First Stiefel-Whitney class of a line bundle over non-orientable surface

Question

Let $\Sigma_g$ be the orientable surface of genus $g$ and $N_{g+1}$ be the non-orientable surface of genus $g+1$. Assume that $\Sigma_g$ is embedded inside $\mathbb{R}^{3}$ such that it admits the antipodal action. We can prove that the quotient of $\Sigma_g$ by the antipodal action is $N_{g+1}$. Let $X=\frac{\Sigma_g\times \mathbb{R}}{(x,t)\sim(-x,-t)}$. Observe that $\mathbb{R}\to X\to N_{g+1}$ is the line bundle.

I have started reading the book on Characteristic classes by Milnor and Stasheff. I am finding difficulties to compute the first Stiefel-Whitney class of this line bundle. Can anyone help me to understand this? If $g=0$, then it is clear that the first Stiefel-Whitney class is the generator of $H^1(\mathbb{R}P^2;\mathbb{Z}_2)\cong \mathbb{Z}_2$.

Answer 1

To save on typing, I'll write for $G$ for $\mathbb{Z}/2\mathbb{Z}$ for the duration of this post.

Let me start with a more general approach. Suppose $p:M\rightarrow N$ is a $2$-fold covering map, where $M$ and $N$ are any reasonably nice topological spaces.

Then one can form the space $E_p:=M\times \mathbb{R}/\sim$ where a point $(m,t)\in M\times \mathbb{R}$ is identified with $(m', -t)$, where $m'$ refers to the unique other point in $p^{-1}(p(m))$. The projection $p$ descends to map (which I'll still write as $p$) $p:E_p\rightarrow N$ giving $E_p$ the structure of a rank $1$ real vector bundle over $N$.

Proposition: Consider the bundle $q:S^\infty\rightarrow \mathbb{R}P^\infty$. Then $E_q = S^\infty\times \mathbb{R}/\sim$ is isomorphic to the tautological bundle over $\mathbb{R}P^\infty$.

Proof: Because $H^1(\mathbb{R}P^\infty;G)\cong G$, there are only two rank $1$ bundles over $\mathbb{R}P^\infty$. Thus, it is enough to show that this bundle is non-trivial. To that end, we consider the sphere bundle inside of it: $S^\infty \times S^0/\sim$. In the trivial bundle, this sphere bundle is disconnected, but in the case of our bundle, it is connected. Indeed, if $\gamma:[0,1]\rightarrow S^\infty$ is any path connecting antipodal points (meaning $-\gamma(0) = \gamma(1)$, then the path $\alpha:[0,1]\rightarrow S^\infty\times S^0/\sim$ given by $\alpha(t) = [(\gamma(t),1)]$ satisfies $\alpha(0) = [(\gamma(0),1)]$ while $\alpha(1) = [(\gamma(1),1)] = [(-\gamma(0),1)] = [(\gamma(0),-1)]$. $\square$

Returning to the more general case, $p$ induces a map $p_\ast: \pi_1(M)\rightarrow \pi_1(N)$. Since $p$ is a $2$-fold covering map, the image $p_\ast(\pi_1(M))\subseteq \pi_1(N)$ is index $2$. Thus, we have a map $\phi:\pi_1(N)\rightarrow \pi_1(N)/(p_\ast(\pi_1(M))) \cong G$.

Proposition. Under the identification $H^1(N;G)\cong \operatorname{Hom}(H_1(N);G)\cong \operatorname{Hom}(\pi_1(N);G)$, the map $\phi$ corresponds to $w_1(E_p\rightarrow N)$.

Proof. The bundle $q:S^\infty\rightarrow \mathbb{R}P^\infty$ is the universal $G$-bundle, so the covering $p:M\rightarrow N$ is classified by a map $i:N\rightarrow \mathbb{R}P^\infty$, so we obtain a commutative diagram \begin{array} A\mathbb{Z}/2\mathbb{Z} & {\longrightarrow} & \mathbb{Z}/2\mathbb{Z} \\ \downarrow{} & & \downarrow{} \\ M & \stackrel{j}{\longrightarrow} & S^\infty\\ \downarrow{} & & \downarrow{} \\ N& \stackrel{i}{\longrightarrow} & \mathbb{R}P^\infty \end{array}

Where the bottom vertical maps are bundle projections.

