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We prove compact perturbation invariance of Fredholm operators using a two-sided inverse modulo compact operators. If $L$ is Fredholm, decomposeThe restriction of $L$ from $(\ker L{)}^{\perp }$ to its closed range is a bounded bijection, so the bounded inverse theorem supplies a bounded inverse there. Extend that inverse by zero on $(\mathrm{ran}L{)}^{\perp }$ to obtain a bounded operator $B$. If $P,Q$ are the orthogonal projections onto $\ker L$ and $(\mathrm{ran}L{)}^{\perp }$, respectively, thenBoth $P$ and $Q$ have finite rank.

Set $T=L+K$. Products of a compact operator and a bounded operator are compact: on one side compactness preserves compact images, and on the other the bounded operator maps the unit ball into a bounded ball. Hencewhere $C,D$ are compact.

Here is why these identities force $T$ to be Fredholm. On $\ker T$, $C=-I$, so part (a) and compactness make $\ker T$ finite-dimensional. There is a positive constant $c$ withOtherwise unit vectors $h_n\perp \ker T$ could satisfy $T{h}_n\to 0$. From $h_n=-C{h}_n+BT{h}_n$ and compactness, a subsequence would converge strongly to a unit vector $h\perp \ker T$ with $Th=0$, which is impossible. The lower bound proves closed range: for a convergent sequence $T{h}_n$, first discard the kernel components, then the remaining $h_n$ are Cauchy and their limit maps to the desired range limit.

For finite codimension, take the adjoint operator of $TB=I+D$, giving $B^{\ast }{T}^{\ast }=I+D^{\ast }$. The operator $D^{\ast }$ is compact: the finite-rank approximation proved in part (e) gives finite-rank adjoints converging in operator norm to $D^{\ast }$, and norm limits of compact operators are compact. On $\ker T^{\ast }$, $D^{\ast }=-I$, so this kernel is finite-dimensional. Finally,Because the range is closed, its cokernel is isomorphic to this finite-dimensional orthogonal complement. This verifies all three Fredholm operator conditions for $L+K$.

For the converse, start with $L+K$ and perturb by the compact operator $-K$. Therefore $L$ is Fredholm if and only if $L+K$ is Fredholm, with no self-adjointness assumption.

Let $u=A^{\ast }h$ be the strict dual certificate for basis pursuit supplied by condition (ii). For $0\ne v\in \ker A$, we haveHere the adjoint operator is the real transpose. The injectivity of $A_S$ ensures $v_{S^c}\ne 0$: otherwise $A_S{v}_S=0$ would force $v=0$. Thus at least one nonzero term lies outside the support of a vector $x$. Since $|u_l|<1$ at every such index,This proves the fixed-sign null space condition, so part (a) gives uniqueness in basis pursuit. If $S^c$ is empty, the injectivity of $A_S=A$ instead means there is no nonzero null space vector, and the feasible set is a singleton. Injective active columns and a strict dual certificate for basis pursuit ensure unique recovery. Both ingredients matter: strictness outside $S$ cannot detect a nonzero null space direction supported entirely inside $S$.

Interpret the angle as the directed subspace angle defined by the infimum of projected unit vectors; it is different from the smallest angle between two subspaces. WriteConsider the bounded linear operator $T:W⟶V^{\perp }$ given by $Tw=P_{V^{\perp }}w$. Its adjoint operator, between these two Hilbert spaces, is $T^{\ast }u=P_{W}u$: for $w\in W$ and $u\in V^{\perp }$, the orthogonal projections give $⟨Tw,u⟩=⟨w,u⟩=⟨w,P_{W}u⟩$. The two positive directed subspace angle cosines yieldThe first bound makes $T$ injective and gives a closed range. Explicitly, if $T{w}_j$ converges, then $\Vert w_j-w_k\Vert \le a^{-1}\Vert T{w}_j-T{w}_k\Vert$, so $(w_j)$ is a Cauchy sequence. The closed subspace of a Hilbert space $W$ is complete, and its limit maps to the proposed range limit. The second bound gives $\ker T^{\ast }=\{0\}$. A vector $u\in V^{\perp }$ orthogonal to the range has $T^{\ast }u=0$, so it must be zero. The range is therefore dense as well as closed in $V^{\perp }$, and $T$ is onto. This is the mechanism of invertibility from lower bounds on an operator and its adjoint.

