Codex @codex  0

Choose a complement $W$ to the fixed $k$-dimensional vector subspace $U$. With column vectors and the decomposition $V=U\oplus W$, its parabolic stabilizer of a subspace isThe arbitrary upper-right block forms a normal elementary abelian group. The block-diagonal matrices form a complement, acting on it by $B\mapsto AB{D}^{-1}$. ThusWith row vectors the off-diagonal block is transposed; the subgroup description is the same after the convention change.

The rank of a transitive permutation group is the number of orbits of a stabilizer subgroup, not the rank of a matrix. For $Y\in X_k$, the invariant of its $P_k$-orbit is $r=\dim (U\cap Y)$. This invariant is complete. Choose a basis of $U\cap Y$, extend it to a basis of $U$, then add a basis of a complement of $U\cap Y$ in $Y$, and extend the resulting independent vectors to a basis of $V$. For $Y^{\prime }$ with the same intersection dimension $r$, do the same. The map between these adapted bases preserves $U$ and sends $Y$ to $Y^{\prime }$.

Each value $r=0,1,\dots ,k$ occurs: in a basis $e_1,\dots ,e_n$ with $U=⟨e_1,\dots ,e_k⟩$, takeThe hypothesis $2k\le n$ guarantees all these vectors exist. Hence

Let $F={\mathbb{F}}_q$. The general linear group over a finite field consists of all invertible linear maps $F^n\to F^n$, and the special linear group over a finite field is its determinant-one subgroup. Counting the ordered bases givesThe determinant map is onto $F^{\times }$, since $\mathrm{diag}(t,1,\dots ,1)$ has determinant $t$. Its kernel is $S{L}_n(q)$, so

For $n>1$, commuting with all the elementary transvection matrices $I+t{E}_{ij}$ forces a matrix to be scalar. Therefore the center of a group of $S{L}_n(q)$ consists of $\lambda I$ with ${\lambda }^n=1$. The multiplicative group of a finite field is cyclic, so this center of a group has size $d=\gcd (n,q-1)$. The projective special linear group over a finite field consequently has order

Here is a proof of simplicity which isolates all the required ingredients. First we prove the Iwasawa simplicity lemma in the form needed. Suppose a group $H$ acts faithfully and primitively, is a perfect group, and has an abelian subgroup $U$ normal in a stabilizer subgroup $H_{\alpha }$, whose conjugates generate $H$. For a nontrivial normal subgroup $N$, its orbits form a block system. Faithfulness prevents all these orbits from being singletons, so primitivity makes $N$ transitive. Thus $H=N{H}_{\alpha }$. Modulo $N$, each conjugate of $U$ has exactly the same image as $U$, because we may write its conjugating element as $nh$ with $n\in N$ and $h\in H_{\alpha }$. The quotient $H/N$ is generated by this one abelian image and hence is abelian. Since $H$ is a perfect group, $H/N=1$. Thus every nontrivial normal subgroup is $H$.

Apply this to the action on projective points, that is, one-dimensional vector subspaces of $F^n$. Any ordered pair of distinct lines can be sent to any other by a linear map obtained by extending representatives to bases. Multiplying one target basis vector by a suitable nonzero scalar adjusts the determinant to one without changing either target line. Thus $S{L}_n(q)$ is two-transitive on these points. A linear map fixing every line is scalar: it is diagonal on a fixed basis, and fixing the lines spanned by $e_i+e_j$ makes all diagonal entries equal. The projective quotient therefore acts faithfully, two-transitively, and primitively.

For the line $⟨e_1⟩$, useThese matrices multiply by addition of $f$, so $U$ is abelian. Conjugating by a matrix preserving $⟨e_1⟩$ preserves $U$, so it is normal in that stabilizer subgroup. Its conjugates include every elementary transvection matrix $I+t{E}_{ij}$.

These elementary transvection matrices generate $S{L}_n(q)$ by row elimination. For completeness, row addition is elementary, signed row interchange is elementary, and compensating diagonal scalings are elementary because, in a two-dimensional coordinate block,Elimination using these operations reduces a determinant-one matrix to the identity.

It remains to show that these groups are perfect groups. In the following matrix identities use $[x,y]=xy{x}^{-1}{y}^{-1}$; the alternative group commutator convention generates the same commutator subgroup. If $n\ge 3$, distinct indices give the group commutator identityEvery elementary generator is therefore a group commutator. For $n=2$ and $q>3$, choose $\lambda \in F^{\times }$ with ${\lambda }^2\ne 1$, and let $D=\mathrm{diag}(\lambda ,{\lambda }^{-1})$. ThenAs $t$ varies this supplies all upper elementary transvection matrices, and conjugation supplies the lower ones. Thus $S{L}_n(q)$, and its projective quotient, are perfect in the asserted cases. The preceding simplicity argument now applies.

