Construction of Stiefel-Whitney Classes

Orientation of Vector Bundles

First we shall understand the geometric meaning of Stiefel-Whitney classes on line bundles.

[Def] (Orientable) Let EXE\rightarrow X be a real vector bundle, with structural group O(k)O(k). Say EE is orientable if there exists a reduction of structural group to SO(k)SO(k).

For linear space VV, define the set of orientation on VV to be

o(V):={(v1,,vk) is a basis for V}/GLk+(R)o(V):=\{(v_1,\cdots,v_k)\text{~is a basis for }V\}/GL_k^+(\mathbb{R})

[Def] (Orientation) An orientation of EE consists of the following material:

  1. xX\forall x\in X, an orientation αxo(Ex)\alpha_x\in o(E_x);

  2. α\alpha is locally constant, namely if local trivialization EUU×RkE_U\cong U\times\mathbb{R}^k, then α\alpha gives

    α:Uo(Rk).\alpha':U\rightarrow o(\mathbb{R}^k).

[Remark] A manifold is orientable if TMTM admits an orientation. The kk-th exterior ΛkE\Lambda^k E of EE, is also a line bundle over XX.

[Prop] EE is orientable \Leftrightarrow ΛkE\Lambda^k E is trivial.

Proof. Global section s:XEs:X\rightarrow E endows ExE_x with an orientation. The inverse also holds. \square


Construction via Orientation

To establish the construction of Stiefel-Whitney class, it requires two important theorem on vector bundles.

[Theorem] (Leray-Hirsch) Let FPXF\rightarrow P\rightarrow X be a fiber bundle, RR is a ring. If there exists u1,,ukH(P;R)u_1,\cdots,u_k\in H^*(P;R), such that

u1F,,ukFu_1|_F,\cdots,u_k|_F

is a basis for H(F;R)H^*(F;R), then

H(P;R)H(X;R)R[u1,,uk]παuiαui\begin{aligned} H^*(P;R)&\cong H^*(X;R)\otimes R[u_1,\cdots,u_k]\\ \pi^*\alpha\cup u_i&\leftrightarrow \alpha\otimes u_i \end{aligned}

Proof. (MV-induction)

  1. If P=X×FP=X\times F, then by Kunneth formula, the theorem establishes.
  2. Let X1,X2XX_1,X_2\subset X​ be subspaces.

Since MM is free, then the sequence is exact. \square

[Theorem] (Splitting principle) Let EXE\rightarrow X be a vector bundle of rank kk and f:YXf:Y\rightarrow X.

  1. f:H(X,Z/2)H(Y,Z/2)f^*:H^*(X,\mathbb{Z}/2)\rightarrow H^*(Y,\mathbb{Z}/2) is injective;
  2. L1,,LkY\exists L_1,\ldots,L_k\rightarrow Y are line bundles, s.t.

fE=L1Lk.f^*E=L_1\oplus\cdots\oplus L_k.

Notice that

w1(E)H1(X,Z/2)Hom(H1(X,Z),Z/2)=Hom(π1(X)ab,Z/2)Hom(π1(X),Z/2){2-cover of X}\begin{aligned} w_1(E)&\in H^1(X,\mathbb{Z}/2)\\ &\cong \mathrm{Hom}(H_1(X,\mathbb{Z}),\mathbb{Z}/2)\\ &=\mathrm{Hom}(\pi_1(X)^{ab},\mathbb{Z}/2)\\ &\cong\mathrm{Hom}(\pi_1(X),\mathbb{Z}/2)\\ &\cong\{2\text{-cover~of~}X\} \end{aligned}

Define XE~:={vΛkE:v=1}ΛkE\widetilde{X_E}:=\{v\in\Lambda^kE:|v|=1\}\subset\Lambda^kE be a 2-cover of XX. Since element in π1(X)\pi_1(X) is the map α:S1X\alpha:S^1\rightarrow X.

[Prop] The following statements are equivalent:

  1. αE\alpha^*E is orientable;
  2. αXE~\alpha^*\widetilde{X_E} is trivial;
  3. α\alpha can be lifted to XE~\widetilde{X_E}.

Thus, [α]π1(X)\forall [\alpha]\in\pi_1(X), define

w1(E)[α]={+1,αE is orientable,1,αE is unorientable.w_1(E)[\alpha] =\left\{ \begin{aligned} &+1,\qquad \alpha^*E\text{~is orientable},\\ &-1,\qquad \alpha^*E\text{~is unorientable}. \end{aligned} \right.

Suppose that EX\boxed{E\rightarrow X} is a vector bundle of rank kk, and L=ΛkEXL=\boxed{\Lambda^kE\rightarrow X}. By the above correspondence,

Define

w1(E):=W(ΛkE)w_1(E):=W(\Lambda^kE)

which means

w1(E)[α]={+1,αE is orientable,1,αE is unorientable.w_1(E)[\alpha]=\left\{ \begin{aligned} &+1,\qquad \alpha^*E\text{~is orientable},\\ &-1,\qquad \alpha^*E\text{~is unorientable}. \end{aligned} \right.

for α:S1X\alpha:S^1\rightarrow X. In fact, w1w_1 gives VectR1(X)H1(X,Z/2)\text{Vect}_\mathbb{R}^1(X)\cong H^1(X,\mathbb{Z}/2), which means that w1w_1 classifies real line bundles. Now we wish to define all Stiefel-Whitney classes.

[Prop] Let L1,L2VectR1(X)L_1,L_2\in\text{Vect}_\mathbb{R}^1(X). Then

w1(L1L2)=w1(L1)+w1(L2).w_1(L_1\otimes L_2)=w_1(L_1)+w_1(L_2).

