Tensor Fields Exterior Algebra and Differential Forms

Fix a list of arguments of the right total type. By [F1], $((R\otimes S)\otimes T)$ evaluates on that list as the product of the three separate values $R(\cdots )S(\cdots )T(\cdots )$.

The same formula [F1] gives exactly the same scalar for $R\otimes (S\otimes T)$ on the same list of arguments. Hence $(R\otimes S)\otimes T=R\otimes (S\otimes T)$.

Again by [F1], $((aR+bS)\otimes T)(\cdots )=(aR+bS)(\cdots )T(\cdots )=a(R\otimes T)(\cdots )+b(S\otimes T)(\cdots )$, and the same computation in the second slot gives bilinearity there as well.

Therefore the tensor product is associative and bilinear in each factor.

Applying $\mathrm{Sym}$ twice gives a double average over $S_k\times S_k$. Reindex by the product permutation $\rho =\sigma \tau$; each $\rho$ occurs exactly $k!$ times, so the second averaging changes nothing. Thus $\mathrm{Sym}(\mathrm{Sym}T)=\mathrm{Sym}T$.

The same reindexing works for $\mathrm{Alt}$, and the sign factors multiply to $\mathrm{sgn}(\sigma \tau )$. Hence $\mathrm{Alt}(\mathrm{Alt}T)=\mathrm{Alt}T$.

If $T$ is symmetric, every summand in [F1] equals $T$, so $\mathrm{Sym}T=T$. Conversely, $\mathrm{Sym}T$ is fixed by every permutation because averaging over the whole group is permutation-invariant. The alternating case is identical, with the sign picked up under permutation.

Therefore $\mathrm{Sym}$ and $\mathrm{Alt}$ are projections onto the symmetric and alternating tensors.

Define the bilinear map $B:V^{\ast }\times V\to \mathbb{R}$ by $B(\beta ,w)=T(\beta ,{\alpha }_2,\dots ,{\alpha }_r,w,v_2,\dots ,v_s)$. Then the two contraction sums are ${\sum }_{i}B(e^i,e_i)$ and ${\sum }_{j}B(b^j,b_j)$.

By [L1], any vector $w$ satisfies $w={\sum }_i{e}^i(w)e_i={\sum }_j{b}^j(w)b_j$, and any covector $\beta$ satisfies $\beta ={\sum }_i\beta (e_i)e^i={\sum }_j\beta (b_j)b^j$. Bilinearity of $B$ therefore gives ${\sum }_{i}B(e^i,e_i)={\sum }_{i,j}{b}^j(e_i)B(e^i,b_j)={\sum }_{j}B({\sum }_i{b}^j(e_i)e^i,b_j)={\sum }_{j}B(b^j,b_j)\mathrm{.}$

The displayed sum is therefore basis-independent, so contraction is intrinsically defined.

Evaluating on $v_1,\dots ,v_{k+\mathrm{\ell }}$ and using [F1] and [F2], $A^{\ast }(S\otimes T)(v_1,\dots ,v_{k+\mathrm{\ell }})=(S\otimes T)(A{v}_1,\dots ,A{v}_{k+\mathrm{\ell }})=S(A{v}_1,\dots ,A{v}_k)T(A{v}_{k+1},\dots ,A{v}_{k+\mathrm{\ell }}),$ which is exactly $(A^{\ast }S\otimes A^{\ast }T)(v_1,\dots ,v_{k+\mathrm{\ell }})$.

Likewise, $A^{\ast }(\sigma \cdot T)(v_1,\dots ,v_{\mathrm{\ell }})=(\sigma \cdot T)(A{v}_1,\dots ,A{v}_{\mathrm{\ell }})=T(A{v}_{\sigma (1)},\dots ,A{v}_{\sigma (\mathrm{\ell })}),$ which equals $(\sigma \cdot A^{\ast }T)(v_1,\dots ,v_{\mathrm{\ell }})$.

Therefore linear pullback respects tensor products and permutations.

