Past exam of the lie Groups and Differentiable Manifolds course of the University of Oxford / 2008 / B (Differentiable Manifolds)

One definition of $T_{x}M$ is the space of derivations at $x$: linear maps $v:C^{\infty }(M)\to \mathbb{R}$ satisfying $v(fh)=f(x)v(h)+h(x)v(f)$. In local coordinates $u^1,\dots ,u^n$, each has the unique form $v={\sum }_i{v}^i\partial /\partial u^i{|}_x$, identifying $T_{x}M$ with ${\mathbb{R}}^n$. Equivalently it is the space of velocities of curves through $x$, identifying curves with the same coordinate derivative.

For a chart $\varphi :U\to {\mathbb{R}}^n$, give $TM{|}_U$ coordinates $(x,v)\mapsto (\varphi (x),D{\varphi }_{x}v)$. If $F=\psi \circ {\varphi }^{-1}$ is a base transition, the induced transition isThis is a smooth bijection with smooth inverse coming from the reverse chart. The resulting bundle atlas makes $TM$ a smooth manifold of dimension $2n$. These local product charts also give the usual Hausdorff and second-countable topology, inherited from a countable base atlas and its fibre coordinates.

For $n\ge 1$, identify nonzero vectors in each fibre up to nonzero real scalar multiplication. Locally this gives $P(TM){|}_U\cong U\times {\mathbb{RP}}^{n-1}$. In a projective chart where $v^j\ne 0$, use the ratios $v^i/v^j$ for $i\ne j$. Under a base change, the fibre line maps by $[v]\mapsto [D{F}_{u}v]$; its projective coordinates are ratios of smooth linear expressions with nonzero denominator on the overlap. This defines a smooth manifold of dimension $n+(n-1)=2n-1$.

For the sphere, a point is a pair $(x,L)$ with $|x|=1$ and an unoriented line $L\subset x^{\perp }$. Choose a unit $v\in L$. The triple $(x,v,x\times v)$ is a positively oriented orthonormal frame. Given a second pair, the unique orientation-preserving linear map between chosen frames lies in $S{O}_3$ and carries the first pair to the second. Thus the action is transitive.

A rotation stabilizing $(x,L)$ fixes $x$, so rotates its tangent plane. To preserve the unoriented line it must act by $v\mapsto v$ or $v\mapsto -v$, giving exactly the identity and a half-turn about $x$. Hence the stabilizer has order two, and $P(T{S}^2)\cong S{O}_3/C_2$.

This is an example of the projectivized tangent bundle.

In a coordinate chart, define the exterior derivative byThe chain rule makes this independent of coordinates. It is real linear, increases form degree by one, agrees with the differential on functions, satisfies $d^2=0$, and obeys the graded Leibniz ruleIt also commutes with pullback by a smooth map. These properties characterize it locally and hence globally.

de Rham cohomology is the quotient of closed forms by exact forms. If $d\alpha =d\beta =0$, Leibniz gives $d(\alpha \wedge \beta )=0$. Changing $\alpha$ to $\alpha +d\eta$ changes the product by $d(\eta \wedge \beta )$; changing $\beta$ to $\beta +d\xi$ changes it by $(-1{)}^{r}d(\alpha \wedge \xi )$. Thusis independent of representatives. This gives the graded-commutative de Rham cohomology ring.

The nonvanishing top-degree form ${\omega }^m$ is a volume form. Declare an ordered tangent basis positive when its value under ${\omega }^m$ is positive. Smooth nonvanishing ensures these choices agree on overlapping oriented charts and vary continuously. Hence a symplectic manifold is orientable; this is its symplectic orientation.

Use the standard convention that the compact manifold is nonempty and has no boundary. Orient it by ${\omega }^m$. Since $d\omega =0$, every power ${\omega }^k$ is closed. If ${\omega }^k=d\eta$ for some $1\le k\le m$, thenagain by Leibniz and closedness. Stokes theorem would give ${\int }_M{\omega }^m=0$. But this integral is strictly positive in the symplectic orientation, a contradiction. ThereforeThe boundaryless hypothesis matters: a compact disk has a symplectic area form but no nonzero ordinary second cohomology class.

Let the two oriented surfaces carry positive area forms $\alpha$ and $\beta$, obtainable from oriented Riemannian metrics. A top-degree form on a surface is automatically closed. On their product setIt is closed, andis a nowhere-vanishing four-form; the two squared terms vanish because each comes from a two-dimensional factor. Thus the product is symplectic, with no compactness assumption needed.

For compact connected oriented $n$-manifolds without boundary, the degree is characterized byfor every smooth top-degree form $\eta$. Equivalently, at any regular value $y$,The preimage is finite by compactness and the inverse function theorem. Taking a form supported in a small neighbourhood of a regular value and applying change of variables proves the signed-count formula. The top de Rham cohomology of a compact connected oriented manifold is one-dimensional, with integration giving its coordinate; pullback and Stokes theorem make the integral characterization independent of the chosen normalized form. Thus this signed count is independent of the regular value. In particular a nonzero degree forces surjectivity, and the number of points in any regular fibre is at least the absolute degree. Degree is invariant under homotopy, as follows by applying Stokes theorem to the pullback on $M\times [0,1]$.

For the polynomial assertion take $k\ge 1$. The map $P:\mathbb{C}\to \mathbb{C}$ extends smoothly to a map of the Riemann sphere fixing infinity: in reciprocal coordinates its target coordinate isThe homotopy $P_t(z)=z^k+t(a_1{z}^{k-1}+\cdots +a_k)$ extends smoothly in the same way. The map $z\mapsto z^k$ has $k$ distinct inverse images of the regular value one, each orientation-preserving, since the real Jacobian of multiplication by a nonzero complex derivative is its squared absolute value. Its degree is $k$, so $\mathrm{deg}P=k\ne 0$. It is onto the sphere and must attain zero at a finite point. This proves every nonconstant complex polynomial has a root, without assuming the result to count roots of a general polynomial.

For a proper smooth $F:{\mathbb{R}}^n\to {\mathbb{R}}^n$, choose a compactly supported $n$-form $\eta$ with $\int \eta =1$ and defineProperness makes its pullback compactly supported. If ${\eta }^{\prime }$ is another choice, the given result gives $\eta -{\eta }^{\prime }=d\beta$ with compact support. The difference of the defining integrals is $\int d(F^{\ast }\beta )=0$, proving independence. Sard's theorem supplies a regular value. A proper map has finitely many preimages of it, and a sufficiently small neighbourhood has no additional inverse sheets outside the chosen inverse-function neighbourhoods: otherwise a sequence of extra preimages would have a convergent subsequence by properness and limit in that fibre. A normalized form there therefore gives the same finite signed-count formula, proving that the degree is an integer.

Finally $x\mapsto x^k$ is proper for $k\ge 1$. For odd $k$, a positive regular value has one inverse image with positive derivative. For positive even $k$, it has two inverse images with opposite derivative signs. ConsequentlyFor $k=0$ it is the constant map one, whose inverse image of the compact singleton $\{1\}$ is all of $\mathbb{R}$. It is not proper, so its degree is not defined by the proper-map construction requested here.

A Riemannian metric is a smooth positive definite symmetric bilinear form $g_q$ on each tangent space. The inverse form on the cotangent space defines the kinetic HamiltonianIt is smooth and satisfies $H(q,ap)=a^{2}H(q,p)$. With the canonical symplectic form $\Omega ={\sum }_{i}d{q}^i\wedge d{p}_i$, define ${\iota }_{X_H}\Omega =dH$. This gives Hamilton's equationsThus $p_i=g_{ij}{\dot{q}}^j$. Differentiating the inverse metric yields ${\dot{p}}_i=\frac{1}{2}{\partial }_i{g}_{jk}{\dot{q}}^j{\dot{q}}^k$. Differentiating $p_i=g_{ij}{\dot{q}}^j$ and substituting givesTherefore the projections of the Hamiltonian integral curves are affinely parametrized geodesics, and lifting a geodesic by its metric covector gives a Hamiltonian integral curve.

For the Killing vector field define the fibrewise linear function $f(q,p)=p(Y_q)=p_i{Y}^i(q)$. Along the lifted geodesic it equals $g(\dot{q},Y)$. Metric compatibility and the geodesic equation giveThus $f$ is conserved. Equivalently the cotangent lift of the isometry flow preserves $H$, and its generator has Hamiltonian $f$, so their Poisson bracket vanishes.

For the final two metrics write $s=\ell (x)+m(y)$, restricting to an open set where the printed tensors are smooth nondegenerate metrics. Their inverse forms giveAlong the $H_1$ flow, $H_1$ is constant, whileHenceThis is the Liouville metric geodesic integral, a quadratic conserved quantity. It does not require a Killing vector field; it arises from the separated metric coefficients. When both printed metrics are Riemannian, their positivity in particular requires $s>0$, $m>0$ and $\ell <0$; the calculation also applies wherever the corresponding inverse tensors are defined.