# Solutions to exercises

## The Möbius strip as a line bundle

In everything that follows we consider the argument map on complex numbers as taking values in the half-open interval $[0,2\pi )$ , and the square root of a complex number $z\in \mathbb {C}$ to be

$z^{\frac {1}{2}}={\sqrt {|z|}}e^{i{\frac {\operatorname {arg} (z)}{2}}}.$ Consider now the unit circle as a subset $S^{1}=\{z\in \mathbb {C} \,|\,|z|=1\}$ of the complex numbers. The topologically obvious way to endow the Möbius strip with the structure of a real line bundle $L$ over $S^{1}$ is to take the direct product of the unit interval and the real line and to glue the two edges in reverse manner to each other, i.e. one defines the total space of $L$ to be

$[0,1]\times \mathbb {R} /\sim ,$ where the equivalence relation $\sim$ is generated by
$(0,y)\sim (1,-y).$ The projection onto $S^{1}$ is then given by the map $(x,y)\mapsto e^{i2\pi x}$ ; it is clear that this is well defined. However since this definition is so artificial it is unfortunately a bit annoying (though not too hard) to construct explicit local trivializations.

A more convenient way of thinking of the Möbius strip as a smooth real line bundle is slightly algebraic: the tautological line bundle over $\mathbb {RP} ^{1}\cong S^{1}$ is the Möbius strip. Inspired by this viewpoint we can define the total space

$L^{\frac {1}{2}}=\left\{(\zeta ,z)\in S^{1}\times \mathbb {C} \,|\,\zeta ^{\frac {1}{2}}z\in \mathbb {R} \right\}.$ The projection $L^{\frac {1}{2}}\to S^{1}$ is the obvious map and is clearly smooth. To see that this is indeed the Möbius strip one can either meditate on it for a while, or one can note that the map $([0,1]\times \mathbb {R} /\sim )\to L^{\frac {1}{2}}$ defined by
$(x,y)\mapsto \left(e^{i2\pi x},ye^{-i\pi x}\right)$ is a homeomorphism that induces an isomorphism of bundles $L\cong L^{\frac {1}{2}}$ . As suggested by the notation let us define another line bundle over $S^{1}$ given by the total space
$L^{1}=\left\{(\zeta ,z)\in S^{1}\times \mathbb {C} \,|\,\zeta z\in \mathbb {R} \right\}.$ After similar meditation as before it should be clear that this is the smooth line bundle corresponding to a strip with a full twist in it (two of the twists involved in the Möbius strip). In this framework it is surprisingly simple to see that $\left(L^{\frac {1}{2}}\right)^{\otimes 2}\cong L^{1}$ : identify fibres of $L^{\bullet }$ as
$L_{\zeta }^{\bullet }=\left\{z\in \mathbb {C} \,|\,\zeta ^{\bullet }z\in \mathbb {R} \right\},$ and define maps $\left(L_{\zeta }^{\frac {1}{2}}\right)^{\otimes 2}\to L_{\zeta }^{1}$ by $z\otimes w\mapsto zw$ . These maps are clearly smooth isomorphisms and extend to smooth isomorphisms of the full bundles. Finally it is also immediate that $L^{1}$ is the trivial line bundle over $S^{1}$ : extend notation further by considering the total space of the trivial bundle as
$L^{0}=\left\{(\zeta ,z)\in S^{1}\times \mathbb {C} \,|\,z\in \mathbb {R} \right\},$ and define maps $L_{\zeta }^{1}\to L_{\zeta }^{0}$ by $z\mapsto \zeta z$ . These are again clearly smooth isomorphisms and extend to give smooth isomorphisms of the bundles. More generally one can say that the tensor product of any real line bundle with itself is trivial. This can be seen by either noting that the transition functions for the tensor product are products (as real valued functions) of the original transition functions, by consulting Stiefel-Whitney classes, or by noting that $\mathbb {RP} ^{\infty }$ is a classifying space for real line bundles and that the composition $\mathbb {RP} ^{\infty }{\overset {\Delta }{\to }}\mathbb {RP} ^{\infty }\times \mathbb {RP} ^{\infty }{\overset {\mu }{\to }}\mathbb {RP} ^{\infty }$ is nullhomotopic, where $\mu$ is the map induced by tensor products.

Finally we can note that the Möbius strip can't be the trivial bundle simply because it is not homeomorphic to a cylinder. An easy way to see that the strip with two half-twists $L^{1}$ can't be deformed to $L^{0}$ in $\mathbb {R} ^{3}$ (even though we have seen them to be homeomorphic) is to note that the boundary of an actual (i.e. not open and not infinite) $L^{1}$ is the Hopf link, and an isotopy from $L^{1}$ to $L^{0}$ would give an ambient isotopy of the Hopf link to the trivial unlink, something that is clearly impossible by intuition or basic arguments involving Link invariants.

Such an isotopy is however possible in four dimensions. For notational convenience let $C$ be the open cylinder $[0,2\pi ]\times (0,1)/\sim$ where the equivalence is generated by $(0,y)\sim (2\pi ,y)$ and consider the embeddings: $i_{0},i_{1}:C\mapsto \mathbb {R} ^{4}$ , defined by

{\begin{aligned}i_{0}(\theta ,t)&=&(cos(\theta ),sin(\theta ),0,t)\\i_{1}(\theta ,t)&=&((1-sin(\theta )t)cos(\theta ),((1-sin(\theta )t)sin(\theta ),cos(\theta )t,0).\end{aligned}} Then $i_{0}$ is the standard cylinder, while $i_{1}$ is the full-twist Möbius band. The map $I:C\times [0,1]\mapsto \mathbb {R} ^{4}$ given by
$I(\theta ,t,s)=((1-sin(\theta )ts)cos(\theta ),((1-sin(\theta )ts)sin(\theta ),cos(\theta )ts,(1-s)t)$ defines an isotopy between the two embeddings.

Exercise 4.5

Show that the tangent bundle $T\mathbb {CP} ^{1}$ of the Riemann Sphere $\mathbb {CP} ^{1}$ is isomorphic to the line bundle ${\mathcal {O}}(2)$ .

A possible way to solve this exercise is using this two theorems on compact Riemann surfaces:

• Every line bundle on the complex projective space $\mathbb {CP} ^{n}$ is of the form ${\mathcal {O}}(d)$ , for some integer $d\in \mathbb {Z}$ , where $d=\deg(L).$ • One of the many consequences of Riemann-Roch Theorem is that $\deg(K_{C})=2g-2$ , where $K_{C}$ is the canonical bundle of $C$ , a compact Riemann Surface of genus $g$ .

If we believe to the two previous statement, the exercise is immediately solved: indeed, $\deg(T_{\mathbb {CP} ^{1}})=\deg(K_{\mathbb {CP} ^{1}}^{-1})=-\deg(K_{\mathbb {CP} ^{1}})=2$ , since $g(\mathbb {CP} ^{1})=0$ . Thus, $T_{\mathbb {CP} ^{1}}\cong {\mathcal {O}}(2)$ .

Exercise 4.6

In the lecture we defined the first Stiefel-Whitney Class in the following way:

1. Prove that this is a characteristic class, i.e. that it is functorial.
2. Calculate the first Stiefel-Whitney class for the tautological bundle over $\mathbb {R} P^{n}.$ For part a), recall the following fact: suppose we have a loop $\gamma :S^{1}\to B$ and a vector bundle $E$ over $B$ . Then the fibre of the loop reverses orientation if and only if $\Lambda ^{n}E$ is non-trivial when restricted to $\gamma .$ Suppose we have a vector bundle $\pi _{2}:E_{2}\to B_{2}$ and a map $f:B_{1}\to B_{2}.$ We now use this fact to prove the required equality, i.e.

$f^{*}w_{1}(E_{2})=w_{1}(f^{*}E_{2}).$ Let $\gamma _{1}$ be a loop in $B_{1}$ , an element of $H_{1}(B_{1},\mathbb {Z} )$ . Then, by definition of the first Stiefel-Whitney class, $w_{1}(f^{*}E_{2})(\gamma _{1})=0$ if and only if $\Lambda ^{n}f^{*}E_{2}$ is trivial when restricted to $\gamma _{1}$ .

On the other hand, $f^{*}w_{1}(E_{2})(\gamma _{1})=w_{1}(E_{2})(f_{*}\gamma _{1}),$ which again equals 0 if and only if $\Lambda ^{n}E_{2}$ is trivial when restricted to $f_{*}\gamma _{1}$ . It is now easy to see that these two conditions are equal directly from the definition of the pullback bundle, since $\Lambda ^{n}f^{*}E_{2}=f^{*}(\Lambda ^{n}E_{2}).$ For part b), we recall that $H^{1}(\mathbb {R} P^{n},\mathbb {Z} /2\mathbb {Z} )=\mathbb {Z} /2\mathbb {Z} .$ Hence we need to figure out whether $w_{1}({\mathcal {O}}_{\mathbb {R} P^{n}}(-1))\in H^{1}(\mathbb {R} P^{n},\mathbb {Z} /2\mathbb {Z} )$ is $0$ or not. We claim that it is the non-zero element.

Let us first try to answer the question for $\mathbb {R} P^{1}$ . We already know from a previous exercise that ${\mathcal {O}}_{\mathbb {R} P^{1}}(-1)$ is isomorphic to the Mobius band and, of course, that $\mathbb {R} P^{1}\cong S^{1}$ . But, the Mobius band is non-orientable. Hence, $w_{1}({\mathcal {O}}_{\mathbb {R} P^{1}}(-1))$ is non-zero.

Now for the general case, we note that linear algebra gives us an embedding $f\colon \mathbb {R} P^{1}\hookrightarrow \mathbb {R} P^{n}$ (just consider the lines in $\mathbb {R} ^{n}$ which lie in a fixed plane) and, since $\mathbb {R} P^{1}$ is isomorphic to $S^{1}$ , it actually defines a loop in $\mathbb {R} P^{n}.$ Since ${\mathcal {O}}_{\mathbb {R} P^{n}}(-1)|_{\mathbb {R} P^{1}}={\mathcal {O}}_{\mathbb {R} P^{1}}(-1)$ , we see that by restricting to $\mathbb {R} P^{1}$ , the first Stiefel-Whitney class of $\mathbb {R} P^{n}$ must, in fact, also be non-zero.