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A Lie algebra representation $V$ is completely reducible when it is a direct sum of irreducible submodules. In finite dimension this is equivalent to saying that every submodule $W\subseteq V$ has an invariant complementary subspace. Indeed, one may split off a minimal nonzero submodule and induct on dimension; conversely, if $V$ is a finite sum of simple submodules, choose a maximal sum $U$ of these with $U\cap W=0$. If $W+U\ne V$, some simple summand $S$ is not contained in $W+U$. Then $S\cap (W+U)=0$ by simplicity, so $U+S$ is a larger sum disjoint from $W$, a contradiction. Thus $V=W\oplus U$.

We prove the required splitting by Haar averaging produces invariant complements. The structural input is the compact real form of a complex semisimple Lie algebra: there is a real semisimple algebra $\mathfrak{k}$ with $L=\mathfrak{k}\oplus i\mathfrak{k}$, whose simply connected integrating Lie group $K$ is compact. This structural fact is independent of the complete-reducibility assertion. One standard construction uses root vectors normalized so that $[e_{\alpha },e_{-\alpha }]=h_{\alpha }$, with real structure constants compatible with the involution $e_{\alpha }\mapsto -e_{-\alpha }$ and $h_{\alpha }\mapsto -h_{\alpha }$. The real spanis closed under brackets, complexifies to $L$, and has negative-definite Killing form. The compactness criterion for real semisimple Lie algebras then gives a compact simply connected $K$. Thus this route uses root structure and compact integration, not the theorem being proved.

Restrict $\rho :L\to {\mathrm{End}}_{\mathbb{C}}(V)$ to $\mathfrak{k}$. The integration of a Lie-algebra representation gives a representation $R:K\to {\mathrm{GL}}_{\mathbb{C}}(V)$, since $K$ is simply connected. Choose any positive-definite Hermitian inner product $⟨,{⟩}_0$ and average using normalized Haar measure:The integral exists by compactness. It is positive-definite because $R(g)v\ne 0$ for $v\ne 0$, and it is $K$-invariant by translation invariance of Haar measure. Equivalently, every $\rho (a)$ with $a\in \mathfrak{k}$ is skew-Hermitian for this inner product.

Let $W$ be an $L$-submodule. It is invariant under $\mathfrak{k}$, hence under the exponentials generating the connected group $K$. For $v\in W^{\perp }$, $w\in W$ and $g\in K$,Consequently the orthogonal complement $W^{\perp }$ is $K$-invariant, and differentiation makes it $\mathfrak{k}$-invariant. It is a complex vector subspace, so it is also invariant under $i\mathfrak{k}$ and hence under $L$. Therefore $V=W\oplus W^{\perp }$ as $L$-modules.

Induction on $\dim V$ now expresses $V$ as a finite direct sum of irreducible submodules. Every finite-dimensional representation of a finite-dimensional complex semisimple Lie algebra is completely reducible. The compact real form need not be the real form originally used to present $L$; in particular the averaging argument does not require the given presentation to be unitary.

The connection difference as an endomorphism-valued one-form follows because the graded Leibniz rule terms cancel:Thus $a=D_1-D$ is a global endomorphism-valued one-form. To define the endomorphism bundle connection, requireThe right side is $C^{\infty }$-linear in $s$, so it is indeed an endomorphism-valued one-form, and replacing $\varphi$ by $f\varphi$ gives $df\otimes \varphi +f\tilde{D}\varphi$. In a frame with connection matrix $A$, its extension to an endomorphism-valued $k$-form $b$ isThe endomorphism-valued exterior product combines the wedge of differential forms with composition in $\mathrm{End}E$. In particular,Expanding $d(A+a)+(A+a)\wedge (A+a)$ proves the curvature difference formula:

The Chern connection is characterized by metric compatibility and $D^{0,1}={\bar{\partial }}_E$. Take Hermitian inner products linear in the first variable. In a holomorphic local frame write $⟨eu,ev⟩=v^{†}Hu$, with $H$ a positive Hermitian matrix. The second condition forces $A^{0,1}=0$, and metric compatibility isIts $(1,0)$ component determines $A$ uniquely:Conversely this formula and its conjugate transpose satisfy the metric equation. Under a holomorphic local frame change, $H^{\prime }=g^{†}Hg$ and $\partial g^{†}=0$, so the displayed formula transforms by $A^{\prime }=g^{-1}Ag+g^{-1}\partial g$. Hence the local matrices patch, proving existence and uniqueness.