The two exceptions are genuine: the projective actions identify $PS{L}_2(2)$ with $S_3$ and $PS{L}_2(3)$ with $A_4$, of orders 6 and 12 respectively. The first contains its normal subgroup of order 3; the second contains the normal Klein four group. We have proved

The Grassmann graph is connected, and its distance can be determined directly. Put $C=A\cap B$, $r=\dim C$, and $s=k-r$. Choose vectors so thatThe vectors of a basis of $C$, followed by all the $a_i$ and $b_i$, are independent: a relation would identify a vector in the two chosen complements modulo $C$, contradicting $A\cap B=C$. For $0\le j\le s$ defineEach successive pair has an intersection of dimension $k-1$. Thus $A_0=A$ and $A_s=B$ are joined by a path of length $s$.

For the lower bound, if two vertices $U,V$ are adjacent, their common vector subspace $D$ has codimension one in each. Both $\dim (U\cap B)$ and $\dim (V\cap B)$ lie between $\dim (D\cap B)$ and $\dim (D\cap B)+1$. Thus one edge can change the dimension of the intersection with $B$ by at most one. It must change from $r$ to $k$ along a path from $A$ to $B$, requiring at least $k-r$ edges. Therefore

If two ordered pairs have equal distance, their intersections have equal dimension. Construct adapted bases for each pair as above and extend both to bases of the ambient vector space. The invertible linear map between them sends the first ordered pair to the second. It preserves intersection dimensions, so it is a graph automorphism. This proves that the Grassmann graph is distance-transitive, with the required simultaneous images. Its diameter is $k$, because $2k\le n$ permits two disjoint $k$-dimensional vector subspaces.

We first prove the algebraic extension step needed for Hahn-Banach separation theorem. Suppose $p$ is a finite sublinear functional on a real vector space and $g$ is a linear functional on a vector subspace $W$, with $g\le p$ there. For $v\notin W$, choose a real number $c$ betweenThis interval is nonempty: for any $w,z\in W$,Taking $w=0$ or $z=0$ also bounds the endpoints by finite numbers. Define $\tilde{g}(w+tv)=g(w)+tc$. For $t>0$, the upper bound for $c$, applied to $w/t$, proves $\tilde{g}(w+tv)\le p(w+tv)$. For $t<0$, writing $t=-s$ and using the lower bound at $w/s$ proves the same inequality. The case $t=0$ is the original domination. Thus a dominated linear functional extends over one further direction. Order all dominated extensions by extension of their domains. A chain has its union as an upper bound, so Zorn's lemma supplies a maximal extension. The one-direction argument shows its domain is the whole vector space. This proves the required algebraic Hahn-Banach theorem, rather than assuming the separation result.

Choose $a\in A$, and put $C=A-a$. This is a radially open convex set containing zero. It is absorbing: along each line through zero, sufficiently small multiples of any vector lie in $C$. Its Minkowski functionalis finite and nonnegative. Positive homogeneity follows by rescaling $t$. If $x\in sC$ and $y\in tC$, convexity gives $x+y\in (s+t)C$, proving subadditivity by taking infima. Moreover,Indeed, $p(x)<1$ gives $x\in sC$ for some $s<1$, hence $x\in C$ by convexity and $0\in C$. Conversely, radial openness at $x\in C$ permits $(1+ϵ)x\in C$ for some $ϵ>0$, giving $p(x)<1$.

Since $A\cap U=\varnothing$, $a\notin U$ and $u-a\notin C$ for every $u\in U$. Define a linear functional on $U+\mathbb{R}a$ byIt is well defined because $a\notin U$. For $t<0$, write $s=-t$; then $p(u+ta)=sp(u/s-a)\ge s=g(u+ta)$. For $t\ge 0$, domination follows from $g(u+ta)\le 0\le p(u+ta)$. The extension step therefore gives $L:V\to \mathbb{R}$ with $L\le p$, $L{|}_U=0$ and $L(a)=-1$. If $x\in A$, thenso $L(x)<0$. ConsequentlyBecause $L(a)\ne 0$, this kernel of a linear map has codimension one and is a hyperplane. This establishes the geometric Hahn-Banach theorem for radially open sets without assuming a norm or a topology on $V$.

For the convex function, use its strict epigraph in $V\times \mathbb{R}$. It is nonempty and convex, and does not contain $(0,0)$. It is also radially open: on every affine line, a finite convex function is continuous. To justify this last fact, convexity orders secant slopes; on a smaller interval the slopes are bounded above and below by slopes to two fixed exterior endpoints, giving a local Lipschitz bound. Thus $t\mapsto f(x+ty)-\beta -ts$ is continuous, and a strict negative value persists near $t=0$.

Apply the separation theorem with the zero subspace. A nonzero separating linear functional has form $L(x,\beta )=g(x)+c\beta$. Its values on the convex epigraph cannot have both signs, since a line segment would then meet its kernel. Choose the sign so that $L>0$ on the epigraph. At $x=0$ and $\beta >0$ this gives $c>0$. Letting $\beta$ decrease to $f(x)$ yields $g(x)+cf(x)\ge 0$. ThereforeThe resulting linear functional need not be nonzero: for example, the zero functional is the only linear minorant of $f(x)=x^2$ on $\mathbb{R}$.

Now suppose the norm and continuity hypotheses hold. Domination at $x$ and $-x$ givesIf $x_n\in \ker l$ and $x_n\to x$ in norm, apply this inequality to $x-x_n$. Since $l(x-x_n)=l(x)$ and both bounding values tend to $f(0)=0$, we obtain $l(x)=0$. The kernel of a linear map is therefore closed.

Finally, a linear functional with closed kernel on a normed vector space is continuous. If $l=0$, this is immediate. Otherwise choose $v$ with $l(v)=1$ and put $M=\ker l$. Closedness gives $d=\mathrm{dist}(v,M)>0$. For $l(x)\ne 0$, $x/l(x)-v\in M$, so $d\le \Vert x\Vert /|l(x)|$. The same resulting bound holds when $l(x)=0$:This proves continuity and explains exactly how closedness of the kernel is used.

False. Take the separable Banach space $E={\ell }^1$, whose continuous dual space is ${\ell }^{\infty }$. We prove that first countability of the weak unit ball would force this dual to be norm separable, producing a contradiction.

Suppose that $B$ had a countable weak neighbourhood base $(U_n)$ at zero. Inside each $U_n$, choose a basic relative weak neighbourhood specified by finitely many continuous linear functionals $F_n$. The union $F={\bigcup }_n{F}_n$ is countable. For any $f\in E^{\prime }$ and $ϵ>0$, some $U_n$ lies in $\{x\in B:|f(x)|<ϵ\}$. On the vector subspacethis implies $\Vert f{|}_{M_n}\Vert \le ϵ$, by scaling vectors into $B$. The Hahn-Banach theorem extends $f{|}_{M_n}$ to $h\in E^{\prime }$ with $\Vert h\Vert \le ϵ$. The functional $f-h$ vanishes on $M_n$ and is therefore a linear combination of $F_n$: it factors through the finite-dimensional coordinate map $x\mapsto (g(x){)}_{g\in F_n}$, and a linear functional on that map's image extends to the finite-dimensional coordinate space. HenceEvery $f\in E^{\prime }$ thus belongs to the norm closure of $\mathrm{span}F$. Taking rational coefficients, or rational real and imaginary parts, gives a countable norm-dense subset of $E^{\prime }$. This proves that weak-ball metrizability requires a norm-separable dual.

For ${\ell }^1$, every bounded sequence $a=(a_j)$ defines $f_a(x)={\sum }_j{a}_j{x}_j$ with $\Vert f_a\Vert =\Vert a{\Vert }_{\infty }$. Conversely, the values $f(e_j)$ of any bounded linear functional determine such a sequence and recover $f$ by density of finitely supported vectors. Thus $E^{\prime }={\ell }^{\infty }$. The uncountable family $\{0,1{\}}^{\mathbb{N}}$ in ${\ell }^{\infty }$ has distance one between distinct members, so this dual is not norm separable: disjoint balls of radius less than $1/2$ would require distinct members of any countable dense set. Since every metrizable space is first countable, the weak unit ball of ${\ell }^1$ is not metrizable.

Write a finite binary word as $s=(s_1,\dots ,s_n)$ and denote its prefix cylinder set byMore generally, a cylinder set fixes finitely many coordinates, or prescribes a subset of their finite coordinate product, leaving all other coordinates free. Each coordinate factor is discrete, so finite-coordinate inverse images are both open and closed. Thus cylinders are clopen. The basic open sets of the product topology restrict only finitely many coordinates, so cylinders form a base. Any finite-coordinate cylinder is a finite union of prefix cylinders of one sufficiently long common length; the prefix cylinders therefore also form a base.

The Bernoulli space here is Cantor space, with compatible metricIt is a compact metric space: from any sequence, choose successively subsequences constant in the first, second and subsequent coordinates, and take the diagonal subsequence. Agreement on the first $n$ coordinates bounds the distance by $2^{-n}$, proving convergence. Compactness also follows from Tychonoff theorem.

Every cylinder indicator function is continuous because the cylinder is clopen. Hence $S(\Omega )$ is a vector subspace of the space of continuous functions on a compact space. To prove density, fix $f\in C_{\mathbb{R}}(\Omega )$ and $\epsilon >0$. By uniform continuity, choose $n$ so that $d(x,y)\le 2^{-n}$ implies $|f(x)-f(y)|<\epsilon$. Choose a point ${\omega }_s$ in each length-$n$ prefix cylinder and putThen $f_n\in S(\Omega )$ and $\Vert f-f_n{\Vert }_{\infty }<\epsilon$. This proves cylinder-function density in Cantor space:

Positivity of $\varphi$ implies monotonicity. The pointwise bounds $-\Vert f{\Vert }_{\infty }1\le f\le \Vert f{\Vert }_{\infty }1$ therefore giveEvaluation on $1$ attains equality, so $\varphi$ is continuous and has norm one. This is the norm of a positive functional on C(K).

The object extended to open sets is the set function $\ell (C)=\varphi (1_C)$, rather than the functional $\varphi$ itself. Positivity gives $\ell (C)\ge 0$, and linearity gives finite additivity on disjoint cylinders. To construct its extension directly, use the disjoint cylinder decomposition of open subsets of Cantor space. For an open $U$, take the shortest prefixes $s$ for which $[s]\subset U$. They are prefix-free, their cylinders are disjoint, and their union is $U$. DefineFor $U=\Omega$, use the empty prefix; for $U=\varnothing$, use the empty sum.

This value is independent of the chosen disjoint prefix-cylinder decomposition. Indeed, compare two such decompositions $(C_i)$ and $(D_j)$. For each fixed $C_i$, its intersections with the $D_j$ give a disjoint open cover of the compact cylinder $C_i$. Compactness reduces this to finitely many nonempty intersections. Each intersection is a prefix cylinder or is empty, so finite additivity gives $\ell (C_i)={\sum }_j\ell (C_i\cap D_j)$. Sum over $i$ and rearrange the nonnegative double sum. Doing the same with each $D_j$ proves equality of the two totals.

For disjoint open sets $U_k$, combine their disjoint cylinder decompositions to obtain one for ${\bigcup }_k{U}_k$. Rearrangement of nonnegative sums proves countable additivity on this family of open sets. The extension is unique because any countably additive extension must have the prescribed sum on every disjoint cylinder decomposition. Open sets are not themselves a sigma-algebra; countable additivity here concerns disjoint open families and their open union.

For the Borel probability measure, let $\mathcal{A}$ be the algebra of finite unions of prefix cylinders. It is also the algebra of clopen sets, because every clopen set is compact and has a finite cylinder cover. Define $\ell (A)=\varphi (1_A)$ on $\mathcal{A}$. If $A={⨆}_{k\ge 1}{A}_k$ with all sets in $\mathcal{A}$, compactness of $A$ gives a finite subcover by the $A_k$. Disjointness makes all remaining members empty. Thus finite additivity already establishes the premeasure condition.

The Caratheodory extension theorem gives a unique measure $P$ on $\sigma (\mathcal{A})$. This is the Borel sigma-algebra, since the cylinders are a countable base; $P(\Omega )=1$. Its values on open sets agree with the extension above. For cylinder simple functions,Both sides are continuous in the supremum norm, so cylinder-function density in Cantor space givesAny other Borel probability measure with this property agrees on every cylinder indicator function, hence on $\mathcal{A}$, and uniqueness in the Caratheodory extension theorem makes it equal to $P$. This proves the Cantor-space representation of positive functionals without assuming the general representation theorem.

Now let $h:\Omega \to X$ be the supplied continuous surjection. The pullbackis a unital isometric embedding: surjectivity gives $\Vert f\circ h{\Vert }_{\infty }=\Vert f{\Vert }_{\infty }$. On the vector subspace $J(C_{\mathbb{R}}(X))$, define $L(Jf)=\varphi (f)$. This is well-defined, has norm one and satisfies $L(1)=1$.

The real Hahn-Banach theorem extends $L$ to $\tilde{L}$ on all of $C_{\mathbb{R}}(\Omega )$ with the same norm. Norm preservation alone does not automatically mean positivity, so verify it. If $0\le g\le 1$, then $\Vert 1-g{\Vert }_{\infty }\le 1$ andScaling proves positivity for every nonnegative $g$. This is the unital contraction positivity criterion, giving a positive extension from a unital subspace of C(K).

Represent $\tilde{L}$ by the already constructed Borel probability measure $P$ on $\Omega$ and take the pushforward measure $Q=h_{\ast }P$. Continuity of $h$ makes its Borel inverse images measurable, and $Q(X)=1$. The pushforward integral identity givesThis is compact-metric representation by Cantor-space pullback. The hypothesis $\varphi (1)=1$ already excludes an empty $X$. The measure is on the Borel sets; no uniqueness claim on an unspecified larger collection of subsets is needed.

The endomorphisms of the finite-dimensional vector space $V$ over the complex numbers form the general linear Lie algebra $L={\mathrm{End}}_{\mathbb{C}}(V)$ with the usual addition and scalar multiplication and Lie bracket $[x,y]=xy-yx$. This commutator is bilinear and antisymmetric, and expanding the six products verifies the Jacobi identity.

For a Lie subalgebra $\mathfrak{g}=L_1$, being an abelian Lie algebra means $[\mathfrak{g},\mathfrak{g}]=0$. A nilpotent Lie algebra has ${\gamma }_1=\mathfrak{g}$, ${\gamma }_{j+1}=[\mathfrak{g},{\gamma }_j]$ eventually zero; a solvable Lie algebra has ${\mathfrak{g}}^{(0)}=\mathfrak{g}$, ${\mathfrak{g}}^{(j+1)}=[{\mathfrak{g}}^{(j)},{\mathfrak{g}}^{(j)}]$ eventually zero. These are the lower central series of a Lie algebra and derived series of a Lie algebra, respectively. Nilpotence is a condition on the Lie bracket, and does not require every member to be a nilpotent endomorphism: a nonzero scalar multiple of the identity spans an abelian Lie algebra.

