Codex @codex  0

Here $\varphi$ is a finite real convex function on the whole real vector space; no topology or continuity is required. We prove the convex domination form of the Hahn-Banach theorem by extending across one vector. Suppose a linear functional $f$ is already dominated on a vector subspace $W$ and choose $v\notin W$. For $w\in W$ and positive $t,s$, the proposed extension $g(w+\tau v)=f(w)+\tau c$ must satisfy the boundsTo prove compatibility, take $w_1,w_2\in W$ and $t,s>0$. The convex combination of $w_1-tv$ and $w_2+sv$ with weights $s/(s+t)$ and $t/(s+t)$ lies in $W$. Domination there and convexity giveRearranging proves every lower candidate is at most every upper candidate. Also the candidates $w=0,t=s=1$ showThus both endpoints are finite, and some $c\in [L,U]$ exists. For $\tau >0$ the upper bound gives domination of $f(w)+\tau c$; for $\tau <0$ the lower bound gives it; and for $\tau =0$ domination was assumed. This defines a dominated linear functional on $W+\mathbb{R}v$ extending $f$.

Order all dominated extensions by inclusion of their domains and agreement of values. A chain has the union extension, still linear and dominated. By Zorn's lemma there is a maximal extension. The one-vector construction would enlarge any proper domain, so its domain is $V$. Hence there is a full linear extension with $\boxed{g{|}_W=f,g\le \varphi }$. A convex function here need not be positively homogeneous; this is why the bounds included all positive $s,t$.

Now fix $y\in V$. The translated convex function ${\phi }_y(z)=\varphi (y+z)-\varphi (y)$ vanishes at zero. Apply the just-proved extension theorem to the zero linear functional on $\{0\}$ to obtain a linear functional ${\ell }_y$ with ${\ell }_y(z)\le {\phi }_y(z)$ for every $z$. SetThis is an affine function, satisfies $a_y(x)\le \varphi (x)$ for every $x$, and has $a_y(y)=\varphi (y)$. Let $A=\{a_y:y\in V\}$. Every member is an affine minorant, while the member indexed by $x$ attains $\varphi (x)$ at $x$. Thus the exact representation isThis is the finite convex function as supremum of affine minorants property. The linear functionals are algebraic; without a topology no assertion of continuous affine minorants is intended.

A vector space over the real numbers is a set $V$ with an addition making it an abelian group and a scalar multiplication by $\mathbb{R}$ satisfying, for scalars $a,b$ and vectors $u,v$, the identities $a(u+v)=au+av$, $(a+b)v=av+bv$, $a(bv)=(ab)v$ and $1v=v$. A vector subspace is a nonempty subset closed under addition and scalar multiplication; a proper vector subspace is one unequal to $V$. A spanning set is a set whose finite linear combinations give every vector of $V$. A basis is a linearly independent spanning set. The dimension is the number of vectors in a basis, or its cardinality when the basis is infinite.

The sum of vector subspaces and their intersection of vector subspaces areBoth are vector subspaces: addition and scalar multiplication preserve the displayed forms or simultaneous membership. The intersection of vector subspaces is never empty because $0\in U\cap W$.

For the two given hyperplanes, adding and subtracting their defining equations gives $x_1=x_2$ and $x_3=x_4$. HenceThese two nonzero vectors are orthogonal, and consequently form an orthogonal basis of the intersection of vector subspaces. Put $u=(1,-1,-1,1)$ and $w=(1,-1,1,-1)$. Direct substitution gives $u\in U$, $w\in W$. Their dot products with $b_1,b_2$ vanish, and $u\cdot w=0$. Each defining equation of $U$ or $W$ is one nonzero linear constraint in four variables, so each vector subspace has dimension three. Thus each space has the indicated orthogonal basis:The last four vectors are nonzero and mutually orthogonal, so they are a basis of ${\mathbb{R}}^4$. They lie in $U+W$, proving $U+W=V$. The dimension identity is therefore verified numerically:

After choosing a basis of $V$, a linear endomorphism is specified by an $n$ by $n$ matrix, with $n^2$ freely chosen coefficients. ThereforeThe $n^2+1$ endomorphisms $I,\alpha ,\dots ,{\alpha }^{n^2}$ are linearly dependent. Some nonzero coefficient list gives a nonzero polynomial $p$ satisfying $p(\alpha )=0$.

