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The monomials $1,x,\dots ,x^n$ are linearly independent and span, so $\boxed{{\dim }_{\mathbb{C}}V=n+1}$. Use the factorial monomial basis for polynomial differentiation $p_j(x)=x^j/j!$, $0\le j\le n$. Direct differentiation gives $e_i(p_j)={\delta }_{ij}$. HenceIndeed every $\varphi \in V$ has the Taylor expansion $\varphi ={\sum }_{j=0}^n{e}_j(\varphi )p_j$; evaluation on the $p_j$ proves independence of the functionals. Derivative orders include zero. If one uses a convention excluding zero from $\mathbb{N}$, $e_0$ must explicitly be included, since all positive-order derivatives annihilate constants and cannot span the dual space.

Differentiation satisfies $D{p}_0=0$ and $D{p}_j=p_{j-1}$ for $j\ge 1$. With column vectors of coefficients, its matrix $A$ has entries $A_{ij}=1$ if $j=i+1$ and zero otherwise, for indices $0\le i,j\le n$:The dual map is $D^{\ast }(\ell )=\ell \circ D$, not a conjugate-linear adjoint. Thus $D^{\ast }{e}_i=e_{i+1}$ for $i<n$ and $D^{\ast }{e}_n=0$. Therefore $\boxed{[D^{\ast }]=[D{]}^T}$ in the corresponding dual basis. For $n=0$ both matrices are the one-by-one zero matrix.

Let $f_1,f_2,f_3$ be the dual basis of $W^{\ast }$, and use $L_1+L_2+L_3=0$. The elementary tensors $e_i\otimes e_j\otimes f_k$ have weight $L_i+L_j-L_k$. Grouping them gives every weight space explicitly:The counts are $6+3\cdot 2+3\cdot 5=27$, so this table exhausts the tensor product. The diagrams below use equilateral coordinates $L_1=(1,0)$, $L_2=(-1/2,\sqrt{3}/2)$, $L_3=(-1/2,-\sqrt{3}/2)$; numbers at points are weight multiplicities.

To decompose the module, first split the first two factors into their symmetric and alternating parts. Put $t_{ij}=(e_i\otimes e_j+e_j\otimes e_i)/2$, including $t_{ii}=e_i\otimes e_i$. The equivariant contractionis onto, since $C(t_{ii}\otimes f_i)=2e_i$. Its kernel $K$ has dimension fifteen and contains the highest-weight vector $t_{11}\otimes f_3$ of weight $2L_1-L_3$.

Here is an explicit weight basis of the trace-free symmetric-square dual tensor module, and thus the requested fifteen-vector basis:Every displayed vector contracts to zero. The first two families occupy distinct one-dimensional weights; at each weight $L_i$ the two vectors in the last family are independent because their respective $t_{ij}\otimes f_j$ terms are distinct. Thus all fifteen are independent and span $K$.

For completeness, $K$ is irreducible, not just an invariant kernel of the right dimension. On elementary tensors the raising action isAt the six weights $2L_i-L_k$, both $E_{12}$ and $E_{23}$ kill the vector only for $(i,k)=(1,3)$. At each weight $-2L_k$, at least one of those operators is nonzero. At weight $L_1$, write a vector as $a(t_{12}\otimes f_2-t_{11}\otimes f_1/2)+b(t_{13}\otimes f_3-t_{11}\otimes f_1/2)$. The two raising equations are $3a+b=0$ and $-a+b=0$, forcing zero. On the two-dimensional weight-$L_2$ space $E_{12}$ is injective, and on the weight-$L_3$ space $E_{23}$ is injective, as substitution in the same action formula shows. Therefore the common raising kernel in $K$ is the single line $\mathbb{C}(t_{11}\otimes f_3)$. By Weyl complete reducibility theorem, every irreducible summand supplies a highest-weight line, so $K$ has only one summand and is irreducible.

The symmetric-factor contraction splits off a defining copy $W$; explicitly $e_i\mapsto {\sum }_j{t}_{ij}\otimes f_j$ is equivariant and its contraction is $4e_i$. For the alternating factor, the invariant volume form identifies ${\Lambda }^{2}W\cong W^{\ast }$, sending $e_1\wedge e_2$ to $f_3$, $e_2\wedge e_3$ to $f_1$ and $e_3\wedge e_1$ to $f_2$. ThereforeThe symmetric dual module is irreducible of dimension six and highest weight $-2L_3$; its symmetric-monomial basis has weights $-2L_i$ and $-L_i-L_j=L_k$, all of multiplicity one. In the original tensor product its highest-weight vector isThe raising action above annihilates it, and its weight is $L_1+L_2-L_3=-2L_3$.

Altogether the Triple tensor decomposition for the defining sl3 representation isThe fifteen-dimensional module has multiplicity one at each $2L_i-L_k$ and $-2L_i$, and multiplicity two at each $L_i$. The six-dimensional module has multiplicity one at $-2L_i$ and $L_i$. Each defining module has the three weights $L_i$, once each. These add to the original table of multiplicities.

The two defining summands are not intrinsically distinguished: other diagonal copies inside their isotypic component have the same three-point weight diagram. Thus the displayed diagrams account for every irreducible submodule type, as well as showing both copies in this decomposition.

The dual space is $V^{\ast }={\mathrm{Hom}}_F(V,F)$, the vector space of all linear functionals with pointwise operations. For finite $d=\dim V$, choose a basis $v_1,\dots ,v_d$ and define its dual basis by ${\epsilon }_j(v_i)={\delta }_{ij}$. Every linear functional is $\ell ={\sum }_j\ell (v_j){\epsilon }_j$, and evaluation on each $v_i$ proves that the ${\epsilon }_j$ are linearly independent. Thus $\dim V^{\ast }=\dim V=d$. This dimension formula is a finite-dimensional assertion.

On the space of polynomials of degree at most $n$, define the evaluation map $E(p)=(p(a_0),\dots ,p(a_n))$. Its kernel is zero: a nonzero degree-at-most-$n$ polynomial cannot have $n+1$ distinct roots. Since domain and codomain both have dimension $n+1$, $E$ is an isomorphism. The evaluations therefore form a basis of the dual space. The linear functional $p\mapsto p^{\prime }(0)$ has a unique expansion in that basis, proving the existence and uniqueness of the requested weights.

Explicitly, use the Lagrange interpolation polynomial basisThe differentiation weights from nodal evaluations are ${\lambda }_j={\ell }_j^{\prime }(0)$. If a node is zero, no singular reciprocal formula is needed: differentiate the finite product directly. For $n=0$ the sole weight is zero.

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 Poincaré-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:

The dual space isthe vector space of linear functionals on $V$. If $e_1,\dots ,e_n$ is a basis of a finite-dimensional $V$, define $e_i^{\ast }(e_j)={\delta }_{ij}$. Every $f\in V^{\ast }$ has the unique expansionso the dual basis $e_1^{\ast },\dots ,e_n^{\ast }$ proves

For $U\le V$, its annihilator of a vector subspace isIf $u_1,\dots ,u_k$ is a basis of $U$ and is extended to a basis $u_1,\dots ,u_n$ of $V$, thenConsequentlyIf $U\ne V$, this dimension is positive, giving a nonzero functional that vanishes on $U$.

For a linear map $\alpha :V\to W$, the dual map isNowsoEvery $g\circ \alpha$ vanishes on $\ker \alpha$, henceThe two spaces have the same dimension, sinceThus

Let $Q:V\to V/U$ be the quotient map. Since $Q$ is onto, $Q^{\ast }$ is injective, and the preceding identity givesThereforeFor the inclusion $J:U\hookrightarrow V$, the map $J^{\ast }:V^{\ast }\to U^{\ast }$ is restriction to $U$. It is onto and has kernelThe first isomorphism theorem now gives the other duals of a subspace and its quotient:

Fix ${\xi }_i\in [t_i,t_{i+k}]$ and choose a knot interval $I=(t_{\ell },t_{\ell +1})$ adjacent to it on which $N_i$ is active. Exactly $k$ B-splines are nonzero on $I$, and their polynomial restrictions form a basis of ${\mathcal{P}}_{k-1}$. Applying the expansion from part (a) to the polynomial piece $N_j{|}_I$ givesUniqueness of coordinates in this local basis forceswhen $N_j$ is active. If $N_j$ vanishes on $I$, all of its local polynomial derivatives vanish and the same equality holds with value zero. At a knot, use either adjacent polynomial piece; the $x$-independence proved in part (a) gives the same coefficient. Henceso the ${\lambda }_i$ form the dual basis to the B-spline basis.

For a basis $(u_i)$ of a finite-dimensional inner product space with conjugate-linear first argument, define $G_{ij}=(u_i,u_j)$ and ${\hat{u}}_k={\sum }_l(G^{-1}{)}_{lk}{u}_l$. Then $(u_i,{\hat{u}}_k)={\delta }_{ik}$ and $({\hat{u}}_j,{\hat{u}}_k)=(G^{-1}{)}_{jk}$. This is the inner-product representation of the dual basis; requiring the dual vectors to lie in the given span is essential.

For two unital associative algebras over a field $k$, their tensor product carries the multiplication in the title, extended by bilinearity, with identity $1\otimes 1$. Choosing a basis $(b_j)$ of the second factor gives the basis of a module $(1\otimes b_j)$ for the resulting left $A$-module; dual basis functionals recover the coefficients uniquely.

For a linear map $T:V\to W$, the dual map $T^{\ast }:W^{\ast }\to V^{\ast }$ sends a linear functional $\phi$ on the output space to the input measurement $T^{\ast }(\phi )=\phi \circ T$. It reverses direction because a measurement of the output can be evaluated on an input by first applying $T$:For $T(x,y)=(x+y,2y)$ and $\phi (u,v)=2u-v$, the input measurement is $T^{\ast }\phi (x,y)=2(x+y)-2y=2x$. The output measurement ignores the part of this transformation contributed by $y$.

If $T$ has matrix $A$ in specified bases, a row representing $\phi$ changes to $\ell A$. Storing the coefficients of a linear functional as a column in the corresponding dual basis gives $A^T{\ell }^T$, explaining why the transpose of a linear map is represented by the matrix transpose. This algebraic operation uses no inner product; it agrees with a matrix transpose even over the complex numbers, rather than automatically introducing complex conjugation. The adjoint operator is defined using an inner product and is a different construction. ^[ref].

In finite dimension, the kernel and image of the dual map are described by annihilators of vector subspaces: