Characteristic Classes of Vector Bundles

This post was machine-translated from the Korean original by Marvin (via Kimi). It may contain errors or awkward phrasing — the Korean original is the source of truth.

Euler Class

So far we have used $\mathbb{Z}/2$-coefficients to effectively sidestep the issue of orientability. We now take orientation into account as well. With $\mathbb{Z}/2$-coefficients we cannot distinguish signs, so every fiber automatically carried a “direction”; but when we pass to $\mathbb{Z}$-coefficients, $1$ and $-1$ become genuinely distinct elements, and the question becomes whether we can consistently assign a direction to each fiber. When this is possible, a more refined invariant appears that lifts the top Stiefel-Whitney class $w_n$ to the integers. This is the Euler class.

Definition 1 An orientation of a rank $n$ vector bundle $p:E\rightarrow B$ is a continuous choice, in each local trivialization, of a generator $u_x$ of $H^n(p^{-1}(x), p^{-1}(x)\setminus 0;\mathbb{Z})\cong\mathbb{Z}$ for each fiber $p^{-1}(x)$. A bundle admitting such an orientation is called an oriented vector bundle.

There are essentially three ways to think about giving an orientation. First, given a vector bundle $E\rightarrow B$, we usually regard $B$ as embedded in $E$ via the zero section $0:B\rightarrow E$. Then in the definition above, the relative cohomology $H^n(p^{-1}(x), p^{-1}(x)\setminus 0;\mathbb{Z})$ is itself the data of a $+$ or $-$ attached to the origin of the fiber, that is, to the corresponding point of the base $B$.

In differential geometry this is interpreted as follows. Removing the origin from the fiber $p^{-1}(x)\cong\mathbb{R}^n$ deformation retracts onto the sphere $S^{n-1}$, so from the long exact sequence of the pair we obtain the isomorphism

\[H^n(p^{-1}(x), p^{-1}(x)\setminus 0;\mathbb{Z})\cong \widetilde{H}^{n-1}(S^{n-1};\mathbb{Z})\cong\mathbb{Z}\]

In differential geometry, the top-dimensional cohomology of a manifold contains the volume form, and this is regarded as determining the orientation, so we can think of orienting a vector bundle as using the orientation of $S^{n-1}$, a space we already understand well.

However, the most familiar way to orient a vector space is to choose a reference ordered basis and declare that another ordered basis is negatively oriented if the change-of-basis matrix to the reference basis has negative determinant. The problem with such a definition is that it contains too much information; what we actually care about is only the sign of the change-of-basis determinant. This viewpoint is closely connected to the Čech cohomology explained earlier in §Stiefel-Whitney Characteristic Classes. That is, we defined an arbitrary vector bundle by a trivializing open cover $\{U_i\}$ and transition functions $g_{ij}: U_{ij}\rightarrow \GL(n;\mathbb{R})$ over $U_{ij}$; choosing one of the signs of $\det$ is then the same as reducing the structure group from $\GL(n;\mathbb{R})$ to $\GL^+(n;\mathbb{R})$. In other words, when passing from one chart to another via the transition functions, determinants that are negative—that is, orientations that are reversed—are no longer allowed, and this restriction filters out the non-orientable vector bundles.

Whether this is possible is determined by $\pi_0(\GL(n;\mathbb{R}))\cong \mathbb{Z}/2$ that we saw earlier. That is, the only information remaining in each transition function related to orientation is the sign $\varepsilon_{ij}=\operatorname{sgn}\det g_{ij}:U_{ij}\rightarrow \{\pm 1\}$ of $\det g_{ij}$, and the class $[\varepsilon_{ij}]\in H^1(B;\mathbb{Z}/2)$ obtained by gluing these together becomes the obstruction to reducing to $\GL^+$. This class is exactly $w_1(E)$ (§Stiefel-Whitney Characteristic Classes, ⁋Definition 5), and we can think of this as the rank-$n$ version of the fact that $H^1(M;\mathbb{Z}/2)$ contained the orientation information of covering spaces. That is, $E$ being orientable is equivalent to $w_1(E)=0$.

Henceforth we assume that all bundles discussed in this section are oriented. Once an orientation is given, the generators $u_x$ scattered over each fiber are assembled into a single cohomology class.

Theorem 2 (Thom isomorphism) For an oriented rank $n$ vector bundle $p:E\rightarrow B$, let $E_0=E\setminus 0(B)$ be the subspace with the zero section removed. Then there exists a unique Thom class $u\in H^n(E, E_0;\mathbb{Z})$ such that for each $x\in B$, the restriction of $u$ to $(p^{-1}(x), p^{-1}(x)\setminus 0)$ is $u_x$. Moreover, the composition of cup product and pullback

\[H^k(B;\mathbb{Z})\xrightarrow{\ \cong\ }H^{k+n}(E, E_0;\mathbb{Z}),\qquad \alpha\longmapsto p^\ast\alpha\smile u\]

is an isomorphism for all $k$.

Proof

Writing only the essentials, when $E$ is the trivial bundle $B\times\mathbb{R}^n$, the pair $(E, E_0)=(B\times\mathbb{R}^n, B\times(\mathbb{R}^n\setminus 0))$, and by the relative version of the Künneth formula

\[H^{k+n}(B\times\mathbb{R}^n, B\times(\mathbb{R}^n\setminus 0))\cong H^k(B)\otimes H^n(\mathbb{R}^n,\mathbb{R}^n\setminus 0)\]

The second factor on the right is $\mathbb{Z}$, and its generator is the fiber orientation $u_x$. In this case it suffices to set $u=1\otimes u_x$, and the general case is obtained by taking a trivializing open cover and gluing these isomorphisms via Mayer-Vietoris. We leave the detailed proof to Chapter 10 of [MS].

