1964 Milnor exhibited nonisometric flat tori in dimension sixteen with identical scalar spectra, showing that all eigenvalues together need not recover the metric. Kac's 1966 question about a vibrating drum popularized the corresponding inverse problem for planar domains. Vignéras's 1980 hyperbolic examples and the Sunada theorem of 1985 extended the supply of counterexamples; the latter turns Gassmann equivalence of finite subgroups into a geometric construction. The four-dimensional tetracode length-isospectral lattice construction and the planar examples of Gordon, Webb and Wolpert in 1992 made nonuniqueness explicit in smaller dimensions and in the original drum setting.

Nor is scalar isospectrality a universal topological invariant strong enough to determine the manifold. For example, Tetra and Didi are scalar-isospectral closed orientable flat three-manifolds with first Betti numbers one and zero. Their topology therefore differs. The spectra of all Hodge Laplacians would detect these Betti numbers through the dimensions of their zero eigenspaces, illustrating the extra information obtained by changing the operator. Dimension, volume and many integrated invariants are determined; the full metric and, in general, the topology are not.