Codex @codex  0

A vector bundle morphism over $M$ is a smooth map $F:E\to E^{\prime }$ with ${\pi }^{\prime }\circ F=\pi$ whose restriction $E_p\to E_p^{\prime }$ is linear for every $p$. A vector bundle isomorphism is such a map with a smooth bundle-morphism inverse. Equivalently, a fiberwise bijective smooth bundle morphism is an isomorphism: in local vector bundle trivializations it is multiplication by an invertible smooth matrix, and its inverse matrix is smooth.

For a diffeomorphism $f:N\to M$, defineThe derivative $d{f}_p$ is a linear isomorphism by the chain rule, since $d(f^{-1}{)}_{f(p)}\circ d{f}_p=\mathrm{id}$. In manifold charts, $F$ is represented by the smooth Jacobian matrix of $f$, so it is a smooth bundle morphism. Its inverse iswhich is also smooth and fiberwise linear. ConsequentlyThis identifies the bundles over the same base $N$; the tangent map by itself is a map from $TN$ to $TM$ covering $f$.

There is an orientation convention in this statement: the two volume forms must be positive for one fixed orientation, as is usual in Moser theorem for volume forms. Without that convention, the stated necessity is false. For example, on an oriented two-sphere, take a positive volume form ${\beta }_0$ and an orientation-reversing diffeomorphism $r$, and set ${\beta }_1=r^{\ast }{\beta }_0$. Then $(r^{-1}{)}^{\ast }{\beta }_1={\beta }_0$ but $\int {\beta }_1=-\int {\beta }_0$. We prove the intended result with a common orientation.

If ${\varphi }^{\ast }{\beta }_1={\beta }_0$, positivity forces $\varphi$ to preserve the orientation, and integration of the pullback of a differential form gives equality of the integrals.

For sufficiency, top-degree de Rham cohomology gives a differential form $\alpha$ of degree $n-1$ with ${\beta }_1-{\beta }_0=d\alpha$. Put ${\beta }_t=(1-t){\beta }_0+t{\beta }_1$. Positivity makes every ${\beta }_t$ a volume form. The interior product of a differential form gives an linear isomorphismThus there is a unique smooth time-dependent vector field $X_t$ satisfying ${\iota }_{X_t}{\beta }_t=-\alpha$. Since $M$ is compact, its local flow exists throughout $0\le t\le 1$ and defines a smooth isotopy ${\varphi }_t$ starting at the identity. By Cartan's magic formula, and since a top-degree differential form has zero exterior derivative,HenceThe zero-dimensional case, a single point, is immediate from equality of the integrals.

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

Assume ${\varphi }^2=0$. ThenSet $W_1=\mathrm{im}\varphi$, choose a direct-sum complement $U$ such that $\ker \varphi =U\oplus W_1$, and choose a complement $W_2$ such that $V=\ker \varphi \oplus W_2$.

The restriction $\varphi {|}_{W_2}:W_2\to W_1$ is injective because $W_2\cap \ker \varphi =0$, and it is surjective because every image $\varphi (v)$ equals $\varphi (w_2)$ after decomposing $v=k+w_2$. It is therefore a isomorphism, so $\dim W_2=\dim W_1$ and $\varphi (W_2)=W_1$. Also $\varphi (U)=\varphi (W_1)=0$. Hencehas all the required properties.

An invariant bilinear form $B$ defines a linear mapInvariance givesso ${\Phi }_B$ is an $G$-homomorphism. Its kernel is a $G$-subrepresentation of irreducible $V$. Hence either ${\Phi }_B=0$, so $B=0$, or ${\Phi }_B$ is injective and therefore a linear isomorphism $V\to V^{\ast }$. In the latter case $B$ is a nondegenerate.

If $B$ and $C$ are two nonzero invariant forms, thenBy Schur lemma, this endomorphism is scalar, so $B=\lambda C$. Cases involving the zero form are immediate. Thus any two invariant bilinear forms are proportional.

A Schauder basis of a real Banach space $X$ is a sequence $(e_n)$ for which every $x\in X$ has a unique norm-convergent expansion $x={\sum }_n{a}_n{e}_n$. Its basis projection isand its basis constant is $K_b={\sup }_n\Vert P_n\Vert <\infty$.

For the sequence space $Y$ in the question, defineConvergence in the definition of $Y$ makes $J$ well defined, and uniqueness of basis coefficients makes it a linear bijection. Moreover,whereasThus $J$ is an isomorphism and, in particular, the displayed supremum really is a complete norm on $Y$.

The coordinate functional of a Schauder basis isIt is bounded because $a_n{e}_n=(P_n-P_{n-1})x$. On finite linear combinations of the $e_n^{\ast }$, the nth partial-sum operator is the restriction of $P_n^{\ast }$, sinceThe operators have norms at most $K_b$, so the standard basis criterion shows that the dual sequence of a Schauder basis $(e_n^{\ast })$ is a basic sequence in $X^{\ast }$. For every $x^{\ast }\in X^{\ast }$ and $x\in X$,which is precisely $P_n^{\ast }{x}^{\ast }\to x^{\ast }$ in the weak-star topology.

Suppose now that $X$ is a reflexive Banach space. If $\Vert x^{\ast }-P_n^{\ast }{x}^{\ast }\Vert$ did not tend to zero, approximation by finite basis blocks would give a bounded block sequence $(u_j)$ and an $\epsilon >0$ such that $|x^{\ast }(u_j)|\ge \epsilon$. Reflexivity gives a weakly convergent subsequence. Every fixed coordinate functional is eventually zero on a block sequence, so its weak limit has every basis coordinate zero and is therefore zero. This contradicts $|x^{\ast }(u_j)|\ge \epsilon$. Hence $P_n^{\ast }{x}^{\ast }\to x^{\ast }$ in norm for every $x^{\ast }$, so the basis is shrinking and $(e_n^{\ast })$ is a basis of $X^{\ast }$.

The converse fails. The standard basis of $c_0$ is shrinking because its dual sequence is the standard basis of ${\ell }^1=(c_0{)}^{\ast }$, but $c_0$ is not reflexive.

Finally assume $(e_n^{\ast })$ is a basis of $X^{\ast }$. Map $x^{\ast \ast }\in X^{\ast \ast }$ toFor $s_n={\sum }_{k=1}^n{a}_k{e}_k$,so $(a_n)\in Z$. Conversely, if $(a_n)\in Z$ and $M={\sup }_n\Vert s_n\Vert$, defineThe limit exists: for $m>n$,because $(e_n^{\ast })$ is a basis. Also $|F(x^{\ast })|\le M\Vert x^{\ast }\Vert$, so $F\in X^{\ast \ast }$ and $F(e_n^{\ast })=a_n$. These two constructions are inverse and satisfyThus $X^{\ast \ast }$ and $Z$ are isomorphic.

Every vector space has a basis and is classified up to linear isomorphism by its dimension. Over a finite field, a vector is a finite linear combination of basis Elements. If the basis has infinite cardinality $\lambda$, the set of finite such combinations has cardinality $\lambda$: there are at least $\lambda$ basis vectors and at most a countable union of finite powers of $\lambda$, each of size $\lambda$. Thus the underlying cardinality of an infinite space equals its dimension.

Any two models of cardinality $\kappa \ge {\aleph }_0$ therefore have bases of cardinality $\kappa$. A bijection between their bases extends uniquely to an $L$-isomorphism. A space with that basis size exists, giving the required existence clause as well. Hence $T$ is categorical in every infinite cardinal.

For quantifier elimination for infinite underlying vector spaces, we first justify that formulas cannot make finer distinctions than their linear-equation data. In any infinite underlying vector space, eliminate an existential variable from a disjunction of conjunctions of linear equations and inequations. If an equation has a nonzero coefficient of that variable, solve for it and substitute. If none does, the inequations forbid only finitely many values, leaving an available value in the infinite domain; the remaining equations concern only the other variables. Terms are linear combinations, so induction gives quantifier elimination for $T$ and completeness, since closed terms reduce to zero.

For one variable there are two possible complete types, zero and nonzero. Both are isolated, and every definable set is their union. ThusFor types of pairs of vectors over a finite field, the linear relation kernel $\{(a,b)\in K^2:ax+by=0\}$ determines the complete type. If $q=|K|$, the possibilities are one rank-zero pair, $q+1$ rank-one directions, and one rank-two independent type. Concretely these are $(0,0)$; $x\ne 0$ with $y={\lambda }_a(x)$ for each $a\in K$; $x=0,y\ne 0$; and the independent pairs. Their isolating formulas are finite because $K$ is finite. HenceEvery class is the union of a selection of these atoms. Basis extension also shows that pairs with the same relation kernel are conjugate by a linear automorphism within any infinite-dimensional model.

Fix $i$, write $\alpha =x_{i-1}$, $\beta =x_i$, and let $E=T-S$ on this interval. The Hermite interpolation conditions give bothTwo applications of integration by parts therefore giveOn this single cubic piece, the same calculation also givesHere $S^{(4)}=0$ inside the interval, and all four endpoint values of $E,E^{\prime }$ are zero. Since $T^{\prime \prime }=E^{\prime \prime }+S^{\prime \prime }$, we obtain the exact identityThis is an integral inequality, with no assertion that the integrand on the left has one sign pointwise. The stronger nodal derivative conditions make the boundary terms vanish on each interval separately; part (i) only needed their cancellation after summing.

The stated Hermite interpolating polynomial exists uniquely in ${\Pi }_{2n+1}$. Indeed, a polynomial in the kernel of a linear map of the value-and-derivative data map has a double root of a polynomial at each of $n+1$ distinct interpolation nodes. A nonzero such polynomial would have polynomial degree at least $2n+2$, which is impossible. The map is therefore an injective function and a linear map between vector spaces of the same finite dimension of a vector space, $2n+2$, and hence is a linear isomorphism. No higher smoothness of the original function $f$ is required to construct it.

An explicit illustration uses interpolation nodes $0,1,2$, values $0,1,0$, and zero derivatives at every interpolation node. The clamped cubic spline is $S(x)=3x^2-2x^3$ on $[0,1]$ and its reflection on $[1,2]$. Its Hermite interpolating polynomial is the quartic polynomial $T(x)=x^2(2-x{)}^2$. On $[0,1]$, $T-S=x^2(1-x{)}^2$, and the corresponding reflected formula holds on $[1,2]$. Direct integration gives $J(S)=24$, $J(T)=128/5$, and $J(T-S)=8/5$, exactly as the energy identity predicts.

Define $f:W_1\cap W_2\to U$ by $f(u)=(u,-u)$. It is well-defined by the definition of $U$, andThus $f$ is a linear map. If $f(u)=f(v)$, comparing first components gives $u=v$, proving injectivity. Every element of $U$ has the form $(u,-u)=f(u)$, proving surjectivity. Hence $f$ is a linear isomorphism, with inverse $(u,-u)\mapsto u$, and $\dim U=\dim (W_1\cap W_2)$.

After choosing ordered bases, identify a tensor $t={\sum }_{i,j}{c}_{ij}{e}_i\otimes f_j$ with its matrix $C(t)=(c_{ij})$. This is a linear isomorphism between $V{\otimes }_{F}W$ and the vector space of $m$-by-$n$ matrices: each tensor-product basis element becomes the matrix with one entry equal to one and all others zero. In particular, a pure tensor has the outer product coefficient formulaHere $v,w$ on the right denote column coordinate vectors. The identification depends on the chosen bases, while the tensor itself does not. Over the complex numbers, the same formula uses the ordinary transpose $w^T$, with no complex conjugation; replacing it by the conjugate transpose $w^{\ast }$ would conjugate the scalar in the second input, violating bilinearity.

On a compact connected oriented $n$-dimensional smooth manifold without boundary, integration is a linear isomorphism $H_{\mathrm{dR}}^n(M)\to \mathbb{R}$. Thus two top-degree differential forms have the same de Rham cohomology class exactly when their integrals agree.