The Thom class can be understood as a fiber-direction cohomology class concentrated near the zero section of the vector bundle $E$. Then the above isomorphism stretches a cohomology class $\alpha$ living on $B$ in the fiber direction of $E$ and multiplies it by $u$, and the claim of the theorem is that this is an isomorphism. Alternatively, from the viewpoint of §Poincaré Duality, ⁋Example 16, $u$ is the (relative) Poincaré dual of the zero section, and the above isomorphism can be thought of as taking the homology class defined by $\alpha$, stretching it along the fiber, and then intersecting it with the zero section to return (but now the homology class lives in the homology group of the total space).

Pulling this Thom class back to the base again gives the Euler class.

Definition 3 For an oriented rank $n$ vector bundle $E\rightarrow B$, the Euler class $e(E)\in H^n(B;\mathbb{Z})$ is defined, for the zero section $0:B\rightarrow E$ and the Thom class $u$ of Theorem 2 (Thom isomorphism), by

\[e(E)=0^\ast\bigl(j^\ast u\bigr)\]

where $j^\ast:H^n(E, E_0)\rightarrow H^n(E)$ is the restriction from the pair to all of $E$.

Then $0^\ast:H^n(E)\rightarrow H^n(B)$ is an isomorphism because $p$ is a homotopy equivalence.

Above we explained that the Thom class is the Poincaré dual of the zero section. Then the Euler class $e(E)$ is its restriction back onto the zero section; that is, viewing it again as a Poincaré dual, it is the self-intersection of the zero section obtained by pushing the zero section slightly to a generic section and taking the intersection with the original zero section, which can be thought of as the vanishing locus of a generic section. Making this intuition precise gives the following.

Proposition 4 The Euler class satisfies the following.

(Naturality) For any $f:B'\rightarrow B$, $e(f^\ast E)=f^\ast e(E)$.
(Whitney) For two oriented bundles, $e(E\oplus F)=e(E)\smile e(F)$.
(Vanishing) If $E$ admits a nowhere-vanishing section then $e(E)=0$. In particular, trivial bundles have $e=0$.
(Mod 2 reduction) The $\mathbb{Z}/2$-reduction of $e(E)$ is the top Stiefel-Whitney class $w_n(E)$ of §Stiefel-Whitney Characteristic Classes, ⁋Definition 5.
(Orientation reversal) Reversing the orientation changes the sign of $e(E)$. Hence if $n$ is odd then $2e(E)=0$.

Proof

(1) follows from the fact that the Thom class is compatible with pullback, and (2) follows from the fact that the external product of the Thom classes of two bundles becomes the Thom class of the Whitney sum ([MS] §9–10).

For (3), suppose there exists a nowhere-vanishing section $s':B\rightarrow E_0$. The straight-line homotopy $t\mapsto t\cdot s'$ is a homotopy in $E$ between the zero section $0$ and $s'$, so $0^\ast=s'^\ast:H^n(E)\rightarrow H^n(B)$. On the other hand, $s'$ factors through $i:E_0\hookrightarrow E$, and from the long exact sequence of the pair the composition $H^n(E, E_0)\xrightarrow{j^\ast}H^n(E)\xrightarrow{i^\ast}H^n(E_0)$ is $0$, so $i^\ast(j^\ast u)=0$, that is, $j^\ast u$ dies on $E_0$. Since $s'$ passes through $E_0$, we have $s'^\ast(j^\ast u)=0$, and therefore $e(E)=0^\ast(j^\ast u)=s'^\ast(j^\ast u)=0$.

(4) is because the $\mathbb{Z}/2$-reduction of the Thom class is exactly the Thom class defining the Stiefel-Whitney class ([MS] §8), and since restriction commutes with reduction, $e(E)\bmod 2=w_n(E)$. For (5), reversing the orientation on each fiber flips the sign of every $u_x$, so $u\mapsto -u$ and hence $e\mapsto -e$. If $n$ is odd, the reflection $v\mapsto -v$ on each fiber is a bundle automorphism with determinant $(-1)^n=-1$ that reverses the orientation, so this automorphism forces $e=-e$ and thus $2e(E)=0$.

All five properties of Proposition 4 are read from the picture we saw earlier, namely that $e(E)$ is the Poincaré dual recording the zero locus of a generic section, including the signs. Apart from the somewhat formal first two conditions, the remaining three are stories of signs and obstructions.

For example, the third condition says that if there is a nowhere-vanishing section, then a generic section can also be chosen without zeros, so the self-intersection disappears and hence $e(E)=0$. A picture worth noting is the trivial line bundle over $S^1$ twisted twice; a generic section of this bundle meets the zero section twice, but the intersection directions are opposite, so they cancel to give $0$.

For the fourth claim, over $\mathbb{Z}$ we count the zero locus of a generic section with signs, but when we reduce to $\mathbb{Z}/2$ we forget those signs. Then the number of zeros counted without signs is exactly the top Stiefel-Whitney class $w_n$, so $e(E)$ is the lift of $w_n$ to the integers that remembers the signs, and the reverse process is taking mod $2$.

For the fifth claim, reversing the orientation flips the sign of all zeros together, giving $e\mapsto -e$, and in particular when $n$ is odd, the reflection $v\mapsto -v$ on each fiber is a bundle automorphism with determinant $(-1)^n=-1$ that reverses the orientation, forcing $e=-e$ and hence $2e(E)=0$; we immediately obtain that the Euler class of an odd-rank oriented bundle is always $2$-torsion.

The name Euler class comes from what it measures. If $M$ is a closed oriented $n$-manifold and $E=TM$ is its tangent bundle, then evaluating $e(TM)$ on the fundamental class $[M]$ of §Poincaré Duality, ⁋Definition 10 gives exactly the Euler characteristic

\[\rchi(M)=\int_{[M]} e(TM)\]

That is, the Euler class is an obstruction measuring whether this bundle admits a nonvanishing section, and if not, how much it is obstructed; in the case of the tangent bundle, the answer appeared as the topological invariant $\rchi(M)$.

Let us verify this concretely on $S^2$. For example, from §Degree of a Map and Brouwer·Lefschetz Fixed Point Theorem, ⁋Theorem 8 we know that any section of $TS^2$ must have a zero somewhere. For instance, suppose $S^2$ is embedded in $\mathbb{R}^3$ by

\[S^2=\{(x,y,z): x^2+y^2+z^2=1\}\]

and consider the height function $z$ defined on it. Then the gradient vector $\nabla z$ of this $z$ is the section of the tangent bundle given by

\[(\nabla z)_{(x,y,z)}=(-xz,-yz,x^2+y^2)\]

and we know that this section vanishes exactly at $(x,y,z)=(0,0,\pm 1)$. We can check that this section meets the zero section transversely and both intersections are positive, so the Euler class must be $2\in H^2(S^2; \mathbb{Z})$. Indeed, using the familiar computation of the Euler characteristic we find $\rchi(S^2)=2$, confirming that this supports our intuition.

Chern Classes

The Stiefel-Whitney class and Euler class we have seen so far were invariants of real vector bundles. We now turn our attention to complex vector bundles. Of course, any complex vector space can be viewed as a real vector space by separating real and imaginary parts, but the natural morphisms between complex vector spaces are in $\GL(n;\mathbb{C})$ rather than $\GL(2n;\mathbb{R})$, and this difference changes many things.

For example, $\GL(2n;\mathbb{R})$ is not connected, but $\GL(n;\mathbb{C})$ is connected, so any complex vector bundle is always orientable. Intuitively it suffices to look at what happens on a single fiber $V\cong\mathbb{C}^n$; choosing a (complex) basis

\[v_1,\ldots,v_n\]

for this fiber, the isomorphism $\mathbb{C}^n\cong \mathbb{R}^{2n}$ naturally gives the real basis

\[v_1,iv_1,\ldots,v_n,iv_n\]

Our claim is that this orientation is preserved even when we choose a different complex basis, which is because when we view the matrix $A\in\GL(n;\mathbb{C})$ connecting two bases as a real linear map, its real determinant is

\[\det\nolimits_{\mathbb{R}}(A)=\lvert\det\nolimits_{\mathbb{C}}(A)\rvert^2>0\]

For example, in the simplest case $n=1$, multiplication by $z=a+bi$ viewed as a real matrix is

\[\begin{pmatrix}a&-b\\ b&a\end{pmatrix}\]

and its determinant is $a^2+b^2>0$; for general $A$ the determinant always becomes positive in this way. That is, change of basis in a complex vector space always preserves orientation, so there is a canonical orientation on $V$ (as a real vector space), and the calculation above is exactly saying that $\GL(n;\mathbb{C})\subset \GL^+(2n; \mathbb{R})$.

In particular, the Euler class is canonically well defined for any complex vector bundle. Moreover, there are additional invariants beyond the Euler class. For example, a complex vector bundle $E$ and its conjugate $\bar{E}$ are the same as underlying real vector bundles, but they are generally different as complex vector bundles, and the Chern classes we will define can distinguish them.

Since the Chern class satisfies $c_n=e(E_\mathbb{R})$ at the top Chern class, it can be thought of as a characteristic class extending the Euler class. There are several ways to define it. In differential geometry one pulls it out from the curvature of a connection via Chern–Weil theory, or one can take an axiomatic approach as we did for Stiefel-Whitney. (Of course, in this case existence must be proved as a separate proposition.) We follow [MS] and define the Chern class by descending step by step from the Euler class, that is, the top Chern class. What is needed in this process is the Gysin exact sequence of Theorem 5 (Gysin exact sequence).

To describe this, let us first define the following. If the base $B$ is paracompact, we can use a partition of unity to put a fiber metric on each fiber of $E$, and then the set of vectors of length $1$

\[S(E)=\{v\in E:\lvert v\rvert=1\}\]

becomes a fiber bundle with fiber $S^{n-1}$. This is called the sphere bundle of $E$, and similarly

\[D(E)=\{v\in E:\lvert v\rvert\leq 1\}\]

is called the disk bundle with fiber the disk $D^n$. We used a metric for convenience, but this is not essential; what matters is that for the pair $(D(E), S(E))$ and the space $E_0=E\setminus 0(B)$ with the zero section removed, this pair is homotopy equivalent to $(E, E_0)$. First, using §Computation of Homology, ⁋Theorem 2 (Excision theorem) for the exterior of $D(E)$, we obtain

\[H^\ast(E, E_0)\cong H^\ast\bigl(D(E), D(E)\setminus 0(B)\bigr)\]

and then applying radial retraction gives

\[(E, E_0)\simeq (D(E), S(E))\]

Then using this we obtain the following.

Theorem 5 (Gysin exact sequence) For the sphere bundle $\pi:S(E)\rightarrow B$ of an oriented rank $n$ vector bundle $E\rightarrow B$, the following long exact sequence exists:

\[\cdots\rightarrow H^{k-n}(B)\xrightarrow{\ \smile e\ }H^k(B)\xrightarrow{\ \pi^\ast\ }H^k(S(E))\xrightarrow{\ \pi_!\ }H^{k-n+1}(B)\rightarrow H^{k+1}(B)\rightarrow\cdots\]

where $e=e(E)$ is the Euler class, $\pi^\ast$ is the pullback, and $\pi_!$ is integration along the fiber.

Proof

Consider the cohomology long exact sequence of the pair $(D(E), S(E))$

\[\cdots\rightarrow H^k(D(E), S(E))\rightarrow H^k(D(E))\rightarrow H^k(S(E))\xrightarrow{\ \delta\ }H^{k+1}(D(E), S(E))\rightarrow\cdots\]

First, the first term is $H^k(D(E), S(E))\cong H^k(E, E_0)\cong H^{k-n}(B)$ by Theorem 2 (Thom isomorphism), and for the second term we obtain $H^k(D(E))\cong H^k(B)$ via the retraction. Through these identifications we obtain the following commutative diagram

$pair cohomology long exact sequence and Gysin exact sequence$

and the maps in the second column are those of the upper exact sequence transported along the vertical isomorphisms. Concretely, let us follow the first map $H^{k-n}(B)\rightarrow H^k(B)$. Lift $\alpha\in H^{k-n}(B)$ to $H^k(E, E_0)$ via the Thom isomorphism $\Phi:\alpha\mapsto p^\ast\alpha\smile u$, and then compose with the first map $j^\ast$ of the upper row to get

\[j^\ast\Phi(\alpha)=j^\ast(p^\ast\alpha\smile u)=p^\ast\alpha\smile j^\ast u\]

where the second equality uses that $j^\ast$ is a homomorphism of relative cohomology rings preserving cup product, that is, $j^\ast p^\ast\alpha=p^\ast\alpha$, which is intuitively obvious because $p^\ast\alpha$ already lives on $H^\ast(E)$. Now we bring this down via the vertical identification $H^k(E)\cong H^k(B)$, which is done through the zero section $0:B\hookrightarrow E$ ($p\circ 0=\mathrm{id}$), so

\[0^\ast(p^\ast\alpha\smile j^\ast u)=0^\ast p^\ast\alpha\smile 0^\ast j^\ast u=\alpha\smile e(E)\]

(Definition 3) Similarly, the second map $H^k(D(E))=H^k(B)\rightarrow H^k(S(E))$ is the restriction, that is, $\pi^\ast$, and the Gysin map $\pi_!$ is the connecting homomorphism $\delta$ transported by the Thom isomorphism $H^{k+1}(D(E), S(E))\cong H^{k-n+1}(B)$.

The third map $\pi_!:H^k(S(E))\rightarrow H^{k-n+1}(B)$ has a somewhat special property. The natural map that a continuous map $\pi:S(E)\rightarrow B$ induces on cohomology is usually the pullback $\pi^\ast:H^\ast(B)\rightarrow H^\ast(S(E))$, which goes in the opposite direction of $\pi$ and preserves degree. On the other hand, $\pi_!$ goes in the same direction as $\pi$ while lowering degree by $(n-1)$; a map that goes against the direction that would naturally be induced by a continuous function is called a wrong-way map, and it is customary to attach the subscript $!$.

The intuition for this map lies in reversing Theorem 2 (Thom isomorphism). If the Thom isomorphism $\alpha\mapsto p^\ast\alpha\smile u$ was a lift in the fiber direction, copying the base class $\alpha$ to each point on the fiber ($p^\ast\alpha$) and then multiplying by the fiber-direction class $u$ to raise the degree, then $\pi_!$ should be thought of as its inverse. That is, viewing a class on $S(E)$ as a base-direction component and a fiber-direction component, the fiber-direction component is integrated out along each fiber $S^{n-1}$ (hence the degree drops by the fiber dimension $n-1$), and the remaining base-direction class is returned as is. The mathematical formulation of this property is the projection formula

\[\pi_!(\pi^\ast\alpha\smile\beta)=\alpha\smile\pi_!\beta,\qquad \alpha\in H^\ast(B), \quad\beta\in H^\ast(S(E))\]

To see this in action, let us look at the tangent bundle $TS^2$ of $S^2$ discussed above. Then in the Gysin sequence

\[\smile e:H^0(S^2)\rightarrow H^2(S^2)\]

is given by the $\times 2$ map, and its cokernel $\mathbb{Z}/2$ appears as torsion in $H^2$ of the sphere bundle $S(TS^2)$. On the other hand, if we had started with the trivial bundle $E$ over $S^2$, this part would have been $\mathbb{Z}$, so the trace of the Euler class pushing the sphere bundle away from the product is contained in this torsion.

Now let us define the Chern class from this. The key fact is that when $k<n-1$

\[H^{k-n}(B)=H^{k-n+1}(B)=0\]

so $\pi^\ast:H^k(B)\rightarrow H^k(S(E))$ is an isomorphism. That is, the cohomology of the sphere bundle coincides with that of the base in low degrees, and after that the Euler class contributes additional terms beyond what comes from the base cohomology.

Consider the deleted total space $E_0=E\setminus 0(B)$ that we have been looking at. A point of $E_0$ is an ordered pair of a point $x\in B$ of the base and a nonzero vector $v\in E_x$ in the fiber of $E$ at that point. Now define the tautological bundle $\pi_0^\ast E$ on $E_0$. This is the vector bundle obtained by pulling back the vector bundle $E\rightarrow B$ along the projection map $\pi_0:E_0\rightarrow B$, and its substance is the vector bundle having fiber $(\pi_0^\ast E)_{(x,v)}= E_x$ at each point $(x,v)\in E_0$. That is, $v$ is also an element of the vector space attached to the point $(x,v)$, and since it is nonzero it defines a $1$-dimensional subspace $\langle v\rangle$ inside this vector space. Now attaching a line in this way at every point of $E_0$ gives a line bundle $L\rightarrow E_0$, and we can consider the quotient $(\pi_0^\ast E)/L\rightarrow E_0$ that this defines inside $\pi_0^\ast E$. This is a canonical complex rank $(n-1)$ bundle over $E_0$ having fiber $E_x/\langle v\rangle$ at each point $(x,v)$, and if we put a Hermitian inner product on each fiber it is also realized as the orthogonal complement $v^\perp\subseteq E_x$ of $v$. (§Complex Inner Product Spaces, ⁋Proposition 4) Since the two realizations are canonically isomorphic, we will denote this rank $(n-1)$ bundle by $L^\perp$ for convenience.

Now let a complex vector bundle $E$ be given, and denote by $E_\mathbb{R}$ its underlying (oriented) real vector bundle. If $E$ has complex dimension $n$ then $E_\mathbb{R}$ has real dimension $2n$. Then $E_0$ is homotopy equivalent to the sphere bundle $S(E_{\mathbb{R}})$ of $E_{\mathbb{R}}$, so by Theorem 5 (Gysin exact sequence)

\[\cdots\rightarrow H^{k-2n}(B)\xrightarrow{\ \smile e\ }H^k(B)\xrightarrow{\ \pi_0^\ast\ }H^k(E_0)\rightarrow H^{k-2n+1}(B)\rightarrow\cdots\]

holds, and as we saw above, when $k\leq 2n-2$ both $H^{k-2n}(B)$ and $H^{k-2n+1}(B)$ are in negative degree and hence $0$, so $\pi_0^\ast:H^k(B)\rightarrow H^k(E_0)$ is an isomorphism.

Definition 6 The Chern class $c_i(E)\in H^{2i}(B;\mathbb{Z})$ of a complex rank $n$ vector bundle $E\rightarrow B$ is defined inductively on the rank $n$ of the vector bundle as follows.

First $c_0(E)=1$, and $c_i(E)=0$ for $i>n$, and we set

\[c_n(E)=e(E_{\mathbb{R}})\in H^{2n}(B;\mathbb{Z})\]

For $0<i<n$, since the $L^\perp$ discussed above is a rank $(n-1)$ vector bundle whose Chern classes are already defined by the inductive hypothesis, we pull these back via the isomorphism $\pi_0^\ast:H^{2i}(B)\rightarrow H^{2i}(E_0)$ and define the (unique) $c_i(E)\in H^{2i}(B)$ satisfying

\[\pi_0^\ast c_i(E)=c_i(L^\perp)\]

to be the $i$-th Chern class of $E$. The sum $c(E)=1+c_1(E)+\cdots+c_n(E)\in H^\bullet(B;\mathbb{Z})$ is called the total Chern class.

As in other situations, what is as important as the definition are the following characteristic properties it satisfies.

Proposition 7 The Chern class satisfies the following.

(Naturality) For any $f:B'\rightarrow B$, $c(f^\ast E)=f^\ast c(E)$.
$c_0(E)=1$, and $c_i(E)=0$ for $i>\rank_{\mathbb{C}}E$.
(Top degree) $c_n(E)=e(E_{\mathbb{R}})$, so if $E$ has a nonzero section then $c_n(E)=0$.

Proof

The second and third conditions follow immediately from the definition using the vanishing of $c_n=e(E_{\mathbb{R}})$ from the third condition of Proposition 4.

The first condition is proved by induction on $n$. The naturality of $c_n$ comes from the naturality of the Euler class. (First condition of Proposition 4) For $0<i<n$, $f$ induces a bundle map $E_0'\rightarrow E_0$ compatible with the deleted space, the complement bundle, and the entire Gysin sequence, and on it $f^\ast(L^\perp)\cong(f^\ast L)^\perp$, so the naturality of $c_i$ follows from the inductive hypothesis and the naturality of $\pi_0^\ast$.

That is, the Chern class satisfies axiomatic properties of a similar kind to the Stiefel-Whitney class. (§Stiefel-Whitney Characteristic Classes, ⁋Definition 5) We showed the existence of Stiefel-Whitney classes by considering the real infinite Grassmannian $\Gr(k,\mathbb{R}^\infty)$, pulling back cohomology classes from there to the original space, and showing that they satisfy the axiomatic conditions for Stiefel-Whitney classes; a similar construction is possible for Chern classes.

Example 8 As the complex analogue of the real tautological line bundle of §Stiefel-Whitney Characteristic Classes, ⁋Example 3, consider the tautological complex line bundle $\gamma$ over $\CP^\infty=\Gr(1,\mathbb{C}^\infty)$. Then the sphere bundle of $\gamma$ is the unit sphere $S^\infty$ in $\mathbb{C}^\infty$, which is contractible¹ so $H^k(S^\infty)=0$ for all $k>0$. Therefore, by Theorem 5 (Gysin exact sequence), $H^1(\CP^\infty)=0$ and

\[\smile c_1(\gamma):H^{k-2}(\CP^\infty)\rightarrow H^k(\CP^\infty)\]

is an isomorphism for $k\geq 2$. Starting from $H^0(\CP^\infty)=\mathbb{Z}$, we obtain that $c_1(\gamma)$ is a generator of $H^2(\CP^\infty;\mathbb{Z})\cong\mathbb{Z}$, and

\[H^\bullet(\CP^\infty;\mathbb{Z})=\mathbb{Z}[c_1(\gamma)]\]

This is, as we saw for real bundles in §Stiefel-Whitney Characteristic Classes, the universal family for complex line bundles. That is, any complex line bundle is obtained uniquely as a pullback of $\gamma$, so the first Chern class gives a bijection

\[\{B\text{ 위의 complex line bundle}\}/\cong\ \xrightarrow{\ c_1\ }\ H^2(B;\mathbb{Z})\]

which is a group isomorphism sending tensor product to addition. Thus all information about a complex line bundle is contained in $c_1$.

More generally, the complex Grassmannian $\Gr(k,\mathbb{C}^\infty)$ takes the place of the real Grassmannian, and its cohomology ring becomes

\[H^\bullet(\Gr(k,\mathbb{C}^\infty);\mathbb{Z})=\mathbb{Z}[c_1,\ldots,c_k]\]

the polynomial ring generated by the Chern classes of the universal bundle, and we will revisit this kind of computation before long.

Meanwhile, just as the Stiefel-Whitney class followed the Whitney sum formula, it is natural to expect that the Chern class satisfies the same formula here as well. The key step to actually prove this is §Projective Bundles and Leray–Hirsch Theorem, ⁋Theorem 5; the proof of this theorem is certainly possible with the discussion so far, but for the sake of narrative flow we defer it to a separate post.

Theorem 9 (Whitney sum formula) For two complex vector bundles $E,E'\rightarrow B$,

\[c(E\oplus E')=c(E)\smile c(E')\]

holds. That is, for all $k$,

\[c_k(E\oplus E')=\sum_{i+j=k}c_i(E)\smile c_j(E')\]

Proof

By §Projective Bundles and Leray–Hirsch Theorem, ⁋Theorem 5, there exists a continuous map $\rho:F(E)\rightarrow B$ such that the pullback $\rho^\ast:H^\bullet(B)\hookrightarrow H^\bullet(F(E))$ is injective and $\rho^\ast E$ splits as a Whitney sum $L_1\oplus\cdots\oplus L_n$ of complex line bundles. By naturality and the injectivity of $\rho^\ast$, it suffices to prove the formula assuming every bundle is a sum of line bundles.

Then it suffices to show

\[c(L_1\oplus\cdots\oplus L_n)=\prod_{i=1}^n\bigl(1+c_1(L_i)\bigr)\]

The key is the two equalities for two line bundles $L,L'$:

\[c_1(L\oplus L')=c_1(L)+c_1(L'),\qquad c_2(L\oplus L')=c_1(L)\smile c_1(L')\]

The second equality is obtained immediately over any base. By Definition 6, the top term of a rank $2$ bundle is

\[c_2(L\oplus L')=e\bigl((L\oplus L')_{\mathbb{R}}\bigr)\]

and by the second result of Proposition 4 this equals $e(L_{\mathbb{R}})\smile e(L'_{\mathbb{R}})=c_1(L)\smile c_1(L')$.

For the first equality, let us first show that for any rank $n$ complex vector bundle $E$ and trivial line bundle $\varepsilon^1$,

\[c(E\oplus\varepsilon^1)=c(E)\]

holds. Let $E'=E\oplus\varepsilon^1$; the section $s(x)=(0,1)$ taking the constant $1$ from the trivial component is nowhere zero, so it gives a section $s:B\rightarrow E'_0$ with $\pi_0\circ s=\mathrm{id}$. Now the orthogonal complement of $(0,1)$ at each point is exactly the fiber of $E$, so $s^\ast(E'^\perp)\cong E$, and hence applying $s^\ast$ to the formula $\pi_0^\ast c_i(E')=c_i(E'^\perp)$ of Definition 6 for $0<i\leq n$ gives, by naturality of Proposition 7,

\[c_i(E')=s^\ast\pi_0^\ast c_i(E')=s^\ast c_i(E'^\perp)=c_i(s^\ast E'^\perp)=c_i(E)\]

The top term $c_{n+1}(E')=e(E'_{\mathbb{R}})$ is $0$ by (3) of Proposition 4 because there exists a nowhere-vanishing section, which matches $c_{n+1}(E)=0$.

On the other hand, as we saw in Example 8, since $\gamma$ is the universal family for complex line bundles, any two line bundles $L,L'$ over a base $B$ are pullbacks $f_1^\ast\gamma$, $f_2^\ast\gamma$ along morphisms $f_1,f_2:B\rightarrow\CP^\infty$ of the base. Now setting

\[f=(f_1, f_2): B\rightarrow \CP^\infty\times\CP^\infty\]

we obtain the following commutative diagram between bases

$classifying map decomposition$

and therefore showing the formula

\[c_1(L\oplus L')=c_1(L)+c_1(L')\]

on $B$ for $L, L'$, that is,

\[c_1(f^\ast(\pi_1^\ast\gamma \oplus \pi_2^\ast\gamma))=c_1(f_1^\ast\gamma)+c_1(f_2^\ast\gamma)\]

is, by the first result of Proposition 7, the same as showing

\[c_1(\pi_1^\ast\gamma\oplus \pi_2^\ast\gamma)=c_1(\pi_1^\ast\gamma)+c_1(\pi_2^\ast\gamma)\]

That is, it suffices to show that the formula holds for any two line bundles $L_1, L_2$ over $\CP^\infty\times\CP^\infty$.

For this, first observe by §Cohomology, ⁋Corollary 10 (Künneth) that

\[H^2(\CP^\infty\times\CP^\infty;\mathbb{Z})\cong H^2(\CP^\infty;\mathbb{Z})\oplus H^2(\CP^\infty;\mathbb{Z})\]

That is, defining restriction maps $j_1^\ast$, $j_2^\ast$ via the inclusions $j_1:z\mapsto(z,q)$, $j_2:z\mapsto(q,z)$, these read off the respective components. On the other hand, we showed above that the desired formula holds for trivial line bundles, and $j_1^\ast L_2$ and $j_2^\ast L_1$ are trivial, so

\[j_1^\ast c_1(L_1\oplus L_2)=c_1(\gamma\oplus\varepsilon^1)=c_1(\gamma)=j_1^\ast\bigl(c_1(L_1)+c_1(L_2)\bigr)\]

and a similar formula holds for $j_2^\ast$. Therefore,

\[c_1(L_1\oplus L_2)=c_1(L_1)+c_1(L_2)\]

and as we saw above, adding naturality makes this hold for arbitrary $L,L'$.

Earlier we said that the Chern class can distinguish a complex vector bundle $E$ and its conjugate $\bar{E}$, that is, the bundle with the same underlying real bundle but with scalar multiplication twisted to $z\cdot v=\bar{z}v$. We can now make this claim precise.

Proposition 10 For the conjugate $\bar{E}$ of a complex vector bundle $E\rightarrow B$,

\[c_i(\bar{E})=(-1)^ic_i(E)\]

holds for all $i$.

Proof

First consider the case of a line bundle $L$. By Definition 6, $c_1(L)=e(L_{\mathbb{R}})$, and $L$ and $\bar{L}$ have the same underlying real bundle but opposite canonical orientations. Indeed, for a nonzero vector $v$ in the fiber, the canonical orientation of $L$ is given by the ordered basis $(v,iv)$, while in $\bar{L}$ since $i$ sends $v$ to $-iv$, the canonical orientation is given by $(v,-iv)$, and the determinant of the change-of-basis matrix between the two bases is $-1$. Therefore by (5) of Proposition 4, $c_1(\bar{L})=-c_1(L)$.

The general case also follows using the splitting principle as in the proof above.

For example, for the tautological bundle $\gamma$ of Example 8, $c_1(\gamma)$ is a generator of $H^2(\CP^\infty;\mathbb{Z})\cong\mathbb{Z}$, so $c_1(\bar{\gamma})=-c_1(\gamma)\neq c_1(\gamma)$, and hence $\gamma\not\cong\bar{\gamma}$. Of course there are limits to this distinction: if all odd Chern classes of a bundle are $2$-torsion or $0$, its conjugate cannot be distinguished by Chern classes alone, but we can still confirm that Chern classes carry richer information than real bundles.

Meanwhile, all examples so far have been line bundles, so let us look at one example showing how Theorem 9 (Whitney sum formula) is used in actual calculations for bundles of higher rank.

Example 11 In this post we compute the total Chern class of the tangent bundle of the finite-dimensional complex projective space $\CP^n=\Gr(1,\mathbb{C}^{n+1})$.

For this, first consider the tautological line bundle $\gamma\subseteq\CP^n\times\mathbb{C}^{n+1}$ defined on it. This is the restriction of the universal line bundle $\gamma$ of Example 8 to $\CP^n\hookrightarrow\CP^\infty$, and since the cell structure has cells only in even dimensions, the restriction $H^k(\CP^\infty;\mathbb{Z})\rightarrow H^k(\CP^n;\mathbb{Z})$ is an isomorphism for $k\leq 2n$. Therefore, setting $\x=c_1(\bar{\gamma})=-c_1(\gamma)$ by Proposition 10, we have

\[H^\bullet(\CP^n;\mathbb{Z})=\mathbb{Z}[\x]/(\x^{n+1})\]

Now consider the tangent bundle. A point $\ell\in\CP^n$ of this space is a line $\ell\subseteq\mathbb{C}^{n+1}$, and fixing a Hermitian inner product, lines near $\ell$ are uniquely represented as graphs of linear maps $\ell\rightarrow\ell^\perp$. That is, writing the bundle of $\mathbb{C}$-linear maps fiberwise as $\Hom$,

\[T\CP^n\cong\Hom(\gamma,\gamma^\perp)\]

Adding the trivial line bundle $\Hom(\gamma,\gamma)$ by Whitney sum gives

\[T\CP^n\oplus\Hom(\gamma,\gamma)\cong\Hom(\gamma,\gamma^\perp\oplus\gamma)\cong\Hom(\gamma,\varepsilon^{n+1})\cong\Hom(\gamma,\varepsilon^1)^{\oplus(n+1)}\]

Here $\varepsilon^{n+1}$ is the rank $n+1$ trivial bundle. Therefore, identifying $\Hom(\gamma, \varepsilon^1)$ with $\overline{\gamma}$, we obtain from Theorem 9 (Whitney sum formula) the formula

\[c(T\CP^n)=c\bigl(T\CP^n\oplus\Hom(\gamma,\gamma)\bigr)=c(\bar{\gamma})^{n+1}=(1+\x)^{n+1}\]

and expanding this, since $H^\bullet(\CP^n)=\mathbb{Z}[\x]/(\x^{n+1})$, we know

\[c(T\CP^n)=(n+1)\x^n+\cdots +1\]

Pontryagin Classes

For real vector bundles as well, $\mathbb{Z}$-coefficient invariants can be obtained via the complex Chern classes.