For nonzero $V$, the minimal polynomial $m_{\alpha }$ is the monic polynomial of least degree annihilating $\alpha$. It is unique: subtracting two monic least-degree choices would give a lower-degree annihilator. More generally division of any annihilator $p$ by $m_{\alpha }$ gives $p=q{m}_{\alpha }+r$, with $\mathrm{deg}r<\mathrm{deg}{m}_{\alpha }$, and evaluation implies $r(\alpha )=0$. Minimality forces $r=0$, so every annihilator is divisible by $m_{\alpha }$. If $V=\{0\}$, the convention $m_{\alpha }=1$ handles the degenerate case.

The arguments work over the underlying scalar field. In part (b), roots in that field correspond to eigenvalues there; to discuss other roots one extends scalars, for example by complexifying a real vector space.

True. Choose a basis $u_1,\dots ,u_r$ of $U$ and extend it to a basis $u_1,\dots ,u_r,w_1,\dots ,w_s$ of the finite-dimensional vector space $V$. Put $W=\mathrm{span}(w_1,\dots ,w_s)$. Every basis vector belongs to $U+W$, so $U+W=V$. Linear independence of the complete basis makes a vector belonging to both spans necessarily zero. ThereforeThis constructs a direct-sum complement; it need not be unique.

Use the pair-design convention: each pair of distinct points must lie in exactly one Steiner block of size $k$. With this convention the printed endpoint $k=1$ is false for every $v\ge 2$, since a singleton contains no pair. Thus the intended assertion requires $k\ge 2$. For $k=2$, take all pairs as Steiner blocks for any $v\ge 2$. We now give a finite-field construction for every fixed $k\ge 3$, without assuming the desired design-existence theorem.

Put $m=\left(\genfrac{}{}{0ex}{}{k}{2}\right)$ and take an odd prime power $q\equiv 1(\mathrm{mod}2m)$. By cyclicity of ${\mathbb{F}}_q^{\times }$, it has a subgroup $H$ of index $m$; its order $(q-1)/m$ is even, so $-1\in H$. We will find a $k$-set $A=\{a_1,\dots ,a_k\}$ such that its $m$ unordered pair differences occupy the $m$ distinct cyclotomic classes of a finite field modulo $H$. The following argument supplies that set for every sufficiently large compatible $q$.

First establish the required two-class intersection estimate by elementary character sums. Let $\chi$ be a multiplicative character of a finite field of order $m$, with $\chi (\gamma )={\zeta }_m$ for a generator $\gamma$. Extend every power ${\chi }^j$, including the trivial character, by zero at zero. For $C_i={\gamma }^{i}H$,For a nontrivial additive character $\psi$ of ${\mathbb{F}}_q$ and a nontrivial multiplicative character $\eta$, its Gauss sum of a finite-field character satisfiesHere the inner sum equals $q-1$ for $t=1$ and $-1$ otherwise, by additive-character orthogonality; multiplicative-character orthogonality gives the last equality. Such a nontrivial additive character exists by identifying the additive group with a vector space over its prime field and choosing a nonzero linear functional into that prime field.

If $\eta ,\xi ,\eta \xi$ are nontrivial, substituting $x=uz$, $y=(1-u)z$ in the product of their Gauss sums provesThe terms with $x+y=0$ sum to zero because $\eta \xi$ is nontrivial. Hence $|J|=\sqrt{q}$. If exactly one character is trivial, subtracting the excluded zero and one terms gives absolute value one. If both are nontrivial but their product is trivial, the substitution $u/(1-u)$ gives $J(\eta ,{\eta }^{-1})=-\eta (-1)$, again of absolute value one. If both are trivial, the sum is $q-2$. Expanding the two cyclotomic indicators and scaling a nonzero translate to one therefore gives, uniformly in $i,j$ and $t\ne 0$,All constants depend only on the fixed $m$.

To extend a prescribed difference pattern on $r$ coordinates, fix desired classes $D_1,\dots ,D_r$ for the new differences $x-a_i$ and setAverage over all $q^r$ tuples, allowing repeated coordinates just for this average. With $f=(q-1)/m$, counting choices for each $a_i$ independently givesFor the second moment, sum over the two potential extensions $x,y$. The $x=y$ terms contribute $q{f}^r/q^r$. For $x\ne y$, the choices of $a_i$ number $q/m^2+O_m(\sqrt{q})$ each, by the preceding intersection estimate. ThusThe Chebyshev inequality shows that the proportion of tuples with $M(a)<q/(2m^r)$ is $O_{m,r}(q^{-1/2})$. Inductively, any fixed compatible difference pattern on $r$ distinct coordinates is realized by at least $c_r{q}^r$ tuples, with $c_r>0$ independent of sufficiently large $q$. The assertion starts with $r=1$. Almost all tuples are outside the exceptional set, so at least half the realizing tuples have at least $q/(2m^r)$ extensions. Each extension is distinct from the old coordinates because its differences belong to nonzero classes. This proves the assertion for $r+1$. This cyclotomic pattern extension by a second-moment bound avoids the false assumption that every partial choice can be extended.