It's not hard to see that $j$ extends to a map $E_p\rightarrow E_q$, so $i$ pulls back $E_q$ to $E_p$. But from our first proposition, $E_q$ is the tautological bundle on $\mathbb{R}P^\infty$. S0, $i$ not only classifies the double cover, but also the vector bundle $E_p$. In particular, the map $i^\ast:H^1(\mathbb{R}P^\infty;G)\rightarrow H^1(N;G)$ has the property that $i^\ast(z) = w_1(E_p\rightarrow N)$, where $z$ generates $H^1(\mathbb{R}P^\infty;G)$. We must now determine $i^\ast(z)$.

Returning to the diagram of fibrations above, we obtain long exact sequence in homotopy groups, a portion of which looks like \begin{array} A0 & \longrightarrow & \pi_1(M) &\longrightarrow & \pi_1(N) & \stackrel{\phi}{\longrightarrow} &\mathbb{Z}/2\mathbb{Z} &\longrightarrow & 0 \\ \downarrow & & \downarrow & & \downarrow{i_\ast} && \downarrow{Id} && \downarrow\\ 0 & \longrightarrow & 0 & \longrightarrow & \mathbb{Z}/2\mathbb{Z} & \stackrel{Id}{\longrightarrow} & \mathbb{Z}/2\mathbb{Z} & \longrightarrow & 0 \end{array}

Commutativity tells us that $Id\circ \phi = Id \circ i_\ast$, so $\phi = i_\ast$. As $H_1$ is the abelianization of $\pi_1$, it also follows that $\phi = i_\ast$ when considered as maps on $H_1$.

Finally, we turn to the universal coefficients theorem for cohomlogy. We obtain a commutative diagram \begin{array} AH^1(\mathbb{R}P^\infty;G) & \stackrel{i^\ast}{\longrightarrow} & H^1(N;G) \\ \downarrow{h} & & \downarrow{h} \\ \operatorname{Hom}(H_1(\mathbb{R}P^\infty;G) & \stackrel{-\circ i_\ast}{\longrightarrow} & \operatorname{Hom}(H_1(N);G) \end{array}

The Ext terms vanish since $H_0$ is free, so the vertical maps in this diagram are isomorphisms. In addition, since $i_\ast = \phi$ on $H_1$, the bottom map can be replaced with $-\circ \phi$.

Then $h(w_1(E_p\rightarrow N)) = h(i^\ast(z)) =h(z) \circ \phi$. But $h(z)$ is the non-trivial element of $\operatorname{Hom}(H_1(\mathbb{R}P^\infty),G)$, i.e., the non-trivial element of $\operatorname{Hom}(G,G)\cong G$, i.e., $h(z) = Id$ Thus, we have identified $w_1(E_p\rightarrow N)$ with $\phi$. $\square$

So, what does this all mean for your specific example? Well, if we pick a simple curve connecting a point on $\Sigma_g$ to its antipodal point, the projection into $N_{g+1}$ gives a closed curve whose whose image in $H_1(N)$ projects to a non-trivial element of $G$ via $\phi$. Then $w_1$ is the cohomology class dual to this curve.

Answer 2

The final answer to your question is $w_1(X) = w_1(TN_{g+1})$. This follows from the isomorphism of real line bundles $X \cong \det(TN_{g+1})$ which is established via the following steps:

1. A real line bundle $L_p$ is associated to a double cover $p : M \to N$.
2. If $p$ is the orientable double cover of a smooth manifold $N$, then it is shown that $L_p \cong \det(TN)$ - this is Proposition $1$.
3. The bundle $E_p$ as in Jason DeVito's answer is considered.
4. Using Lemmata $1$ and $2$, it is shown that $E_p \cong L_p$ - this is Proposition $2$.
5. Propositions $1$ and $2$ are combined to conclude that $X = E_p \cong L_p \cong \det(TN_{g+1})$.

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First note that there is a bijective correspondence between isomorphism classes of double covers of a space $N$ and isomorphism classes of real line bundles on $N$. In one direction, we view a double cover $p : M \to N$ as a principal $\mathbb{Z}_2$-bundle (the $\mathbb{Z}_2$ action interchanges the points in each fiber), so we can use the natural representation $\mathbb{Z}_2 \to GL(1, \mathbb{R}^*)$ to form the associated line bundle, which we denote by $\pi : L_p \to N$. In the opposite direction, given a line bundle $\pi : L \to N$, there is an associated double cover given by $p : S(L) \to N$, where $S(L)$ denotes the sphere bundle of $L$ (the elements of norm $1$ with respect to a fixed bundle metric) and $p = \pi|_{S(L)}$. Equivalently, we could instead take the orthonormal frame bundle of $L$ with respect to a fixed bundle metric - this description makes it clear that the two directions are inverses of one another, but the former description will be more useful for what follows.

By this correspondence, a double cover $p : M \to N$ is isomorphic to the double cover $p : S(L_p) \to N$; in particular, $p$ is trivial if and only if $L_p$ is trivial. In general, this description can be used to show that the bundle $p^*L_p \to S(L_p)$ is trivial. To see this, note that

$$p^*L_p = \{(v, w) \in S(L_p)\times L_p \mid p(v) = \pi(w)\} = \{(v, w) \in S(L_p)\times L_p \mid \pi(v) = \pi(w)\}$$

where we use that fact that $p = \pi|_{S(L_p)}$. The map $p^*L_p \to S(L_p)$ is given by projection onto the first factor, so $\sigma : S(L_p) \to p^*L_p$ defined by $\sigma(v) = (v, v)$ is a nowhere-zero section of $p^*L_p$ and hence $p^*L_p$ is trivial. This property can be used to characterise $L_p$.

Lemma $1$: Let $p : M \to N$ be a double cover where $N$ is connected, and let $\ell$ be a real line bundle on $N$. Then $p^*\ell$ is trivial if and only if $\ell$ is trivial or $\ell \cong L_p$.

Proof: The Gysin sequence for $p : M \to N$ reads

$$\dots \to H^0(N; \mathbb{Z}_2) \xrightarrow{\cup\ w_1(L_p)} H^1(N; \mathbb{Z}_2) \xrightarrow{p^*} H^1(M; \mathbb{Z}_2) \to \dots \tag{$\ast$}$$

Note that $p^*\ell$ is trivial if and only if $w_1(p^*\ell) = 0$. As $w_1(p^*\ell) = p^*w_1(\ell)$, it follows from exactness that vanishing occurs if and only if $w_1(\ell) \in \operatorname{span}\{w_1(L_p)\}$. So either $w_1(\ell) = 0$, in which case $\ell$ is trivial, or $w_1(\ell) = w_1(L_p)$, in which case $\ell \cong L_p$. $\square$

The connectedness assumption is necessary in the above statement. In the disconnected case, the line bundle $\ell$ could be trivial on some connected components, and isomorphic to $L_p$ on others - unless $p$ is the trivial covering, then we need not have $\ell \cong L_p$. If $p$ and $\ell$ are non-trivial on each connected component, then $p^*\ell$ is trivial if and only if $\ell \cong L_p$.

Proposition $1$: Let $p : M \to N$ be the orientation double cover of a connected smooth manifold $N$. Then $L_p \cong \det(TN)$.