Definition 12 The Pontryagin class $p_i(E)\in H^{4i}(B;\mathbb{Z})$ of a real vector bundle $E\rightarrow B$ is defined from the Chern class of the complexification $E\otimes_{\mathbb{R}}\mathbb{C}$ by

\[p_i(E)=(-1)^i c_{2i}(E\otimes_{\mathbb{R}}\mathbb{C})\]

The complexification $E\otimes_{\mathbb{R}}\mathbb{C}$ is isomorphic to its conjugate $\overline{E\otimes\mathbb{C}}$ via $v\otimes z\mapsto v\otimes\bar{z}$. Then by Proposition 10, $c_{2i+1}(E\otimes\mathbb{C})=-c_{2i+1}(E\otimes\mathbb{C})$, that is, all odd Chern classes become $2$-torsion ($2c_{2i+1}=0$) and essentially carry no meaningful information. For this reason we define the $i$-th class using only the even-position (signed) Chern classes, and since Chern classes live in cohomology of degree twice their index, Pontryagin classes end up in $H^{4i}(B;\mathbb{Z})$. Intuitively this is bringing what the Stiefel-Whitney class did over $\mathbb{Z}/2$ to $\mathbb{Z}$-coefficients (without passing to complex vector bundles), or bringing what the Chern class did for complex vector bundles down to real vector bundles.

The basic properties also descend from the Chern classes along complexification. The total Pontryagin class is written $p(E)=1+p_1(E)+p_2(E)+\cdots$.

Proposition 13 For real vector bundles $E,F\rightarrow B$, the following hold.

(Naturality) For any $f:B'\rightarrow B$, $p(f^\ast E)=f^\ast p(E)$.
(Whitney) $2\bigl(p(E\oplus F)-p(E)\smile p(F)\bigr)=0$. In particular, if there is no $2$-torsion in $H^\bullet(B;\mathbb{Z})$ then $p(E\oplus F)=p(E)\smile p(F)$.
For a complex vector bundle $E$, $E_{\mathbb{R}}\otimes_{\mathbb{R}}\mathbb{C}\cong E\oplus\bar{E}$, so $p_i(E_{\mathbb{R}})$ is a polynomial in the Chern classes of $E$. For example, $p_1(E_{\mathbb{R}})=c_1(E)^2-2c_2(E)$.

Proof

(1) follows immediately because complexification commutes with pullback and from the naturality of Proposition 7. (2) also follows by applying Theorem 9 (Whitney sum formula) to $(E\oplus F)\otimes\mathbb{C}\cong(E\otimes\mathbb{C})\oplus(F\otimes\mathbb{C})$; as observed below Definition 12, all odd Chern classes are $2$-torsion, so the terms involving them vanish when multiplied by $2$, and the remaining even terms give $p(E)\smile p(F)$.

Only (3) requires a small calculation. When we complexify $E_{\mathbb{R}}\otimes\mathbb{C}$, the complex structure appears as $J\in \End(E)$ with eigenvalues $\pm i$. Extending this $\mathbb{C}$-linearly, the $\pm i$ eigenspace decomposition gives $E_{\mathbb{R}}\otimes\mathbb{C}\cong E\oplus\bar{E}$. Then by Theorem 9 (Whitney sum formula) and Proposition 10, $c_2(E_{\mathbb{R}}\otimes\mathbb{C})=c_2(E\oplus\bar{E})=2c_2(E)-c_1(E)^2$, and bringing this to the Pontryagin class gives the desired result.

References

[MS] J. W. Milnor and J. D. Stasheff, Characteristic Classes, Annals of Mathematics Studies 76, Princeton University Press, 1974.

[BT] R. Bott and L. W. Tu, Differential Forms in Algebraic Topology, Springer, 1982.

[Hat] A. Hatcher, Vector Bundles and K-Theory, online notes, 2017.

View $S^\infty$ as the unit sphere in $\mathbb{C}^\infty=\bigcup_n\mathbb{C}^n$. For the shift map $T(x_1,x_2,\ldots)=(0,x_1,x_2,\ldots)$, normalize vectors by $v\mapsto v/\lvert v\rvert$; the two straight-line homotopies $x\mapsto\bigl((1-t)x+tT(x)\bigr)/\lvert(1-t)x+tT(x)\rvert$ and $x\mapsto\bigl((1-t)T(x)+te_1\bigr)/\lvert(1-t)T(x)+te_1\rvert$ connect the identity to $T$, and $T$ to the constant map $x\mapsto e_1=(1,0,\ldots)$, respectively. Neither denominator vanishes: for the first, the coordinates of $x$ and $T(x)$ are shifted by one so $(1-t)x+tT(x)=0$ forces $x=0$; for the second, the first coordinate of the sum is $t$, so for it to be $0$ we need $t=0$, but then $T(x)=0$, hence $x=0$. Joining these two, $S^\infty$ contracts to a point. ↩

Share on

Twitter Facebook LinkedIn

K

Euler Class

Chern Classes

Pontryagin Classes

Share on

댓글남기기