Codex @codex  0

For an algebraic dual space $X^{\ast }={\mathrm{Hom}}_F(X,F)$, the linear map $J:X\to X^{\ast \ast }$ sends $x$ to evaluation at $x$. It is injective because linear functionals separate nonzero vectors: extend any nonzero vector to a basis and use its coordinate functional. If $X$ has a finite basis $e_i$ with dual basis $e_i^{\ast }$, every $B\in X^{\ast \ast }$ equals $J({\sum }_{i}B(e_i^{\ast })e_i)$, so $J$ is an isomorphism. For infinite-dimensional spaces, surjection does not follow from this finite sum argument.

Choose a base of a root system and its positive roots, so that the triangular decomposition of a Lie algebra is $L={\mathfrak{n}}^{-}\oplus H\oplus {\mathfrak{n}}^{+}$. A primitive element of a Lie algebra representation of weight $\omega \in H^{\ast }$ is a nonzero vector $v$ such thatEquivalently it is a highest-weight vector. Being primitive depends on the chosen positive system; no integrality condition is part of this definition.

Construct the Verma module for an arbitrary $\omega$. The subalgebra $\mathfrak{b}=H\oplus {\mathfrak{n}}^{+}$ has a one-dimensional module ${\mathbb{C}}_{\omega }$ on which $H$ acts by $\omega$ and ${\mathfrak{n}}^{+}$ acts by zero. This is a representation because $[\mathfrak{b},\mathfrak{b}]\subseteq {\mathfrak{n}}^{+}$. Setwhere $U$ denotes the universal enveloping algebra. The Poincare-Birkhoff-Witt theorem, applied with negative-root vectors first, then Cartan vectors and positive-root vectors, givesas vector spaces. In particular $v_{\omega }=1\otimes 1$ is nonzero, generates $M(\omega )$, and is primitive of weight $\omega$.

The same ordered monomials show that $M(\omega )$ is a direct sum of weight spaces, with weights of the formand that its weight-$\omega$ space is exactly $\mathbb{C}{v}_{\omega }$. Each vector has only finitely many weight components. If a submodule contains a vector with nonzero weight-$\omega$ component, choose $h\in H$ that separates $\omega$ from the finitely many other weights of that vector. Polynomial interpolation in the action of $h$ extracts a nonzero multiple of $v_{\omega }$. Such a submodule is all of $M(\omega )$, because $v_{\omega }$ generates it. Consequently every proper submodule has zero component in weight $\omega$.

Let $N$ be the sum of all proper submodules. Every finite sum still has zero weight-$\omega$ component, so $N$ is proper; it contains every proper submodule and is therefore the unique maximal proper submodule. Its quotientis irreducible, and the nonzero image of $v_{\omega }$ is primitive of weight $\omega$. This proves existence for every $\omega \in H^{\ast }$ and gives the irreducible quotient of a Verma module. It also proves uniqueness up to isomorphism among irreducible modules generated by a primitive vector of that weight: the defining relations induce a surjection from $M(\omega )$, whose proper kernel must be $N$.

These modules are not generally finite-dimensional. For example, in an ${\mathfrak{sl}}_2$ root subalgebra, a primitive vector with $hv=\lambda v$ satisfiesIf its irreducible highest-weight module is finite-dimensional, the weight-lowering sequence eventually stops. For the first $m$ with $f^{m+1}v=0$ and $f^{m}v\ne 0$, the identity forces $\lambda =m\in {\mathbb{Z}}_{\ge 0}$. In general a finite-dimensional highest-weight module therefore requires nonnegative integral values on all simple coroots; arbitrary $\omega$ in the request must allow infinite-dimensional representations. This is consistent with the finite-dimensional qualification made in Question 2.

