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Local coordinates $x^{\mu }$ induce the coordinate-basis tangent vectors ${\partial }_{\mu }=\partial /\partial x^{\mu }$. Under a coordinate change they transform by the chain rule.

For nonzero real $a_i$, scale the ambient coordinates by $q_i=a_i{r}_i$. On the half where $\epsilon q_i>0$, omitting $q_i$ is a manifold chart onto the open unit ball; its inverse restores $q_i=\epsilon \sqrt{1-{\sum }_{j\ne i}{q}_j^2}$. These $2(n+1)$ charts form a smooth atlas of the ellipsoid, including for negative $a_i$.

The cotangent space at $p$ is the dual vector space $T_p^{\ast }M=(T_{p}M{)}^{\ast }$. Its elements are covectors on the tangent space. In a manifold chart $x$, the basis $d{x}^i{|}_p$ is dual to ${\partial }_i{|}_p$. Under a coordinate change to $y$, the coefficients of $\alpha =a_{i}d{x}^i=b_{j}d{y}^j$ obey $b_j={\sum }_i{a}_i\partial x^i/\partial y^j$. The disjoint union of these spaces is the cotangent bundle.

A topological manifold has dimension $n$ when its manifold charts have coordinate domains open in ${\mathbb{R}}^n$. For a smooth manifold, each tangent space has dimension $n$, and a tangent bundle has vector bundle rank $n$.

On the unit sphere, use $\zeta =(x+iy)/(1-z)$ away from the north pole and $\eta =(x-iy)/(1+z)$ away from the south pole. Their overlap relation $\eta =1/\zeta$ is a holomorphic map, giving a holomorphic atlas for the Riemann sphere. The opposite sign of $iy$ in the second manifold chart ensures a holomorphic rather than antiholomorphic transition.

For a smooth differential one-form $\alpha ={\alpha }_i(x)d{x}^i$ with $d\alpha =0$, take a manifold chart identifying a neighborhood with a ball centered at zero, and set $\Phi (x)={\int }_0^1{x}^i{\alpha }_i(sx)ds$. Closedness means ${\partial }_j{\alpha }_i={\partial }_i{\alpha }_j$. ConsequentlyThus $d\Phi =\alpha$ in any dimension. This is the one-form case of the Poincare lemma, and applies to spacetime as well as three-dimensional vector calculus. A global potential need not exist on a domain with nonzero periods.

The matrix logarithm near the identity maps $Sp(n)=\{A\in U(2n):AJ{A}^t=J\}$ onto an open set of its real compact symplectic Lie algebra $\{X:X^{\ast }=-X,XJ+J{X}^t=0\}$. Its inverse is the matrix exponential. These statements follow by applying the logarithm to $A^{\ast }=A^{-1}$ and $A^t=J^{-1}{A}^{-1}J$, and differentiating $e^{tX}J{e}^{t{X}^t}$. Left translations provide manifold charts at every group element. In block form $X=\left(\begin{array}{cc}P& Q\\ -\bar{Q}& \bar{P}\end{array}\right)$ with $P^{\ast }=-P$ and $Q^t=Q$, giving real dimension $n^2+n(n+1)=n(2n+1)$.

On an oriented Riemannian manifold, the metric volume form is the volume form taking value one on positively oriented orthonormal frames. In an oriented manifold chart it has the displayed expression. Under an orientation-preserving coordinate change with Jacobian $A$, $\det g$ acquires the factor $(\det A{)}^2$, precisely matching the transformation of the top exterior product. Thus the expressions agree on chart overlaps. Orientation is needed to make a globally positive form, though the volume density exists without it.

An $n$-dimensional smooth manifold is a Hausdorff space with a countable topological base, equipped with a smooth atlas of homeomorphisms from open subsets onto open subsets of ${\mathbb{R}}^n$, whose overlap maps are smooth diffeomorphisms. The smooth structure is the maximal smooth atlas compatible with these manifold charts. Here manifolds have no boundary unless specified otherwise.