By [F1], $\alpha \wedge \beta$ is obtained by applying the alternation operator to $\alpha \otimes \beta$. Alternation produces an alternating multilinear form, so $\alpha \wedge \beta \in {\mathrm{Alt}}^{k+\mathrm{\ell }}(V)$.

Both tensor product and alternation are linear in each argument, so [F1] gives $(a\alpha +b{\alpha }^{\mathrm{\prime }})\wedge \beta =a(\alpha \wedge \beta )+b({\alpha }^{\mathrm{\prime }}\wedge \beta ),$ and similarly in the second slot.

Therefore the wedge product is alternating and bilinear.

Using [F1] twice and bilinearity from [L1], both $(\alpha \wedge \beta )\wedge \gamma$ and $\alpha \wedge (\beta \wedge \gamma )$ are the full alternation of the multilinear tensor $\alpha \otimes \beta \otimes \gamma$ with the same normalization factor. Hence they are equal.

In the shuffle formula of [F1], swapping the $k$ inputs destined for $\alpha$ with the $\mathrm{\ell }$ inputs destined for $\beta$ contributes the sign of the block permutation, namely $(-1{)}^{k\mathrm{\ell }}$. Therefore every term of $\alpha \wedge \beta$ matches the corresponding term of $(-1{)}^{k\mathrm{\ell }}\beta \wedge \alpha$.

Therefore the wedge product is associative and graded commutative.

Let $\omega \in {\mathrm{Alt}}^k(V)$. Expanding each input vector in the basis and using multilinearity shows that $\omega$ is determined by its values on basis $k$-tuples. Because $\omega$ is alternating, every tuple with a repeated index vanishes and every tuple with distinct indices reduces, up to sign, to one with increasing indices. Thus $\omega$ is a linear combination of the displayed wedges from [F1].

Suppose ${\sum }_I{c}_I{e}^I=0$, where $I=(i_1<\cdots <i_k)$ and $e^I:=e^{i_1}\wedge \cdots \wedge e^{i_k}$. Evaluating at $(e_{j_1},\dots ,e_{j_k})$ with $j_1<\cdots <j_k$, [L1] gives $e^I(e_{j_1},\dots ,e_{j_k})=1$ when $I=(j_1,\dots ,j_k)$ and $0$ otherwise. Hence every $c_I=0$.

Therefore the displayed wedges form a basis of ${\mathrm{Alt}}^k(V)$.

Choose a basis $e_1,\dots ,e_n$ of $V$. By [L1], ${\mathrm{Alt}}^k(V)$ has one basis vector for each strictly increasing $k$-tuple $(i_1,\dots ,i_k)$ from $\{1,\dots ,n\}$.

Such tuples are the same thing as $k$-element subsets of an $n$-element set, so [F1] counts them by $\left(\genfrac{}{}{0px}{}{n}{k}\right)$. Therefore $\dim {\mathrm{Alt}}^k(V)=\left(\genfrac{}{}{0px}{}{n}{k}\right)$. If $k>n$, there are no such tuples, so the basis is empty and ${\mathrm{Alt}}^k(V)=0$.

This is exactly the claimed dimension formula and vanishing statement.

Choose a basis $e_1,\dots ,e_n$ of $V$. Define $\tilde{A}$ on the spanning set of decomposable wedges by $\tilde{A}(e_{i_1}\wedge \cdots \wedge e_{i_k}):=A(e_{i_1},\dots ,e_{i_k})$ for $i_1<\cdots <i_k$, and extend linearly. This is possible because [L1] gives a basis indexed by those increasing tuples.

For arbitrary $v_1,\dots ,v_k$, expand each $v_j$ in the chosen basis. Multilinearity of $A$ and of the wedge, together with alternation on both sides, reduce both expressions to the same signed sum over increasing $k$-tuples. Hence $\tilde{A}(v_1\wedge \cdots \wedge v_k)=A(v_1,\dots ,v_k)$.