For simple coroots $h_1,\dots ,h_r$ forming a basis of $H$, the fundamental weights are the dual basis ${\omega }_1,\dots ,{\omega }_r\in H^{\ast }$:For the special linear Lie algebra ${\mathfrak{sl}}_3(\mathbb{C})$, take the diagonal trace-zero Cartan subalgebra and positive roots corresponding to upper triangular matrix units. Write ${\epsilon }_i(\mathrm{diag}(t_1,t_2,t_3))=t_i$. The simple roots are ${\alpha }_1={\epsilon }_1-{\epsilon }_2$, ${\alpha }_2={\epsilon }_2-{\epsilon }_3$, and the simple coroots areThe fundamental weights of sl3 are thereforeChecking these on $h_1,h_2$ gives the two coordinate vectors $(1,0)$ and $(0,1)$. In the Euclidean trace-zero plane they can also be writtenThe Fundamental representations of sl3 make these weights concrete: ${\mathbb{C}}^3$ has primitive vector $e_1$ of weight ${\omega }_1$, and ${\Lambda }^2{\mathbb{C}}^3$ has primitive vector $e_1\wedge e_2$ of weight ${\omega }_2$. Every finite-dimensional irreducible highest weight is $a{\omega }_1+b{\omega }_2$ with nonnegative integers $a,b$.

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 dual space here is the algebraic dual space $X^{\ast }={\mathrm{Hom}}_{\mathbb{R}}(X,\mathbb{R})$, consisting of every linear functional; no topology or boundedness is imposed. For a basis $e_1,\dots ,e_n$, define $e_i^{\ast }({\sum }_j{x}_j{e}_j)=x_i$. These are linear functionals satisfying $e_i^{\ast }(e_j)={\delta }_{ij}$. Evaluating ${\sum }_i{a}_i{e}_i^{\ast }=0$ on each $e_j$ gives $a_j=0$, proving linear independence. Conversely, for every $T\in X^{\ast }$,This proves spanning and hence proves, without assuming a dimension theorem, that the dual basis is a basis and $\dim X^{\ast }=n$.

The canonical bidual evaluation map isIts definition makes no choice of basis and it is linear. If $x\ne 0$, one coordinate functional is nonzero on $x$, so $J$ is injective. For any $B\in X^{\ast \ast }$, choose $x={\sum }_{i}B(e_i^{\ast })e_i$. Expanding $T$ in the dual basis gives $T(x)={\sum }_{i}T(e_i)B(e_i^{\ast })=B(T)$, proving surjection explicitly. Thus $J$ is an isomorphism.

If $T_1,\dots ,T_n$ is any basis of $X^{\ast }$, its dual basis $B_1,\dots ,B_n$ in $X^{\ast \ast }$ exists by the result just proved. Put $v_j=J^{-1}(B_j)$. These vectors form a basis of $X$, and $T_i(v_j)=B_j(T_i)={\delta }_{ij}$. Every basis of the finite-dimensional dual is a dual basis, of a uniquely determined basis of $X$.

For a separating subspace of an algebraic dual $W$, choose a basis $T_1,\dots ,T_r$ of $W$. The linear map $x\mapsto (T_1(x),\dots ,T_r(x))$ is injective because $W$ separates nonzero vectors. The rank-nullity theorem implies $n\le r$; but $W\subseteq X^{\ast }$ implies $r\le n$. HenceThe zero-dimensional case also has this conclusion.

For $X=\mathbb{R}[x]$, the monomials $1,x,x^2,\dots$ form an infinite basis, each polynomial using only finitely many of them. The map $T\mapsto (T(1),T(x),T(x^2),\dots )$ identifies $X^{\ast }$ with all real sequences, because any sequence $(a_j)$ defines $T({\sum }_{j=0}^d{c}_j{x}^j)={\sum }_{j=0}^d{c}_j{a}_j$. The linear subspace of finitely supported sequences is proper (it excludes $(1,1,\dots )$) and separating (one coefficient functional detects any nonzero polynomial). Thus

Use the PDF's covector-first convention: its space ${\mathcal{T}}_l^k$ consists of tensor fields inThis reverses the order in which some texts list tensor type. For a finite-dimensional real vector space $V$, define the tensor contraction first on decomposable tensors:The hats mean omission, with the other factors left in their original order. The formula is a multilinear map of its individual factors, so the universal property of a tensor product gives a unique linear map with this formula.

Apply it with $V=T_{p}M$ at every $p$. Evaluation of a covector on a vector is basis independent: under a change of frame, one factor transforms by a matrix and the other by its inverse transpose, and the matrices cancel in the pairing. Thus the fiber maps agree on overlapping vector bundle trivializations. In a local dual basis, the coefficients of the contracted tensor are finite sums of coefficients with the selected covariant and contravariant indices set equal. They remain smooth. Henceis a globally defined smooth contraction for $k,l\ge 1$. It is also linear over $C^{\infty }(M)$.