Differentiating $H^{-1}$ givesTherefore the $(2,0)$ vector-bundle curvature is zero. There is no $(0,2)$ component, andThe last equality is ${\bar{\partial }}^{2}A=0$ in holomorphic local frames, so is intrinsic. This is Dolbeault closedness of Chern curvature. The induced endomorphism bundle connection is itself the Chern connection for the Hilbert-Schmidt metric on $\mathrm{End}E$: the tensor product of a metric-compatible connection and its dual preserves the induced metric, and its $(0,1)$ part is the induced holomorphic Dolbeault operator. These two properties and uniqueness suffice; no further matrix calculation is needed.

For two metrics, both Chern connections have the same $(0,1)$ part, so $a\in A^{1,0}(\mathrm{End}E)$. The $(1,1)$ part of the curvature difference formula is just ${\bar{\partial }}_{\mathrm{End}E}a$; the other two terms have type $(2,0)$, which must cancel because both vector-bundle curvatures have type $(1,1)$. ThusBy Dolbeault theorem, this proves that the curvature Dolbeault classis independent of the metric.

Endomorphism composition and wedge product of differential forms define the powers in $A^{k,k}(\mathrm{End}E)$, rather than in a tensor power of the endomorphism bundle. The Dolbeault operator is a derivation, so $\bar{\partial }({\Theta }^k)=0$. The noncommutative telescoping identity gives the stronger explicit independence statementAll powers of vector-bundle curvature have even total degree, so the derivation introduces no extra signs before $a$. Consequentlyis metric independent. For $k>n$, the form space and the resulting class are zero. This proves metric independence of curvature Dolbeault powers without assuming that endomorphism-valued forms commute.

Let ${\mathcal{A}}^{p,q}$ denote the sheaf of differential forms of type (p, q) printed in the PDF. Take the pointwise Hermitian inner product to be linear in its first argument, and use $\mathrm{vol}={\omega }^n/n!$. The bidegrees in the question specify the conjugate-linear Hodge star: it is uniquely characterized byThus $\ast (c\beta )=\bar{c}\ast \beta$, and $\ast$ maps bidegree $(p,q)$ to $(n-p,n-q)$. If $S$ is the complex-linear extension of the real Hodge star operator, then $\ast =S\circ \text{complex conjugation}$. The operator $S$ is real and $S^2=(-1{)}^{k(2n-k)}$ on degree $k=p+q$, soUsing the complex-linear star instead would give a different bidegree, $(n-q,n-p)$; keeping the convention explicit prevents that ambiguity.

The Hermitian Lefschetz operator is exterior multiplication by $\omega$. Define $\Lambda$ pointwise byIt lowers bidegree by $(1,1)$. In a unitary real coframe with $\omega ={\sum }_j{e}^j\wedge f^j$, it is $\Lambda ={\sum }_j{\iota }_{f_j}{\iota }_{e_j}$, the corresponding sum of interior product of a differential form operators. This gives its existence and identifies it as a smooth operator. On a compact manifold, integrate the pointwise equality against $\mathrm{vol}$ to obtainNo integration by parts is needed for this order-zero operator. Thus $\Lambda$ is both the pointwise and the global formal adjoint.

Write $[A,B]=AB-BA$, and let $k$ denote the degree of the input form. The given Lefschetz commutator is $[L,\Lambda ]=(k-n)I$. For $r=1$ the required formula is exactly this identity. If it holds for $r$, the commutator derivation identity givesThe second commutator acts on degree $k+2r$, so the coefficient on the right isThis proves the commutator formula for powers of the Lefschetz operator

For injectivity of powers of the Lefschetz operator, $r=0$ is immediate. Suppose $r\ge 1$ and $r+k\le n$. If $k=0$ or $1$, then $\Lambda \psi =0$, and $L^r\psi =0$ impliesThe scalar is nonzero, since $k+r\le n$ makes $k-n+r-1\le -1$. Repeating with the smaller power shows $\psi =0$. For $k\ge 2$, induct on $r+k$, treating all $r=0$ cases as already established. The same commutator calculation yields, with $c=r(k-n+r-1)\ne 0$,The induction hypothesis applies to $L^{r-1}$ on degree $k$, since $(r-1)+k<r+k$ and $(r-1)+k\le n$. Hence $\psi =L\alpha$ with $\alpha =c^{-1}\Lambda \psi$ of degree $k-2$. Moreover $L^{r+1}\alpha =L^r\psi =0$. Since $(r+1)+(k-2)=r+k-1$, induction also makes $L^{r+1}$ injective on that degree, so $\alpha =0$ and $\psi =0$. All operators preserve restriction to open sets; the argument applies on every open set. Therefore $L^r$ is an injective morphism of sheaves whenever $r+k\le n$.

