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Use the convention that the Hermitian inner product is conjugate-linear in its first argument and linear in its second. In orthonormal coordinates, $⟨v,w⟩={\sum }_i\overline{v_i}{w}_i$. It satisfies conjugate symmetry, sesquilinearity and positive definiteness. Expanding in the orthonormal basis givesApplying the Hermitian inner product and $⟨e_i,e_j⟩={\delta }_{ij}$ yieldsThis identifies exactly where complex conjugation enters; replacing the first factor by $⟨e_i,v⟩$ would generally be wrong under this convention.

Adopt the convention that an inner product is linear in its second slot. Over $\mathbb{R}$, it is a symmetric bilinear form with $⟨v,v⟩\ge 0$, equality only for $v=0$. Over $\mathbb{C}$, a Hermitian inner product obeysThese imply conjugate linearity in the first slot: $⟨au+bv,w⟩=\overline{a}⟨u,w⟩+\overline{b}⟨v,w⟩$. The strict positivity condition distinguishes an inner product from a general Hermitian form.

The characteristic polynomial factors as $\det (\lambda I-A)=(\lambda -1)({\lambda }^2-1)$. Thus $\boxed{\lambda =1\text{ twice},\lambda =-1\text{ once}}$ in algebraic multiplicity. An orthonormal basis of corresponding eigenvectors isDirect multiplication verifies $A{v}_1=v_1$, $A{v}_2=v_2$, $A{v}_3=-v_3$. Their mutual Hermitian inner products vanish and each has norm one, which proves linear independence. For the repeated eigenvalue, Gaussian elimination givesHence $\boxed{\mathrm{rank}(A-I)=1,\dim \ker (A-I)=2}$. The second equality follows from the rank-nullity theorem and agrees with the two-dimensional eigenspace. A repeated eigenvalue alone would not imply a two-dimensional eigenspace for an arbitrary matrix; here the Hermitian matrix is diagonalizable.

For real orthonormal column vectors, $u^{T}u=v^{T}v=1$ and $u^{T}v=v^{T}u=0$. Expand using associativity of matrix multiplication:HenceMore precisely, it is the orthogonal projection onto the span of $u,v$: it fixes both vectors and annihilates every vector orthogonal to both.

The transpose convention presumes real vectors, or the explicitly stated bilinear identities above. For complex vectors orthonormal in the Hermitian inner product, the correct orthogonal projection is $u{u}^{†}+v{v}^{†}$. The printed transpose formula need not work in that setting: $u=(i,0{)}^T$ and $v=(0,1{)}^T$ are Hermitian-orthonormal, but $u{u}^T+v{v}^T=\mathrm{diag}(-1,1)$ has square $I_2$, which is different from itself.

Let $Av=\lambda v$ with $v\ne 0$, using the standard complex Hermitian inner product. ThenTaking the complex conjugate and using $A^{†}=-A$ yields $\overline{\lambda }{v}^{†}v=v^{†}{A}^{†}v=-\lambda v^{†}v$. Since $v^{†}v>0$,Thus every eigenvalue of an anti-Hermitian matrix lies on the imaginary axis, including the possible zero eigenvalue; each nonzero eigenvalue is purely imaginary.

Now take $Au=\lambda u$ and $Av=\mu v$. On one hand $u^{†}Av=\mu u^{†}v$. On the other handConsequently $(\mu -\lambda )u^{†}v=0$. Distinct eigenvalues therefore implywhich is the required orthogonality of the eigenvectors.

For a compact Lie group $G$, the complex representation ring of a compact group $R(G)$ is the Grothendieck group of finite-dimensional continuous complex group representations, with the relations $[M\oplus N]=[M]+[N]$. Multiplication is $[M][N]=[M\otimes N]$, and the unit is the trivial one-dimensional representation. The ring of class functions iswith pointwise addition and multiplication. The character of $M$ is ${\chi }_M(g)=\mathrm{tr}{\rho }_M(g)$. The trace identities for direct sums and tensor products makea well-defined ring homomorphism $\chi :R(G)\to c\ell (G)$.

