Tangent Cotangent and the Differential — Examples

In the standard chart on ${\mathbb{R}}^n$, the tangent basis is the standard coordinate basis by [L1].

The vector $u$ defines both the derivation ${\sum }_i{u}^i{\mathrm{\partial }}_i{\mathrm{\mid }}_a$ and the velocity of the straight line $t\mapsto a+tu$, and [L2] identifies these two descriptions.

Thus $T_a{\mathbb{R}}^n$ is canonically the usual vector space ${\mathbb{R}}^n$.

The Jacobian of $(r,\theta )\mapsto (x,y)$ is $\left(\begin{array}{cc}\cos \theta & -r\sin \theta \\ \sin \theta & r\cos \theta \end{array}\right)$.

Applying [L1] with this Jacobian gives the displayed formulas for ${\mathrm{\partial }}_r$ and ${\mathrm{\partial }}_{\theta }$ in the Cartesian basis.

This is the standard basis-change example.

In stereographic coordinates, the map is $u\mapsto 2u\mathrm{/}(1-u^2)$.

Differentiating this rational function gives $2(1+u^2)\mathrm{/}(1-u^2{)}^2$, so [L1] identifies this scalar as the matrix of the differential in the chosen one-dimensional bases.

Hence the differential is explicitly computed by the coordinate formula.

If $v$ is the velocity of a curve $\gamma$ in $S^n$ with $\gamma (0)=p$, then differentiating $\gamma (t)\cdot \gamma (t)=1$ at $0$ gives $p\cdot v=0$.

Conversely, if $p\cdot v=0$, the normalized curve $\gamma (t):=(p+tv)\mathrm{/}\mathrm{\parallel }p+tv\mathrm{\parallel }$ lies in $S^n$, satisfies $\gamma (0)=p$, and has velocity $v$ at $0$.

Contact-equivalent curves have the same coordinate, hence ambient, velocity, so the displayed map is well defined. Conversely, choose an index $k$ with $p_k\mathrm{\ne }0$ and the standard sphere chart on the hemisphere where the sign of the $k$th coordinate is fixed, obtained by deleting that coordinate. If two curves have the same ambient derivative, their derivatives after this coordinate deletion agree, so they are contact equivalent. The ambient-derivative map is therefore injective. Steps 1.1-1.2 show that its image is exactly $p^{\perp }$, and [L1] identifies its domain with the derivation tangent space.

The curve ${\gamma }_{p,a}$ lies in $S^1$, starts at $p$, and has ambient derivative $a(-y,x)$ at $0$. By [L2] it therefore determines a tangent derivation at $p$.

Let $p_0\in S^1$ and choose a local angle chart $\theta \mapsto p(\theta )=(\cos \theta ,\sin \theta )$ around $p_0$. Its coordinate tangent vector has ambient velocity $(-\sin \theta ,\cos \theta )$, so [L2] makes every tangent derivation over this arc uniquely $a{\mathrm{\partial }}_{\theta }$. In the induced chart of [L3], the displayed map is therefore $(\theta ,a)⟼(\theta ,a)\mathrm{.}$ Thus it and its inverse are smooth on every such bundle-chart domain.

Hence $T{S}^1$ is a cylinder.

The standard global chart on ${\mathbb{R}}^n$ induces a global tangent-bundle chart $T{\mathbb{R}}^n\to {\mathbb{R}}^n\times {\mathbb{R}}^n$.

Because there is only one chart, there are no nontrivial transition maps, so this induced chart is a global diffeomorphism.

Therefore $T{\mathbb{R}}^n$ is trivial.

The coordinate function $u\circ F$ is $x^2$, so $F^{\ast }(du)=d(x^2)=2xdx$.

Likewise $v\circ F=x+y$, so $F^{\ast }(dv)=d(x+y)=dx+dy$.

Thus coordinate one-forms pull back by the expected substitution rule.

For any target germ $[g]$, the composite $g\circ F$ is constant near every point of $M$.

Every derivation annihilates constant germs, so $d{F}_p(v)([g])=0$ for every $v\in T_{p}M$.

Hence $d{F}_p=0$.

At the origin, the angle coordinate is not defined, and for $r>0$ the same point admits many values of $\theta$ differing by multiples of $2\pi$.

Therefore the polar description is not a single-valued homeomorphism on any neighbourhood containing the origin, so it cannot be a chart there by [F1].

This refutes the statement.

Let $v={\mathrm{\partial }}_x{\mathrm{\mid }}_0$. In the $x$-chart its coordinate tuple is $1$, while [L1] gives $v=2{\mathrm{\partial }}_y{\mathrm{\mid }}_0$, so in the $y$-chart its coordinate tuple is $2$.

The same intrinsic tangent vector therefore has different coordinate tuples in different charts.

Hence the tuple itself is chart dependent and not the intrinsic tangent vector.