On a Kähler manifold, the Kähler identities, with the positive form convention used above, areTaking formal adjoints conjugates the scalar and reverses the order in the commutator. Since ${\Lambda }^{\ast }=L$, this givesTo derive the Kähler Laplacian identity, put $P=\partial$ and $B=\bar{\partial }$, so $P^2=B^2=0$ and $PB+BP=0$. Write $\{A,B\}=AB+BA$ for an anticommutator. The Kähler identities say $B^{\ast }=-i[\Lambda ,P]$ and $P^{\ast }=i[\Lambda ,B]$. Thereforebecause expansion leaves only terms containing $P^2$ or $B^2$. Furthermore,Expanding these expressions and using $PB=-BP$ makes them equal. Since $d=P+B$ and $d^{\ast }=P^{\ast }+B^{\ast }$, the mixed anticommutators already vanish, and consequentlyClosedness of the Kähler form implies $[L,P]=[L,B]=0$. Using the adjoint identity above,The equality of the three Laplacians therefore proves each Laplacian commutes with $L$.

For a compact Kähler manifold, the Dolbeault Hodge decomposition is the orthogonal decompositionHere harmonicity for the Dolbeault Laplacian and for the Hodge Laplacian agrees by the preceding identity. If $\bar{\partial }\alpha =0$ and $\alpha =h+\bar{\partial }u+{\bar{\partial }}^{\ast }v$, then $\bar{\partial }{\bar{\partial }}^{\ast }v=0$. Taking its inner product with $v$ gives $\Vert {\bar{\partial }}^{\ast }v{\Vert }^2=0$. Thus $\alpha$ is cohomologous to $h$. Conversely, a harmonic form is $\bar{\partial }$-closed, and a harmonic exact form $h=\bar{\partial }u$ satisfies $\Vert h{\Vert }^2=({\bar{\partial }}^{\ast }h,u)=0$. This proves existence and uniqueness of the harmonic representative and henceThese spaces are finite dimensional by the ellipticity of the Dolbeault Laplacian on the compact manifold.

Complex conjugation commutes with the real operator ${\Delta }_d$ and interchanges the bidegrees $(p,q)$ and $(q,p)$. It is therefore a conjugate-linear bijection of the harmonic spaces, proving Hodge symmetry. The real Hodge star operator commutes with ${\Delta }_d$, as does complex conjugation; hence their composition, our conjugate-linear Hodge star, sends harmonic $(p,q)$-forms bijectively to harmonic $(n-p,n-q)$-forms. Its square is the nonzero scalar established above. This proves Hodge duality, and gives

For the final hard Lefschetz isomorphism on Dolbeault cohomology, take $k=p+q\le n$ and $r=n-k$. This restriction is necessary to make the displayed power nonnegative. The Hermitian Lefschetz operator raises bidegree by $(1,1)$, and the Lefschetz operator preserves harmonic forms; thereforeis well defined. It is injective by injectivity of powers of the Lefschetz operator, since $r+k=n$. Hodge symmetry and Hodge duality give $h^{n-q,n-p}=h^{q,p}=h^{p,q}$, so it is a bijection between finite-dimensional spaces of equal dimension. Because $[L,\bar{\partial }]=0$, the map on harmonic representatives agrees with exterior multiplication by $[\omega {]}^{n-k}$ on Dolbeault cohomology. We concludeFor $p+q>n$, the corresponding valid statement is the inverse of the positive-power isomorphism from bidegree $(n-q,n-p)$ to $(p,q)$; a negative exterior-multiplication power is not defined.

View a possible complex eigenvector $v$ using the Hermitian inner product. For a real symmetric matrix, $A^{†}=A$, so $v^{†}Av$ is real. Since $v^{†}Av=\lambda v^{†}v$ and $v^{†}v>0$, every eigenvalue is real.