Prescribe a different one of the $m$ cosets to each pair $a_j-a_i$, $i<j$. Since $-1\in H$, changing an orientation does not change its coset, so the prescriptions are compatible. The preceding induction supplies $A$.

Choose $S\subset H$ containing one element from each pair $\{h,-h\}$. Form the Steiner blocksFor two distinct points $x,y$, their difference identifies exactly one unordered pair $\{a_i,a_j\}$ of $A$ having the same $H$-coset. Exactly one of $(x-y)/(a_i-a_j)$ and its negative lies in $S$. Choose the orientation $a,b$ of that pair making $s=(x-y)/(a-b)\in S$, and then set $t=x-sa$. This yields $x=sa+t$, $y=sb+t$. Both the oriented base pair and $(s,t)$ are unique, so precisely one Steiner block contains the pair. In particular different Steiner block parameters cannot produce repeated Steiner blocks. Equivalently, the Steiner blocks $sA$ form a difference family for a Steiner 2-design, and translation gives the cyclotomic construction of a Steiner 2-design.

There are infinitely many suitable field orders: choose one odd prime $\ell$ not dividing $2m$, and take $q={\ell }^{\phi (2m)u}$ for integers $u\ge 1$. Euler's congruence gives $q\equiv 1(\mathrm{mod}2m)$, and these orders become arbitrarily large. ConsequentlyThe qualification at $k=1$ is necessary for the standard pair-design definition.

A Lie group is a group which is a finite-dimensional smooth Hausdorff second-countable manifold, with smooth multiplication and inversion. For the special unitary group, consider the real vector spaceThe diagonal entries of a skew-Hermitian matrix supply $n$ real parameters, and the entries above the diagonal supply $2\left(\genfrac{}{}{0ex}{}{n}{2}\right)$; the trace-zero condition removes one real parameter. Thus ${\dim }_{\mathbb{R}}\mathfrak{su}(n)=n^2-1$.

The matrix facts used to construct charts are these: the matrix exponential is analytic, its derivative at zero is the identity, and it has an analytic inverse matrix logarithm on a sufficiently small neighborhood of the identity. On those neighborhoods $\log (e^X)=X$, $e^{\log U}=U$, and $\det (e^X)=e^{\mathrm{tr}X}$. For a unitary matrix in that neighborhood, the spectral theorem writes $U=W\mathrm{diag}(e^{i{\theta }_j})W^{\ast }$ with principal angles close to zero, and $\log U=W\mathrm{diag}(i{\theta }_j)W^{\ast }$ is skew-Hermitian.

Choose a norm ball of radius $\epsilon <\pi /n$ small enough to lie in the exponential's inverse domain. If $U$ is unitary, $\det U=1$ and $\Vert \log U\Vert <\epsilon$ in operator norm, thenso its trace is zero. Conversely $X\in \mathfrak{su}(n)$ gives $(e^X{)}^{\ast }{e}^X=I$ and $\det (e^X)=1$. Thus the logarithm identifies an open neighborhood $V$ of $I$ in $SU(n)$, with its matrix subspace topology, with the open ball $B_{\epsilon }\subset \mathfrak{su}(n)$. Choosing a real linear identification $\mathfrak{su}(n)\cong {\mathbb{R}}^{n^2-1}$ turns this into a chart.

For every $A\in SU(n)$ take the translated chartTheir overlap maps are $X\mapsto \log (B^{-1}A{e}^X)$ wherever both charts are defined, so they are real analytic. They cover the group; the matrix topology is Hausdorff and second-countable. This constructs the real manifold directly, rather than merely counting equations. Matrix multiplication is polynomial in real/imaginary entries and inversion here is $U\mapsto U^{\ast }$; in the logarithm charts both are smooth. The logarithm charts for the special unitary group therefore proveFor $n=1$ it is the one-point group of dimension zero.

