On the Number of Holes in a Common Drinking Straw

Sketch. The map $p\colon\mathbb{R}\to S^1,\ t\mapsto e^{2\pi i t}$ is the universal covering, with fibre $\mathbb{Z}$. The path-lifting property gives a unique lift $\widetilde\ell$ of each loop $\ell\colon[0,1]\to S^1$ with $\widetilde\ell(0)=0$, and the difference $\widetilde\ell(1)-\widetilde\ell(0)\in\mathbb{Z}$ — the winding number — depends only on the homotopy class of $\ell$. The resulting map $\pi_1(S^1)\to\mathbb{Z}$ is a bijection.

For $n\geq 2$, the covering map induces an isomorphism $\pi_n(\mathbb{R})\xrightarrow{\sim}\pi_n(S^1)$, and $\pi_n(\mathbb{R})=0$ because $\mathbb{R}$ is contractible. Hence $\pi_1$ has a single free generator (the loop $\gamma$ winding once through the bore) and all higher $\pi_n$ vanish — one hole's worth of loops.▢

Sketch. Equip $S^1$ with the minimal CW structure: one $0$-cell $e^0$ and one $1$-cell $e^1$ with both endpoints attached to $e^0$. The cellular chain complex is $$0\longrightarrow \mathbb{Z}\langle e^1\rangle \xrightarrow{\;\partial\;} \mathbb{Z}\langle e^0\rangle\longrightarrow 0,$$ and $\partial e^1=e^0-e^0=0$ since the two attaching endpoints cancel with opposite orientations. The homology of this complex is $\mathbb{Z}$ in degrees $0$ and $1$ and zero elsewhere. The single $\mathbb{Z}$ in degree one is generated by the fundamental class $[\gamma]$ of the core circle — the unique non-bounding $1$-cycle — and this is the hole.▢

Sketch. Since $H_*(X;\mathbb{Z})$ is free, the Universal Coefficient Theorem gives $H^n(X;\mathbb{Z})\cong\mathrm{Hom}(H_n(X;\mathbb{Z}),\mathbb{Z})$, so $H^0=H^1=\mathbb{Z}$ and $H^n=0$ for $n\geq 2$. For the ring structure, the cup product $\smile\colon H^1\otimes H^1\to H^2$ lands in the zero group, hence $\alpha\cup\alpha=0$ for any $\alpha\in H^1$. Therefore $H^*$ is generated by the single class $\alpha$ dual to $[\gamma]$, with $\alpha^2=0$ — one generator, one hole, no further structure.▢

Sketch. The angular one-form $d\theta$ is globally well-defined on $S^1\cong\mathbb{R}/2\pi\mathbb{Z}$ even though $\theta$ is only a local coordinate, and it is closed because every $1$-form on a $1$-manifold is. It is not exact: if $d\theta=df$ for some smooth $f$, then by the fundamental theorem of calculus $\oint_\gamma d\theta=0$, contradicting the explicit computation $\oint_\gamma d\theta=2\pi$.

By the de Rham theorem, $H^*_{\mathrm{dR}}(X;\mathbb{R})\cong H^*(X;\mathbb{R})$, so $H^1_{\mathrm{dR}}$ has rank one and must be generated by $[d\theta]$. The class $[d\theta]$ is precisely the obstruction to a closed $1$-form on the straw being exact, and this obstruction is the hole.▢

Sketch. The group $\mathbb{Z}$ admits the short free resolution $$0\longrightarrow \mathbb{Z}[\mathbb{Z}] \xrightarrow{\,t-1\,} \mathbb{Z}[\mathbb{Z}] \xrightarrow{\;\varepsilon\;} \mathbb{Z}\longrightarrow 0$$ as $\mathbb{Z}[\mathbb{Z}]=\mathbb{Z}[t,t^{-1}]$-modules, where $\varepsilon$ is the augmentation. Applying $-\otimes_{\mathbb{Z}[\mathbb{Z}]}\mathbb{Z}$ collapses $t-1$ to the zero map, leaving $\mathbb{Z}\xrightarrow{0}\mathbb{Z}$ and so $H_0(\mathbb{Z};\mathbb{Z})=H_1(\mathbb{Z};\mathbb{Z})=\mathbb{Z}$, $H_n=0$ otherwise. Equivalently, $S^1=B\mathbb{Z}$, so the group (co)homology of $\mathbb{Z}$ is, by definition, the (co)homology of $S^1$. The hole of the straw is the rank-one freeness of the integers as a group.▢

Sketch. A complex vector bundle on $S^1$ is determined by its clutching function, an element of $\pi_0\, GL_n(\mathbb{C})$; since $GL_n(\mathbb{C})$ deformation-retracts onto $U(n)$ and $\pi_0\,U(n)=0$ for all $n$, every complex bundle on $S^1$ is trivial. Hence $\widetilde K^{\,0}(X)=0$ and $K^0(X)=\mathbb{Z}$ records rank only.

