Tensor Fields Exterior Algebra and Differential Forms — Examples

For vectors $(a_1,a_2)$ and $(b_1,b_2)$, [L1] gives $(S\otimes T)((a_1,a_2),(b_1,b_2))=S(a_1,a_2)T(b_1,b_2)=a_1{b}_2,$ so $S\otimes T=e^1\otimes e^2$.

Again by [L1], $\mathrm{ctr}(R)=R(e^1,e_1)+R(e^2,e_2)=1+1=2.$

This computes the announced tensor product and contraction.

The displayed formula is linear in $(a,b)$ and in $(c,d)$ separately, so $g$ is bilinear.

By [F1], bilinearity is exactly the requirement for a type $(0,2)$ tensor. Hence $g$ is such a tensor.

Therefore the Euclidean dot product is a concrete type $(0,2)$ tensor.

For fixed $v$, the map $\alpha \mapsto \alpha (Av)$ is linear because evaluation of a fixed vector is linear on $V^{\ast }$. For fixed $\alpha$, the map $v\mapsto \alpha (Av)$ is linear because $A$ and $\alpha$ are linear. Thus $T$ is bilinear on $({\mathbb{R}}^2{)}^{\ast }\times {\mathbb{R}}^2$.

By [F1], this means $T$ is a type $(1,1)$ tensor.

Therefore an endomorphism gives a concrete type $(1,1)$ tensor through evaluation.

The tensor associated to ${\mathrm{id}}_{{\mathbb{R}}^n}$ is $T(\alpha ,v)=\alpha (v)$. Therefore [F1] gives $\mathrm{ctr}(T)={\sum }_{i=1}^n{e}^i(e_i)=n\mathrm{.}$

In the chosen basis, the matrix of the identity endomorphism is the identity matrix, so [F2] gives trace $n$. This agrees with step 1.1, and neither value depends on the basis.

Thus the identity endomorphism has coordinate-independent trace $n$.

Since $dx$ and $dy$ are $1$-forms, [L1] gives $dx\wedge dy=-dy\wedge dx$ and $dx\wedge dx=-dx\wedge dx$, hence $dx\wedge dx=0$.

By [L2], the unique increasing triple wedge $dx\wedge dy\wedge dz$ forms a basis of ${\mathrm{Alt}}^3({\mathbb{R}}^3)$.

These are the standard wedge-product identities in the dual basis.

The matrix with entries $e^i(v_j)$ is exactly the coordinate matrix $[v_1\cdots v_n]$ of the ordered $n$-tuple $(v_1,\dots ,v_n)$.

Applying [L1] to ${\alpha }_i=e^i$ gives $⟨e^1\wedge \cdots \wedge e^n,v_1\wedge \cdots \wedge v_n⟩=\det (e^i(v_j))=\det [v_1\cdots v_n]\mathrm{.}$

By [L2], this scalar determines the decomposable top wedge relative to the standard volume form.

Thus the determinant is the top-degree pairing against $e^1\wedge \cdots \wedge e^n$.

The coefficients of $g$ in the standard coordinates are constant, so $g$ is smooth.

For vectors $u,v$, one has $g(u,v)=g(v,u)$, so each fibre value $g_p$ is symmetric. Hence [F1] places $g$ in the symmetric fibrewise part, and [L1] identifies that part as a smooth subbundle.

Therefore the Euclidean metric is a symmetric smooth two-tensor.

By [F1], $F^{\ast }dx=\cos \theta dr-r\sin \theta d\theta ,F^{\ast }dy=\sin \theta dr+r\cos \theta d\theta \mathrm{.}$

Using [L1] and bilinearity of the wedge product, $F^{\ast }(dx\wedge dy)=F^{\ast }dx\wedge F^{\ast }dy=(\cos \theta dr-r\sin \theta d\theta )\wedge (\sin \theta dr+r\cos \theta d\theta )=rdr\wedge d\theta \mathrm{.}$

Thus the area form becomes $rdr\wedge d\theta$ in polar coordinates.

Along the curve, $x\circ c=\cos t$, $y\circ c=\sin t$, so $c^{\ast }x=\cos t,c^{\ast }y=\sin t,c^{\ast }dx=-\sin tdt,c^{\ast }dy=\cos tdt\mathrm{.}$

Therefore $c^{\ast }\eta =-(\sin t)(-\sin tdt)+(\cos t)(\cos tdt)=({\sin }^{2}t+{\cos }^{2}t)dt=dt\mathrm{.}$

Thus the angular form pulls back to the standard parameter form on the circle.

The map $F$ is noninjective and has differential $d{F}_t=0$ for every $t\in \mathbb{R}$.

A vector field $X$ as in the false claim would satisfy $d{F}_t(X_t)=Y_{F(t)}$ at each point. Step 1.1 makes that impossible, because $d{F}_t(X_t)=0$ for every $X_t\in T_t\mathbb{R}$, whereas $Y_{F(t)}\mathrm{\ne }0$.

Hence $Y$ has no pullback along $F$, giving the announced counterexample and agreeing with [L1].

The reflection satisfies $R^{\ast }dx=dx$ and $R^{\ast }dy=d(-y)=-dy$.

By [L1], $R^{\ast }\omega =R^{\ast }(dx\wedge dy)=R^{\ast }dx\wedge R^{\ast }dy=dx\wedge (-dy)=-dx\wedge dy\mathrm{.}$

Therefore the top-degree form changes sign under the reflection, so the claimed invariance is false.

Write $v=a{\mathrm{\partial }}_x+b{\mathrm{\partial }}_y+c{\mathrm{\partial }}_{{\xi }_1}+d{\mathrm{\partial }}_{{\xi }_2}$ at $(x,y,{\xi }_1,{\xi }_2)$. Then $d\pi (v)=a{\mathrm{\partial }}_x+b{\mathrm{\partial }}_y$, so the covector $\xi ={\xi }_{1}dx+{\xi }_{2}dy$ gives $\xi (d\pi (v))={\xi }_{1}a+{\xi }_{2}b$.

The form $\lambda ={\xi }_{1}dx+{\xi }_{2}dy$ takes the same value on $v$, namely $\lambda (v)={\xi }_{1}a+{\xi }_{2}b$. Thus ${\lambda }_{(p,\xi )}(v)=\xi (d\pi (v))$. Its coefficients are smooth coordinate functions, so [F1] makes it a smooth $1$-form.

Therefore the canonical one-form is a concrete covariant tensor on the cotangent bundle.