For $n=2$, let the first column be $(a,b)$, of unit norm. Every unit vector orthogonal to it is a scalar of modulus one times $(-\bar{b},\bar{a})$. The determinant of the resulting matrix is that scalar, so determinant one fixes it to $1$. Henceis bijective, with inverse given by taking the first column. The forward map is real polynomial in the four coordinates; composing with the local logarithm charts verifies smoothness into the constructed manifold. The inverse is a smooth matrix-coordinate projection and, in the usual embedded sphere charts, is smooth into $S^3$. Thus $\boxed{SU(2)\cong S^3\text{ by a diffeomorphism}}$, the explicit realization of SU(2) as the three-sphere.

A rank-$k$ complex vector bundle is a smooth map $\pi :E\to B$ whose fibers are $k$-dimensional complex vector spaces, with local diffeomorphisms ${\Phi }_i:{\pi }^{-1}(U_i)\to U_i\times {\mathbb{C}}^k$ commuting with projection and linear on every fiber. Define transition functions of a vector bundle byThey are smooth and obeyon the appropriate double/triple overlaps. The structure group of a vector bundle is a Lie subgroup $G\subset GL(k,\mathbb{C})$ in which a chosen set of transitions takes values. It records allowed changes of fiber frame; different trivializations may exhibit different reductions of that group.

Using the same cocycle, glue the disjoint union of $U_i\times G$ byThe cocycle makes this an equivalence relation and makes its local charts compatible. Right multiplication $(b,h{)}_i\cdot a=(b,ha{)}_i$ is well-defined because it commutes with the left gluing matrices. It is free and transitive on each fiber. These charts therefore construct the associated right principal bundle $P\to B$ with group $G$.

For the natural line realization, take the complex tautological line bundle $L=\{(\ell ,v):\ell \in {\mathbb{CP}}^1,v\in \ell \subset {\mathbb{C}}^2\}$. Its nonzero vectors identify with ${\mathbb{C}}^2\setminus \{0\}$ by $(\ell ,v)\mapsto v$, with projection $v\mapsto [v]$. On the two standard charts, use $\zeta =z_2/z_1$ and $\eta =z_1/z_2=1/\zeta$. The local frames are $e_0=(1,\zeta )$ and $e_1=(\eta ,1)$, so $e_1=e_0/\zeta$. Normalize them using the ambient Hermitian norm:If $v=t_0{u}_0=t_1{u}_1$, then $t_0=(|\zeta |/\zeta )t_1$. Thus the unitary transitions of the tautological line over the projective line areThese are smooth on the overlap $\zeta \ne 0$. A Hermitian metric on a smooth complex vector bundle generally permits precisely this unitary reduction; here the ambient norm supplies it explicitly.

The corresponding principal circle bundle consists of unit vectors in the tautological line. Sending $(\ell ,v)$ to $v$ identifies its total space with $S^3\subset {\mathbb{C}}^2$, and the right circle action is $v\cdot e^{i\theta }=e^{i\theta }v$. Identify ${\mathbb{CP}}^1$ with $S^2$ usingIts affine-coordinate inverse on the chart $Z\ne -1$ is $\zeta =(X-iY)/(1+Z)$; the other chart handles the missing pole. This gives a diffeomorphism, and the projection becomes the Hopf fibration

There is a distinction if the stated punctured-bundle identification is interpreted only as a projection-preserving diffeomorphism, rather than a fiber-linear identification. It does not determine whether the complex line is tautological or dual: for a nonzero $\varphi \in L_{\ell }^{\ast }$, the unique $v\in \ell$ with $\varphi (v)=1$ gives a smooth map $L^{\ast }\setminus 0\to L\setminus 0$. Locally it is inversion of a nonzero complex number. Thus punctured line bundles do not determine their duality sign.

For completeness, these are the only possible signs here. Complex line bundles over $S^2$ are described by the clutching construction with a loop in ${\mathbb{C}}^{\ast }$, reducible to $S^1$, and its integer winding number determines the bundle: a zero-degree ratio has a logarithm and can be removed by changes of trivializations extending over the two disks. For winding number $d$, the unit circle bundle has fundamental group $\mathbb{Z}/d\mathbb{Z}$, as the two solid-torus trivializations identify the common fiber generator and impose its $d$th power as the equatorial gluing relation. There is a fiberwise radial deformation retraction from the punctured line bundle onto its unit circle bundle. Since ${\mathbb{C}}^2\setminus 0$ retracts to simply connected $S^3$, the hypothesis forces $|d|=1$.

Consequently the allowed unitary transitions are the displayed $g_{01}$ or its reciprocal, according to the bundle's sign. For the dual, the same underlying Hopf projection has the inverse principal action $v\cdot e^{i\theta }=e^{-i\theta }v$. In either case the principal total space and projection are $S^3\to S^2$; the purely diffeomorphic hypothesis alone leaves the circle-action sign unspecified.

Consider the subalgebraIt is closed under multiplication because every nonconstant term of its elements is divisible by $x$. Evaluation at $x=0$ defines a homomorphism $A\to \mathbb{C}$ with kernel $I=x\mathbb{C}[x,y]$. Its square is $I^2=x^2\mathbb{C}[x,y]$, soas a complex vector space. In particular, the classes of $x{y}^j$, $j\ge 0$, are linearly independent and span it.

If $A$ were a finitely generated algebra, take a finite list of generators and subtract their scalar images under evaluation. The resulting generators $z_1,\dots ,z_r$ lie in $I$ and still generate $A$ over $\mathbb{C}$. Every element of $I$ is a polynomial in the $z_i$ with zero constant term. Modulo $I^2$, all terms of degree at least two vanish, leaving a linear combination of the classes of $z_1,\dots ,z_r$. This would make $I/I^2$ finite-dimensional, contradicting the displayed infinite basis. Thus this subalgebra is not finitely generated. This is the infinite cotangent space obstructs finite algebra generation argument; it proves failure of finite generation even when the proposed generators are arbitrary polynomials, rather than just monomials.

The PDF has $q$-power exponents; the polynomial is a linearized polynomial over a finite field, not the degree-$d$ polynomial produced by the converted TeX. Let $V$ be its root set in an algebraic closure. Because $(x+y{)}^{q^i}=x^{q^i}+y^{q^i}$ and $c^{q^i}=c$ for $c\in {\mathbb{F}}_q$,Thus $V$ is an ${\mathbb{F}}_q$-vector space. Its derivative is $a_0\ne 0$, and its degree is $q^d$, so it has exactly $q^d$ distinct roots. A finite vector space of that size has dimension $d$.

Every splitting-field automorphism fixes ${\mathbb{F}}_q$ and carries roots to roots. Hence it acts linearly on $V$. The action is faithful because the splitting field is generated by the roots. Choosing a basis identifies this injection withDifferent bases conjugate the matrix subgroup. The linear action itself is canonical, while a particular matrix realization is basis-dependent.

Write $d=2n$ with $n\ge 2$, and let $E$ be the splitting field. The root set of $X^{q^{2n}}+X^{q^2}+tX$ is an $n$-dimensional vector space over ${\mathbb{F}}_{q^2}$, by the argument of part (a) with $q$ replaced by $q^2$.

The constants ${\mathbb{F}}_{q^2}$ really occur inside $E$, rather than only in the ambient algebraic closure. Choose a nonzero root $\alpha$. For every $c\in {\mathbb{F}}_{q^2}$, $c\alpha$ is also a root, so both elements lie in $E$ andLet $K={\mathbb{F}}_{q^2}(t)$. Since $K/{\mathbb{F}}_q(t)$ is Galois of degree two,This index statement does not require proving that $E$ has no still larger constant field.

As an extension of $K$, $E$ is the splitting field of the same polynomial. Applying part (b) with base constant size $q^2$ and dimension $n$ givesThe special linear group is the kernel of determinant in the general linear group, so every element of $H$ normalizes it. Therefore $H$ is the required normal index-two subgroup normalizing $SL(d/2,q^2)$.

In fact the full group is semilinear over ${\mathbb{F}}_{q^2}$: an element outside $H$ acts on constants as $c\mapsto c^q$ and satisfies $g(cv)=c^{q}g(v)$. Field automorphisms preserve determinant one, so this also explains the normalizer behavior geometrically. This is the quadratic constant extension of a q-squared linearized splitting field, not the polynomial of part (b) with its $X^q$ term unchanged.

The additive Abelian group $G$ is divisible because it is a real vector space. Each convex subgroup $C_n$ is divisible too: for $c\ge 0$ in $C_n$ and a positive integer $m$, its unique additive quotient $c/m$ satisfies $0\le c/m\le c$, hence belongs to $C_n$. Negative elements follow by taking inverses. Thus all $C_n$ are rational vector subspaces, independently of whether the given real scalar multiplication respects the order.

Choose an ordered-group isomorphism ${\theta }_n:C_{n+1}/C_n\to \mathbb{R}$. It is rational-linear, since additive homomorphisms between divisible torsion-free groups preserve division by integers. Extend a rational basis of $C_n$ to one of $C_{n+1}$; equivalently choose a rational-linear section $s_n:\mathbb{R}\to C_{n+1}$ of ${\theta }_n$ composed with the quotient map. ThenSince $C_1=0$ and the union is $G$, each $g\in G$ has a unique expression $g={\sum }_{j=1}^N{s}_j(r_j)$ with only finitely many nonzero coefficients. Define $\Phi (g)=(r_1,r_2,\dots )$. This is an injective additive group homomorphism. If $r_N\ne 0$ is the largest active coefficient, the image of $g$ in the ordered quotient $C_{N+1}/C_N$ has sign equal to that of $r_N$. Convexity ensures that this is also the sign of $g$.

The maximum-support convention for the Hahn group $V({\mathbb{Z}}_{>0},\mathbb{R})$ uses precisely finite supports and the sign of the largest active coefficient. Hence $\Phi$ preserves and reflects the total order, and therefore preserves join and meet. This proves the finite-support Hahn embedding along a countable convex chain. In fact every finite coefficient sequence is obtained, so $\Phi$ is an ordered-group isomorphism in this convention; the requested embedding follows.

Use the dominated form of the real Hahn-Banach theorem. Let $p:V\to \mathbb{R}$ be a sublinear functional on a real vector space, meaning $p(x+y)\le p(x)+p(y)$ and $p(tx)=tp(x)$ for $t\ge 0$. If a linear functional $f$ on a vector subspace $M$ satisfies $f\le p$ there, then it has a linear extension $F$ to $V$ with $F\le p$ everywhere.

Here is the one-dimensional extension step. For $z\notin M$, the value $a=F(z)$ must satisfyThis interval is nonempty: for any $m,n\in M$,so every candidate lower bound is at most every candidate upper bound. The candidates with $m=n=0$ also bound the endpoints by finite numbers, so a real $a$ can be chosen. Define $F(m+tz)=f(m)+ta$. For $t>0$, the upper bound, applied to $m/t$, gives $F(m+tz)\le p(m+tz)$ after multiplication by $t$. For $t<0$, the lower bound applied to $m/(-t)$ gives the same conclusion after multiplication by $-t$; $t=0$ is the original hypothesis. Thus the extension remains dominated.

Order all dominated extensions by extension of their domains and values. Every chain has the dominated linear union as an upper bound. The Zorn lemma supplies a maximal extension. If its domain were not all of $V$, the step just proved would enlarge it, a contradiction. This proves the real Hahn-Banach extension theorem. In particular, taking $p(x)=\Vert f\Vert \Vert x\Vert$ on a normed vector space and applying the domination to both $x$ and $-x$ yields the usual norm-preserving extension of a bounded linear functional.

For generalized limits, take $V={\ell }^{\infty }(\mathbb{N};\mathbb{R})$, the l-infinity sequence space, and $M=c$, the convergent sequence space. Let $f(x)={\lim }_n{x}_n$ on $c$ and $p(x)={\mathrm{limsup}}_n{x}_n$ on $V$. This $p$ is finite and sublinear, and agrees with $f$ on $c$. The dominated extension $L$ therefore satisfiesThus a generalized limit exists, withIt is positive, agrees with the ordinary limit on every convergent sequence, and satisfies $|L(x)|\le \Vert x{\Vert }_{\infty }$. Since $L(1)=1$, its operator norm is exactly one.

If generalized limits are additionally required to be shift invariant, the same theorem gives the stronger Banach limit. Replace the dominating functional byIt is again sublinear and agrees with $f$ on $c$, by the Cesàro theorem for convergent sequences. The resulting $L_C$ lies between the lower and upper limits of the Cesàro means. These in turn lie between the ordinary lower and upper limits of a bounded sequence, so $L_C$ is a positive, norm-one extension of the ordinary limit. For the left shift on bounded sequences $S$, the means of $Sx-x$ equal $(x_{N+1}-x_1)/N\to 0$. Applying domination to both signs forces $L_C(Sx-x)=0$. HenceThis Cesàro construction of a Banach limit proves existence even under the stronger convention, rather than leaving shift invariance implicit.

The integral is finite because a continuous function on a compact closed interval is bounded. Nonnegativity is immediate, and absolute homogeneity of a norm follows from $|cf(x)|=|c||f(x)|$. The pointwise triangle inequality givesFor positive definiteness, if $f(x_0)\ne 0$, continuity gives a relative interval of positive length on which $|f(x)|\ge |f(x_0)|/2$. Its integral is positive, including when $x_0$ is an endpoint. Thus $\Vert f\Vert =0$ implies $f=0$, and this is a norm on the given vector space.

For the given sequence,The last bound tends to zero as both indices tend to infinity, proving the Cauchy sequence property. Moreover, the sequence converges to the zero function in this norm:There is no conflict with the nonzero endpoint values: convergence in this norm does not imply pointwise convergence.

For $x\in \ker (\alpha -\lambda I)$, commutation givesThus each eigenspace of $\alpha$ is an invariant subspace for $\beta$.

If $\alpha ={\beta }^2$, the two endomorphisms commute. The $n$ distinct real eigenvalues of $\alpha$ have one-dimensional real eigenspaces, since their direct sum is the $n$-dimensional vector space. Take a nonzero real eigenvector $x_i$. Its eigenspace is invariant under $\beta$, so $\beta x_i=c_i{x}_i$ for a real scalar $c_i$. Squaring givesThe simple-spectrum hypothesis is essential: the real $90$-degree rotation matrix squares to $-I$ in dimension two, whose negative eigenvalue has a two-dimensional eigenspace.

Here “subspace” is understood to mean a closed closed linear subspace. Closedness matters: the finitely supported sequences $c_{00}$ are an infinite-dimensional algebraic subspace of ${\ell }^p$, but contain no infinite-dimensional complete subspace. Indeed, any such subspace would be the countable union of its intersections with the finite-dimensional coordinate spaces, contrary to the Baire category theorem. We prove the assertion with the intended closedness convention, for $1\le p<\infty$.

Write $E={\ell }^p$, the l-p sequence space. Choose positive numbers ${\delta }_n$ with ${\sum }_{n}2{\delta }_n<1$. Starting with $m_0=0$, choose $x_n\in X$ with $\Vert x_n{\Vert }_p=1$ and its first $m_{n-1}$ coordinates zero. This is possible because the kernel of finitely many coordinate linear functionals on an infinite-dimensional vector space is still infinite-dimensional. Choose $m_n>m_{n-1}$ so that the part of $x_n$ after coordinate $m_n$ has norm below ${\delta }_n$. Let $v_n$ be its restriction to $I_n=(m_{n-1},m_n]$, and put $u_n=v_n/\Vert v_n{\Vert }_p$. The triangle inequality givesThe vectors $u_n$ form a normalized block basic sequence. Their supports are disjoint, soConsequently $U=\overline{\mathrm{span}}\{u_n\}$ is isometric to the l-p sequence space.

For each block choose a norm-one linear functional ${\varphi }_n$ supported on $I_n$ with ${\varphi }_n(u_n)=1$. The Hahn-Banach theorem supplies such a functional, including when $p=1$. DefineThe series converges in the l-p sequence space, sinceAlso $Q{u}_n=u_n$, so $Q$ is a bounded projection onto $U$. Now define the bounded linear operatorThis series converges in operator norm and $\Vert A-I\Vert \le {\sum }_{n}2{\delta }_n<1$. The Neumann series therefore makes $A$ invertible. Disjoint support gives $A{u}_n=x_n$. Thus $Y=AU=\overline{\mathrm{span}}\{x_n\}$ lies in the closed subspace $X$, is isomorphic to the l-p sequence space, and is the range of the bounded projection $AQ{A}^{-1}$ on the whole space $E$. This proves the required copy of ${\ell }^p$ is complemented in the ambient ${\ell }^p$, not just in $X$.

Now let $Z$ be an infinite-dimensional complemented subspace of $E$. The construction supplies $Y\subset Z$ with $Y\cong E$ and a bounded projection $R:E\to Y$. Its restriction to $Z$ gives the topological direct sumIf $P:E\to Z$ is a bounded projection, then $(I-R)P$ is a bounded projection onto $W$: its range lies in $W$, and it fixes every element of $W$. Hence $W$ is a complemented subspace of $E$, so $E\cong W\oplus V$ for some closed $V$.

We give the absorption argument in the Pełczyński decomposition method explicitly. Reindexing coordinates by a bijection between ${\mathbb{N}}^2$ and $\mathbb{N}$ yields $E\cong ({⨁}_{n\ge 1}E{)}_p$. Applying the fixed Banach space isomorphism $E\cong W\oplus V$ in every coordinate givesThe extra $W$ is absorbed by shifting the countably many $W$ summands. The maps are bounded with bounded inverses because the same two-space decomposition is used uniformly in each summand; we may equip each finite direct sum with its equivalent $p$-sum norm. Since $Z\cong E\oplus W$, $Z\cong {\ell }^p$.

We work over an algebraically closed field $k$, so the assertion concerns a geometric projective line. Let $V=k^4$ and let $W$ be the vector space of homogeneous cubic forms on $V$. It has dimension $\left(\genfrac{}{}{0ex}{}{6}{3}\right)=20$, so the parameter space of cubic equations up to nonzero scalar is $\mathbb{P}(W)={\mathbb{P}}^{19}$. The Grassmannian $G=\mathrm{Gr}(2,V)$ parametrizes lines in ${\mathbb{P}}^3$. Its standard graph charts are four-dimensional affine spaces, so its dimension is four. The Plücker embedding realizes it as a closed quadric in ${\mathbb{P}}^5$, proving projectivity. The general linear group ${\mathrm{GL}}_4$ acts transitively on two-planes and is an irreducible open subset of the affine space of matrices; its orbit map proves that $G$ is irreducible.

Consider the incidence varietyRestriction of a cubic to a fixed line is surjective onto the four-dimensional space of binary cubic forms: after choosing coordinates with that line given by $x_2=x_3=0$, extend any binary cubic using $x_0,x_1$ alone. Thus its kernel has dimension sixteen. On each Grassmannian chart these restriction maps depend algebraically on the line parameters and have constant rank four. Their kernels form a rank-sixteen vector bundle, and $\mathcal{I}$ is its projective bundle, with fibre ${\mathbb{P}}^{15}$. In particular, $\mathcal{I}$ is irreducible of dimension $4+15=19$. It is closed in the product, since restriction vanishes precisely when four polynomial coefficients vanish.

The projection $\pi :\mathcal{I}\to \mathbb{P}(W)$ is a projective morphism, hence its image is closed. It remains to prove that this image has dimension nineteen. Equal dimensions alone would not suffice: we must control a fibre locally.

TakeIn the open Grassmannian chart around ${\ell }_0$, every line is uniquely a graphRestriction of $F_0$ to that line isThe four monomials are linearly independent over every field. Hence the only line in this chart lying on $F_0$ has $a_0=a_1=b_0=b_1=0$. The fibre ${\pi }^{-1}([F_0])$ therefore has local dimension zero at $({\ell }_0,[F_0])$. We do not need $F_0$ to define a smooth surface, nor do we claim its entire fibre is finite.

The fiber dimension theorem, in its local dimension inequality, givesThus the closed image has dimension at least nineteen. A proper closed subset of the irreducible ${\mathbb{P}}^{19}$ has smaller dimension, so $\pi (\mathcal{I})={\mathbb{P}}^{19}$. Every cubic equation consequently has a line in its zero set. In particular, every nonsingular cubic surface contains a line. The proof in fact establishes existence for arbitrary cubic surfaces over an algebraically closed field; it does not assert a rational line over an arbitrary nonclosed ground field.

A central simple algebra over $K$ is a finite-dimensional associative $K$-algebra with no nonzero proper two-sided ideals and with center $K$. The Artin–Wedderburn theorem writes it uniquely as $M_r(D)$ with $D$ a central division algebra. After a finite separable splitting field extension it becomes $M_d$, so its dimension is $d^2$, and $d$ is its degree of a central simple algebra. Passing from $D$ to a matrix algebra over $D$ does not change its Brauer class.

The Brauer group consists of these classes, with product induced by tensor product. Extending to a common splitting field shows that a tensor product of central simple algebras is again central simple; its two-sided ideals and center descend from the corresponding full matrix algebra. The neutral class is $[K]$, and the inverse of $[A]$ is $[A^{\mathrm{op}}]$, becauseThe displayed map becomes the standard isomorphism for matrices over a splitting field, so it is already an isomorphism over $K$. Thus Brauer classes measure precisely the obstruction to being a split matrix algebra.