For the odd part, $K^1(X)=[X,U]_*=\pi_1(U)=\mathbb{Z}$ by the Bott periodicity theorem [Bot59]; the generator is the clutching function $z\mapsto z$, which represents the same loop as the core circle $\gamma$. The single $\mathbb{Z}$ in $K^1$ is the hole, this time recorded as a one-parameter family of inequivalent automorphisms over the bore.▢

Sketch. Real $n$-plane bundles on $S^1$ are classified by their clutching function $S^0\to O(n)$, i.e. by $\pi_0\, O(n)=\mathbb{Z}/2$ — the orientation component. The trivial element gives the trivial bundle; the nontrivial element gives the (rank-$n$) Möbius bundle. Stabilising in $n$ yields $\widetilde{KO}^{\,0}(S^1)=\mathbb{Z}/2$. Equivalently, by Bott periodicity for $KO$ (period $8$, with $\pi_*KO=\mathbb{Z},\mathbb{Z}/2,\mathbb{Z}/2,0,\mathbb{Z},0,0,0$), one has $\widetilde{KO}^{\,0}(S^1)=KO^{-1}(\mathrm{pt})=\pi_1(KO)=\mathbb{Z}/2$.

The group is cyclic on one generator: the same hole, but with strictly more information attached than complex K-theory or singular cohomology can carry. In particular, the straw and the Möbius strip — homotopy-equivalent and so indistinguishable to $H_*$ and $K^*$ — represent the trivial and nontrivial classes respectively, certifying the cylinder under consideration as a bona fide untwisted straw.▢

Sketch. The Atiyah–Hirzebruch spectral sequence for the generalised homology theory $\Omega_*^{SO}$ has $E^2_{p,q}=H_p(X;\Omega_q^{SO}(\mathrm{pt}))\Rightarrow\Omega_{p+q}^{SO}(X)$ [Sto68, Ch. VI], [MS74, App. A]. With $\Omega_0^{SO}=\mathbb{Z}$ and $\Omega_1^{SO}=0$, the relevant entries are $E^2_{1,0}=H_1(S^1;\mathbb{Z})=\mathbb{Z}$ and $E^2_{0,1}=0$; no differentials are possible into or out of these positions in total degree $\leq 1$, so $\Omega_1^{SO}(X)=\mathbb{Z}$.

A geometric representative for the generator is the inclusion $\gamma\colon S^1\hookrightarrow X$ of the core circle; the bordism class of a map $f\colon M^1\to S^1$ from a closed oriented $1$-manifold is detected by its total degree, $\Omega_1^{SO}(S^1)\xrightarrow{\sim}\mathbb{Z}$. The single $\mathbb{Z}$ is the hole, this time witnessed by oriented manifolds mapping into the bore.▢

Sketch. The suspension isomorphism $\widetilde\pi_n^{\,s}(\Sigma Y)\cong\widetilde\pi_{n-1}^{\,s}(Y)$ is the defining shift property of the sphere spectrum [Ada74, §3]. Since $S^1\cong \Sigma S^0$, we obtain $\widetilde\pi_n^{\,s}(S^1)\cong\widetilde\pi_{n-1}^{\,s}(S^0)=\pi_{n-1}^{\,s}$, the $(n{-}1)$th stable stem. The first values are classical computations via the Adams spectral sequence and the $J$-homomorphism: $\pi_0^s=\mathbb{Z}$ (degree), $\pi_1^s=\mathbb{Z}/2\langle\eta\rangle$, $\pi_2^s=\mathbb{Z}/2\langle\eta^2\rangle$, $\pi_3^s=\mathbb{Z}/24\langle\nu\rangle$, $\pi_4^s=\pi_5^s=0$, $\pi_6^s=\mathbb{Z}/2\langle\nu^2\rangle$, $\pi_7^s=\mathbb{Z}/240\langle\sigma\rangle$ [Rav86, §1.1].

The hole shows up cleanly in degree $1$: $\widetilde\pi_1^{\,s}(X)=\pi_0^s=\mathbb{Z}$, generated by the inclusion of $\gamma$. Every other class is a fact about the geometry of exotic spheres which, by an accident of stable algebra, also pertains to the straw.▢

Sketch. For a compact, locally contractible $K\subset S^n$, Alexander duality asserts $\widetilde H_i(S^n\!\setminus\! K;\mathbb{Z})\cong\widetilde H^{n-i-1}(K;\mathbb{Z})$ [Hat02, Thm 3.44]. Taking $K=S^1$ and $n=3$, $i=1$, gives $\widetilde H_1(S^3\setminus S^1)\cong\widetilde H^{\,1}(S^1)=\mathbb{Z}$. The duality is implemented by the linking-number pairing $\mathrm{lk}\colon H_1(K)\otimes H_1(S^3\setminus K)\to\mathbb{Z}$, which with one generator on each side is the identity matrix.

Translated to the kitchen: there is exactly one independent way to thread a closed loop through the straw, and its threading is detected by the linking number with the core circle $\gamma$. This is the invariant one computes physically by passing a piece of string through the bore — fingers and Alexander agree.▢