Normalize Haar measure by ${\int }_{G}dg=1$. Averaging a positive definite Hermitian inner product giveswhich is positive definite and $G$-invariant. This proves unitarization of a compact-group representation. In a unitary representation the orthogonal complement of an invariant subspace is invariant. Induction on dimension therefore gives complete reducibility of compact-group representations. Hence every element of $R(G)$ is a finite integral linear combination of irreducible classes.

We establish the needed character orthogonality for compact groups carefully. For irreducible unitary representations $V,W$, let $G$ act on ${\mathrm{Hom}}_{\mathbb{C}}(W,V)$ byIts average $P={\int }_G(g\cdot )dg$ is a projection onto ${\mathrm{Hom}}_G(W,V)$: averaging makes every image invariant, and it fixes every intertwiner. The trace of this action is ${\chi }_V(g)\overline{{\chi }_W(g)}$, since the inverse of a unitary matrix has conjugate trace. Taking the trace of the projection givesFor completeness, Schur's lemma follows here from invariance of the kernel and image of an intertwiner. A nonzero intertwiner between irreducibles is an isomorphism. An endomorphism of an irreducible complex representation has an eigenvalue $\lambda$; the noninvertible intertwiner $A-\lambda I$ must vanish. Thus the last dimension is one for isomorphic irreducibles and zero for inequivalent ones.

If a virtual character ${\sum }_V{n}_V{\chi }_V$ vanishes, integrating it against $\overline{{\chi }_W}$ gives $n_W=0$ for every irreducible $W$. The irreducible multiplicities therefore distinguish the classes, andThis also proves that two finite-dimensional representations with the same character are isomorphic.

For SU(2), the irreducibles are $V_k={\mathrm{Sym}}^k({\mathbb{C}}^2)$, $k\ge 0$, of dimension $k+1$. One can see this classification through the usual ${\mathfrak{sl}}_2$ operators $H,E,F$: a highest-weight vector $v$ has $Hv=kv$, $Ev=0$, and the relations $[H,E]=2E$, $[H,F]=-2F$, $[E,F]=H$ give, by induction,The torus weights are integers. If $F^{r}v\ne 0$ but $F^{r+1}v=0$, the identity at $j=r+1$ forces $k=r\ge 0$. The vectors $v,Fv,\dots ,F^{k}v$ span an invariant irreducible module, which in an irreducible representation is the whole space. This is the symmetric-power module. Conversely its successive monomial weight vectors are linked by $E,F$ with nonzero coefficients, proving its irreducibility. Thus the classification of finite-dimensional representations of SU2 gives every irreducible, rather than just a list of examples.

On the maximal torus $T=\{\mathrm{diag}(z,z^{-1}):|z|=1\}$, their characters areWith $x={\chi }_1=z+z^{-1}$, multiplication gives ${\chi }_{k+1}=x{\chi }_k-{\chi }_{k-1}$, starting with ${\chi }_0=1$. Hence each ${\chi }_k$ is a monic integral polynomial of degree $k$ in $x$. For example ${\chi }_2=x^2-1$ and ${\chi }_3=x^3-2x$. They form a basis over $\mathbb{Z}$, soThe character map realizes this ring as the finite integral polynomials in $z+z^{-1}$, interpreted as continuous class functions. Its image is not the whole infinite-dimensional ring of continuous class functions; injectivity is the assertion being proved.

Use the finite-dimensional complex-module setting of the course. We prove the Weyl complete reducibility theorem by Haar averaging produces invariant complements, defining its constructions rather than invoking complete reducibility itself.

Choose a compact real form $\mathfrak{k}\subset \mathfrak{g}$: it is a real Lie subalgebra with $\mathfrak{g}=\mathfrak{k}\oplus i\mathfrak{k}$, and its simply connected Lie group $K$ is compact. Existence of this form and compact group is a standard fact for complex semisimple Lie algebras. Restrict the representation to $\mathfrak{k}$; integration of a Lie-algebra representation gives a group representation $R:K\to GL(V)$. Since $K$ is connected and generated by exponentials, the $\mathfrak{g}$-invariant subspace $W$ is also $K$-invariant.

Choose any positive-definite Hermitian inner product on $V$. With the Haar measure of $K$ normalized to mass one, defineThis is a Hermitian inner product, and it is positive-definite: for $v\ne 0$ every $R(k)v$ is nonzero, so the continuous positive integrand has positive integral. Right invariance of Haar measure gives $⟨R(a)v,R(a)w{⟩}_K=⟨v,w{⟩}_K$ for $a\in K$.

Let $W^{\prime }=W^{\perp }$ for this averaged inner product. Finite-dimensional inner-product linear algebra gives $V=W\oplus W^{\prime }$. If $v\in W^{\prime }$ and $w\in W$, thenbecause $R(a^{-1})w\in W$. Thus $W^{\prime }$ is $K$-invariant. Differentiating makes it $\mathfrak{k}$-invariant, and since it is a complex vector subspace, $\mathfrak{g}=\mathfrak{k}+i\mathfrak{k}$ also preserves it. ThereforeThe zero or whole-space submodule cases are included. This proves the required complement theorem without circularly assuming it.

The finite-dimensional hypothesis is essential to the statement. The nonsplit zero-weight Verma extension for sl2 is a counterexample for unrestricted modules: $M(0)$ has basis $v_n$ with $h{v}_n=-2n{v}_n$, $f{v}_n=v_{n+1}$ and $e{v}_n=n(1-n)v_{n-1}$. Its submodule spanned by $v_n$ for $n\ge 1$ has trivial one-dimensional quotient. A complement would be trivial and contained in the zero-weight line $\mathbb{C}{v}_0$, yet $f{v}_0=v_1\ne 0$. Hence no such complement exists in that infinite-dimensional case. The paper's finite-dimensional representation setting supplies the necessary hypothesis.

Interpret $\theta$ as a continuous finite-dimensional complex group representation, so its values lie in ${\mathrm{GL}}_{\mathbb{C}}(V)$, and interpret approximation in the uniform norm. Assume the usual Hausdorff convention for the compact topological group. The following argument proves the required form of the Peter-Weyl theorem without first assuming that $G$ is a Lie group.

Start with normalized Haar measure. It is both left- and right-invariant on a compact group. Average any Hermitian inner product to make a finite-dimensional representation unitary. The left regular action $L_{a}h(x)=h(a^{-1}x)$ is already unitary on $L^2(G)$.

Choose a continuous nonnegative function $k$ supported in a small symmetric identity neighbourhood, with $k(z^{-1})=k(z)$ and $\int k=1$. Such functions are obtained from a nonzero continuous bump near the identity, symmetrizing and normalizing. DefineThe convolution operator $T_k$ commutes with left translations. Its continuous kernel is Hermitian, hence it is self-adjoint. It is compact on $L^2(G)$: approximate the continuous kernel uniformly by finite sums of separated functions of $x$ and $y$, using the Stone-Weierstrass theorem on the compact product. The corresponding operators have finite rank and converge in operator norm. It also maps into $C(G)$ and satisfiesby the Cauchy-Schwarz inequality; continuity follows from the uniform continuity of the translated kernel.

The spectral theorem for compact self-adjoint operators decomposes the orthogonal complement of its kernel into finite-dimensional eigenspaces with nonzero eigenvalues. Every such eigenspace is left-translation invariant, and all its elements are continuous, since $h=T_{k}h/\lambda$ there. If $W\subset C(G)$ is one of these invariant spaces, write $\rho (a)=L_a{|}_W$. Evaluation at the identity givesThus $h$ is a matrix coefficient of the dual representation. This identifies each finite spectral sum with finite-dimensional representation data.

