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I need help for solving Ex. 7C from 'Characteristic Classes' by Milnor/Stasheff: The exercise asks to find a formula for the (total) Stiefel-Whitney class of $\xi^m\otimes\eta^n$ over a paracompact base space in terms of the the SW classes of $\xi^m$ and $\eta^n$. (superscript for dimension of the vector bundle). The book tells me to start off by

  1. computing it for $n=m=1$ as $w_1(\xi^1\otimes\eta^1)=w_1(\xi^1)+w_1(\eta^1)$, where I'm already struggling. My guess would be to use the uniqueness of SW classes over paracompact spaces, but I don't know how to exactly

  2. Compute cohomology of $G_m\times G_n$ by the Künneth formula. This is more or less clear, but I don't understand the role that it plays for the whole exercise

  3. As hint: We can first compute the formula in the special case that both bundles are Whitney sums of line bundles

  4. Establish the formula $w(\xi^m\otimes\eta^n)=p_{m,n}(w_1(\xi^m),\dots,w_m(\xi^m),w_1(\eta^n),\dots,w_n(\eta^n))$ with $p_{m,n}$ being characterized as $p_{m,n}(\sigma_1,\dots,\sigma_m,\tau_1,\dots,\tau_n)=\prod_i \prod_j (1+s_i+t_j)$, where $\sigma_i$ are the elementary symmetric functions associated to $s_1,\dots,s_m$ and likewise $\tau_j$ and $t_1\dots,t_n$

Can someone please provide stepwise guidance as to how to solve this problem, i.e more or less take me by the hand.

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1 Answer 1

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I provided a solution to this exercise in my note here; I have copied the proof below. Note, this may not be the exact proof Milnor had in mind.


Lemma: Let $L_1$ and $L_2$ be real line bundles over a paracompact space $B$. Then $w_1(L_1\otimes L_2) = w_1(L_1) + w_1(L_2)$.

Proof: Let $\pi_i : \mathbb{RP}^{\infty}\times\mathbb{RP}^{\infty}$ denote projection onto the $i^{\text{th}}$ factor and let $\mu : \mathbb{RP}^{\infty}\times\mathbb{RP}^{\infty} \to \mathbb{RP}^{\infty}$ be a classifying map for $\pi_1^*\gamma\otimes\pi_2^*\gamma$. By the Künneth theorem, $\pi_1^*w_1(\gamma)$ and $\pi_2^*w_1(\gamma)$ form a basis for $H^1(\mathbb{RP}^{\infty}\times\mathbb{RP}^{\infty}; \mathbb{Z}_2)$, so $w_1(\pi_1^*\gamma\otimes\pi_2^*\gamma) = a\pi_1^*w_1(\gamma) + b\pi_2^*w_1(\gamma)$ for some $a, b \in \mathbb{Z}_2$.

If $\sigma : \mathbb{RP}^{\infty}\times\mathbb{RP}^{\infty} \to \mathbb{RP}^{\infty}\times\mathbb{RP}^{\infty}$ is the map which interchanges factors, then $\pi_1\circ\sigma = \pi_2$ and $\pi_2\circ\sigma = \pi_1$, so $\sigma^*\mu^*w_1(\gamma) = a\pi_2^*w_1(\gamma) + b\pi_1^*w_1(\gamma)$, but $\sigma\circ\mu$ classifies $\pi_2^*\gamma\otimes\pi_1^*\gamma \cong \pi_1^*\gamma\otimes\pi_2^*\gamma$ so $\sigma\circ\mu$ is homotopic to $\mu$. Therefore

$$a\pi_2^*w_1(\gamma) + b\pi_1^*w_1(\gamma) = (\sigma\circ\mu)^*w_1(\gamma) = \mu^*w_1(\gamma) = a\pi_1^*w_1(\gamma) + b\pi_2^*w_1(\gamma),$$

which implies $a = b$. So either $w_1(\pi_1^*\gamma\otimes\pi_2^*\gamma) = \pi_1^*w_1(\gamma) + \pi_2^*w_1(\gamma)$, or $w_1(\pi_1^*\gamma\otimes\pi_2^*\gamma) = 0$.

Now let $f_i : B \to \mathbb{RP}^{\infty}$ be a classifying map for $L_i$. Then

\begin{align*} (f_1, f_2)^*(\pi_1^*\gamma\otimes\pi_2^*\gamma) &\cong ((f_1, f_2)^*\pi_1^*\gamma)\otimes((f_1, f_2)^*\pi_2^*\gamma)\\ &\cong (\pi_1\circ(f_1, f_2))^*\gamma\otimes (\pi_2\circ(f_1, f_2)^*)\gamma\\ &\cong f_1^*\gamma\otimes f_2^*\gamma\\ &\cong L_1\otimes L_2. \end{align*}

As $w_1(L_1\otimes L_2) = w_1((f_1, f_2)^*(\pi_1^*\gamma\otimes\pi_2^*\gamma) = (f_1, f_2)^*w_1(\pi_1^*\gamma\otimes\pi_2^*\gamma)$, if $w_1(\pi_1^*\gamma\otimes\pi_2^*\gamma) = 0$, then $w_1(L_1\otimes L_2) = 0$. This is clearly false, just take $L_1$ to be non-trivial and $L_2$ to be trivial. Therefore $w_1(\pi_1^*\gamma\otimes\pi_2^*\gamma) = \pi_1^*w_1(\gamma) + \pi_2^*w_1(\gamma)$ and so

\begin{align*} w_1(L_1\otimes L_2) &= (f_1, f_2)^*w_1(\pi_1^*\gamma\otimes\pi_2^*\gamma)\\ &= (f_1, f_2)^*(\pi_1^*w_1(\gamma) + \pi_2^*w_1(\gamma))\\ &= (f_1, f_2)^*\pi_1^*w_1(\gamma) + (f_1, f_2)^*\pi_2^*w_1(\gamma)\\ &= (\pi_1\circ(f_1, f_2))^*w_1(\gamma) + (\pi_2\circ(f_1, f_2))^*w_1(\gamma)\\ &= f_1^*w_1(\gamma) + f_2^*w_1(\gamma)\\ &= w_1(f_1^*\gamma) + w_1(f_2^*\gamma)\\ &= w_1(L_1) + w_1(L_2). \end{align*}

$$\tag*{$\square$}$$

With this lemma in hand, we can move on to the general case thanks to the splitting principle.

Theorem: Let $E$ and $F$ be real vector bundles over a paracompact space $B$. Let $m = \operatorname{rank} E$ and $n = \operatorname{rank} F$. Then $w(E\otimes F) = p_{m, n}(w_1(E), \dots, w_m(E), w_1(F), \dots, w_n(F))$ where $p_{m, n}$ is the unique polynomial which satisfies

$$p_{m, n}(\sigma_1, \dots, \sigma_m, \tau_1, \dots, \tau_n) = \prod_{i=1}^m\prod_{j=1}^n(1 + x_i + y_j)$$

where $\sigma_k = \sigma_k(x_1, \dots, x_m)$ and $\tau_k = \tau_k(y_1, \dots, y_n)$ are the $k^{\text{th}}$ elementary symmetric polynomials in $m$ and $n$ variables respectively.

Proof: By the splitting principle, there is a paracompact space $Y$ and a map $g : Y \to B$ such that $g^*E \cong \ell_1'\oplus\dots\oplus\ell_m'$ and $g^* : H^*(Y; \mathbb{Z}_2) \to H^*(B; \mathbb{Z}_2)$ is injective. Again by the splitting principle, there is a paracompact space $X$ and a map $f : X \to Y$ such that $f^*g^*F \cong \eta_1\oplus\dots\oplus\eta_n$, and $f^* : H^*(X; \mathbb{Z}_2) \to H^*(Y; \mathbb{Z}_2)$ is injective. Letting $\ell_i = f^*\ell_i'$, we have $f^*g^*E \cong \ell_1\oplus\dots\oplus\ell_m$. So

$$f^*g^*(E\otimes F) \cong (f^*g^*E)\otimes(f^*g^*F) \cong (\ell_1\oplus\dots\oplus\ell_m)\otimes(\eta_1\oplus\dots\oplus\eta_n) \cong \bigoplus_{i=1}^m\bigoplus_{j=1}^n\ell_i\otimes\eta_j.$$

Therefore,

\begin{align*} w(f^*g^*(E\otimes F)) &= w\left(\bigoplus_{i=1}^m\bigoplus_{j=1}^n\ell_i\otimes\eta_j\right)\\ &= \prod_{i=1}^m\prod_{j=1}^nw(\ell_i\otimes\eta_j)\\ &= \prod_{i=1}^m\prod_{j=1}^n(1 + w_1(\ell_i\otimes\eta_j))\\ &= \prod_{i=1}^m\prod_{j=1}^n(1 + w_1(\ell_i) + w_1(\eta_j))\\ &= \prod_{i=1}^m\prod_{j=1}^n(1 + x_i + y_j) \end{align*}

where the penultimate equality uses the lemma and $x_i := w_1(\ell_i)$, $y_j := w_1(\eta_j)$.

Denote the final expression above by $q(x_1, \dots, x_m, y_1, \dots, y_n)$. Note that $q$ is a polynomial which is symmetric in the $x_i$ and the $y_j$ separately, so by the fundamental theorem of symmetric polynomials, there is a unique polynomial $p_{m,n}$ such that

$$q(x_1, \dots, x_m, y_1, \dots, y_m) = p_{m,n}(\sigma_1, \dots, \sigma_m, \tau_1, \dots, \tau_n).$$

Now note that $\sigma_i(x_1, \dots, x_m) = w_i(\ell_1\oplus\dots\oplus\ell_m) = w_i(f^*g^*E) = f^*g^*w_i(E)$ and likewise $\tau_j(y_1, \dots, y_n) = f^*g^*w_j(F)$, so

\begin{align*} f^*g^*w(E\otimes F) &= w(f^*g^*(E\otimes F))\\ &= q(x_1, \dots, x_m, y_1, \dots, y_n)\\ &= p_{m,n}(\sigma_1, \dots, \sigma_m, \tau_1, \dots, \tau_n)\\ &= p_{m,n}(f^*g^*w_1(E), \dots, f^*g^*w_m(E), f^*g^*w_1(F), \dots, f^*g^*w_n(F))\\ &= f^*g^*p_{m,n}(w_1(E), \dots, w_m(E), w_1(F), \dots, w_n(F)). \end{align*}

By the injectivity of $f^*$ and $g^*$, we have $w(E\otimes F) = p_{m,n}(w_1(E), \dots, w_m(E), w_1(F), \dots, w_n(F))$.

$$\tag*{$\square$}$$

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