For eigenvectors $v,w$ with distinct eigenvalues $\lambda ,\mu$, symmetry givesso $v^{†}w=0$. In particular, real eigenvectors are orthogonal in the real dot product.

Use the permitted diagonalizability assumption to take bases of the real eigenspaces whose union spans ${\mathbb{R}}^n$. Apply the Gram-Schmidt process within each eigenspace. Its linear combinations remain eigenvectors of that same eigenvalue; vectors in different eigenspaces are already orthogonal. The resulting union is an orthonormal basis of eigenvectors, giving an orthogonal diagonalization of a real symmetric matrix.

Let $e_1,\dots ,e_n$ be the coordinate vectors and chooseThese vectors lie in the sum-zero standard representation of the symmetric group. They are mutually orthogonal for the usual Hermitian inner product, with $\Vert v_j{\Vert }^2=j(j+1)$, so they form a basis of its $(n-1)$-dimensional space.

If $i\le j$, every transposition in $X_i$ exchanges two coordinates equal to $1$, so each fixes $v_j$. If $i=j+1$, summing the $j$ transpositions gives coordinate $-1$ in each of positions $1,\dots ,j$ and coordinate $j$ in position $j+1$, hence $X_{j+1}{v}_j=-v_j$. Finally, if $i>j+1$, the $i$th coordinate of $v_j$ is zero. In the sum of the $i-1$ transpositions, any coordinate in its support retains its original value in $i-2$ terms and becomes zero in the remaining term; the $i$th output coordinate is the sum of the original first $i-1$ coordinates, which is zero.

Thus all actions are diagonal in this basis:This supplies an explicit Gelfand–Tsetlin basis of the standard representation.

The joint spectrum $\mathrm{Spec}(n)$ consists of the tuples of simultaneous eigenvalues of $X_1,\dots ,X_n$ occurring in irreducible complex $S_n$-modules. A Gelfand–Tsetlin basis is a Gelfand–Tsetlin basis: its vectors lie in the one-dimensional joint eigenspaces selected by successive restrictions. The weight of $v$ is $\alpha =(a_1,\dots ,a_n)$ when $X_{j}v=a_{j}v$ for every $j$.

Define $\mathrm{Cont}(n)$ to be the integer tuples satisfying the content vector of a standard Young tableau conditions: $a_1=0$; for $j>1$, at least one of $a_j-1,a_j+1$ occurs earlier; and between any two occurrences of the same integer $a$, both $a-1$ and $a+1$ occur. We will justify their tableau interpretation in part (iv).

Here is the local spectral rules for Young–Jucys–Murphy elements calculation used in this question. Average a positive definite Hermitian inner product over $S_n$, making the representation unitary. The $X_j$ are commuting self-adjoint operators because transpositions are self-adjoint involutions. Put $s_i=(ii+1)$ and $d=a_{i+1}-a_i$. The group-algebra relations areFor a unit weight vector $v$, the first relation gives $d⟨v,s_{i}v⟩=1$. Thus $d\ne 0$, $|d|\ge 1$, andhas the weight obtained by swapping $a_i,a_{i+1}$, with $\Vert w{\Vert }^2=1-d^{-2}$. If $d=\pm 1$, then $w=0$ and $s_{i}v=\pm v$. If $|d|>1$, then $w\ne 0$, providing an admissible spectral swap in the same irreducible module.

For the proposed three consecutive weights $(a,a+1,a)$, these rules give $s_{i}v=v$ and $s_{i+1}v=-v$. But the braid relation in a Coxeter group would implya contradiction. Hence the proposed tuple is not in $\mathrm{Spec}(n)$. The same argument, with the signs reversed, excludes $(a,a-1,a)$ as well.

For a standard Young tableau $T$, let $c_T(r)=j-i$ when its entry $r$ lies in cell $(i,j)$. Its content vector of a standard Young tableau is $C(T)=(c_T(1),\dots ,c_T(n))$. We use the following sign convention for axial distance in a Young tableau:This is the axial distance from $j$ to $i$. In particular, write $d=c_T(r+1)-c_T(r)$ for the adjacent transposition $s_r=(rr+1)$.

Here is Young orthogonal form, with one orthonormal basis vector $w_T$ for each standard Young tableau of a fixed shape:If $s_{r}T$ is not standard, omit that term. In this case $r,r+1$ occupy adjacent cells in one row or column, so $d=1$ or $d=-1$, respectively, and the scalar action is $+1$ or $-1$. On a standard pair $T,R=s_{r}T$, $|d|\ge 2$ and the matrix isThis is a real orthogonal involution.