To get uniform rather than just $L^2$ approximation, let $S_N$ project onto increasing finite sums of the nonzero eigenspaces. For $f\in C(G)$, $T_{k}f$ is orthogonal to the kernel, so $S_N{T}_{k}f\to T_{k}f$ in $L^2$. Applying the displayed smoothing bound givesMeanwhile, choosing the support of $k$ sufficiently small makes $T_{k}f$ uniformly close to $f$, by uniform continuity under small translations. Since $\Vert T_k{\Vert }_{C\to C}\le 1$, it also makes $T_k^{2}f$ uniformly close to $f$. Every $T_k{S}_N{T}_{k}f$ belongs to a finite sum of invariant eigenspaces, hence is a finite sum of matrix coefficients. This completes the convolution proof of uniform Peter-Weyl approximation.

A coefficient $\ell (\theta (g)v)$ equals $\mathrm{tr}(\alpha \theta (g))$ for the rank-one endomorphism $\alpha =v\otimes \ell$. Finite sums of coefficients can be combined using a direct-sum representation and a block-diagonal $\alpha$. Therefore the conclusion has exactly the requested form:

For a class function $f$, apply conjugation averaging on a compact group,It fixes $f$ and is a contraction in the uniform norm, so averaging an approximation preserves its error bound. A trace coefficient averages to $\mathrm{tr}(\overline{\alpha }\theta (g))$, whereThis endomorphism commutes with the representation. A finite-dimensional unitary representation splits into irreducibles by repeatedly taking invariant orthogonal complements. On each irreducible diagonal block, Schur lemma makes its average a scalar multiple of the identity, namely its trace divided by the block dimension. Off-diagonal blocks do not contribute to the total trace. Thus the averaged coefficient is a finite linear combination of irreducible characters. Consequently irreducible complex characters span a uniformly dense subspace of the continuous class functions.

Use the standard boundaryless-manifold convention, so $M$ is a closed manifold. Its orientation and Riemannian metric determine a unique positive Riemannian volume form ${\mu }_g$ taking value one on every positively oriented orthonormal frame. In an oriented atlas it isFor the coordinate invariance of the Riemannian volume form, on an overlap put $J=\partial y/\partial x$. Then $g_x=J^T{g}_{y}J$ and $d{y}^1\wedge \cdots \wedge d{y}^n=(\det J)d{x}^1\wedge \cdots \wedge d{x}^n$. The oriented transition has $\det J>0$, so the square root of the metric determinant and the coordinate volume transform together. The local forms therefore agree and define ${\mu }_g$ globally.

For real $r$-forms the Hodge star operator is defined byThe pointwise wedge pairing is nondegenerate, so this uniquely defines a smooth map $\ast :{\Omega }^r(M)\to {\Omega }^{n-r}(M)$. On an oriented orthonormal coframe, it sends a basis wedge to the signed complementary wedge. Applying it twice gives

For Hodge star eigenvalues on graded differential forms, extend $\ast$ complex-linearly to the direct sum of all form degrees. This interpretation matters: on a fixed degree it is an endomorphism only in middle degree. In positive odd dimension, $r(n-r)$ is even for every $r$, so ${\ast }^2=I$ and only $\pm 1$ can occur. Both occur, with eigenforms $1+{\mu }_g$ and $1-{\mu }_g$.

In positive even dimension, ${\ast }^2$ is $+I$ on even-degree forms and $-I$ on odd-degree forms. Hence ${\ast }^4=I$ and the only possible eigenvalues are $\pm 1,\pm i$. The first two occur as above. For the latter, choose a smooth local orthonormal frame and multiply its first dual differential one-form by a nonzero smooth bump function supported in the chart. This gives a form $\eta$ for which $\eta$ and $\ast \eta$ are independent. For $\lambda =\pm i$,Thus all four occur. Consequently, for $n\ge 1$,On degree $m$ in dimension $2m$, the Hodge star has eigenvalues $\pm 1$ for even $m$ and $\pm i$ for odd $m$; signed complementary middle-degree wedges show both occur for $m\ge 1$. In zero dimensions it is multiplication by the signed unit volume at each point, so only the signs actually present occur.