Write a local frame as $e_a$, its dual basis as ${ϵ}^a$, and set $D{e}_a=A^b{}_a{e}_b$. The one-form definition gives $D{ϵ}^a=-A^a{}_b{ϵ}^b$. Repeated indices are summed. Every tensor field has a unique local expansionDefine $DT$ by applying the scalar operator to its coefficient and applying $D$ to one frame factor at a time, adding all these terms. In components this isIn each term only the indicated slot is replaced. This formula is real-linear and satisfies the Leibniz rule for a tensor product: differentiating a coefficient product uses the scalar product rule, and the list of differentiated frame factors splits into the two factors' lists.

To check that it is intrinsic, write $e$ as the row of frame fields and $ϵ$ as the column of dual fields. Change frame by $e^{\prime }=eP$, where $P$ is an invertible smooth matrix. ThenDifferentiating the inverse matrix gives $Z(P^{-1})=-P^{-1}(ZP)P^{-1}$; it cancels the extra frame-change terms in the dual factors. The same cancellation in each tensor slot makes the two local formulas agree. Thus they glue to a global tensor derivation.

In particular, for arbitrary tensor types, not necessarily the same type,The resulting type is the sum of their two covariant counts and their two contravariant counts. This also supplies the required real-linearity for every type, including functions as type $(0,0)$.

The dual vector space is $V^{\ast }={\mathrm{Hom}}_{\mathbb{R}}(V,\mathbb{R})$, the vector space of all linear functionals on $V$. For a vector subspace $U\le V$, its annihilator of a vector subspace is

Given a basis $x_1,\dots ,x_n$, define $x_i^{\ast }$ by $x_i^{\ast }(x_j)={\delta }_{ij}$ and extend linearly. If ${\sum }_i{a}_i{x}_i^{\ast }=0$, evaluation at $x_j$ gives $a_j=0$, so the $x_i^{\ast }$ are linearly independent. Every $\varphi \in V^{\ast }$ satisfiesso they also span $V^{\ast }$ and therefore form its dual basis.

Write $\varphi =a{x}_1^{\ast }+b{x}_2^{\ast }+c{x}_3^{\ast }+d{x}_4^{\ast }$. The condition $\varphi \in U^0$ isGaussian elimination gives $b=-2c-3d$ and $a=c+2d$. Hence

Use the decomposition into Lagrangian subspaces $V=V_{+}\oplus V_{-}$ from part (a). The bilinear form $\Omega$ pairs $V_{+}$ and $V_{-}$ nondegenerately: a vector in either summand annihilating the other already annihilates its own summand and hence all of $V$, so is zero.

Choose a basis $e_1,\dots ,e_n$ of $V_{+}$. The linear mapis an linear isomorphism. Let $f_j$ correspond to the dual basis of the $e_i$. Then $\Omega (e_i,f_j)={\delta }_{ij}$, and the pairings within each summand vanish. Thus

Write $\alpha =\sqrt[n]{m}$ and $\zeta =e^{2\pi i/n}$. The $n$ field embeddings of $L$ into $\mathbb{C}$ send $\alpha$ to ${\zeta }^j\alpha$, so the field trace isFor the trace pairing $T(x,y)={\mathrm{Tr}}_{L/\mathbb{Q}}(xy)$, the basis ${\beta }_i={\alpha }^i$ therefore has dual basis

Every element of $\mathbb{Z}[\alpha ]$ is an algebraic integer. If $x\in {\mathcal{O}}_L$ and $z\in \mathbb{Z}[\alpha ]$, then $xz$ is an algebraic integer and its trace is an integer. Thus $x$ belongs to the trace-dual lattice of $\mathbb{Z}[\alpha ]$, which is generated over $\mathbb{Z}$ by the displayed dual basis. Since $1/n=m/(nm)$, every such linear combination lies in $(nm{)}^{-1}\mathbb{Z}[\alpha ]$. Hence

Let $({\theta }^1,\dots ,{\theta }^n)$ be the dual basis of the positively oriented orthonormal basis $(v_1,\dots ,v_n)$. By the definition of the Riemannian volume form,The interior product of a differential form with the outward unit normal isThe vectors $(v_2,\dots ,v_n)$ form a positive orthonormal frame of $T\partial X$ by the outward-normal-first boundary orientation. Consequently the pullback of the last display is the positive unit boundary volume form: