The De Rham Complex Homotopy and Mayer Vietoris — Examples

A smooth function is uniquely its value at $p$, so ${\mathrm{\Omega }}^0(M)=\mathbb{R}$. There are no nonzero cotangent vectors, hence no positive-degree forms, and $d=0$.

The degree-zero quotient has no boundaries, and evaluation at $p$ sends $[a][b]$ to $ab$ and $[1]$ to $1$. Thus it is the algebra $\mathbb{R}$; every other group is zero.

For any closed $k$-form with $k\ge 1$, the radial primitive is ${\eta }_x(v_1,\dots ,v_{k-1})={\int }_0^1{t}^{k-1}{\omega }_{tx}(x,v_1,\dots ,v_{k-1})dt$, and $d\eta =\omega$. Thus every positive-degree class is zero.

Euclidean space is nonempty and connected, so its degree-zero classes are constant functions with their ordinary multiplication. Negative degrees are zero by the complex convention. When $n=0$, the only form space is the constants on a point, giving the same ring.

A function is a tuple $(a_1,\dots ,a_m)$; every function is locally constant, and each tangent space is zero, so $df=0$ and all positive-degree form spaces vanish. The quotient has $B^0=0$.

Restriction to the finite disjoint points is the direct-sum identification. For tuples $a,b$, $(ab{)}_j=a_j{b}_j$, so multiplication is coordinatewise. For $m=0$ this is the zero algebra and for $m=1$ it is $\mathbb{R}$.

Stereographic projection from the omitted point makes each of $U,V$ diffeomorphic to $\mathbb{R}$. Their overlap has two open-arc components $W_0,W_1$, each diffeomorphic to an open interval and hence contractible by linear contraction in that coordinate. F3 makes their positive-degree groups vanish, while F2 identifies their degree-zero groups with constants on components. Thus the initial Mayer–Vietoris segment is $0\to H^0(S^1)\to {\mathbb{R}}^2\stackrel{s}{\to }{\mathbb{R}}^2\to H^1(S^1)\to 0$, where $s(a,b)=(b-a,b-a)$.

The kernel is $\{(a,a)\}$ and the image is the diagonal. The functional $(c,d)\mapsto d-c$ has exactly that kernel and is onto, so the cokernel is $\mathbb{R}$. Exactness computes both groups. Degrees above one and negative degrees have zero form spaces.

The form is smooth and closed because two-forms on a one-manifold vanish. For $\gamma (t)=(\cos t,\sin t)$, $0\le t\le 2\pi$, substitution gives ${\gamma }^{\ast }\alpha =dt\mathrm{/}(2\pi )$, hence ${\int }_{S^1}\alpha =1$.

The circle is compact, oriented, boundaryless and embedded, and the form is closed, so its nonzero period obstructs exactness. Since the sphere theorem gives $\dim H^1(S^1)=1$, this nonzero class is a basis.

Apply F1 with $n=2$. It gives $H^0(S^2)=H^2(S^2)=\mathbb{R}$ and $H^1(S^2)=0$, with all remaining degrees zero.

The same theorem with $n=2$ gives the constant-function group in degree zero and the zero group in degree one. A product of two degree-two classes has degree four and is zero, so the ring is $\mathbb{R}[u]\mathrm{/}(u^2)$ with $\mathrm{\mid }u\mathrm{\mid }=2$.

For tangent vectors $v_1,\dots ,v_n$, expansion along the first column gives ${\omega }_x(v_1,\dots ,v_n)=\det (x,v_1,\dots ,v_n)$. The outward orientation is precisely the convention that this determinant is positive on positive tangent bases. Since $x$ is a nonzero normal to the tangent space, the determinant is nonzero on every tangent basis; thus $\omega$ is smooth, positive and nowhere zero.

The sphere is nonempty, compact, oriented and boundaryless, so the positive-top-form clause of the integral obstruction theorem makes $\omega$ nonexact. It is closed by top degree. The sphere computation gives a one-dimensional $H^n$, and its nonzero class therefore generates it.

The radial map $r(x)=x\mathrm{/}\mathrm{\mid }x\mathrm{\mid }$ retracts onto $S^2$. The homotopy $F(x,t)=((1-t)+t\mathrm{/}\mathrm{\mid }x\mathrm{\mid })x$ has norm $(1-t)\mathrm{\mid }x\mathrm{\mid }+t>0$; it begins at $x$, ends at $r(x)$ and fixes points of $S^2$.

This is the $n=3$ instance of the punctured-space theorem, so its only nonzero groups are $H^0=H^2=\mathbb{R}$. Products of positive-degree classes vanish because their degree is at least four.

The homotopy $F(x,t)=((1-t)+t\mathrm{/}\mathrm{\mid }x\mathrm{\mid })x$ has radius $(1-t)\mathrm{\mid }x\mathrm{\mid }+t$, between $\mathrm{\mid }x\mathrm{\mid }$ and $1$, hence strictly between $a$ and $b$. It is smooth, fixes the unit circle, and connects the identity to $ir$; also $ri=\mathrm{id}$.

Thus $i^{\ast },r^{\ast }$ are inverse graded algebra homomorphisms by smooth homotopy invariance. The sphere computation gives one generator $u$ in degree one and the unit in degree zero; $u^2=0$ because $H^2(S^1)=0$. The annulus has the same multiplication, so its ring is the exterior algebra on one degree-one generator.

The denominator is positive. Writing its coefficients as $a=-y\mathrm{/}(x^2+y^2)$ and $b=x\mathrm{/}(x^2+y^2)$ gives ${\mathrm{\partial }}_{x}b={\mathrm{\partial }}_{y}a=(y^2-x^2)\mathrm{/}(x^2+y^2{)}^2$. Hence $d\alpha =({\mathrm{\partial }}_{x}b-{\mathrm{\partial }}_{y}a)dx\wedge dy=0$.

On $\gamma (t)=(\cos t,\sin t)$, ${\gamma }^{\ast }\alpha =dt$ and its integral is $2\pi$. If $\alpha =df$, then the chain rule and the fundamental theorem would make this integral $f(\gamma (2\pi ))-f(\gamma (0))=0$. Thus the closed witness is not exact.

All functions are locally constant, hence identify with their $H^0$ classes. At the point $2$, $F^{\ast }{G}^{\ast }u=u\circ G\circ F$ has value $u(1)=1$, whereas $G^{\ast }{F}^{\ast }u=u\circ F\circ G$ has value $u(3)=0$.

The actual functoriality identity is $(G\circ F{)}^{\ast }=F^{\ast }{G}^{\ast }$, so the two displayed values refute the proposed covariant ordering even when all sources and targets coincide.

Take a subordinate partition ${\rho }_U+{\rho }_V=1$. The local forms $\eta {\mathrm{\mid }}_U=d(-{\rho }_V\omega )$ and $\eta {\mathrm{\mid }}_V=d({\rho }_U\omega )$ glue by the connector formula. Explicitly $\eta =0$ on $W_0$ and $\eta =d{\rho }_U=-d{\rho }_V$ on $W_1$, with the prescribed smooth zero extensions at the removed points. Thus $[\eta ]=\delta [(0,1)]$.

The degree-zero groups on $U,V$ are constants, whose difference on the two overlap components is $(b-a,b-a)$. The vector $(0,1)$ is not diagonal, since equality to $(c,c)$ would require $c=0$ and $c=1$. Exactness of the Mayer–Vietoris sequence defining this connector therefore gives $\delta [(0,1)]\mathrm{\ne }0$.