My last post discussed zero sections of vector bundles, and showed some examples where the zero sections are determined by a particular cohomology class called the Euler class. In this post, I'll go through two constructions of the Euler class and work through calculations of Euler classes using both constructions. The first construction—using Chern classes—is geometric and a bit more concrete, but obscures the fact that the Euler class depends only on the topology of our vector bundle. The second construction—using the Thom class—is purely topological but fairly abstract.

## Geometric construction via the first Chern class

For complex line bundles, the Euler class is equal to the first Chern class (and on general complex vector bundles, it is equal to the top nonzero Chern class). This relation allows us to take a more geometric view of the Euler class.

Although the Chern class is a topological invariant of a complex vector bundle, independent of any choice of geometry, it is convenient to define the Chern class using a connection on the vector bundle. It turns out that the resulting Chern class is well-defined, independent of the particular connection used in this definition. So for now, suppose that we have a complex vector bundle $$\CC^d \to E \to B$$. This notation means that we have a vector bundle $$E$$ over a base space $$B$$, where above every point of $$x \in B$$ we have a $$d$$-dimensional complex vector space $$\CC_x^d$$. Suppose also that we have a connection $$\nabla$$ on $$E$$ which allows us to parallel transport vectors between the different space $$\CC_x^d$$. The curvature of this connection is a matrix-valued 2-form $$\Omega$$. The $$d \times d$$ matrix $$\Omega(X, Y)$$ tells us how a vector in our line bundle gets rotated if we use the connection to parallel transport the vector around an infinitesimal loop in the $$X, Y$$ plane. The first Chern class is then defined to be the 2-form $$\tr \Omega$$, where $$\tr \Omega(X, Y)$$ gives the trace of the matrix $$\Omega(X, Y)$$. (In general, the $$k$$th Chern class is given by $$\tr \Omega^{\wedge k}$$, the trace of the $$k$$th wedge power of $$\Omega$$). On a complex line bundle, $$\Omega$$ is a $$1 \times 1$$ matrix, so the first Chern class is equal to the 2-form $$\Omega$$ itself.

Since Chern classes only exist for complex vector bundles, we cannot use this approach to understand the simpler vector bundles like the cylinder and Möbius strip which I talked about last time. But we can compute the Euler class for the tangent bundle of the sphere $$TS^2$$. And the computation is a lot quicker than the Thom space construction which we will see next: if we take $$\nabla$$ to be the Levi-Civita connection, then the usual Gaussian curvature 2-form $$\Omega$$ is our first Chern class, and hence also our Euler class.

Since $$H^2(S^2) \cong \ZZ$$, $$\Omega$$ is cohomologous to a scalar multiple of the normalized area form. And to compute the scalar, we can simply integrate $$\Omega$$ over our domain $$S^2$$. By the Gauss-Bonnet theorem, $$\int_S^2 \Omega = \chi(S^2) = 2$$. Hence, we conclude again that our Euler class is given by the element $$2 \in \ZZ \cong H^2(S^2)$$, and we see that this value of $$2$$ comes directly from the Euler characteristic $$\chi(S^2)$$.

## Topological construction via the Thom class

The Euler class is often described by constructing an equally-mysterious cohomology class called the Thom class and pulling it pack through a sequence of maps. In this section, we'll try to unpack the definitions and work through some examples. Formally, the Thom class is a certain relative cohomology class in the unit disk bundle associated to our vector bundle, and the Euler class is the restriction of the Thom class to our base space, viewed as the zero section of the unit disk bundle. We will stat by reviewing some facts about relative homology and cohomology.

### Relative homology

Given a topological space $$X$$, the $$k$$th absolute homology group $$H_k(X)$$ consists of $$k$$-dimensional boundary-free subsets which are not themselves the boundary of any $$(k+1)$$-dimensional subset. For any subset $$A \subset X$$, the $$k$$th homology group relative to $$A$$, denoted $$H_k(X, A)$$, relaxes the boundary-free condition and instead consists of $$k$$-dimensional subsets whose boundaries lie in $$A$$, and do not form the boundary of any $$(k+1)$$-dimensional subset even when combined with any portion of $$A$$.