To prove the formula, explicitly use the permitted Young seminormal form theorem: the Specht module has a tableau basis $v_T$ whose vectors are joint eigenvectors of the Young–Jucys–Murphy elements, and whose actions, on an admissible pair oriented so that tableau length increases from $T$ to $R$, areNonadmissible swaps have the scalar actions just described. The theorem supplies these as one globally consistent representation, not as independently chosen two-dimensional blocks.

Average a positive definite Hermitian inner product over $S_n$. Distinct tableau lines are orthogonal because their commuting self-adjoint Young–Jucys–Murphy elements have distinct joint weights. Also $s_r$ is self-adjoint, so the two seminormal equations implyUpon setting $w_T=v_T/\Vert v_T\Vert$, the coefficient from $w_T$ to $w_R$ becomes $\Vert v_R\Vert /\Vert v_T\Vert =\sqrt{1-d^{-2}}$; the reverse coefficient is the same. All phases are consistent because the seminormal basis was global, and the group relations remain true under this change of basis. This proves Young orthogonal form.

For $\lambda =(n)$, there is exactly one standard Young tableau, the row $1,2,\dots ,n$. Its content vector of a standard Young tableau isEvery adjacent distance is $1$, so $s_r{w}_T=w_T$ for every $r$. Thus $\boxed{S^{(n)}\cong {\mathbb{C}}_{\mathrm{triv}}}$ is the trivial representation.

Here is an intrinsic definition that also proves well-definedness. For a complex line bundle $L$, define its First Chern class to be the Euler class of its canonically oriented underlying real rank-two bundle: pull its integral Thom class back along the zero section after forgetting relative support. The complex orientation fixes the sign, so this construction makes no arbitrary choice of generator.

For a rank-$r>0$ complex vector bundle $E$, let $\pi :\mathbb{P}(E)\to X$ be its projective bundle of lines, let $S\subset {\pi }^{\ast }E$ be the tautological bundle, and put $h=-c_1(S)=c_1(S^{\ast })$. On every fiber ${\mathbb{CP}}^{r-1}$, $h$ is the positive degree-two generator. The Leray-Hirsch theorem says that if globally defined cohomology classes restrict to a free basis of the cohomology of every fiber, then multiplication by those classes identifies the cohomology of the total space with a free module over the base. Applied here, it givesThe finite trivializing cover in the question is sufficient for this application: the assertion holds on each trivializing open set by the Künneth theorem, since the fiber has finite free cohomology, and the Mayer–Vietoris sequence and the Five lemma glue it over the finite cover. The same local argument constructs the oriented Thom class used for line bundles. It does not require a choice of classifying map.

There are therefore unique classes $c_i(E)\in H^{2i}(X;\mathbb{Z})$ such thatDefine $c_0(E)=1$ and $c_i(E)=0$ for $i>r$. For rank zero the Total Chern class is $1$. This is the projective bundle definition of Chern classes. Existence and uniqueness follow by expressing $h^r$ in the displayed free module basis, with degrees determining each coefficient. The projective bundle and tautological bundle are intrinsic to $E$, and the line Thom class is uniquely fixed by orientation. Hence the resulting Chern classes do not depend on trivializations or other auxiliary choices. For a line bundle the relation is $h+c_1(E)=0$, agreeing with the original normalization. Pulling back this unique relation also proves naturality. This standard construction and the sum theorem are treated in Vector Bundles and K-Theory, Section 3.1.

The requested result is the Whitney sum formula for Chern classes:Here $c(E)=1+c_1(E)+\cdots +c_r(E)$, and the formula holds for complex vector bundles over a common base. In particular, a trivial bundle has total Chern class $1$.

Now take $B={\mathbb{CP}}^{n-1}$ and let $L\subset B\times {\mathbb{C}}^n$ be its tautological bundle. The standard Hermitian inner product gives the rank-$(n-1)$ complex bundle $Q=L^{\perp }$, with $L\oplus Q\cong {\underline{\mathbb{C}}}^n$. The orthogonal complex line flag manifold in the question is precisely $\mathbb{P}(Q)$: over a first line $\ell$, the second line is any line in ${\ell }^{\perp }$. Local orthonormal frames give this identification as a fiber bundle, with fiber ${\mathbb{CP}}^{n-2}$.

Let $x=c_1(L^{\ast })$ on $B$, also writing $x$ for its pullback to $\mathbb{P}(Q)$. By part 3(a), $H^{\ast }(B;\mathbb{Z})=\mathbb{Z}[x]/(x^n)$. The Whitney sum formula for Chern classes givesLet $S_2$ be the second tautological line on $\mathbb{P}(Q)$, and set $y=c_1(S_2^{\ast })$. Thus $x,y$ are exactly the pullbacks of the positive hyperplane classes from the two projective factors. The projective bundle definition of Chern classes for $Q$ givesTogether with $x^n=0$, this gives a surjective graded ring mapThere are no additional relations. Indeed the second relation is monic of degree $n-1$ in $y$, so polynomial division makes its source free over $\mathbb{Z}[x]/(x^n)$ on $1,y,\dots ,y^{n-2}$. The projective bundle formula for complex vector bundles gives exactly the same free basis on the target. The map sends each basis element to its corresponding basis element, and is therefore an isomorphism. ConsequentlyThis is the cohomology ring of the orthogonal complex line flag manifold. Its additive basis is $x^i{y}^j$ with $0\le i<n$, $0\le j<n-1$. As a symmetry check, multiplying the second relation by $y-x$ gives $y^n-x^n=0$, so $y^n=0$ as expected from the second projection. For $n=2$, every line has a unique orthogonal line, and the relations become $x^2=0$, $x+y=0$, giving the ring of ${\mathbb{CP}}^1$. If $n=1$, the space is empty and the second relation is $1=0$, so the printed formula still gives the zero cohomology ring.

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.

Let $\omega$ be the Kähler form associated with the metric. The Lefschetz operator of a Kähler manifold and adjoint Lefschetz operator areHere the adjoint is taken with respect to the pointwise Hermitian inner product on complexified forms, or equivalently the global $L^2$ inner product. It lowers degree by two and bidegree by $(1,1)$. In terms of the Hodge star operator, $\Lambda ={\ast }^{-1}L\ast$.

Since $d\omega =0$, $dL=Ld$, so $L$ takes closed differential forms to closed differential forms and exact differential forms to exact differential forms. This already gives its action on de Rham cohomology. For both operators together, use the Kähler identities: $L$ commutes with the Hodge Laplacian, and taking adjoints shows that $\Lambda$ does too. Thus both preserve harmonic differential forms.

When $M$ is compact, the Hodge decomposition theorem identifies $H^k(M,\mathbb{C})$ with its space of harmonic $k$-forms. If $h_{\alpha }$ is the unique harmonic differential form representing $[\alpha ]$, defineThe resulting forms are harmonic and therefore closed, and uniqueness of $h_{\alpha }$ makes these definitions independent of the initial representative. The cohomological definition of $\Lambda$ uses harmonic differential forms as representatives: $\Lambda$ need not take an arbitrary closed form to a closed form.

Work over $\mathbb{R}$ or $\mathbb{C}$, in finite dimension and characteristic zero. The Killing form of a Lie algebra is the trace form of its Adjoint representation:It is bilinear and symmetric by the cyclic property of the trace. It is an invariant bilinear form on a Lie algebra, since the Jacobi identity gives ${\mathrm{ad}}_{[x,y]}=[{\mathrm{ad}}_x,{\mathrm{ad}}_y]$ and henceEquivalently, $\kappa ([z,x],y)+\kappa (x,[z,y])=0$. A Lie algebra automorphism preserves the form, because it conjugates the adjoint matrices. Its radical of a bilinear form ${\mathfrak{r}}_{\kappa }=\{x:\kappa (x,\mathfrak{g})=0\}$ is an ideal of a Lie algebra by invariance. The center of a Lie algebra lies in this radical, so the form vanishes for an abelian algebra. On a direct sum of ideals the summands are orthogonal and the form restricts to their own Killing forms.

A short argument proves the requested implication without assuming the Cartan criterion for semisimplicity. Let $\mathfrak{a}$ be an abelian ideal of a Lie algebra, $x\in \mathfrak{a}$, and $y\in \mathfrak{g}$. The map ${\mathrm{ad}}_x$ takes $\mathfrak{g}$ into $\mathfrak{a}$ and vanishes on $\mathfrak{a}$, while ${\mathrm{ad}}_y$ preserves $\mathfrak{a}$. Therefore ${\mathrm{ad}}_x{\mathrm{ad}}_y$ has image in $\mathfrak{a}$ and is zero on $\mathfrak{a}$. In a basis extending a basis of $\mathfrak{a}$, both diagonal blocks are zero, so its trace vanishes. This proves that abelian ideals lie in the radical of the Killing form.

If the solvable radical $\mathfrak{r}$ were nonzero, its derived series of a Lie algebra would have a last nonzero term $\mathfrak{a}$. The Jacobi identity makes every derived term an ideal of $\mathfrak{g}$, and the last one is abelian. Thus $0\ne \mathfrak{a}\subseteq {\mathfrak{r}}_{\kappa }$, contradicting nondegeneracy. We obtainThe converse holds as well in characteristic zero; together these implications are the Cartan criterion for semisimplicity.

For a complex simple Lie algebra, that criterion gives nondegeneracy on $\mathfrak{g}$. Let $\mathfrak{h}$ be a Cartan subalgebra. Use the standard root-space decompositionFor $h,h^{\prime }\in \mathfrak{h}$ and a root vector $e_{\alpha }$, choose $t\in \mathfrak{h}$ with $\alpha (t)\ne 0$. Invariance givesThus $\mathfrak{h}$ is orthogonal to every nonzero root space. If $h\in \mathfrak{h}$ is also orthogonal to $\mathfrak{h}$, it is orthogonal to all of $\mathfrak{g}$ and hence is zero. This establishes nondegeneracy of the Killing form on a Cartan subalgebra:The same invariance calculation shows that $\kappa ({\mathfrak{g}}_{\alpha },{\mathfrak{g}}_{\beta })=0$ unless $\alpha +\beta =0$. Opposite root spaces are therefore paired nondegenerately. Tracing the adjoint action on the root-space decomposition yieldssince the root spaces of a complex semisimple algebra are one-dimensional and its adjoint action on $\mathfrak{h}$ is zero.

The relevant Euclidean subspace of a Cartan subalgebra iswhere the $h_{{\alpha }_i}$ are the simple coroots, normalized by ${\alpha }_i(h_{{\alpha }_i})=2$. The standard Euclidean property is that $\kappa$ is real and positive definite on this space. The displayed trace formula explains it: $\kappa (h,h)$ is a sum of real squares, and the roots span ${\mathfrak{h}}^{\ast }$, so all squares vanish only for $h=0$. Via the nondegenerate form, the roots can consequently be regarded as vectors in a real Euclidean normed vector space. The induced dual inner product makes root reflections orthogonal and allows root lengths and angles to be encoded by Cartan integers and the Dynkin diagram.

This positivity is on a specified real subspace; the complex Killing form is a bilinear form, not a positive Hermitian inner product. On a compact real form the Killing form is instead negative definite. For example the Killing form of the special linear Lie algebra gives $\kappa (X,Y)=4\mathrm{tr}(XY)$ for ${\mathfrak{sl}}_2$, so $\kappa (L_0,L_0)=8$ but $\kappa (i{L}_0,i{L}_0)=-8$.

A Hermitian metric on a holomorphic vector bundle $E\to X$ is a smoothly varying family of positive-definite Hermitian forms $h_x$ on its fibers. A Chern connection is a connection on a vector bundle $\nabla$ that is compatible with $h$ and whose $(0,1)$ part is the bundle's Dolbeault partial connection, ${\nabla }^{0,1}={\bar{\partial }}_E$.

Choose a local holomorphic frame $e$ and write $h=h(e,e)>0$. The connection form in this frame isIf $e^{\prime }=ge$ for a nowhere-zero holomorphic function $g$, then $h^{\prime }=|g{|}^{2}h$ andwhich is exactly the connection-form transformation law. The local formulas therefore define a global connection.

Identify $\mathcal{O}(-1)$ with the tautological bundle over Complex projective space. The standard Hermitian inner product of ${\mathbb{C}}^{n+1}$ restricts to each tautological line. On the affine chart $U_i=\{Z_i\ne 0\}$, put $w_j=Z_j/Z_i$ for $j\ne i$ and use the holomorphic frameThenThe curvature form of a connection isConsequently Chern-Weil theory gives the closed representativewhere ${\omega }_{FS}$ is the Fubini-Study form.