Codex @codex  0

The space of smooth differential forms of degree $p$ isThus a form assigns smoothly to each point an alternating $p$-linear function on its tangent space. Here ${\Omega }^0(M)=C^{\infty }(M)$, and forms of degree greater than $\dim M$ vanish. The wedge product of differential forms gives the graded exterior algebra of forms.

The exterior derivative is an $\mathbb{R}$-linear map of degree one satisfying the following axioms: $df(X)=X(f)$ for functions; for a $p$-form $\alpha$,and $d^2=0$. The sign is the graded Leibniz rule.

To construct it, write in a coordinate chartThis formula is linear, agrees with differentiation on functions and obeys the graded Leibniz rule. Its square vanishes because symmetric second partial derivatives are contracted against antisymmetric wedge products.

The construction is independent of the chart. For coordinates $y^j=y^j(x)$, the chain rule gives $d{y}^j={\sum }_i({\partial }_i{y}^j)d{x}^i$, and the coordinate formula gives $d(d{y}^j)=0$ by equality of mixed partial derivatives. Consequently, expressing a form as ${\sum }_I{b}_{I}d{y}^I$ and applying the $x$-coordinate formula gives exactly ${\sum }_{I}d{b}_I\wedge d{y}^I$, the $y$-coordinate formula. The local constructions agree on overlaps and therefore define a global exterior derivative.

For uniqueness, first derive locality from the axioms. If $\alpha$ vanishes near $p$, choose a smooth bump function $\chi$ supported in that neighborhood and equal to one near $p$. Then $\chi \alpha =0$, andimplies $(d\alpha {)}_p=0$. Thus the value depends only on the local germ of $\alpha$. Local coordinate functions can be extended using bump functions, so the same axioms apply to their germs. Since $d(d{x}^i)=d^2{x}^i=0$, the graded Leibniz rule forces precisely the coordinate formula above. The exterior derivative exists and is unique, as expressed by uniqueness of the exterior derivative from its axioms.

For a CW complex $X$, let $X^k$ be its $k$-skeleton. The cellular chain complex isThe isomorphism uses excision and $H_k(D^k,S^{k-1})\cong \mathbb{Z}$, after choosing an orientation for each cell. In degree zero, take $X^{-1}=\varnothing$ and the free group on the vertices. The differential is the compositeThe long exact sequences of skeleton pairs imply $d_{k-1}{d}_k=0$. The cellular boundary formula describes its coefficient at a $(k-1)$-cell as the mapping degree of the attaching sphere after all other $(k-1)$-cells are collapsed. In degree one this is the signed difference of the two endpoints. The cellular homology theorem identifies the homology of this complex with singular homology: relative homology of successive skeletons is concentrated in their cell dimension, and the exact skeleton sequences leave precisely these kernels modulo images. This is the construction, rather than merely a count of cells.

The filtration ${\mathbb{CP}}^0\subset {\mathbb{CP}}^1\subset {\mathbb{CP}}^2$ gives one cell in each of dimensions zero, two and four, since ${\mathbb{CP}}^k\setminus {\mathbb{CP}}^{k-1}\cong {\mathbb{C}}^k$. There are no odd-dimensional cells, so every cellular differential vanishes. ThusThe complex projective line has one zero-cell and one two-cell. Its product with itself has one zero-cell, two two-cells and one four-cell. Its cellular chain complex again has zero differentials, givingA homeomorphism induces isomorphisms on homology groups. The second groups have different ranks, and therefore

Now orient the Complex projective plane by its complex coordinates. The inclusion of its two-skeleton ${\mathbb{CP}}^1$ identifies the fundamental class of a projective line with a generator $h$ of $H_2({\mathbb{CP}}^2;\mathbb{Z})\cong \mathbb{Z}$: the cellular generator has no incoming three-cell boundary and no outgoing boundary.

Take the projective lines $L_0=\{z_2=0\}$ and $L_1=\{z_1=0\}$. A path of unitary coordinate transformations takes one to the other, so their oriented homology classes are both $h$. They meet in the single point $[1:0:0]$. In its affine chart, let $w=z_1/z_0$ and $v=z_2/z_0$. Their tangent spaces are the complex $w$-axis and complex $v$-axis, so the intersection is a transverse intersection. The ordered real basesconcatenate to the complex orientation of the ambient four-manifold. Thus the local sign is $+1$, andThis computes the intersection form using distinct representatives of the same class, avoiding an ill-defined attempt to count a line intersecting itself as a set. Bilinearity now givesEquivalently, the map $H_2\to \mathrm{Hom}(H_2,\mathbb{Z})$ sending $a$ to $\lambda (a,-)$ is an isomorphism. This is exactly a unimodular intersection pairing, and proves projective lines generate a unimodular intersection form. Reversing the ambient orientation would change the matrix to $(-1)$ and would still be unimodular.