If $L:{\bigwedge }^{k}V\to W$ is another linear map with the same property, then $L$ and $\tilde{A}$ agree on every decomposable basis wedge, hence on all of ${\bigwedge }^{k}V$ by linearity and [L1]. So $\tilde{A}$ is unique.

Therefore every alternating $k$-linear map factors uniquely through ${\bigwedge }^{k}V$.

The map $(v_1,\dots ,v_k)\mapsto A{v}_1\wedge \cdots \wedge A{v}_k$ is alternating and $k$-linear in $v_1,\dots ,v_k$. By [L1], it therefore factors uniquely through a linear map ${\bigwedge }^{k}A:{\bigwedge }^{k}V\to {\bigwedge }^{k}W$ with the stated action on decomposable wedges.

The identity map and the composite $({\bigwedge }^{k}B)\circ ({\bigwedge }^{k}A)$ have the expected values on every decomposable wedge: ${\bigwedge }^k({\mathrm{id}}_V)(v_1\wedge \cdots \wedge v_k)=v_1\wedge \cdots \wedge v_k,$ and $(({\bigwedge }^{k}B)\circ ({\bigwedge }^{k}A))(v_1\wedge \cdots \wedge v_k)=BA{v}_1\wedge \cdots \wedge BA{v}_k\mathrm{.}$ The same formula holds for ${\bigwedge }^k(B\circ A)$.

By uniqueness in [L1], the maps in step 2.1 must agree. Therefore exterior powers preserve identities and composition.

Hence $V\mapsto {\bigwedge }^{k}V$ and $A\mapsto {\bigwedge }^{k}A$ define a functor.

By [F2], $({\alpha }_1\wedge \cdots \wedge {\alpha }_k)(v_1,\dots ,v_k)$ is the alternating sum over permutations of ${\prod }_i{\alpha }_i(v_{\sigma (i)})$, which is exactly the determinant of the matrix $({\alpha }_i(v_j))$. By [F1], this is the value of the pairing on the displayed decomposable elements.

Choose a basis $e_1,\dots ,e_n$ of $V$ with dual basis $e^1,\dots ,e^n$. By [L1], the wedges $e^I$ form a basis of ${\bigwedge }^k{V}^{\ast }$ and the wedges $e_I$ form a basis of ${\bigwedge }^{k}V$, and step 1.1 shows $⟨e^I,e_J⟩={\delta }_{IJ}$. Therefore the pairing matrix in these bases is the identity, so the pairing is nondegenerate.

Thus the canonical exterior-power pairing is bilinear, has the determinant formula on decomposables, and is nondegenerate.

By [L1], the space ${\bigwedge }^n{V}^{\ast }={\mathrm{Alt}}^n(V)$ is one-dimensional.

Since ${\bigwedge }^{n}V$ is the dual of ${\bigwedge }^n{V}^{\ast }$ by definition, or equivalently because [L2] identifies it with the dual through a nondegenerate pairing, it has the same dimension. Hence $\dim {\bigwedge }^{n}V=1$.

Therefore both top exterior powers are one-dimensional.

Evaluate both sides on $(v_2,\dots ,v_{k+\mathrm{\ell }})$. By [F2], the terms in $(\alpha \wedge \beta )(v,v_2,\dots ,v_{k+\mathrm{\ell }})$ split into two groups: those where $v$ lands among the $k$ arguments sent to $\alpha$, and those where it lands among the $\mathrm{\ell }$ arguments sent to $\beta$.

The first group is exactly $({\iota }_v\alpha )\wedge \beta$ by [F1]. To move $v$ into the first slot of $\beta$ in the second group, it must cross the $k$ slots occupied by $\alpha$, which contributes the sign $(-1{)}^k$; that group is therefore $(-1{)}^k\alpha \wedge {\iota }_v\beta$.

Summing the two groups gives the claimed formula, so interior product is a graded antiderivation.