One common example which helps to motivate these definitions is the homology group $$H_1(M, \partial M)$$ of a surface $$M$$ relative to its boundary $$\partial M$$. If we have a connected surface of nontrivial topology, a basis for $$H_1(M)$$ provides a maximal collection of loops which we can cut along without disconnecting the surface.

A basis for the first homology group of the double torus.

However, this property no longer holds for surfaces with boundary. Even for an annulus, the loops in the first homology group now disconnect our surface.

The first homology group $$H_1(A)$$ of an annulus $$A$$ is generated by a loop around the center, which cuts the annulus into two pieces!

There are two problems here: (1) nontrivial loops in the first homology group may disconnect the surface. Even though they cut the surface into two pieces, they are not technically equal to the boundary of either piece, since the pieces may also have boundary components arising from the global boundary $$\partial M$$. (2) there are cuts which we can make without disconnecting the surface—cuts in the radial direction. But they are not included in the first homology group since they are not loops.

The definition of relative homology groups solves both problems: in $$H_1(M, \partial M)$$, we say that a set is the boundary of a region if it can be combined with part of $$\partial M$$ to form the boundary of a region. And rather than restricting the elements of our homology group to be loops, we allow any subset whose boundary is entirely contained in $$\partial M$$.

Under mild technical conditions on the subset $$A$$, the relative homology group $$H_k(X, A)$$ is equal to the absolute homology group of the quotient space $$H_k(X / A)$$ where we collapse all of $$A$$ to a single point.

#### Long exact sequence of homology

If we know the homology groups of $$X$$ and $$A$$, we can use them to calculate the relative homology groups of the pair $$(X, A)$$. The key tool is a long exact sequence involving all three sets of homology groups: $\cdots \xrightarrow{} H_{n+1}(X, A) \xrightarrow{\partial} H_n(A) \xrightarrow{i_*} H_n(X) \xrightarrow{j_*} H_n(X, A) \xrightarrow{\partial} H_{n-1}(A) \xrightarrow{} \cdots$ Here the function $$i_*$$ is the map on homology induced by the inclusion $$i : A \to X$$, the function $$j_*$$ is the map on homology induced by the quotient map $$j : X \to X/A$$, and the function $$\partial$$ is the standard boundary map (which maps from $$H_n(X, A) \to H_{n-1}(A)$$ since the boundary of a relative homology class must always lie in $$A$$).

### Relative cohomology

The whole relative story can be dualized to relative cohomology. Although we will mostly need cohomology with $$\ZZ$$ or $$\ZZ_2$$ coefficients later on, I will briefly discuss relative de Rham cohomology here, since I find that it helps to highlight about the differences between homology and cohomology.

On a manifold $$X$$ without boundary, the elements of the de Rham cohomology group $$H^k(X)$$ are differential $$k$$-forms $$\eta$$ whose exterior derivative $$d\eta$$ vanishes. Two differential forms $$\eta, \omega$$ are said to be equal if their difference $$\eta - \omega$$ is itself the exterior derivative of some $$(k-1)$$ form. On a manifold with boundary, the $$k$$-forms in $$H^k(X)$$ are constrained to lie tangent to the boundary $$\partial X$$. On the other hand, the elements of the relative cohomology group $$H^k(X, \partial X)$$ are given by $$k$$-forms whose restrictions to the boundary are zero.

Returning to the annulus example, the absolute cohomology group $$H^1(A)$$ is generated by harmonic the 1-form $$\eta$$ which circulates around the annulus counterclockwise, whereas the relative cohomology group $$H^1(A, \partial A)$$ is generated by the harmonic 1-form $$\rho$$ which flows in the radial direction from the inner boundary to the outer boundary.