A flag of a vector space is an increasing chain of vector subspaces. The flag we construct is a complete flag, $0=V_0\subset V_1\subset \cdots \subset V_d=V$, where $\dim V_i=i$, and each $V_i$ is an invariant subspace for $\mathfrak{g}$. We first prove the common-eigenvector assertion in the Lie theorem, by induction on $\dim \mathfrak{g}$; the zero algebra is immediate. For nonzero solvable $\mathfrak{g}$, its derived algebra is proper, so there is a codimension-one ideal of a Lie algebra $\mathfrak{k}$ containing $[\mathfrak{g},\mathfrak{g}]$. Write $\mathfrak{g}=\mathfrak{k}\oplus \mathbb{C}x$. By induction there are $v\ne 0$ and a linear functional $\lambda$ on $\mathfrak{k}$ such that $kv=\lambda (k)v$ for every $k\in \mathfrak{k}$.

Let $W$ be the cyclic subspace spanned by $v,xv,x^{2}v,\dots$. The commutator derivation identity and $[\mathfrak{k},x]\subseteq \mathfrak{k}$ show inductively thatThus $W$ and all its initial cyclic spans are $\mathfrak{k}$-invariant. If $m=\dim W$, the first $m$ cyclic vectors form a basis, $W$ is also $x$-invariant, and ${\mathrm{tr}}_{W}k=m\lambda (k)$. For $k\in \mathfrak{k}$, the trace of a matrix commutator givesHence $\lambda ([k,x])=0$, since the field has characteristic zero. The nonzero common weight spaceis $x$-invariant: $k(xw)=x(kw)+[k,x]w=\lambda (k)xw$. The restriction of $x$ to $V_{\lambda }$ has an eigenvector, because $\mathbb{C}$ is an algebraically closed field. This is a common eigenvector for $\mathfrak{g}$. Its line is invariant, and repeating the argument on the quotient vector space gives the complete invariant flag. Equivalently, this proves simultaneous triangularization of a Lie algebra representation.

In a basis adapted to this complete flag, every member of $\mathfrak{g}$ is upper triangular, so every member of its derived algebra is strictly upper triangular. Products of $d$ strictly upper triangular matrices vanish, and each iterated Lie bracket of $d$ such matrices is a sum of these products. Consequently the derived algebra is a nilpotent Lie algebra. We may therefore take $L_2=[L_1,L_1]$: it is an ideal, and the quotient Lie algebra $L_1/L_2$ is abelian. This also covers $\dim V=0$.

A linear map $\alpha$ is a nilpotent endomorphism if ${\alpha }^q=0$ for some positive integer $q$, and a semisimple endomorphism if it is diagonalizable over $\mathbb{C}$. Decompose $V$ into its generalized eigenspaces $V_{\lambda }=\ker (\alpha -\lambda I{)}^d$, where $d=\dim V$. Define ${\alpha }_s$ on $V_{\lambda }$ to be $\lambda I$, and set ${\alpha }_n=\alpha -{\alpha }_s$. Then ${\alpha }_s$ is diagonalizable, ${\alpha }_n$ is nilpotent, and both preserve these vector subspaces and commute. ThusFor uniqueness, suppose $\alpha =S+N$ with $S$ semisimple, $N$ nilpotent and $SN=NS$. Both commute with $\alpha$, so preserve each $V_{\lambda }$. On an eigenspace of $S$ of eigenvalue $\mu$ inside $V_{\lambda }$, the map $\alpha$ is $\mu I+N$ and has only the eigenvalue $\mu$. Since the same subspace lies in $V_{\lambda }$, $\mu =\lambda$. Hence $S=\lambda I$ on $V_{\lambda }$, proving $S={\alpha }_s$ and $N={\alpha }_n$. This is the additive Jordan–Chevalley decomposition.

We need a polynomial consequence of this decomposition. Hermite interpolation supplies a polynomial $p$ with $p(\alpha )={\alpha }_s$, by prescribing $p(t)\equiv \lambda (\mathrm{mod}(t-\lambda {)}^d)$ for each eigenvalue. For any endomorphism $T$, its semisimple part can likewise be expressed as a polynomial in $T$ with zero constant term: if zero is an eigenvalue, its interpolation condition already forces this; if not, add the independent condition $p(0)=0$.

On $\mathrm{End}(V)$, the maps $\mathrm{ad}{\alpha }_s$ and $\mathrm{ad}{\alpha }_n$ commute. The first is diagonalizable, with eigenvalues $\lambda -\mu$ on $\mathrm{Hom}(V_{\mu },V_{\lambda })$. The second is nilpotent, sincewhich vanishes for $k\ge 2q-1$ when $N^q=0$. Uniqueness therefore proves adjoint compatibility of additive Jordan decomposition: $\mathrm{ad}{\alpha }_s=(\mathrm{ad}\alpha {)}_s$.

Now assume $\alpha \in M$. The condition $[\alpha ,W]\subseteq U\subseteq W$ implies that both $W$ and $U$ are invariant under $\mathrm{ad}\alpha$, and every polynomial in $\mathrm{ad}\alpha$ with zero constant term maps $W$ into $U$. Define $\beta$ to be multiplication by $\overline{\lambda }$ on $V_{\lambda }$. It commutes with $\alpha$. On $\mathrm{Hom}(V_{\mu },V_{\lambda })$, $\mathrm{ad}\beta$ acts by $\overline{\lambda -\mu }$. Polynomial interpolation on the finite set of differences gives $\mathrm{ad}\beta =q(\mathrm{ad}{\alpha }_s)$ with $q(0)=0$. The preceding paragraph then expresses $\mathrm{ad}\beta$ as a polynomial in $\mathrm{ad}\alpha$ with zero constant term. Consequently $[\beta ,W]\subseteq U$, so $\beta \in M$.

The assumed trace orthogonality nilpotence lemma now follows directly. On $V_{\lambda }$, $\alpha =\lambda I+{\alpha }_n$ and $\beta =\overline{\lambda }I$, while the matrix trace of the nilpotent restriction of ${\alpha }_n$ is zero. HenceEvery summand is nonnegative, so every eigenvalue of $\alpha$ is zero. Its Jordan–Chevalley decomposition therefore has ${\alpha }_s=0$, and $\alpha$ is nilpotent. Notice that neither $U$ nor $W$ was required to be a Lie subalgebra.

True. We prove weak-star topology on an entire infinite-dimensional Banach dual is not metrizable. Suppose instead that $E^{\prime }$ has a countable local base $(W_n)$ at zero. Choose a basic weak-star topology neighbourhood $U_n\subseteq W_n$, with its conditions involving a finite set $S_n\subseteq E$. The $U_n$ still form a local base.

For any $x\in E$, the set $\{\varphi :|\varphi (x)|<1\}$ is a neighbourhood of zero, so some $U_n$ is contained in it. Every functional annihilating $S_n$, and every scalar multiple of that functional, belongs to $U_n$. Consequently every such functional also annihilates $x$. This forcesIndeed a finite-dimensional span is norm closed, and the Hahn-Banach theorem supplies a bounded linear functional vanishing on it and nonzero at any point outside it.

It follows that $E={\bigcup }_n\mathrm{span}{S}_n$. Each span is a proper finite-dimensional closed vector subspace and has empty interior in the infinite-dimensional Banach space $E$. This contradicts the Baire category theorem. HenceThe uniform norm bound that made the metric work on $B^{\prime }$ is absent on the whole dual. The answers to (i), (ii), (iii) and (iv) are therefore all true.

We prove Kantorovich duality theorem here by a positive-functional extension argument; it produces the minimizing transport plan at the same time as equality of the values. Let $Y=X\times X$ and $c(x,y)=d(x,y)$. This is a bounded continuous function because $X$ is a compact metric space. Write $C(Y)$ for the real space of continuous functions on a compact space, and letOn this vector subspace define $\ell (s)=\int fd\mathbf{P}+\int gd\mathbf{Q}$. It is well-defined: two representations differ by a constant in the first variable and the opposite constant in the second, and both measures have mass one. It is positive, since $s\ge 0$ implies $\min f+\min g\ge 0$ and hence $\ell (s)\ge 0$. Also $\ell (1)=1$.

Define the majorant envelope, which is a sublinear functional:Constants majorize every $h$, and positivity gives $\min h\le p(h)\le \max h$, so this quantity is finite. Taking approximate minimizing majorants proves subadditivity; rescaling majorants proves positive homogeneity. Moreover $p(s)=\ell (s)$ for $s\in S$, and subtracting $s$ from a majorant provesThe dual value in the question is exactlySubadditivity at $c+(-c)=0$ gives $m_d\le p(c)$.

On $S+\mathbb{R}c$, assign the value $\ell (s)+t{m}_d$ to $s+tc$. If $c\notin S$ the representation is unique. If $c\in S$, then $m_d=\ell (c)$, so the assignment is still well-defined. It is dominated by $p$: for $t\ge 0$, use $t{m}_d\le tp(c)=p(tc)$; for $t<0$, positive homogeneity gives $p(tc)=(-t)p(-c)=t{m}_d$. Together with the translation identity, these verify domination in both cases.

The real Hahn-Banach theorem extends this functional to $T:C(Y)\to \mathbb{R}$ with $T\le p$. If $h\ge 0$, then $p(-h)\le \ell (0)=0$, so $T(h)\ge 0$. Thus $T$ is a positive linear functional, with $T(1)=1$ and $T(c)=m_d$. Positivity also gives $|T(h)|\le \Vert h{\Vert }_{\infty }$, so $T$ is continuous. The Riesz-Markov-Kakutani representation theorem supplies a Borel probability measure ${\pi }_0$ on the compact space $Y$, satisfying $T(h)={\int }_{Y}hd{\pi }_0$.

For every $f\in C(X)$, its pullback $f(x)$ belongs to $S$, so $\int f(x)d{\pi }_0=\int fd\mathbf{P}$. Likewise $\int g(y)d{\pi }_0=\int gd\mathbf{Q}$. Uniqueness in the Riesz-Markov-Kakutani representation theorem shows that these are exactly the two marginal distributions. Finally, every feasible pair $f+g\le c$ gives a lower bound for the cost of every transport plan, by integration. Our constructed plan achieves that bound:This Kantorovich duality by positive extension establishes the requested existence and equality without assuming an optimal plan in advance.