On a chart domain $U$, the coordinate bases $\mathrm{\partial }\mathrm{/}\mathrm{\partial }{x}^i$ and [F1, given, construct] $d{x}^i$ identify each fibre in [F1] with the fixed finite-dimensional vector space $\mathrm{Mult}((({\mathbb{R}}^m{)}^{\ast }{)}^r\times ({\mathbb{R}}^m{)}^s,\mathbb{R})$ of type $(r,s)$ tensors on ${\mathbb{R}}^m$. This gives local trivializations $T_s^{r}M{\mathrm{\mid }}_U\cong U\times \mathrm{Mult}((({\mathbb{R}}^m{)}^{\ast }{)}^r\times ({\mathbb{R}}^m{)}^s,\mathbb{R})\mathrm{.}$

On an overlap, [L1] shows that each contravariant slot picks up one inverse [L1, step 1.1, algebra] Jacobian factor and each covariant slot picks up one Jacobian factor. Hence the tensor-coordinate change map is fibrewise linear, smooth in the base point, and satisfies the cocycle law because Jacobians and inverse Jacobians do.

Therefore [L2] applies to these local transition maps and produces a smooth [F1, L2, step 2.1] vector bundle. By construction its fibre over $p$ is the tensor space from [F1].

Thus the tensor transition laws define the smooth tensor bundle $T_s^{r}M$. [step 3.1]

By [F1] and [L1], a tensor field is a section of a smooth vector bundle whose local bundle coordinates are exactly the tensor coefficients relative to the chart bases $\mathrm{\partial }\mathrm{/}\mathrm{\partial }{x}^i$ and $d{x}^i$.

Applying [L2] to those trivializations shows that the section is smooth exactly when each local coefficient function is smooth.

Therefore smoothness of a tensor field is equivalent to smoothness of its coordinate components.

In any chart, [L1] identifies $S$, $T$, and $R$ with families of smooth coefficient functions. By [L2], the coefficients of $S\otimes T$ are finite sums of products of the coefficients of $S$ and $T$. Those are smooth.

In the same chart, [L2] writes each contracted coefficient of $R$ as a finite sum of coordinate coefficients of $R$. Because the contraction formula is basis-independent, these chartwise definitions glue. The resulting coefficient functions are smooth by [L1].

Therefore tensor products and the contraction defined above preserve smoothness.

In local coordinates, [F1] expresses each coefficient of $F^{\ast }T$ as a finite sum of the coefficients of $T\circ F$ multiplied by partial derivatives of $F$. Those are smooth, so [L1] implies that $F^{\ast }T$ is smooth.

The identity map has identity differential, so [F1] gives $({\mathrm{id}}_M{)}^{\ast }T=T$.

For $v_1,\dots ,v_k\in T_{p}M$, [F1] and [L2] give $((G\circ F{)}^{\ast }T{)}_p(v_1,\dots ,v_k)=T_{G(F(p))}(d{G}_{F(p)}d{F}_p{v}_1,\dots ,d{G}_{F(p)}d{F}_p{v}_k),$ which is exactly $(F^{\ast }(G^{\ast }T){)}_p(v_1,\dots ,v_k)$.

Therefore pullback of covariant tensors is smooth and functorial.

The field $Y$ is a type $(1,0)$ tensor field, so the false claim requires a vector field $X$ on $\mathbb{R}$ with $d{F}_t(X_t)=Y_{F(t)}$ for every $t$.

The differential of the constant map is zero at every point, so $d{F}_t(X_t)=0$ for every possible $X_t\in T_t\mathbb{R}$, whereas $Y_{F(t)}\mathrm{\ne }0$. Thus the required equality is impossible.

Thus a general mixed tensor field does not have a pullback by every smooth map, even though [L1] shows that purely covariant tensors do.

By [L1], symmetrization and alternation act fibrewise on $T_k^{0}M$ as smooth bundle endomorphisms over ${\mathrm{id}}_M$. Their fibres are the usual linear projections onto the symmetric and alternating tensors.

Because a projection has constant rank equal to the dimension of its image, the fibre ranks of these bundle maps are constant on $M$. Therefore [L2] shows that their images are smooth vector subbundles.