For any $f\in H$, choose the unique $w\in W$ with $Tw=P_{V^{\perp }}f$. Then $P_{V^{\perp }}(f-w)=0$, so $f-w\in V$. Moreover, if $w\in W\cap V$, then $Tw=0$ and hence $w=0$. Every vector has a unique decomposition, andThe direct sum is a topological one as well: the component $w=T^{-1}{P}_{V^{\perp }}f$ depends boundedly on $f$, with operator norm at most $a^{-1}$.

For precision, the quoted equality of the norms of complementary oblique projections needs both summands nonzero. For example, with $H=\mathbb{R}$, $V=H$ and $W=\{0\}$, the oblique projection is $I$, so $\Vert I\Vert =1$ but $\Vert I-I\Vert =0$. The secant function has value one here, so the second equality in the quoted formula fails. With nonzero complementary summands its intended version is valid. The proof above does not use that formula. The angle itself is undefined on a zero source space because it has no unit vectors; expressing the hypotheses as the two lower bounds handles zero spaces without ambiguity.

It is enough to prove the stronger planar Kakeya maximal function boundWe first work with a $\delta$-separated net of unoriented directions $e_1,\dots ,e_M$, with $M\asymp {\delta }^{-1}$. Write ${\alpha }_{ij}$ for angular distance modulo $\pi$. Choose arbitrary length-one, width-$\delta$ rectangles $T_j$ in these directions. The permitted rectangle intersection fact isThe first alternative includes parallel rectangles. Their locations are arbitrary; only separation of their directions matters.

Define $Af(j)={\delta }^{-1}{\int }_{T_j}f$, and put $\Vert a{\Vert }_{{\ell }_{\delta }^2}^2=\delta {\sum }_j|a_j{|}^2$. With these weights the adjoint operator is $A^{\ast }a={\sum }_j{a}_j{1}_{T_j}$. Since a separated angular net has only a bounded number of directions at each distance scale $k\delta$ from $e_i$, its overlap matrix satisfiesThe diagonal term is of size $\delta$. Using $|a_i{a}_j|\le (|a_i{|}^2+|a_j{|}^2)/2$ and symmetry gives the Schur test estimateBy duality of Lp spaces, $\Vert Af{\Vert }_{{\ell }_{\delta }^2}\lesssim \sqrt{\log (2/\delta )}\Vert f{\Vert }_2$. This is uniform over every choice of the translated rectangles. For each direction choose a rectangle approaching the supremum for $|f|$, then take the limit. That gives the same estimate for the discretized Kakeya maximal function.

To recover all directions, partition the direction circle into arcs of length comparable to $\delta$, each with a net direction. A tube in an arc is contained in a rectangle in its net direction with width $C\delta$ and length at most two. A bounded subdivision in the length direction reduces this to the same averaging operators; the wider tubes obey the same overlap estimate with fixed-factor changes. ThereforeFinally $\sqrt{\log (2/\delta )}\le C_{\epsilon }{\delta }^{-\epsilon }$ for every $\epsilon >0$. The planar Kakeya maximal estimate therefore has the required arbitrary small power loss.

Interpret the printed Euclidean norm literally on the full vector array $X^2$. It is a single global norm, not the sum of pixelwise gradient lengths in discrete isotropic total variation. Set $A=\nabla$ and let $A^{\ast }$ be its exact adjoint operator for the chosen difference and boundary conventions. The closed dual ball and its image signal areThe set $C$ is nonempty, convex and compact, hence closed, because it is a linear image signal of a compact ball in finite dimensions. Euclidean norm duality givesthe support function of $C$.

The metric projection onto a closed convex set $w=P_{C}g$ uniquely minimizes $\Vert g-w{\Vert }^2$ over $C$. Its variational characterization of convex projection saysPut $u_{\ast }=g-w$. This inequality says precisely that ${\sigma }_C(u_{\ast })=⟨u_{\ast },w⟩$. For any $v\in X$, the primal energy satisfiesCompleting the square shows that the right side is uniquely minimized by $v=g-w$. At that point the inequality is equality, soThis also follows from the proximal operator of a support function and Moreau decomposition: the convex conjugate of ${\sigma }_C$ is the indicator functional of a constraint set for $C$. The squared fidelity is strictly convex and coercive, so the primal minimizer exists and is unique.

The convex projection in this formula is the removed component $g-u_{\ast }$, not generally $u_{\ast }$ itself. A metric convex projection onto one fixed closed convex set is idempotent. On a right singular vector direction with positive singular value of any nonzero $A$, this denoising map reduces to soft thresholding with a positive threshold, and applying it twice shrinks again. It therefore cannot be such a convex projection for all data. This qualifies the printed convex projection wording while giving the required global gradient-norm projection residual.

Compute $P_{C}g$ by solving the convex dual least-squares problemIts gradient is $\nabla f(p)=A(A^{\ast }p-g)$ and has Lipschitz constant $L=\Vert A{\Vert }^2$. Projected gradient descent givesFor fixed $0<\tau <2/L$, the finite-dimensional projected-gradient convergence theorem ensures that $p^n$ converges to a dual minimizer. Reconstruct $u^n=g-A^{\ast }{p}^n$; its limit is the unique primal solution even if dual minimizers are nonunique. If $A=0$, the primal solution is simply $g$ and no iteration is needed. For unit-grid forward differences with periodic or zero-difference boundaries, $\Vert A{\Vert }^2\le 8$, so $0<\tau <1/4$ is sufficient. Grid-spacing factors or other boundary stencils change this bound.

The normalization here is global. If a pixelwise sum of gradient lengths had instead been intended, the dual feasible set would be a product of pixelwise balls and convex projection would normalize each block separately; it is a different regularizer and should not be silently substituted.

For the discrete Hessian-norm denoising variant, write $H={\nabla }^2:X\to X^4$ and useThis is again a compact convex image signal of a Euclidean ball, now under the second-difference adjoint. In suitable conventions $H^{\ast }$ is a discrete double divergence, but its exact boundary adjoint is what defines the set. It lies in $(\ker H{)}^{\perp }$, so components in $\ker H$ are preserved by denoising. Interior second derivatives annihilate affine image signals; whether all such image signals remain in the kernel depends on the boundary convention. With a pixelwise Hessian norm, the corresponding balls would instead be four-component blocks.

Use real differential forms and the usual convention that a smooth manifold has no boundary unless one is specified. On a noncompact manifold the formal-adjoint identity below uses compactly supported forms. On a compact manifold it has no boundary contribution; if a boundary were allowed, extra boundary conditions would be necessary.

The Riemannian metric gives the dual inner product on $T_x^{\ast }M$ and hence the inner product on exterior powers of a cotangent space:Extend this bilinearly. Equivalently, the increasing wedge products of an orthonormal coframe are an orthonormal basis of ${\Lambda }^p{T}_x^{\ast }M$. The Hodge star operator is the unique map $\ast :{\Lambda }^p{T}_x^{\ast }M\to {\Lambda }^{n-p}{T}_x^{\ast }M$ satisfying $\alpha \wedge \ast \beta =⟨\alpha ,\beta ⟩{\mathrm{vol}}_g$. If $e^1,\dots ,e^n$ is a positively oriented orthonormal coframe, $\ast e^I=\epsilon (I,I^c)e^{I^c}$. Moving the $p$ factors of $I$ across the $n-p$ factors of its complement changes the sign by $(-1{)}^{p(n-p)}$, soThe basis formula also proves that $\ast$ is an isometry.

For $\beta \in {\Omega }^{p-1}(M)$ and $\alpha \in {\Omega }^p(M)$ with compactly supported product, the Stokes theorem and the Leibniz rule for the exterior derivative giveThus $(d\beta ,\alpha {)}_{L^2}=(-1{)}^p{\int }_M\beta \wedge d\ast \alpha$. The formal adjoint operator $\delta$ satisfies $\ast \delta \alpha =(-1{)}^{p}d\ast \alpha$. Applying the inverse star and its square on degree $p-1$ gives the codifferentialThe exponent is congruent to $p+(p-1)(n-p+1)$ modulo two, so this convention is consistent in every degree. Define the Hodge Laplacian by $\Delta =d\delta +\delta d$ and a harmonic differential form by $\Delta \alpha =0$. On a compact manifold,Both nonnegative terms vanish for a harmonic form, giving $d\alpha =\delta \alpha =0$. Conversely these two equations imply harmonicity.

Now let $n=4$. The printed eigenvalue relation omits the subscript on its right-hand side; the intended equation is $\ast {\alpha }_{\pm }=\pm {\alpha }_{\pm }$. On two-forms ${\ast }^2=1$, soThese are respectively self-dual two-forms and anti-self-dual two-forms, and sum to $\alpha$. Their uniqueness follows because the $+1$ and $-1$ eigenspaces have zero intersection. Star is an orthogonal involution here, hence is self-adjoint and its opposite eigenspaces are orthogonal.

For closed two-forms define $Q([\alpha ],[\beta ])={\int }_N\alpha \wedge \beta$. Adding $d\xi$ to $\alpha$ changes the integral by ${\int }_{N}d(\xi \wedge \beta )=0$; adding an exact form to $\beta$ has the same effect. This is the real de Rham intersection form in dimension four. The wedge product of differential forms in degrees two and two is symmetric because $(-1{)}^{2\cdot 2}=1$; hence $Q$ is a symmetric bilinear form on $H_{\mathrm{dR}}^2(N;\mathbb{R})$.

In four dimensions $\delta =-\ast d\ast$ in every degree. If $h$ is a harmonic two-form, $\delta h=0$ gives $d\ast h=0$, and $dh=0$ gives $\delta \ast h=-\ast dh=0$. Thus $\ast h$ is harmonic, and ${\mathcal{H}}^2={\mathcal{H}}^{+}\oplus {\mathcal{H}}^{-}$ is the harmonic star-eigenspace decomposition. Use the permitted unique harmonic representative for each de Rham cohomology class. For harmonic $h,k$,The mixed terms vanish by orthogonality. The first restriction is positive definite and the second negative definite. In particular a nonzero harmonic $h$ has $Q(h,\ast h)=\Vert h{\Vert }_{L^2}^2>0$, proving nondegeneracy. Hence the signature of the intersection form from harmonic duality isThe finite dimensions and the harmonic description also follow from the Hodge decomposition theorem.

For completeness, the Hodge decomposition theorem on a compact oriented Riemannian manifold without boundary states the finite-dimensional harmonic space and the $L^2$-orthogonal direct sumThe three summand components are unique, although their potentials need not be. Given an exact three-form $\beta =d\omega$ on $N$, decompose its smooth two-form potential as $\omega =h+d\xi +\delta \psi$, with $\psi \in {\Omega }^3$. Since $dh=0$ and $d^2=0$, $\beta =d\delta \psi$. Put $a=\delta \psi$. In dimension four, $a=-\ast d\ast \psi$, and ${\ast }^2=1$ on two-forms gives $\ast a=-d\ast \psi$, so $d\ast a=0$. ConsequentlyThis constructs a self-dual primitive of an exact three-form. The sum contains no factor $1/2$: it is twice the self-dual projection of the coexact potential. Using the unscaled projection alone would give $d\sigma =\beta /2$. This proves the final unheaded request.

Consider the operators on all of $C_0(X)$ or $L^2(\mu )$, extending the displayed positive-function formula linearly. Here $C_0(X)$ is the space of continuous functions vanishing at infinity. The multiplier has modulus at most one, so $\Vert P_{t}f\Vert \le \Vert f\Vert$, and multiplication of exponentials gives $P_{s+t}=P_s{P}_t$ and $P_0=I$. For strong continuity on $C_0(X)$, choose a compact $K$ outside which $|f|$ is small. On $K$ the continuous $k$ is bounded, making $e^{-tk}\to 1$ uniform; outside $K$, $|(e^{-tk}-1)f|\le |f|$. For $L^2(\mu )$, use pointwise convergence and the dominated convergence theorem with bound $|(e^{-tk}-1)f{|}^2\le |f{|}^2$. Thus both are contraction semigroups, even when $k$ is unbounded.

The generator of a multiplication semigroup isIndeed pointwise difference quotients converge to $-kf$. A norm limit therefore forces the displayed domain condition, using an almost-everywhere convergent subsequence for the $L^2$ case. Conversely $|(e^{-tk}-1)/t|\le k$, so the dominated convergence theorem proves the $L^2$ limit when $kf\in L^2$. In $C_0(X)$ write the error as $kf$ times $1-(1-e^{-tk})/(tk)$, with the value at $k=0$ supplied by continuity. That second factor tends uniformly to zero where $k$ is bounded and has modulus at most one; the compact-set and small-tail argument applied to $kf$ proves uniform convergence. To prove directly that $L$ is a closed operator, take $f_n\to f$ and $L{f}_n\to g$. Pointwise limits in $C_0$, or a common almost-everywhere convergent subsequence in $L^2$, give $g=-kf$. Thus $f$ is in the appropriate generator domain and $Lf=g$. In $L^2$ the domain is dense because $f{1}_{\{k\le n\}}\to f$; in $C_0$ compactly supported continuous functions are a dense subspace contained in the domain.

A real multiplication operator on its maximal $L^2$ domain is an unbounded self-adjoint operator. One direct verification is to test the adjoint relation against functions supported on $\{k\le n\}$. If $h$ is in the adjoint operator domain with representing vector $g$, those tests force $g=-kh$ on each such set, hence $kh\in L^2$. The converse follows by integration. On the sigma-finite measure spaces used below, the spectrum of a real multiplication operator is its essential range; spectral values are detected by unit vectors supported where the multiplier is arbitrarily close to that value, while outside the essential range its reciprocal is a bounded resolvent multiplier.

For the heat semigroup on $L^2(\mathbb{R},dx)$, take the normalization $L_H=f^{\prime \prime }$. The unitary Fourier transform converts it to multiplication by $-{\xi }^2$, and converts the semigroup to multiplication by $e^{-t{\xi }^2}$. ConsequentlyThe Sobolev space domain is exactly $\{f:{\xi }^2\hat{f}\in L^2\}$. The continuous spectrum comes from the full essential range of $-{\xi }^2$, not from square-integrable Fourier eigenvectors.

For the Ornstein-Uhlenbeck semigroup on standard Gaussian measure, the normalized Probabilists' Hermite polynomials $h_n={\mathrm{He}}_n/\sqrt{n!}$ form an orthonormal basis. The construction below gives $L_{OU}=D^2-xD$ and $L_{OU}{h}_n=-n{h}_n$, so the Hermite coefficient map turns it into a real multiplication operator on ${\ell }^2({\mathbb{N}}_0)$. ThereforeFinally, the heat semigroup on the whole real line has no positive spectral gap, whereas the standard Ornstein-Uhlenbeck semigroup has gap one. For the first claim, take a nonzero smooth compactly supported $\varphi$ and set $f_R(x)=R^{-1/2}\varphi (x/R)$. Then $\Vert f_R{\Vert }_2^2=\Vert \varphi {\Vert }_2^2$, but $\Vert f_R^{\prime }{\Vert }_2^2=R^{-2}\Vert {\varphi }^{\prime }{\Vert }_2^2$. This disproves a uniform whole-line Poincare inequality $\Vert f{\Vert }_2^2\le C\Vert f^{\prime }{\Vert }_2^2$; one can also choose $\int \varphi =0$. Lebesgue measure here is infinite, so it has no normalized probability mean to subtract. For Gaussian measure, the Hermite expansion instead gives the sharp Gaussian Poincaré inequalityEquality holds for affine functions. The identity on the right is interpreted on the energy-form domain, which is larger than the full operator domain.

For noisy data, exact equality $Ax=y$ may be impossible: the data can have a component outside the operator range. A least-squares solution of a linear inverse problem instead minimizesFor any variation $h\in X$, the adjoint operator givesThus statistical homogeneity in every direction is the normal equation for a linear inverse problem, $A^{\ast }Ax=A^{\ast }y$. It is also sufficient for a global minimum: when it holds,The residual is orthogonal to the closure of the range, since it lies in $\ker A^{\ast }$. A normal-equation solution therefore fits the component of the data representable by the forward map, while leaving the perpendicular component unfitted. If there is more than one such solution, select the minimum-norm least-squares solution, which lies in $(\ker A{)}^{\perp }$.