The metric structure additionally gives the Kantorovich–Rubinstein theorem formulation. For a feasible pair define $h(x)={\inf }_y\{d(x,y)-g(y)\}$. The triangle inequality for $d$ makes $h$ one-Lipschitz continuous, $h\ge f$, and $h(y)\le -g(y)$. Hence the dual objective is at most $\int hd\mathbf{P}-\int hd\mathbf{Q}$. Conversely $(h,-h)$ is feasible for every one-Lipschitz $h$. ThereforeNormalizing $h(x_0)=0$ for a fixed $x_0$ gives a uniformly bounded equicontinuous family; it is closed and compact in the uniform topology by the Arzelà-Ascoli theorem. The objective is uniformly continuous on this family, so the supremum is attained as well. In particular, $m_d$ is the first Wasserstein distance between the two measures.

The algebraic dual space is $V^{\ast }={\mathrm{Hom}}_{\mathbb{R}}(V,\mathbb{R})$, the vector space of all real linear functionals. For a finite basis $e_1,\dots ,e_n$, define the dual basis by $e_i^{\ast }({\sum }_j{a}_j{e}_j)=a_i$, so $e_i^{\ast }(e_j)={\delta }_{ij}$. Every linear functional satisfiesThis equality follows by evaluating both sides on each vector of the basis. It proves spanning directly. If ${\sum }_i{b}_i{e}_i^{\ast }=0$, evaluation on $e_j$ gives $b_j=0$, proving linear independence without assuming a theorem on dual dimensions.

For a vector subspace $U$, its annihilator of a vector subspace isExtend a basis $e_1,\dots ,e_r$ of $U$ to a basis of $V$. Then $U^{\circ }$ has basis $e_{r+1}^{\ast },\dots ,e_n^{\ast }$, so $\boxed{\dim U^{\circ }=n-\dim U}$.

The dual map of $\alpha :V\to W$ is ${\alpha }^{\ast }:W^{\ast }\to V^{\ast }$, $\lambda \mapsto \lambda \circ \alpha$. In the corresponding dual bases its matrix is the transpose of the matrix for $\alpha$. A matrix and its transpose have the same rank, giving $\mathrm{rank}{\alpha }^{\ast }=\mathrm{rank}\alpha$. Directly,Thus $\boxed{\ker {\alpha }^{\ast }=(\mathrm{im}\alpha {)}^{\circ }}$. Also every $\lambda \circ \alpha$ vanishes on $\ker \alpha$, so $\mathrm{im}{\alpha }^{\ast }\subseteq (\ker \alpha {)}^{\circ }$. Both spaces have dimension $\mathrm{rank}\alpha$ by the preceding annihilator of a vector subspace formula and rank-nullity theorem, provingAlternatively, a linear functional vanishing on the kernel descends to the image of $\alpha$ and can be extended to $W$ by extending a basis.

For the infinite-dimensional polynomial ring, the coefficient functionals $L_i$ are linearly independent: evaluate a finite relation on each monomial. However, they do not span the algebraic dual space. The linear functional $p\mapsto p(1)$ takes value one on every monomial, while any finite combination of the $L_i$ vanishes on every sufficiently high-degree monomial. Thus the $L_i$ do not form a basis of $V^{\ast }$. An infinite formal sum of coefficient functionals is not a finite linear combination.

The Weyl complete reducibility theorem states that every finite-dimensional Lie algebra representation of a finite-dimensional complex semisimple Lie algebra is a direct sum of Irreducible Lie algebra representations. Equivalently, every invariant vector subspace has an invariant complement.

We use the permitted Casimir operator properties in the following precise form. There is a central quadratic operator $C$ commuting with the action on every module; it is zero on the trivial Lie algebra representation, and on every nontrivial finite-dimensional Irreducible Lie algebra representation it is a nonzero scalar $c$. This follows from the Schur lemma and the Casimir eigenvalue $c_{\lambda }=(\lambda ,\lambda +2\rho )$, using the Killing form normalization. We also use the permitted one-dimensional-representation fact: a complex semisimple Lie algebra has only trivial one-dimensional representations. Equivalently, it is a perfect Lie algebra, $[\mathfrak{g},\mathfrak{g}]=\mathfrak{g}$, so a character $\mathfrak{g}\to \mathbb{C}$ annihilating brackets must vanish. Neither fact assumes complete reducibility of the module being proved reducible.

First prove that every finite-dimensional short exact sequencewith trivial quotient splits. If $W$ is nontrivial irreducible, the Casimir operator $C_E$ has image in $W$ and restricts to $c{I}_W$ there. Consequently $\ker C_E$ is a one-dimensional invariant complement to $W$. If $W$ is trivial irreducible, $E$ has a basis in which every action is $\left(\begin{array}{cc}0& a(x)\\ 0& 0\end{array}\right)$. The Lie algebra representation identity makes $a([x,y])=0$, so the one-dimensional-representation fact gives $a=0$ and again the sequence splits.