For Hodge integration by parts in arbitrary degree, let $\alpha \in {\Omega }^{r-1}(M)$ and $\beta \in {\Omega }^r(M)$. The graded Leibniz rule and Stokes theorem on the closed manifold giveUsing the given codifferential and the square of the Hodge star,Thereforewhich proves that $\delta$ is the formal adjoint of $d$. The complex version inserts conjugation in the usual Hermitian inner product.

The Hodge decomposition theorem gives an $L^2$-orthogonal decompositionwhere ${\mathcal{H}}^k(M)=\ker \Delta$ is finite dimensional and consists of the harmonic differential forms. Each de Rham cohomology class has a unique harmonic representative. Equivalently, with harmonic projection $P_{\mathcal{H}}$ there is a smooth Green operator of the Hodge Laplacian $G$ satisfying $\Delta G=G\Delta =I-P_{\mathcal{H}}$.

The solvability condition for the Hodge Poisson equation isNecessity follows from $⟨\Delta \alpha ,h⟩=⟨\alpha ,\Delta h⟩=0$. Conversely the orthogonality condition gives $P_{\mathcal{H}}\beta =0$, so ${\alpha }_0=G\beta$ solves the Poisson equation for differential forms.

The nonnegativity of the Hodge Laplacian identityidentifies its kernel with closed and coclosed forms. Thus the affine space of solutions of the Hodge Poisson equation isFor completeness, if a closed form is decomposed as $\omega =h+d\xi +\delta \zeta$, then $d\delta \zeta =0$, and $\Vert \delta \zeta {\Vert }^2=⟨\zeta ,d\delta \zeta ⟩=0$. Hence its cohomology class is represented by $h$. An exact harmonic form $h=d\xi$ must vanish, since $\Vert h{\Vert }^2=⟨\delta h,\xi ⟩=0$. This proves the canonical isomorphism ${\mathcal{H}}^k(M)\cong H_{\mathrm{dR}}^k(M)$, so the solution set is an affine space whose translation vector space is isomorphic to the requested de Rham cohomology.

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.

Given a one-dimensional character $\chi$ of $Z$, form the nonzero space of functions $f:G\to \mathbb{C}$ satisfying $f(zg)=\chi (z)f(g)$. Right translation $[\rho (h)f](g)=f(gh)$ defines a representation of dimension $[G:Z]$. Centrality gives $\rho (z)f(g)=f(gz)=\chi (z)f(g)$, so $Z$ acts by $\chi$ on the entire space. This representation has an irreducible summand: average any positive Hermitian inner product over the finite group, so invariant subspaces have invariant orthogonal complements, and decompose inductively by dimension. Every irreducible summand has the same central one-dimensional character $\chi$. Thus every irreducible representation of $Z$ occurs by restriction in the requested way.

Let $g$ be the Riemannian metric underlying the Hermitian metric and $J$ its compatible complex structure. The fundamental form of a Hermitian manifold is $\omega (u,v)=g(Ju,v)$. It is a real alternating form of type $(1,1)$. At a point choose an orthonormal real coframe $x^1,y^1,\dots ,x^n,y^n$ adapted to $J$, with $x^j+i{y}^j$ of type $(1,0)$. ThenThe orientation here is the complex orientation. In coordinates for which $g={\sum }_j((d{x}^j{)}^2+(d{y}^j{)}^2)$, the same convention reads $\omega =(i/2){\sum }_{j}d{z}^j\wedge d{\bar{z}}^j$.