For a space with all the requested topological properties but no such atlas, take the topological tripod: three closed intervals joined at one endpoint. It is a compact connected subspace of the plane with its induced metric. A countable base of planar rational balls restricts to a countable base, the metric makes it Hausdorff, and compactness gives a finite subcover of every open cover, hence a locally finite refinement and paracompactness.

At an interior point of any arm, arbitrarily small neighborhoods are intervals. A coordinate ball in dimension at least two would remain connected after deleting its center; an interval does not. Dimension zero would make the space discrete. Thus any possible connected manifold structure would have dimension one. But a sufficiently small neighborhood of the junction, minus the junction, has three components, whereas an interval chart has two. This contradiction rules out even a topological manifold structure, and hence any smooth one.

The product smooth structure uses manifold charts $(\varphi ,\psi ):U\times V\to \varphi (U)\times \psi (V)\subset {\mathbb{R}}^{m+n}$. Transition maps act separately in the two coordinate blocks and are smooth with smooth inverses. Products of countable bases give a countable base, and the product remains Hausdorff. Therefore $\boxed{\dim (M\times N)=\dim M+\dim N}$ with this natural smooth structure.

Define the tangent space by point derivations: a tangent vector at $x$ is an $\mathbb{R}$-linear map $D$ on germs of smooth functions at $x$, satisfying $D(ab)=a(x)D(b)+b(x)D(a)$. Addition and scalar multiplication preserve this rule, so these derivations form a vector space. In coordinates $u^1,\dots ,u^n$, the local identityfollows by integrating the derivative of $h$ along the coordinate line segment. Derivations annihilate constants, so it gives $Dh={\sum }_{i}D(u^i){\partial }_{i}h(u(x))$. The coordinate derivations ${\partial }_i{|}_x$ are independent since they evaluate the coordinate functions as ${\delta }_i^j$. ThusFor a smooth map, define the differential of a smooth map intrinsically by $(f_{\ast }D)(h)=D(h\circ f)$. It is again linear and satisfies the derivation rule at $f(x)$, so it maps $T_{x}M$ into $T_{f(x)}N$. Its coordinate matrix is the Jacobian of the coordinate expression of $f$, independently of the charts by the intrinsic definition and chain rule.

Finally, a zero differential forces each target coordinate function to have all derivatives zero on a small connected source coordinate ball mapping into one target chart. Integration on straight segments makes those functions constant there. Hence $f$ is locally constant. Each nonempty fiber is both open and closed, so connectedness gives the zero-differential constancy theorem, $\boxed{f\text{ is constant}}$.

A Riemannian metric is a smooth field of positive-definite symmetric bilinear forms $g_x$ on the tangent spaces, equivalently a smooth section of ${\mathrm{Sym}}^2{T}^{\ast }M$ with that positivity. Its musical isomorphism isNondegeneracy makes every fiber map an isomorphism. In coordinates the map has matrix $g_{ij}$, and its inverse ${♯}_g$ has matrix $g^{ij}$. Smoothness of the inverse follows from the inverse-matrix formula and the nonvanishing determinant. Both maps cover the identity on $M$, giving a smooth vector bundle isomorphism.

For local flattening of a Riemannian metric, choose a manifold chart neighborhood $V$ around $p$ with closure contained in $U$, and let $h={\varphi }^{\ast }\delta$ on $V$. Choose a smooth bump function $0\le \chi \le 1$, with compact support in $V$ and equal to one on a smaller neighborhood $W$ of $p$. SetThe formulas glue smoothly because the modification is supported strictly inside $V$. A convex combination of positive-definite forms is positive definite. On $W$ the metric is exactly $h$, so $\varphi$ gives an isometry to a Euclidean open subset. Outside $U$ the original metric remains unchanged.

Geodesic completeness means that every geodesic with arbitrary initial point and tangent vector extends for every real affine parameter. On each connected component, the Hopf-Rinow theorem equates this with completeness of the Riemannian distance.

The Euclidean-end claim needs a compact-core interpretation that is absent from its literal hypotheses. Indeed, $M=\{x\in {\mathbb{R}}^n:|x|>1\}$ with $n\ge 2$, $X=\varnothing$ and $g=\delta$ satisfy those hypotheses. The geodesic $\gamma (t)=(2-t)e_1$ reaches the missing unit sphere at $t=1$ and cannot continue within $M$. Thus the printed assumptions alone do not prove completeness.

Here is the intended compact-core completeness for a Euclidean end. Write $\Phi :M\setminus X\to \{|x|>1\}$ for the end coordinates and assume additionally thatis compact for every sufficiently large $R$. Choose such an $R$ beyond the metric transition and large enough to include the initial point of a geodesic. Its speed $s$ is constant. On every segment outside $K_R$, the Euclidean radius changes at a rate at most $s$, so over a finite parameter interval it cannot exceed $R+sT$. The full segment therefore stays in the compact set $K_{R+sT}$. Its velocities also stay in a compact subset of $TM$, since their metric norm is fixed and the base set is compact. The smooth geodesic ordinary differential equation consequently extends past any supposed finite endpoint. The reversed-time argument is identical. This proves completeness under the compact-core condition, and explains exactly what the exterior-of-a-ball counterexample lacks.

For upward stability of Riemannian completeness, lengths of all curves satisfy $L_{\tilde{g}}\ge L_g$, hence $d_{\tilde{g}}\ge d_g$. A $d_{\tilde{g}}$-Cauchy sequence is therefore $d_g$-Cauchy and has a limit $p$ because $g$ is complete. Smooth positive-definite metrics induce the manifold topology. More explicitly, on a small coordinate ball around $p$, $\tilde{g}$ has a bounded largest matrix eigenvalue, so the coordinate straight segment gives $d_{\tilde{g}}(p,q)\le C|\varphi (q)-\varphi (p)|$. Thus convergence to $p$ is also in $d_{\tilde{g}}$. That distance is complete, and Hopf-Rinow yieldsFor disconnected $M$, apply this argument on each component.

There is a small domain issue in the printed notation. If ${\mathbb{R}}^3/{\mathbb{R}}^2$ means a quotient vector space by the horizontal plane, the displayed formula does not descend to that quotient vector space: $(0,0,0)$ and $(1,0,0)$ represent the same class but give different values. The intended construction is stereographic projection, restricted to the unit sphere with the north pole removed. Interpreting the slash as removal of the plane $z=1$ also supplies a suitable ambient domain. The holomorphic stereographic atlas of the sphere is obtained as follows.

Write $N=(0,0,1)$ and $S=(0,0,-1)$. On $U_N=S^2\setminus \{N\}$ use $\zeta =(x+iy)/(1-z)$. Its inverse, with $\zeta =u+iv$, isThese formulas give a smooth manifold chart from $U_N$ onto $\mathbb{C}$. On $U_S=S^2\setminus \{S\}$ choose the second manifold chartIts inverse is $x=2\mathrm{Re}\eta /(1+|\eta {|}^2)$, $y=-2\mathrm{Im}\eta /(1+|\eta {|}^2)$ and $z=(1-|\eta {|}^2)/(1+|\eta {|}^2)$, so this is also a smooth manifold chart onto $\mathbb{C}$. The conjugation in this second stereographic projection is essential. On the overlap, $x^2+y^2=1-z^2$, soBoth directions of this transition are holomorphic maps on ${\mathbb{C}}^{\times }$, with nonzero derivative. The two manifold charts cover the sphere, hence define a holomorphic atlas, giving precisely the Riemann sphere. If one instead used $x+iy$ in both manifold charts, the transition would be $1/\bar{\zeta }$ and would not be holomorphic.

Orient the sphere by this holomorphic atlas. The given volume form is smooth at infinity: replacing $\zeta$ by $1/\eta$ in its exterior product gives the same expressionMoreover, $id\zeta \wedge d\bar{\zeta }=2du\wedge dv$, so the normalization of this volume form isThus the printed volume form has half the area of the standard round unit sphere; replacing its integral by $4\pi$ would introduce an erroneous factor of two.