In de Rham cohomology for manifolds without boundary, a cohomology class $$[\eta] \in H^{n-k}(X)$$ is dual to some homology class $$[\gamma] \in H_k(X)$$ if the integral of $$\eta$$ over each $$(n-k)$$-dimensional subset is equal to the number of intersections between $$\gamma$$ and that subset. On manifolds with boundary, one should pair absolute homology classes in $$H_k(X)$$ with relative cohomology classes in $$H^{n-k}(X, \partial X)$$ and vice versa. Note, for instance, how the generator $$\eta$$ of $$H^1(A)$$ integrates to 1 the generator of $$H_1(A)$$, even though two concentric curves in the annulus may not intersect each other! On the other hand, the generator $$\rho$$ of $$H^1(A, \partial A)$$ integrates to 1 along radial segments, which always intersect the generator loop of $$H_1(A)$$ once.

#### Long exact sequence of cohomology

Just as for homology, the relative cohomology groups of a pair $$(X, A)$$ fit into a long exact sequence alongside the cohomology groups of $$X$$ and $$A$$: $\cdots \xrightarrow{} H^{n-1}(A) \xrightarrow{\delta} H^n(X, A) \xrightarrow{j^*} H^n(X) \xrightarrow{i^*} H^n(A) \xrightarrow{\delta} H^{n+1}(X, A) \xrightarrow{} \cdots$ The maps here are the duals of the maps in the long exact sequence for homology: the function $$i^*$$ is the map on cohomology induced by the inclusion $$i : A \to X$$, the function $$j^*$$ is the map on cohomology induced by the quotient map $$j : X \to X/A$$, and the function $$\delta$$ is the codifferential (the dual of the boundary map $$\partial : H^{n+1}(X, A) \to H^n(A)$$).

### The Thom space

To define the Thom class and Thom space, we start with an oriented vector bundle $$\RR^d \to E \to B$$ over a base space $$B$$. An orientation of $$E$$ amounts to a continuous choice of orientation of the vector spaces $$\RR^d_x$$ at each point $$x$$. We then build two associated bundles. First, we build the disk bundle $$D^d \to D(E) \to B$$, whose fiber at a point $$x \in B$$ is the unit disk in our vector space $$\RR^n_x$$ above $$x$$. Then we build the sphere bundle $$S^{d-1} \to S(E) \to B$$ whose fiber at a point $$x \in B$$ is the boundary of the disk above $$x$$ in $$D(E)$$.

The Thom class is a cohomology class $$c \in H^d(D(E), S(E))$$ whose restriction to any fiber $$(D^d_x, S^{d-1}_x)$$ yields the positively-oriented generator of $$H^d(D^d_x, S^{d-1}_x) \cong \ZZ$$. And the Euler class is a cohomology class $$e \in H^d(B)$$ obtained by restricting $$c$$ to the zero section of $$E$$, which we can identify with $$B$$.

The Thom class $$c$$ provides us with an isomorphism $$\Phi^k : H^k(B) \to H^{k+d}(D(E), S(E))$$ given by taking the cup product with $$c$$. Explicitly, the mapping is given by $$\Phi^k(x) = p^*(x) \smile c$$, where $$p : E \to B$$ denotes the canonical projection onto the base space. (In de Rham cohomology, we can equivalently write $$\Phi^k(\eta) = p^* \eta \wedge c$$).

### Thom classes for bundles on $$S^1$$

We'll start by computing the Thom classes for two simple line bundles: the cylinder and the Möbius trip.

#### The Thom class of the cylinder

First, let's look at how the Thom class is constructed on the cylinder $$S^1 \times \RR$$. In this case, our disk bundle $$D(E) = S^1 \times I$$ is simply the annulus $$A$$, and our sphere bundle $$S(E) = S^1 \times S^0$$ is the pair of disjoint circles lying on the boundary $$\partial A$$ of $$D(E)$$. Hence, the Thom class lies in the relative cohomology group $$H^1(A, \partial A)$$ which we investigated earlier. In particular, the Thom class is precisely the generator $$\rho$$ which we identified above. Since $$\rho$$ is orthogonal to the centerline of the annulus, the Euler class $$e$$ is zero.

