Computing $\Omega_1^{\text{Spin}}\cong \Bbb Z_2$

Question

I'm trying to understand why $\Omega_1^{\text{Spin}}\cong \Bbb Z_2$. I know it's a pretty standard computations but I'd like to have an explicit description (and explanation) of what's going on.

As far as I understood, we are considering framing of 3 dimensional vector bundles, since $\text{Spin}(n)$ is the double cover of $SO(n)$ for $n\geq 3$. In order to identify the bounding spin structure on $S^1$ I first consider a framing of $TD^2\oplus \Bbb R$ and think of it as a section of the frame bundle $F\to D^2$, then we consider the double cover $\tilde{F}\xrightarrow{\cdot 2} F$ and see what's the preimage of the induced framing on $S^1$. Following what's explained in this question it's somehow clear that the induced framing on $S^1$ is lifted to the connected double cover of $S^1$. In fact the framing induced on the boundary is given by $S^1\to \Bbb C^2$ $s\mapsto (s, is)$ (since it's isomorphic to the constant one on $D^2$ restricted to $S^1=\partial D^2$)

Hence this is the bounding spin structure on $S^1$.

What's unclear to me is how to prove that the disconnected double cover of $S^1$ is the non-bounding spin structure of $S^1$. Can someone explain it to me or point some reference where it's proven?

Answer 1

I've been wanting to see details for this for a couple of years, because they state it in Lawson-Michelson but give no details. I think I finally came up with an argument: we can see that the disconnected spin structure on $Fr(TS^1)\cong S^1$ does not bound by showing that the connected spin structure is the ONLY one that bounds. Specifically you can show that if $\Sigma_{g,1}$ is an oriented surface of genus $g\geq 0$ with $1$ boundary component, then for any $g$ and any spin structure on $Fr(T\Sigma_{g,1})$ the induced spin structures on $S^1\cong \partial \Sigma_{g,1}$ are all the same, and therefore the other spin structure must not bound. In particular if $g =0$ then $\Sigma_{0,1} \cong D^2$ and your question assumes that we've already established that the induced spin structure is the connected one.

Some generalities

For me a Spin structure on a principal $SO(n)$ bundle $P_{SO}$ over $X$ is a principal $Spin(n)$ bundle $P_{Spin}$ over $X$ and a double cover $P_{Spin} \to P_{SO}$ which is equivariant with respect to the usual double cover $Spin(n) \to SO(n)$.

Note that $Spin(n)$ is the pullback of $Spin(n+1)$ along the inclusion $\iota\colon SO(n) \to SO(n+1)$, so if $E$ is a principal $SO(n)$ bundle over $X$ which has an equivariant embedding $\tilde{f}\colon E \to \bar{E}$ into a principal $SO(n+1)$ bundle over $Y$ covering a map $f\colon X\to Y$, then a spin structure $\bar{P}\to\bar{E}$ pulls back to a spin structure $P = \tilde{f}^*\bar{P}$ on $E$.

If $E\to X$ is a principal $SO(n)$ bundle for some $n$, let $\mathfrak{Spin}(E)$ denote the set of isomorphism classes of spin structures on $E$. The classification of spin structures says that if $\mathfrak{Spin}(E)$ is non-empty then it admits a free, transitive action of the cohomology group $H^1(X;\mathbb{Z}/2)$, and if $\tilde{f}\colon E\to \bar{E}$ is an embedding of principal bundles as above then it induces a function $$ \tilde{f}^*\colon\mathfrak{Spin}(\bar{E}) \to \mathfrak{Spin}(E) $$

which is equivariant wrt the map $f^*\colon H^1(Y;\mathbb{Z}/2) \to H^1(X;\mathbb{Z}/2)$.

For example if $M$ is an oriented $n$-manifold and $N = \partial M$ is oriented according to some convention, then we get a canonical embedding of frame bundles $Fr(TN) \to Fr(TM)$, as follows. Over a point $x\in N$ the fibre $Fr(TN)_x$ is the space $Iso(\mathbb{R}^{n-1}, T_xN)$ of linear isomorphisms with determinant $1$ from standard euclidean space to the tangent space at $x$, and analogously for $M$. Given an $x\in N$ and a $\varphi\in Fr(TN)_x$ we can define an isomorphism $\tilde{\varphi}\colon \mathbb{R}^n \to T_xM\cong T_xN\oplus\mathbb{R}$ via $\tilde{\varphi}(e_i) = \varphi(e_i)$ for $i< n$, and set $\tilde{\varphi}(e_n)$ as the unique $v$ in $T_xM$ so that $\{\varphi(e_1),\dots,\varphi(e_{n-1}), v\}$ is an orthogonal basis with positive orientation. This embedding is used to define a spin structure on $TN$ from one on $TM$. The map $H^1(M;\mathbb{Z}/2) \to H^1(N;\mathbb{Z}/2)$ is often computable for manifolds, so we can understand a lot about $\mathfrak{Spin}(TM) \to \mathfrak{Spin}(TN)$.

Our specific case:

An embedding $S^1 \to \Sigma_{g,1}$ as the oriented boundary induces an embedding $Fr(TS^1) \to Fr(T\Sigma_{g,1})$ of a principal $SO(1)\cong 1$ bundle into a principal $SO(2)\cong S^1$ bundle. As $g$ varies any choice of collar $U_g$ of $\partial\Sigma_{g,1}$ will be diffeomorphic, and by choosing particular models (as submanifolds of $\mathbb{R}^3$ for example) we can pick a single $U$ which is a collar for all $g$. If we denote by $f_g$ the inclusion of $U$ into $\Sigma_{g,1}$ then the induced map on cohomology

$$ f_g^* \colon H^1(\Sigma_{g,1};\mathbb{Z}/2) \to H^1(U;\mathbb{Z}/2) \cong H^1(S^1;\mathbb{Z}/2)$$

will be zero (this is essentially because the attaching map of the top cell for the closed surface $\Sigma_g$ crosses each $1$-cell twice, once in each direction). Since $U\subset \Sigma_{g,1}$ has co-dimension $0$ then $Fr(TU) = Fr(T\Sigma_{g,1})|_U$ i.e. we get an embedding of frame bundles for free, and by equivariance of the $H^1$ actions the map on spin structures

$$ \mathfrak{Spin}(T\Sigma_{g, 1}) \to \mathfrak{Spin}(TU) $$

is constant. The manifold $\Sigma_{g,1}$ is parallelizable for every $g$, so each manifold admits in particular a trivial spin structure $\Sigma_{g,1} \times Spin(2)$ which induces the trivial spin structure on $Fr(TU)$; therefore every spin structure on $Fr(T\Sigma_{g,1})$ induces the trivial one on $Fr(TU)$. As a result, any bounding spin structure on $Fr(TS^1)$ must be induced by the trivial spin structure on the collar. By embedding $U$ into $\mathbb{R}^2$ as $D^2\setminus\{0\}$, we see that it is the same as the spin structure induced by the disk, and it is assumed we already understand this to be the connected spin structure.