For $k\ge 1$, the holomorphic map $f(\zeta )={\zeta }^k$ extends over infinity, since the target reciprocal coordinate is ${\eta }^k$ when the source reciprocal coordinate is $\eta$. Its pullback of a differential form isUsing $s=r^{2k}$ givesThe degree of a map between oriented manifolds is thereforeFor $k=0$, the formula on the finite manifold chart is the constant $1$. Its unique continuous extension is also $1$ at infinity, rather than an undefined expression ${\infty }^0$. Its pullback of a differential form is zero and its degree of a map between oriented manifolds is zero.

For the preimage calculation when $k\ge 1$, choose a regular value $w\in {\mathbb{C}}^{\times }$. There are exactly $k$ distinct roots of ${\zeta }^k=w$. At each root the real Jacobian determinant is $|k{\zeta }^{k-1}{|}^2>0$, so every local contribution to the degree of a map between oriented manifolds is $+1$. Their sum is $k$, agreeing with the integral. The exceptional values $0$ and infinity are avoided because they are branch values when $k>1$. For the constant map, any $w\ne 1$ is a regular value with no preimages, giving the same answer zero. This establishes the degree of a power map of the Riemann sphere for every allowed $k$.

The exterior derivative is an $\mathbb{R}$-linear map $d:{\Omega }^p(M)\to {\Omega }^{p+1}(M)$, agrees with the differential $df$ of a smooth function, satisfies the graded Leibniz rule $d(\alpha \wedge \beta )=d\alpha \wedge \beta +(-1{)}^p\alpha \wedge d\beta$ for $\alpha \in {\Omega }^p(M)$, and has $d^2=0$. These properties determine it locally and hence globally. To see locality directly, if a differential form $\alpha$ vanishes near $x$, choose a smooth bump function $\chi$ equal to one near $x$ and supported where $\alpha =0$. The identity $d(\chi \alpha )=d\chi \wedge \alpha +\chi d\alpha$ gives $(d\alpha {)}_x=0$. Thus global forms can be computed using local extensions in a manifold chart.

In coordinates write $\alpha ={\sum }_I{a}_{I}d{x}^{i_1}\wedge \cdots \wedge d{x}^{i_p}$. Since $d(d{x}^i)=d^2{x}^i=0$, the graded Leibniz rule forcesThis is the uniqueness of the exterior derivative from its axioms. The coordinate formula also establishes existence: it has the stated properties, and the chain rule shows that coordinate changes give the same operator.

The de Rham cohomology is the real quotient vector spaceThus a cohomology class records a closed differential form modulo an exact differential form. In degree zero there are no exact forms; closed functions are locally constant. The Poincare lemma says that every closed form of positive degree on a star-shaped open subset of ${\mathbb{R}}^n$ is exact, and hence that positive-degree closed forms are locally exact on a smooth manifold.

For the first de Rham cohomology of the two-sphere, let $U=S^2\setminus \{N\}$ and $V=S^2\setminus \{S\}$. Stereographic projection identifies each with ${\mathbb{R}}^2$, so a closed one-form $\alpha$ has primitives $f_U,f_V$. On the connected overlap $U\cap V$, the derivative of $f_U-f_V$ is zero, so this difference is a constant. Subtracting that constant from $f_U$ makes the primitives agree. They glue to a global smooth primitive. Therefore $H_{\mathrm{dR}}^1(S^2)=0$.

Let $q:S^2\to {\mathbb{RP}}^2$ be the double covering map and $a$ the antipodal map. The pullback of a differential form identifies forms downstairs with $a$-invariant forms upstairs. Averaging $(\eta +a^{\ast }\eta )/2$ commutes with $d$. If an invariant form is exact upstairs, averaging its primitive proves it exact downstairs. Conversely an invariant cohomology class has an invariant representative by the same averaging. This proves the de Rham cohomology of a finite quotient identification $H_{\mathrm{dR}}^p({\mathbb{RP}}^2)\cong H_{\mathrm{dR}}^p(S^2{)}^{a^{\ast }}$. In degree zero the sphere is connected and $a^{\ast }$ fixes constants; degree one is zero; in degree two the granted action is multiplication by $-1$, whose invariant real subspace is zero. Forms of degree greater than two vanish. HenceThis is real de Rham cohomology; it does not detect the integral two-torsion of the real projective plane.