#### The Thom class of the Möbius strip

Now, what about the Möbius strip? We can still build the Thom space in exactly the same way: our disk bundle $$D(E)$$ is a Möbius strip of unit width, and the sphere bundle $$S(E)$$ is the boundary of $$D(E)$$, which is topologically a circle. However, the Möbius strip is not orientable, so we run into trouble when we try to define the Thom class. The solution is to switch from considering standard cohomology with integer coefficients to use $$\ZZ_2$$ coefficients instead. Working mod 2 essentially ignores signs, and allows us to treat all surfaces as if they were oriented. In this case, the Thom class is a cohomology class $$c \in H^d(D(E), S(E);\ZZ_2)$$ whose restriction to any fiber yields the nontrivial element of $$H^d(D_x^d, S_x^{d-1};\ZZ_2)\cong \ZZ_2$$.

In the case of the Möbius strip, then, we are interested in the relative cohomology group $$H^1(M, \partial M; \ZZ_2)$$, where $$M$$ is the unit-width Möbius strip. Now that we're using $$\ZZ_2$$ coefficients, we cannot use de Rham cohomology any more, so we will do some explicit calculations on a particular cell decomposition of the Möbius strip, which includes the zero section in its 1-skeleton.

A cell decomposition of the Möbius strip. Note that the two sides are identified with the same edges and are given opposite orientations.

Our cell decomposition has three vertices, five edges, and two faces, so $$C^0 = \ZZ_2^3$$, $$C^1 = \ZZ_2^5$$ and $$C^2 = \ZZ_2^2$$. The coboundary maps $$\delta_0 : C^0 \to C^1$$ and $$\delta_1 : C^1 \to C^2$$ are given by $\delta_0 = \begin{pmatrix} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & 1 \\ 0 & 0 & 0 \\ 1 & 0 & 1 \end{pmatrix}, \quad \text{and} \quad \delta_1 = \begin{pmatrix} 1 & 1 & 1 & 1 & 0\\ 1 & 1 & 0 & 1 & 1 \end{pmatrix}.$ Hence, our absolute cohomology groups are \begin{aligned} H^0(M;\ZZ_2) &= \ker \delta_0 = \langle(1, 1, 1)\rangle \cong \ZZ_2,\\ H^1(M;\ZZ_2) &= \ker \delta_1 / \im \delta_0 = \langle(0, 1, 0, 1, 0), (1, 1, 0, 0, 0), (1, 0, 1, 0, 1)\rangle / \im \delta_0 = \ZZ_2,\\ H^2(M;\ZZ_2) &= C^2 / \im \delta_1 = 0. \end{aligned}

The relative cohomology is a little easier, since, we quotient each space of cochains by the space of boundary cochains. So all of our spaces get a little smaller: $$C^0(M, \partial M) = \ZZ_2^3 / \ZZ_2^2 = \ZZ_2$$, $$C^1(M, \partial M) = \ZZ_2^5 / \ZZ_2^2 = \ZZ_2^3$$, and $$C^2(M, \partial M) = \ZZ_2^2$$. And we obtain relative boundary maps $\delta_0' = \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}, \quad \text{and} \quad \delta_1' = \begin{pmatrix} 1 & 1 & 1 \\ 1 & 1 & 1 \end{pmatrix}.$ Our resulting relative cohomology groups are \begin{aligned} H^0(M, \partial M;\ZZ_2) &= \ker \delta_0' = 0,\\ H^1(M, \partial M;\ZZ_2) &= \ker \delta_1' / \im \delta_0' = \langle(1, 1, 0), (1, 0, 1)\rangle / \langle (1, 1, 0)\rangle \cong \ZZ_2,\\ H^2(M, \partial M;\ZZ_2) &= C^2 / \im \delta_1' = \ZZ_2. \end{aligned} The most important result for us is that the first relative cohomology group $$H^1(M, \partial M;\ZZ_2)$$ has one nontrivial element $$c$$, which is represented by the relative cochain with a 1 on edges $$e_1$$ and $$e_4$$ and zero everywhere else. This element $$c$$ is the Thom class of the Möbius strip.

The nontrivial element $$c \in H^1(M, \partial M;\ZZ_2)$$ is the Thom class of the Möbius strip.

Since the Möbius strip is nonorientable, it does not technically have an Euler class. But restricting $$c$$ to the centerline of the Möbius strip pulls $$c$$ back onto the base space $$S^1$$, yielding a sort of "mod 2" Euler class $$e \in H^1(S^1; \ZZ_2)$$. In this case, $$e$$ is the single nontrivial class in $$H^1(S^1;\ZZ_2)$$. In particular, $$e$$ is nonzero, and thus distinct from the Euler class for the cylinder which we computed earlier. (Formally, this "mod 2" Euler class is known as the top Stiefel-Whitney class, and is equal to the Euler class taken mod 2 on orientable vector bundles where the Euler class is defined)

### The Thom class for $$TS^2$$

Now, let's get a little more ambitious and try to compute the Thom class for the tangent bundle $$TS^2$$ of the two-dimensional sphere. For this example, we will have to use more long exact sequences and other algebraic machinery, since I do not know how to compute the relevant cohomology groups directly.

The sphere bundle $$S(TS^2)$$ is the set of all unit tangent vectors on the sphere, and can be identified with the group $$SO(3)$$ of $$3 \times 3$$ rotation matrices: any point $$x \in S^2$$ is a unit vector in $$\RR^3$$, and a unit tangent vector at $$x$$ provides a second orthogonal vector. We can take these two vectors to be the first two columns of a rotation matrix, and they uniquely determine the last column, which must be their cross product. The group $$SO(3)$$ can also be identified with the real projective space $$\RP^3$$, as both can be expressed as the quotient of the set of unit quaternions modulo $$\pm 1$$. If we look up the cohomology groups of $$\RP^3$$, we find that they are $H^0(\RP^3) = \ZZ, \quad H^1(\RP^3) = 0, \quad H^2(\RP^3) = \ZZ_2, \quad H^3(\RP^3) = \ZZ.$

As far as I know, the disk bundle $$D(TS^2)$$ does not have such a nice description, but its topology is quite simple: since each disk is contractible, $$D(TS^2)$$ deformation retracts onto the sphere $$S^2$$, and in particular its cohomology groups are all isomorphic to the corresponding cohomology groups of $$S^2$$.

Since we know the cohomology groups for $$S(TS^2) \cong \RP^3$$ and $$D(TS^2)$$, we can use the long exact sequence for cohomology to work out the relative cohomology group $$H^2(D(TS^2), S(TS^2))$$ which the Thom class $$c$$ lives in. In particular, we have get an exact sequence $H^{1}(\RP^3) \xrightarrow{\delta} H^2(D(TS^2), \RP^3) \xrightarrow{j^*} H^2(D(TS^2)) \xrightarrow{i^*} H^2(\RP^3) \xrightarrow{\delta} H^{3}(D(TS^2), \RP^3)$ The first term $$H^1(\RP^3)$$ is equal to zero. The last term is also equal to zero, since the Thom isomorphism tells us that $$H^3(D(TS^2), \RP^3) \cong H^1(S^2) = 0.$$ So we obtain a short exact sequence $0 \xrightarrow{} H^2(D(TS^2), \RP^3) \xrightarrow{j^*} H^2(D(TS^2)) \xrightarrow{i^*} H^2(\RP^3) \xrightarrow{} 0$ Since $$D(TS^2)$$ deformation retracts onto $$S^2$$, the central term $$H^2(D(TS^2))$$ is isomorphic to $$H^2(S^2) \cong \ZZ$$. And the right-hand term $$H^2(\RP^2)$$ is equal to $$\ZZ_2$$. Hence, our short exact sequence is really $0 \xrightarrow{} H^2(D(TS^2), \RP^3) \xrightarrow{j^*} \ZZ \xrightarrow{i^*} \ZZ_2 \xrightarrow{} 0$ and thus we must have $$H^2(D(TS^2), S(TS^2)) \cong \ZZ$$, and its generator is the Thom class of $$TS^2$$. Furthermore, our short exact sequence shows that this generator maps to the element $$2 \in \ZZ \cong H^2(S^2)$$. Hence, the Euler class of the sphere is twice the generator of $$H^2(S^2)$$, reflecting the fact that the Euler characteristic $$\chi(S^2) = 2$$.