Use the complex Hodge star operator, the complex-linear extension of the real Hodge star operator. With the pointwise Hermitian inner product linear in its first variable, its defining identity isfor forms of the same total degree. It maps type $(p,q)$ to $(n-q,n-p)$ and satisfies ${\ast }^2=(-1{)}^{p+q}$ in real dimension $2n$. It commutes with complex conjugation. These conventions distinguish it from the conjugate-linear Hodge star $\alpha \mapsto \ast \bar{\alpha }$.

Write a top holomorphic-type form as $\eta =a{\zeta }^1\wedge \cdots \wedge {\zeta }^n$, where ${\zeta }^j=x^j+i{y}^j$. On each oriented two-plane, $\ast x^j=y^j$ and $\ast y^j=-x^j$, so the two-dimensional star of ${\zeta }^j$ is $-i{\zeta }^j$. In taking the star of their product, regrouping complementary one-form factors across the $n$ two-planes contributes $(-1{)}^{n(n-1)/2}$. Thus the Hodge star on top holomorphic forms isThis is pointwise and requires neither holomorphicity of the coefficient $a$ nor closedness of $\omega$.

To compute the formal adjoint, take compactly supported forms $\alpha$ of degree $k-1$ and $\beta$ of degree $k$. The Stokes theorem givesSince the degree of $d(\ast \bar{\beta })$ is $2n-k+1$, the identity for ${\ast }^2$ makes the right side equal to ${\int }_X\alpha \wedge \ast \overline{(-\ast d\ast \beta )}$. Consequently the formal adjoint of $d$ in even real dimension is $d^{\ast }=-\ast d\ast$. Splitting by type, $-\ast \partial \ast$ lowers antiholomorphic degree, while $-\ast \bar{\partial }\ast$ lowers holomorphic degree. Orthogonality of distinct types therefore yieldsThe argument applies locally with compact support, and globally without support restrictions on a compact manifold without boundary.

The Dolbeault Laplacian is ${\Delta }_{\bar{\partial }}=\bar{\partial }{\bar{\partial }}^{\ast }+{\bar{\partial }}^{\ast }\bar{\partial }$. A form is $\bar{\partial }$-harmonic when it lies in its kernel. On a compact Hermitian manifold, the adjoint identities implyThus harmonicity is equivalent to both first-order equations vanishing. The Dolbeault Hodge decomposition on a compact Hermitian manifold states that the harmonic space ${\mathcal{H}}_{\bar{\partial }}^{p,q}$ is finite-dimensional and that smooth forms have the orthogonal direct-sum decomposition

To deduce the identification with Dolbeault cohomology, let $\bar{\partial }\alpha =0$ and decompose $\alpha =h+\bar{\partial }u+{\bar{\partial }}^{\ast }v$. The closedness of $\alpha$ makes it orthogonal to the entire image of ${\bar{\partial }}^{\ast }$. Taking its inner product with ${\bar{\partial }}^{\ast }v$ and using the orthogonality of the decomposition gives $\Vert {\bar{\partial }}^{\ast }v{\Vert }^2=0$. Hence $\alpha =h+\bar{\partial }u$, so each cohomology class has a harmonic representative. If a harmonic form equals $\bar{\partial }u$, its squared norm is $⟨{\bar{\partial }}^{\ast }h,u⟩=0$, so the representative is unique. We have proved

For the complementary types, define $C\alpha =\ast \bar{\alpha }$. It maps $(p,q)$ to $(n-p,n-q)$ and is conjugate-linear. If $\alpha$ is harmonic, then $\bar{\partial }\alpha =0$ and $\partial (\ast \alpha )=0$ by the adjoint formula. Conjugating the latter equation gives $\bar{\partial }(\ast \bar{\alpha })=0$. Also, with $k=p+q$,since $\partial \bar{\alpha }=\overline{\bar{\partial }\alpha }$. Therefore $C$ carries harmonic forms to harmonic forms, and $C^2=(-1{)}^k$ makes it a bijection. This proves complementary Dolbeault harmonic types on a Hermitian manifold:The metric supplies a natural conjugate-linear isomorphism. If a complex-linear isomorphism is desired, choose a basis $h_j$ of the first space and send it to the basis $C{h}_j$ of the second, extending linearly. This last choice is not canonical; no Kähler assumption is needed for either the dimension equality or the harmonic bijection.

For a compact Lie group, take any positive-definite Hermitian inner product $h_0$ on $V$ and average using probability Haar measure:It remains positive definite, because the integrand is positive for a nonzero vector, and translation invariance makes it $G$-invariant. If $W\subset V$ is an invariant subspace, then $W^{\perp }$ is invariant too: $h(gv,w)=h(v,g^{-1}w)=0$ when $v\in W^{\perp }$. Thus $V=W\oplus W^{\perp }$. Induction on dimension proves complete reducibility. This uses the usual continuous finite-dimensional representations of the compact group.

For a connected complex reductive algebraic group, representations are understood to be rational representations. Use a compact real form $K\subset G$ with $\mathrm{Lie}(G)=\mathrm{Lie}(K){\otimes }_{\mathbb{R}}\mathbb{C}$. The structural construction takes a compact real form of the semisimple derived factor and the unit circles in the central torus, then passes through their finite central quotient. This is the compact-form structure of a complex reductive group, rather than an assumption that its given representation is already unitary.

The group $K$ is Zariski dense: the Lie algebra of its Zariski closure contains the complex span of $\mathrm{Lie}(K)$, hence all of $\mathrm{Lie}(G)$; connectedness then makes that closure $G$. Average over $K$ as above. The orthogonal complement of a $G$-submodule is $K$-invariant. Its stabilizer is a Zariski-closed subgroup, since preserving a fixed subspace is a system of algebraic matrix conditions. The stabilizer contains dense $K$, so it is all of $G$. This again gives an invariant complement to every submodule and provesIf reductivity is defined to allow finitely many components, first obtain a $G^0$-equivariant projection $P:V\to W$. Average its conjugates over the finite component group. Since $P$ commutes with $G^0$, the average is independent of coset representatives; it is $G$-equivariant and remains the identity on $W$. Its kernel supplies the required complement. Thus the result holds in that convention too.

A vector in a rational representation is an unstable vector in a rational representation when $0\in \overline{Gv}$, with Zariski closure. For $G=S{L}_n(\mathbb{C})$, the Hilbert-Mumford criterion for the affine null cone isEvery algebraic one-parameter subgroup is conjugate to $\mathrm{diag}(t^{a_1},\dots ,t^{a_n})$ with integer $a_j$ summing to zero. This follows by decomposing the standard representation into characters of the multiplicative group, then imposing determinant one. In its weight decomposition, a limit of zero means that only strictly positive weights occur.

The reverse implication in the criterion is immediate: the subgroup lies in $G$ and supplies points of the orbit tending to zero. For the harder implication, the complex orbit is locally closed and Zariski open in its closure. A nonempty Zariski-open subset of a complex irreducible variety is dense also in the usual topology. Thus choose $g_{j}v\to 0$ in the ordinary vector-space topology. Singular value decomposition, with determinant phases adjusted, givesUse an $S{U}_n$-invariant Hermitian inner product on $V$. By compactness, after a subsequence ${\ell }_j\to \ell$. Put $w=\ell v$, and decompose it into the weights of the diagonal torus, $w={\sum }_{\chi }{w}_{\chi }$. These weight spaces are orthogonal because the unitary diagonal torus preserves the inner product. HenceFor every $\chi$ with $w_{\chi }\ne 0$, its final norm is bounded away from zero for large $j$, forcing $\chi (b_j)\to -\infty$. There are only finitely many such weights, so choose a single real vector $b$ with sum zero and $\chi (b)<0$ for all of them. The inequalities are strict with integer coefficients. A nearby rational vector still satisfies them; multiply its negative by a common denominator to obtain integers $a_r$ of sum zero with $\chi (a)>0$ throughout this support. Then $\mu (t)=\mathrm{diag}(t^{a_r})$ takes $w$ to zero. Conjugating gives $\lambda (t)={\ell }^{-1}\mu (t)\ell$ taking $v$ to zero, completing the proof. The zero vector itself satisfies the criterion trivially.

For degree-six binary forms, a nontrivial algebraic one-parameter subgroup of $S{L}_2$ is, after conjugation and choosing its direction, $\mathrm{diag}(t^a,t^{-a})$ with $a>0$. On $x^{6-j}{y}^j$ the weight is $a(6-2j)$. Strictly positive weights permit only $j=0,1,2$, so the form is divisible by $x^4$. Conversely every such form tends to zero under that subgroup. Changing basis replaces $x$ by any linear factor. The root-multiplicity criterion for unstable binary forms therefore givesA triple root alone is insufficient: $x^3{y}^3$ has weight zero and is not unstable. Nor does merely having a repeated root suffice.

For $\ell \in V^{\ast }$ define $(g\ell )(v)=\ell (g^{-1}v)$. The representing matrix is the transpose of the inverse matrix on $V$, so the dual character isThe last equality follows by averaging a Hermitian inner product to make the finite-group group representation unitary. Complex finite-group representations are isomorphic exactly when their characters agree. Hence $V$ is self-dual exactly when its character is real-valued.

The invariant tensor identification $(V\otimes V{)}^G\simeq {\mathrm{Hom}}_G(V^{\ast },V)$ and Schur lemma show that, for irreducible $V$, the trivial multiplicity is zero unless $V^{\ast }\simeq V$, and one in that case. Dually, a self-duality isomorphism gives a nonzero invariant bilinear form $B$ on $V$. Its radical is an invariant subspace, so irreducibility makes $B$ nondegenerate. The space of invariant bilinear forms is one dimensional. Since the transpose $B^T$ is another such form, $B^T=cB$; transposing again gives $c^2=1$. Therefore $B$ is symmetric or alternating: in the negative-transpose case, $B(v,v)=-B(v,v)$ forces $B(v,v)=0$. Both types cannot coexist, since proportional nonzero forms cannot have both transpose signs.

In odd dimension an alternating matrix has determinant zero: $\det B=\det (-B^T)=(-1{)}^n\det B=-\det B$. Thus only the symmetric case is possible. Over $\mathbb{C}$ a nondegenerate symmetric bilinear form has a basis with matrix $I$: diagonalize it by congruence and rescale each nonzero diagonal entry. In that basis invariance says $\rho (g{)}^T\rho (g)=I$. Consequently

Let $g$ be the real Riemannian metric underlying the Hermitian metric, and let $J$ be the complex structure. The fundamental form of a Hermitian manifold isThe identity $g(Ju,Jv)=g(u,v)$ implies $g(Ju,v)=-g(u,Jv)$, so this is an alternating real two-form. Moreover $\omega (Ju,Jv)=\omega (u,v)$. A two-form with this invariance has type $(1,1)$: on two vectors of type $(1,0)$, applying $J$ to both multiplies its value by $i^2=-1$, forcing that value to vanish; the same argument applies to type $(0,1)$. Hence $\omega$ is a real $(1,1)$-form.

Use the complex-linear convention for the complex Hodge star operator, extending the real star. The complex orientation and metric define $dV={\omega }^n/n!$, andHere the Hermitian inner product is linear in the first argument. With this convention, on the even-dimensional underlying real manifold the formal adjoints are $d^{\ast }=-\ast d\ast$, ${\bar{\partial }}^{\ast }=-\ast \partial \ast$ and ${\partial }^{\ast }=-\ast \bar{\partial }\ast$. A conjugate-linear star is another convention, but then the adjoint formulas must change accordingly.