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Submanifolds $A,B\subseteq M$ intersect cleanly when $A\cap B$ is an embedded submanifold andat every $x\in A\cap B$. A transverse intersection is the special case in which $T_{x}A+T_{x}B=T_{x}M$.

Pulling the external product of two normally complex submanifolds back along the diagonal gives their transverse intersection. Its normal complex structure is the direct sum of the restricted normal complex structures. Multiplication of Thom classes for a direct sum identifies this geometric representative with the cup product in complex cobordism. Empty intersection represents zero.

// Target: geometry-and-topology.bigb

For a transverse intersection, the quotient map $TM{|}_L\to {\nu }_1{|}_L\oplus {\nu }_2{|}_L$ is onto and has kernel $TL$. This gives the displayed isomorphism of actual normal bundles. Complex structures on the two normal bundles therefore induce a complex structure on the intersection's normal bundle.

// Target: geometry-and-topology.bigb

Parametrize one ellipse as $\gamma (t)=c+A(\cos t,\sin t)$ and substitute it into the quadratic equation $Q=0$ of the other. At a transverse intersection, $g(t)=Q(\gamma (t))$ has a simple zero, so its sign changes. Periodicity makes the number of zeros even. The function is a trigonometric polynomial of degree at most two, so multiplication by $z^2$, with $z=e^{it}$, gives a nonzero polynomial of degree at most four. Hence there are at most four distinct intersection points. Coincident ellipses are excluded by transversality.

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.

The degree and its local signs. Orient $S^n$ and let $[S^n]$ generate its top reduced homology ${\tilde{H}}_n(S^n;\mathbb{Z})\cong \mathbb{Z}$. The mapping degree is the integer characterized byFor $n\ge 1$ this is the usual top homology definition using the fundamental class; using reduced homology also covers $S^0$.

Suppose $f$ is smooth and $p$ is a regular value. Each $x\in f^{-1}(p)$ has invertible tangent map $D{f}_x$, so the inverse function theorem makes $f^{-1}(p)$ discrete. It is also closed in the compact sphere, hence finite. Choose disjoint small neighborhoods of these points on which $f$ is a local diffeomorphism. The Excision theorem identifies the source local homology with the direct sum of one copy of $\mathbb{Z}$ for each inverse image. The induced local map is multiplication by $+1$ or $-1$ according as $D{f}_x$ preserves or reverses orientation. The map from the global fundamental class to these local orientation classes therefore proves the degree as a sum of local degrees formula:Here the determinant is computed in positively oriented tangent bases. Thus the mapping degree counts inverse images with signs, rather than just their cardinality.

The quotient map. Put $m=n+1$ and write $f=F{|}_{S^{m-1}}$. Give $D^m$ its standard orientation and its boundary the induced orientation. The connecting homomorphismis an isomorphism: the disk has zero positive reduced homology. It sends the relative fundamental class $[D^m,S^{m-1}]$ to $[S^{m-1}]$. If the map on relative homology induced by $F$ multiplies this class by $d$, naturality givesThus $d=\mathrm{deg}f$. Collapsing the boundary gives a sphere $D^m/S^{m-1}\cong S^m$, and the quotient map identifies its top reduced homology with the top relative homology of the disk pair. Give this quotient sphere the orientation determined by that identification. The relation $qF=\tilde{F}q$ now showsThis is the quotient-sphere degree identity.

The graph intersection. Orient $D^m\times D^m$ by the product orientation, orient $N=D^m\times \{0\}$ by its first factor, and orient the graph of a function $\Gamma$ by $x\mapsto (x,F(x))$. An intersection is precisely a zero of $F$, and no such zero lies on the boundary because $F(S^{m-1})\subset S^{m-1}$. At an intersection, transverse intersection means that $D{F}_x$ is surjective, hence invertible. The zeros are consequently isolated and finite.

For the smooth intersection number use the ordered tangent spaces $T_{x}N$ first and $T_{(x,F(x))}\Gamma$ second. Relative to the product basis their concatenated basis has matrixIts determinant is $\det D{F}_x$, so the intersection sign is ${ϵ}_x=\mathrm{sgn}\det D{F}_x$. By the same local Excision theorem argument, now in relative homology at the interior point $0$, the sum of these signs equals the multiplier of $F_{\ast }$ on $H_m(D^m,S^{m-1};\mathbb{Z})$. Therefore the graph intersection formula for mapping degree isIf the tangent spaces are ordered $\Gamma$ first and $N$ second, every sign changes by $(-1{)}^m$; the order above specifies the appropriate convention.

The one-dimensional model $F(x)=2x^3-x$ on $[-1,1]$ illustrates the graph intersection formula for mapping degree: its three zeros have signs $+1,-1,+1$, and its endpoint map has mapping degree $1$ on reduced homology.

Choose $Q\in SO(n+1)$ with $Q{e}_{n+1}=v$. Conjugation by $Q$ identifieswithIt follows that the unit-vector stabilizer in a special orthogonal group is an embedded submanifold diffeomorphic to $SO(n)$, andAt $R\in S_v$, its tangent space is

For any $v\ne w$, both $S_v$ and $S_w$ contain the identity. At the identity,Identify $\mathfrak{so}(n+1)$ with the exterior square ${\Lambda }^2{\mathbb{R}}^{n+1}$. The orthogonal complement of $T_I{S}_v={\Lambda }^2(v^{\perp })$ is $v\wedge v^{\perp }$. If $v$ and $w$ are linearly independent, the nonzero bivector $v\wedge w$ belongs to the orthogonal complements of both tangent spaces. Their sum is therefore a proper subspace of $T_{I}SO(n+1)$, so the intersection is not transverse. If $w=-v$, then $S_w=S_v$, whose tangent space is proper, so they are again not transverse. Consequentlyas summarized by distinct unit-vector stabilizers are not transverse.

A submanifold is a manifold sitting inside another manifold through an immersion or embedding. An embedded submanifold has the subspace topology and local coordinates in which it is a coordinate plane. Smooth transverse intersections are governed by the tangent spaces of the participating submanifolds.