For general $W$, induct on $\dim W$. Choose an irreducible submodule $W_1\subseteq W$. The induced sequence with kernel $W/W_1$ and middle term $E/W_1$ splits by induction. The inverse image $E_1\subseteq E$ of its invariant complement is a submodule fitting into $0\to W_1\to E_1\to \mathbb{C}\to 0$. The irreducible-kernel case gives an invariant line in $E_1$ mapping isomorphically to the quotient. It is also an invariant complement to $W$ in $E$. The case $W=0$ starts this induction. This proves splitting of a trivial quotient for a semisimple Lie algebra, including kernels that are not assumed completely reducible.

Now let $U\subseteq V$ be any invariant subspace. On the Hom representation $\mathrm{Hom}(V,U)$ the action isLet $E$ consist of the maps whose restriction to $U$ is a scalar multiple of $I_U$. This is a submodule, and restriction givesThe right-hand map is surjective because an ordinary linear projection $V\to U$ exists; its quotient action is trivial because a commutator with $I_U$ is zero. The splitting just proved supplies an invariant $T\in E$ with $T{|}_U=I_U$. Thus $T$ intertwines the actions, $T^2=T$, andThe cases $U=0$ and $U=V$ are immediate. Choosing an irreducible submodule and repeating this complement construction proves the Weyl complete reducibility theorem. This last step is invariant complement from an equivariant projection.

Dropping finite dimensionality gives an example with the complex simple Lie algebra ${\mathfrak{sl}}_2$. Its Verma module of highest weight zero has a basis $v_0,v_1,\dots$ withwhere $e{v}_0=0$. These actions obey $[h,e]=2e$, $[h,f]=-2f$, $[e,f]=h$. The span of $v_1,v_2,\dots$ is a proper submodule and the quotient is trivial. It has no invariant complement: such a complement would be a trivial line, while $f$ is injective on the entire module. Thus a simple Lie algebra can have an infinite-dimensional representation that is not completely reducible, even over $\mathbb{C}$.

Let ${\Pi }_G$ be the orthogonal projection onto the good vector subspace $G$, and let the normalized input be $|\psi ⟩$. Set $p=\Vert {\Pi }_G\psi {\Vert }^2={\sin }^2\theta$ with $0<\theta <\pi /2$, and defineAssuming coherent access to the two reflections, the amplitude amplification theorem states thatpreserves the plane spanned by $|g⟩,|b⟩$ and acts as a rotation, givingHence the good-outcome probability is ${\sin }^2((2j+1)\theta )$. When $p$ is known, choose the nearest nonnegative integer to $\pi /(4\theta )-1/2$. The resulting angle is within $\theta$ of $\pi /2$, so the good probability is at least ${\cos }^2\theta =1-p$. For small $p$ this is close to one and requires $O(1/\sqrt{p})$ iterations. Exact success occurs when $(2j+1)\theta =\pi /2$. Known $p$ also allows exact amplitude amplification by ancilla qubit dilution or selective phase adjustment when ordinary integer iterations would overshoot.

If $|\psi ⟩=A|0⟩$ has a known coherent preparation, its reflection is implemented with $A$, $A^{†}$ and a zero-state phase flip. A coherent membership test supplies the reflection about $G$. Merely possessing an unknown copy of $|\psi ⟩$ does not automatically supply its reflection. For $p=1$ the state is already good; for $p=0$ this two-reflection construction cannot generate a good component. These cases delimit the theorem's algorithmic assumptions.

Use the Hausdorff convention for a locally convex space: a real vector space $X$ is equipped with a separating family $\mathcal{P}$ of seminorms. Thus each $p$ is nonnegative, subadditive and satisfies $p(tx)=|t|p(x)$, and for every $x\ne 0$ some $p\in \mathcal{P}$ has $p(x)>0$. The topology has a neighborhood basis at $a\in X$ consisting ofEquivalently, it is the coarsest vector-space topology making all these seminorms continuous. The separating condition is precisely what makes this topology Hausdorff. If the Hausdorff requirement were omitted, continuous linear functionals could not distinguish points in the common kernel of the seminorms.

The continuous dual space $X^{\ast }$ consists of all continuous linear functionals $X\to \mathbb{R}$. We first prove the needed Hahn-Banach theorem. Let $p$ be a real-valued sublinear function on $X$, meaning $p(a+b)\le p(a)+p(b)$ and $p(ta)=tp(a)$ for $t\ge 0$, and let $g$ be linear on a vector subspace $M$, with $g(m)\le p(m)$. To extend across $v\notin M$, writeThe necessary bounds on $c$ areThey are compatible becauseTaking one variable equal to zero also shows that $L,U$ are finite. Choose $c\in [L,U]$. The upper bound proves domination when $t>0$, after dividing $m+tv$ by $t$; the lower bound proves it when $t<0$, after dividing by $-t$. Thus $\tilde{g}\le p$ on $M+\mathbb{R}v$. This is the one-dimensional dominated extension of a real linear functional.