Proof. For line bundle, the tensor product of unorientable bundle is orientable. \square

Whitney product formula: Notice that Λk+l(EF)=ΛkEΛlF\Lambda^{k+l}(E\oplus F)=\Lambda^kE\otimes\Lambda^lF. Then

w1(EF)=w1(Λk+l(EF))=w1(ΛkEΛlF)=w1(E)+w1(F).\begin{aligned} w_1(E\oplus F) &=w_1(\Lambda^{k+l}(E\oplus F))\\ &=w_1(\Lambda^kE\otimes\Lambda^lF)\\ &=w_1(E)+w_1(F). \end{aligned}

Let EXE\rightarrow X be a real vector bundle of rank kk. Define

P(E)x:={lines through the origin of Ex}=P(Ex)\begin{aligned} \mathbb{P}(E)|_x:&=\{\text{lines through the origin of }E_x\}\\ &=\mathbb{P}(E_x) \end{aligned}

If EUU×RkE_U\cong U\times\mathbb{R}^k, then P(E)UE×RPk1\mathbb{P}(E)|_U\cong E\times\mathbb{R}P^{k-1} is a vector bundle.

Denote the tautological line bundle over P(E)\mathbb{P}(E) by

ηx:={(l,u)P(Ex)×ExuL}\eta_x:=\{(l,u)\in\mathbb{P}(E_x)\times E_x|u\in L\}

Define

LE:={(l,u)P(E)×Eπ(l)=p(u),ul}πE\begin{aligned} L_E:&=\{(l,u)\in\mathbb{P}(E)\times E|\pi(l)=p(u), u\in l\}\\ &\subset \pi^*E \end{aligned}

Then LP(E)\boxed{L\rightarrow\mathbb{P}(E)} is a line bundle. Consider the tautological line bundle over RPn\mathbb{R}P^n:

ηnRPn\eta_n\rightarrow \mathbb{R}P^n

Non-zero: since η1RP1\eta_1\rightarrow\mathbb{R}P^1 is a Mobius line bundle, which is unorientable, then w(η1)=a0w(\eta_1)=a\neq 0. With the natural embedding i:RP1RPni:\mathbb{R}P^1\rightarrow \mathbb{R}P^n, we have

w1(ηn)=a0H1(RPn,Z/2).w_1(\eta_n)=a\neq 0\in H^1(\mathbb{R}P^n,\mathbb{Z}/2).

Pullback: Let f:XYf:X\rightarrow Y and α:S1X\alpha:S^1\rightarrow X be a loop.

By direct calculation:

w1(fE)[α]={+1,(αf)E is orientable;1,(αf)E is unorientable.w_1(f^*E)[\alpha]=\left\{ \begin{aligned} &+1, \qquad (\alpha^*f^*)E\text{~is~orientable};\\ &-1, \qquad (\alpha^*f^*)E\text{~is~unorientable}. \end{aligned} \right.

(fw1(E))[α]={+1,(fα)E is orientable;1,(fα)E is unorientable.(f^*w_1(E))[\alpha]=\left\{ \begin{aligned} &+1, \qquad (f\circ\alpha)^*E\text{~is~orientable};\\ &-1, \qquad (f\circ\alpha)^*E\text{~is~unorientable}. \end{aligned} \right.

Hence w1(fE)=fw1(E)w_1(f^*E)=f^*w_1(E). Then {1,w1(ηn),,w1(ηn)n1}\{1,w_1(\eta_n),\cdots,w_1(\eta_n)^{n-1}\} is a basis for H(RPn;Z/2)H^*(\mathbb{R}P^n;\mathbb{Z}/2), and

{1,w1(LE),,w1(LE)k1}\{1,w_1(L_E),\cdots,w_1(L_E)^{k-1}\}

is a basis for H(P(Ex);Z/2)=H(RPk;Z/2)H^*(\mathbb{P}(E_x);\mathbb{Z}/2)=H^*(\mathbb{R}P^{k};\mathbb{Z}/2). By Leray-Hirsch Theorem,

H(P(E);Z/2)H(X)Z/2{1,w1(L),,w1(L)k1}παw1(L)sαw1(L)s\begin{aligned} H^*(\mathbb{P}(E);\mathbb{Z}/2) &\cong H^*(X)\otimes \mathbb{Z}/2\{1,w_1(L),\ldots,w_1(L)^{k-1}\}\\ \pi^*\alpha\cup w_1(L)^s&\leftarrow \alpha\otimes w_1(L)^s \end{aligned}

where π:H(X;Z/2)H(P(E);Z/2)\pi^*:H^*(X;\mathbb{Z}/2)\rightarrow H^*(\mathbb{P}(E);\mathbb{Z}/2) is injective.

The injectivity is by π(H(X;Z/2))=H(X)Z/2\pi^*(H^*(X;\mathbb{Z}/2))=H^*(X)\otimes\mathbb{Z}/2​.

Thus,

w1(LE)k=(πa1)w1(LE)k1++(πak1)w1(LE)+πak.w_1(L_E)^k=(\pi^*a_1)w_1(L_E)^{k-1}+\cdots+(\pi^*a_{k-1})w_1(L_E)+\pi^*a_k.

Define the ii-th Stiefel-Whitney class by

wi(E):=ai,1ik.w_i(E):=a_i,\qquad 1\leq i\leq k.


Respecting Axioms

We are now in a position to show that: the Stiefel-Whitney classes defined above satisfies Axioms. In the previous part, we have verified the nontrivality, pullback of w1w_1 and Whitney product formula for line bundle.