Proof: If $p: M \to N$ is a smooth double covering, then $TM \cong p^*TN$, so $0 = w_1(TM) = w_1(p^*TN) = p^*w_1(TN)$. Since $w_1(TN)$ is in the kernel of $p^*$, it follows from the Gysin sequence $(\ast)$ that $w_1(TN) \in \operatorname{span}\{w_1(L_p)\}$. If $w_1(TN) = 0$, then $N$ is orientable so $p$ is trivial and hence so is $L_p$. As $w_1(\det(TN)) = w_1(TN) = 0$, we see that $\det(TN)$ is also trivial and hence $L_p \cong \det(TN)$. On the other hand, if $w_1(TN) = w_1(L_p)$, then $w_1(L_p) = w_1(TN) = w_1(\det(TN))$, so $L_p \cong \det(TN)$. $\square$

Let $p : M \to N$ be a double cover and let $\kappa$ denote the non-trivial deck transformation. Consider the line bundle $\rho: E_p \to N$ defined in Jason DeVito's answer where $E_p = (M\times\mathbb{R})/\sim$ where $(m, t) \sim (\kappa(m), -t)$, and $\rho$ is the map induced by projection onto the first factor, i.e. $\rho([(m, t)]) = p(m)$.

Lemma $2$: If $p : M\to N$ is a double covering with $N$ connected, then $E_p$ is trivial if and only if $p$ is.

Proof: If $p$ is the trivial double covering, it admits a section $\varphi : N \to M$. Then $\sigma : N \to E_p$ given by $\sigma(n) = [(\varphi(n), 1)]$ is a nowhere-zero section of $E_p$, and hence $E_p$ is trivial.

If $p$ is non-trivial, consider $S(E_p) = (M\times S^0)/\sim$, the sphere bundle of $E_p$. As $p$ is non-trivial, $M$ is path-connected, so for any $m \in M$, there is a path $\gamma : [0, 1] \to M$ with $\gamma(0) = m$ and $\gamma(1) = \kappa(m)$. Then the path $\alpha : [0, 1] \to S(E_p)$ given by $\alpha(t) = [(\gamma(t), 1)]$ has endpoints $\alpha(0) = [(\gamma(0), 1)] = [(m, 1)]$ and $\alpha(1) = [(\gamma(1), 1)] = [(\kappa(m), 1)] = [(m, -1)]$. Note that $[(m, 1)]$ and $[(m, -1)]$ are the two preimages of $p(m)$ under the map $S(E_p) \to N$. As they are connected by a path, we see that $S(E_p) \to N$ is a non-trivial double cover, and hence $E_p$ is not trivial. $\square$

Given a double covering $p : M \to N$, we now have two real line bundles, $L_p$ and $E_p$, which are trivial if and only if $p$ is trivial. It is natural to ask whether they are related.

Proposition $2$: If $p : M \to N$ is a double covering with $N$ connected, then $L_p \cong E_p$.

Proof: If $p$ is trivial, then both $L_p$ and $E_p$ are trivial, so $L_p\cong E_p$. Suppose now that $p$ is a non-trivial double covering. Note that

\begin{align*} p^*E_p &= \{(m', [(m, t)]) \in M\times E_p \mid p(m') = \rho([(m, t)])\}\\ &= \{(m', [(m, t)]) \in M\times E_p \mid p(m') = p(m)\}\end{align*}

where the map $p^*E_p \to M$ is given by projection onto the first factor. Then $\sigma: M \to p^*E_p$ given by $\sigma(m) = (m, [(m, 1)])$ is a nowhere-zero section of $p^*E_p$ and hence $p^*E_p$ is trivial. As $p$ is a non-trivial covering, the line bundle $E_p$ is non-trivial by Lemma $2$, so $E_p \cong L_p$ by Lemma $1$. $\square$

For the specific case you asked about, namely the smooth orientable double cover $p : \Sigma_g \to N_{g+1}$, the bundle you denoted $X$, which above is denoted $E_p$, is isomorphic to $L_p$ by Proposition $2$, which in turn is isomorphic to $\det(TN_{g+1})$ by Proposition $1$. In particular, $w_1(X) = w_1(\det(TN_{g+1})) = w_1(TN_{g+1})$.