### The Thom class for a twisted line bundle over the torus

In my last post, I discussed a twisted line bundle over the torus. We built a real line bundle by specifying that as you go around one generator loop, the fibers sweep out an annulus, whereas if you go around the other generator loop the fibers sweep out a Möbius strip. Let us denote this bundle by $$\RR \to E \to T^2$$. We will compute the Thom class of $$E$$.

We start by drawing the torus as a square with opposite edges identified. To make the disk bundle associated to our line bundle, we can thicken the square to obtain a cube. Now we just need to identify the opposite sides of the cube properly.

We only index the cells which do not lie on the boundary, since only these cells are used to compute the relative cohomology group.

Note that our bundle has annuli sitting above the loops which run parallel to edge $$e_1$$, and has Möbius strips sitting above the loops which run parallel to edge $$e_2$$. We have relative cochain groups are $$C^0(E, \partial E) = \ZZ_2^1, C^1(E, \partial E) = \ZZ_2^4, C^2(E, \partial E) = \ZZ_2^5, C^3 = \ZZ_2^2$$. The coboundary maps are given by $\delta_0 = \begin{pmatrix} 0 \\ 0 \\ 1 \\ 1 \end{pmatrix},\quad \delta_1 = \begin{pmatrix} 0 & 0 & 0 & 0\\ 0 & 1 & 1 & 1\\ 0 & 1 & 1 & 1\\ 1 & 0 & 0 & 0\\ 1 & 0 & 0 & 0 \end{pmatrix}, \quad \delta_2 = \begin{pmatrix} 1 & 0 & 0 & 1 & 1\\ 1 & 0 & 0 & 1 & 1 \end{pmatrix}$ Our resulting relative cohomology groups are \begin{aligned} H^0(E, \partial E;\ZZ_2) &= \ker \delta_0 = 0,\\ H^1(E, \partial E;\ZZ_2) &= \ker \delta_1 / \im \delta_0 = \frac{\langle(0, 1, 1, 0), (0, 0, 1, 1)\rangle}{\langle (0, 0, 1, 1)\rangle} \cong \ZZ_2,\\ H^2(E, \partial E;\ZZ_2) &= \ker \delta_2 / \im \delta_1 = \frac{\langle(1, 0, 0, 0, 1), (0, 1, 0, 0, 0), (0, 0, 1, 0, 0), (0, 0, 0, 1, 1)\rangle}{\langle (0, 0, 0, 1, 1), (0, 1, 1, 0, 0)\rangle} \cong \ZZ_2^2,\\ H^3(E, \partial E;\ZZ_2) &= C^3 / \im \partial_2 \cong \ZZ_2 \end{aligned} The Thom class of $$E$$ is the nontrivial element of $$H^1(E, \partial E; \ZZ_2)$$, which is represented by a cochain supported on edges $$e_2$$ and $$e_3$$. Restricting to the torus $$T^2$$ lying in the center, we find that our mod-2 Euler class is represented by the cocycle supported on edge $$e_2$$—the edge whose fiber was twisted into a Möbius strip.