These facts explain the formulation, but do not make the inverse problem well posed. A compact operator of infinite rank typically has nonclosed operator range, so even the least-squares infimum need not be attained. A least-squares solution exists exactly when the orthogonal projection of $y$ onto $\overline{\mathrm{Ran}A}$ belongs to $\mathrm{Ran}A$. In a singular system of a compact operator, this becomes the Picard criterion. Also, the nonzero eigenvalues of $A^{\ast }A$ are the squares of the singular values of $A$: solving the operator normal equation directly can worsen numerical conditioning. The operator normal equation expresses least squares; it does not itself regularize the inverse problem. The subsequent finite Landweber iteration and its stopping rule provide the inverse-problem regularization.

Use an inner product linear in the first variable. For each $y$, the bounded linear functional $x\mapsto ⟨Tx,y⟩$ is represented uniquely by a vector $T^{\ast }y$ under the Riesz representation theorem, so $⟨Tx,y⟩=⟨x,T^{\ast }y⟩$. Uniqueness and the conjugate-linearity in the second argument imply $T^{\ast }(ay+bz)=a{T}^{\ast }y+b{T}^{\ast }z$. FurthermoreThus the adjoint operator exists, is linear and bounded, and is unique because a vector orthogonal to every $x$ is zero.

A normal operator satisfies $T^{\ast }T=T{T}^{\ast }$. The unilateral unilateral shift operator $S(x_1,x_2,\dots )=(0,x_1,x_2,\dots )$ on ${\ell }^2$ is bounded but not normal: $S^{\ast }S=I$, whereas $S{S}^{\ast }=I-P_{e_1}$. For general $T$,Normality implies equality of norms. Conversely equality for every $x$ makes this self-adjoint operator's quadratic form zero; the polarization identity then gives $T^{\ast }T-T{T}^{\ast }=0$.

An approximate eigenvalue $\lambda$ has unit vectors $x_n$ with $\Vert (T-\lambda I)x_n\Vert \to 0$. It is necessarily in the spectrum, since a bounded inverse would prevent this. Conversely suppose $\lambda$ is not an approximate eigenvalue. Then $A=T-\lambda I$ is bounded below, $\Vert Ax\Vert \ge c\Vert x\Vert$. It is injective and has closed range: a convergent sequence $A{x}_n$ forces $x_n$ to be Cauchy. The operator $A$ is normal, so $\Vert A^{\ast }x\Vert =\Vert Ax\Vert$ and $\ker A^{\ast }=0$. The identity $(\mathrm{ran}A{)}^{\perp }=\ker A^{\ast }$ makes its range dense. Closed and dense means surjective, and its inverse is bounded by $1/c$. Hence $\lambda$ is in the resolvent. This proves $\boxed{\sigma (T)={\sigma }_{\mathrm{ap}}(T)}$.

Apply the adjoint operator identity of part (c) with $\chi =\hat{A}\varphi$ and use part (a):For a nonzero eigenfunction, positivity of the norm in particular forces $E\ge \hslash \omega /2$. Hence the requested norm isIt vanishes exactly at the ground-state energy.

We prove the finite-dimensional spectral theorem for normal operators by induction on dimension. A complex linear operator has an eigenvector $v$ with eigenvalue $\lambda$, by the fundamental theorem of algebra. For a normal operator $T$, the operator $T-\lambda I$ is also normal, andfor every $w$. Applying this to $v$ shows $T^{\ast }v=\overline{\lambda }v$.

Normalize $v$. Its orthogonal complement $W=v^{\perp }$ is invariant under both $T$ and $T^{\ast }$: the adjoint operator identities pair $Tw$ with $v$ through $T^{\ast }v$, and $T^{\ast }w$ with $v$ through $Tv$. The restricted operator on $W$ remains normal, and its adjoint is the restriction of $T^{\ast }$. Induction supplies an orthonormal eigenbasis of $W$, which together with $v$ gives one for $V$.

Conversely, in an orthonormal basis of eigenvectors, $T$ is diagonal and $T^{\ast }$ is the diagonal matrix of the conjugate eigenvalues. The two diagonal operators commute. Thus normality is equivalent to an orthonormal eigenbasis. The zero-dimensional case is immediate.

The algebraic dual map is $\delta ={\alpha }^{\vee }:W^{\ast }\to V^{\ast }$, defined by $\delta (\phi )=\phi \circ \alpha$. The annihilator of a vector subspace $U\subseteq V$ isEvery functional $\phi \circ \alpha$ vanishes on $\ker \alpha$, so $\mathrm{im}\delta \subseteq (\ker \alpha {)}^{\circ }$. Conversely, let $\lambda \in (\ker \alpha {)}^{\circ }$. Define a functional on $\mathrm{im}\alpha$ by $\tilde{\lambda }(\alpha v)=\lambda (v)$. It is well defined because two preimages differ by an element of $\ker \alpha$. Extend a basis of $\mathrm{im}\alpha$ to one of $W$ and extend the functional arbitrarily, for instance by zero on the extra basis vectors. The resulting $\phi \in W^{\ast }$ satisfies $\delta \phi =\lambda$.

This is the dual image and kernel annihilator identity. ThereforeThe basis extension theorem also gives $\dim U^{\circ }=\dim V-\dim U$. By the rank-nullity theorem,Apply rank-nullity theorem to both maps, using $\dim W^{\ast }=\dim W$, to obtainOver $\mathbb{C}$, this is the algebraic dual, without complex conjugation; the adjoint operator in part (b) uses the inner products instead.

The adjoint operator ${\alpha }^{\ast }:W\to V$ is uniquely characterized byLet $A=\beta {\beta }^{\ast }+{\gamma }^{\ast }\gamma :V\to V$. Its quadratic form is nonnegative:If $Av=0$, both norms vanish. In particular $v\in \ker \gamma =\mathrm{im}\beta$, while ${\beta }^{\ast }v=0$ means $v$ is orthogonal to $\mathrm{im}\beta$, by the defining adjoint operator identity. A vector belonging to a vector subspace and its orthogonal complement must be zero. Hence $A$ is injective, and finite dimensionality makes it surjective.

ThusThe same calculation shows it is a positive-definite operator when $V\ne \{0\}$.

The form attached to $V$ is $B_V(x,y)={\mathrm{tr}}_V(xy)$. More generally, for a Lie algebra representation $\rho$, the Trace form of a Lie algebra representation isThe unqualified Killing form is the special case of the Adjoint representation,The distinction matters: a Trace form of a Lie algebra representation can be degenerate even when $\mathfrak{g}$ is semisimple, for example on the trivial Lie algebra representation.

The Trace form of a Lie algebra representation is bilinear and symmetric, because $\mathrm{tr}(AB)=\mathrm{tr}(BA)$. It is an invariant bilinear form on a Lie algebra:This follows by expanding both commutators and cyclically permuting factors under the matrix trace. Equivalently,Its radical of a bilinear form is an ideal of a Lie algebra, since if $B_V(y,-)=0$, then $B_V([x,y],z)=-B_V(y,[x,z])=0$. The Killing form is also preserved by every automorphism of a Lie algebra, because the corresponding adjoint operators are conjugate. On a complex finite-dimensional Lie algebra, the Cartan criterion for semisimplicity says that the Killing form is nondegenerate exactly when the Lie algebra is semisimple. The Cartan solvability criterion says that $\mathfrak{g}$ is solvable exactly when $B(\mathfrak{g},[\mathfrak{g},\mathfrak{g}])=0$.

We next construct the sl2 subalgebra associated with a root. Use the root-space decompositionFor $u\in {\mathfrak{g}}_{\beta }$, $v\in {\mathfrak{g}}_{\gamma }$, invariance of the Killing form givesThus $B({\mathfrak{g}}_{\beta },{\mathfrak{g}}_{\gamma })=0$ unless $\beta +\gamma =0$, and $B({\mathfrak{g}}_{\beta },\mathfrak{h})=0$ for nonzero $\beta$. Nondegeneracy of $B$ on $\mathfrak{g}$ now implies that $-\alpha$ is a root and that $B$ pairs ${\mathfrak{g}}_{\alpha }$ and ${\mathfrak{g}}_{-\alpha }$ nondegenerately.

Nondegeneracy of $B{|}_{\mathfrak{h}}$ defines a unique $t_{\alpha }\in \mathfrak{h}$ byChoose $x\in {\mathfrak{g}}_{\alpha }$ and $y\in {\mathfrak{g}}_{-\alpha }$ with $B(x,y)=1$. Their Lie bracket lies in the zero root space, namely $\mathfrak{h}$, andTherefore $[x,y]=t_{\alpha }$.

The essential nonisotropic root lemma is that $\alpha (t_{\alpha })\ne 0$. Suppose instead that it vanished. Then $[t_{\alpha },x]=[t_{\alpha },y]=0$, so $⟨x,y,t_{\alpha }⟩$ would be a solvable Lie algebra with derived algebra $\mathbb{C}{t}_{\alpha }$. Apply the Lie theorem to its action on $\mathfrak{g}$ by the Adjoint representation. The commutator $\mathrm{ad}{t}_{\alpha }=[\mathrm{ad}x,\mathrm{ad}y]$ is strictly upper triangular in a suitable basis, hence nilpotent. But $t_{\alpha }\in \mathfrak{h}$, so the root-space decomposition makes $\mathrm{ad}{t}_{\alpha }$ diagonalizable. A diagonalizable nilpotent linear map is zero. Thus $t_{\alpha }$ is central in $\mathfrak{g}$. The center of a Lie algebra of a semisimple Lie algebra is zero; equivalently a central element lies in the radical of the Killing form. This forces $t_{\alpha }=0$, contradicting $\alpha \ne 0$.

Writing $c=\alpha (t_{\alpha })$, defineThe root-space decomposition and $[x,y]=t_{\alpha }$ giveThe three vectors are linearly independent because they lie in the distinct summands ${\mathfrak{g}}_{\alpha }$, $\mathfrak{h}$, and ${\mathfrak{g}}_{-\alpha }$. Their span is therefore a copy of the sl2 Lie algebra.

The weight lattice consists of the functionals integral on all coroots. With $H_{\alpha }$ the coroot above, the weight lattice iswhere the fundamental weights satisfy ${\omega }_i(H_{{\alpha }_j})={\delta }_{ij}$ for the simple roots ${\alpha }_j$. Here $P$ lies in the real span of the roots, viewed inside ${\mathfrak{h}}^{\ast }$.

The classification of finite-dimensional sl2 representations says that every finite-dimensional complex sl2 Lie algebra representation is a direct sum of irreducibles $L_m$, $m\in {\mathbb{Z}}_{\ge 0}$, on which the standard $H$ has eigenvalues $m,m-2,\dots ,-m$. Restrict any finite-dimensional Lie algebra representation of $\mathfrak{g}$ to each sl2 subalgebra associated with a root. If $v$ has weight $\lambda$, then $H_{\alpha }v=\lambda (H_{\alpha })v$, so $\lambda (H_{\alpha })$ is an integer. Thus every weight lies in $P$. The same restrictions show that the commuting simple coroots act diagonalizably, justifying the simultaneous weight-space decomposition.

For $\mathfrak{sp}(4)$, the roots are $\pm 2{\epsilon }_1$, $\pm 2{\epsilon }_2$, and $\pm {\epsilon }_1\pm {\epsilon }_2$. Work on ${\mathbb{C}}^4$ with the alternating bilinear form having matrixThe symplectic Lie algebra isUsing the matrix units $E_{ij}$, take the Cartan subalgebraDefine ${\epsilon }_i(a_1{h}_1+a_2{h}_2)=a_i$. A regular diagonal element of $\mathfrak{h}$ has centralizer precisely $\mathfrak{h}$, and every element of $\mathfrak{h}$ acts diagonalizably. Thus it is a Cartan subalgebra. The requested Cartan decomposition is the root-space decompositionChoose positive roots ${\epsilon }_1-{\epsilon }_2$, $2{\epsilon }_2$, ${\epsilon }_1+{\epsilon }_2$, $2{\epsilon }_1$. The symplectic root sl2 triple are given explicitly byFor the negative root spaces, use the corresponding $Y_{\alpha }$. These eight root vectors, together with $h_1,h_2$, form a basis: the block description above has dimension $4+3+3=10$, and the ten listed vectors are independent. Finally, the matrix unit identityverifies $[X_{\alpha },Y_{\alpha }]=H_{\alpha }$ for every row. The diagonal differences verify $[H_{\alpha },X_{\alpha }]=2X_{\alpha }$ and $[H_{\alpha },Y_{\alpha }]=-2Y_{\alpha }$. Thus each row supplies a basis of the required sl2 subalgebra associated with a root.