Homology groups of Klein bottle's unit tangent bundle.

Let $$K$$ denote Klein bottle and $$T^1K$$ its unit tangent bundle. I want to compute homology group of $$T^1K$$, I've seen this discussion: Homology groups of unit tangent bundle, I don't understand much of what's written there, but it seems, that homology of $$T^1K$$ must be the same as the homology of $$T^1T$$ (unit tangent bundle of torus), because torus and Klein bottle have the same Euler characteristic, but how to compute them explicitly? Maybe $$T^1K$$ is homeomorphic to $$T^1T$$?

I understand, that $$T^1K$$ is oriented 3-manifold, because tangent bundle is always orientable, so its third homology must be $$\mathbb{Z}$$.

I also tried to use Mayer-Vietoris sequences: to cut this bundle in two bundles over Mobius strips, but I couldn't do much with it.

• It's important to note that the discussion in the post you have linked relies on the fact that the base space is orientable, which cannot be used here since $K$ is not. Dec 18, 2022 at 23:02

I'm going to use your Mayer-Vietoris idea for computing the integer cohomology of $$T^1 K$$, but with a twist.

To that end, let $$\pi:T^1 K\rightarrow K$$ be the natural projection map. I'll use $$U'$$ and $$V'$$ for the two Mobius band halves of $$K$$, which overlap in the boundary circle of each Mobius band. Let $$U = \pi^{-1}(U')$$ and $$V = \pi^{-1}(V')$$. Then $$U$$ and $$V$$ obviously form an open cover of $$T^1 K$$, so we can use them for Mayer-Vietoris. We just need to understand the topology of $$U$$, $$V$$, and $$U\cap V$$.

Claim 1: Both $$U$$ and $$V$$ deformation retract to copies of $$K$$.

For $$U'$$ deformation retracts to its core circle, and $$\pi^{-1}$$ of this deformation gives a deformtion of $$U$$ to $$\pi^{-1}(S^1_c)$$, with $$S^1_c$$ denoting the core circle in a Mobius band.

So, let's figure out what $$\pi^{-1}(S^1_c)$$ is. First, restricting the tangent bundle of $$K$$ to a Mobius band half gives the tangent bundle of the Mobius band. This restricts to a non-trivial $$\mathbb{R}^2$$ bundle over $$S^1_c$$ (just draw a picture!), so $$\pi^{-1}(S^1_c)$$ is a non-trivial $$S^1$$ bundle over $$S^1$$. That is, it's a Klein bottle as claimed. $$\square$$

Claim 2: The space $$U\cap V$$ deformation retracts to $$T^2$$.

The space $$U'\cap V'$$ deformation retracts to a circle winding around $$S^1_c$$ two times. I'll denote this by $$S^1_d$$ (where $$d$$ is for "double") Thus, $$\pi^{-1}(U\cap V)$$ deformation retracts to $$\pi^{-1}(S^1_d)$$. Then the tangent bundle of $$K$$ restrcits to the tangent bundle of $$M$$, which restricts to a trivial bundle over $$S^1_d$$ (again, just draw a picture). Thus, $$\pi^{-1}(S^1_d)$$ is a trivial $$S^1$$ bundle, so is $$T^2$$ as claimed. $$\square$$

Claim 3:: With respect to the deformations in Claim 1 and 2, the inclusion maps $$U\cap V\rightarrow U,V$$ are homotopic to the double covering $$T^2\rightarrow K$$.

For, if we imagine homotoping $$S^1_d$$ to the core circle $$S^1_c$$, we get a double cover map $$S^1_d\rightarrow S^1_c$$. And this double cover map acts as the identity on the $$S^1$$-fibers because coverings are local diffeomorphisms. $$\square$$

We can now compute $$\pi_1(T^1 K)$$ using Seifert-van Kampen. To that end, we'll write $$\pi_1(T^2) = \langle a,b| aba^{-1}b^{-1}\rangle$$, and $$\pi_1(K) = \langle s,t| sts^{-1}t\rangle$$. Then $$\pi_1(K)$$ has a unique index $$2$$ abelian subgroup, generated by $$s^2$$ and $$t$$, so we may assume the double cover $$T^2\rightarrow K$$ maps $$a$$ to $$s^2$$ and $$b$$ to $$t$$. Now, Seifert-van Kampen immediately gives:

Claim 4: We have $$\pi_1(T^1 K)\cong \langle s,t, u,v| sts^{-1}t, uvu^{-1}v, s^2 u^{-2}, tv^{-1} \rangle$$.

By the way, Claim 4 establishes that $$\pi_1$$ is not abelian, since there is a surjective map from $$\pi_1$$ to the order $$8$$ quaternion group $$Q_8 = \{\pm 1, \pm i, \pm j, \pm k\}$$ given by sending $$(s,t,u,v)\mapsto (i,j,k,j)$$. In particular, $$T^1 K$$ and $$T^1 T^2$$ are not homotopy equivalent.

Since $$H_1$$ is the abelianization of $$\pi_1$$, we can now compute $$H_1$$.

Claim 5: We have $$H_1(T^1 K) \cong \mathbb{Z}\oplus (\mathbb{Z}/(2))^2$$

For, if we assume all the variables commute, every element in $$H_1(T^1 K)$$ can be written in the form $$s^\alpha t^\beta u^\gamma v^\delta$$. But the relation $$sts^{-1}t$$ now implies that $$t$$ has order $$2$$, so we may assume $$\beta \in \{0,1\}$$. Likewise, $$\delta \in\{0,1\}$$. In addition, the relation $$s^2u^{-2}$$ means that we may assume $$\gamma \in \{0,1\}$$.

This gives a map from $$H_1(T^1 K)$$ to $$\mathbb{Z}\oplus (\mathbb{Z}/(2))^2$$, sending $$s^\alpha t^\beta u^\gamma v^\delta$$ to $$(\alpha+\gamma,\beta + \delta,\gamma)$$. We claim that this map is an isomorphism.

To see that it's well defined, we just need to check that each relation in $$\pi_1(T^1K)$$ is sent to the identity. But, e.g., $$sts^{-1} t = t^2\mapsto (0,2,0) = (0,0,0)$$, etc. Surjectivity is obvious. And injectivity is also easy to verify. $$\square$$

The calculation of $$H_1$$, together with your prior knowledge that $$T^1 K$$ is orientable, allows us to compute all homology groups.

Claim 6: The non-zero homology of $$T^1 K$$ is given by $$H_n(T^1 K)\cong \begin{cases}\mathbb{Z} & n=0, 2,3\\ \mathbb{Z}\oplus(\mathbb{Z}/(2))^2 & n=1 \end{cases}.$$

We obviously have $$H_0(T^1 K) \cong \mathbb{Z}$$, you've already noted that $$H_3(T^1 K)\cong \mathbb{Z}$$, and we just computed $$H_1(T^1 K)$$. So, all that remains is $$H_2(T^1 K)$$. We will compute this via a combination of Poincare duality and universal coefficients.

We start with the torsion subgroup of $$H_2(T^1 K)$$. By Poincare duality, this is the same as the torsion subgroup of $$H^1(T^1 K)$$. But by universal coefficients, $$H^1(T^1 K)$$ has torsion given by an Ext term involving $$H_0(T^1 K)$$. Since $$H_0(T^1 K)$$ is free, this Ext term vanishes, so $$H^1(T^1 K)$$ is torsion free. Thus, so is $$H_2(T^1 K)$$.

To compute the free part of $$H_2(T^1 K)$$, we proceed analogously. By universal coefficients, the free part of $$H_2(T^1 K)$$ is isomorphic to the free part of $$H^2(T^1 K)$$, which, via Poincare duality, is isomorphic to the free part of $$H_1(T^1 K)$$, which is isomorphic to $$\mathbb{Z}$$. $$\square$$.

• In case you're wondering why I pivoted from homology to fundamental group, it was because I immediately saw how to compute the induced map $\pi_1(T^2)\rightarrow \pi_1(K)$, but not how to immediately do it for $H_1$. Dec 19, 2022 at 13:18
• Thank you for your answer! I can't understand why in claims 4 and 5 you don't use the last relation $t^{-1}v$? Shouldn't $t$ and $v$ be the same? Dec 19, 2022 at 14:27
• You're right that you can use that relation just as you suggested. But I don't think it significantly shortens the proof of claim 5. Dec 19, 2022 at 15:27
• Yes, I see it now, thank you for your solution! Dec 20, 2022 at 10:35

All cohomology groups in this post will be with coefficients in $$\mathbb{Z}/(2)$$. We can calculate the $$\mathbb{Z}/(2)$$ coefficient cohomology using the unoriented Gysin sequence, which gives a long exact sequence for the deleted space $$E_0$$ of a rank $$k$$ vector bundle $$E\to M$$. Consider the long exact sequence induced by the inclusion of $$E_0\hookrightarrow E$$. $$\cdots\to H^i(E,E_0)\to H^i(E)\to H^i(E_0)\to H^{i+1}(E,E_0)\to \cdots$$ The Thom isomorphism theorem tells us that there exists a class $$u\in H^{k}(E, E_0)$$ such that $$\smile u: H^{i-k}(E)\to H^{i}(E,E_0)$$ is an isomorphism. Replacing $$H^i(E,E_0)$$ by $$H^{i-k}(E)$$ yields the following short exact sequence: $$\cdots\to H^{i-k}(E)\to H^i(E)\to H^i(E_0)\to H^{i-k+1}(E)\to \cdots$$ where the map $$g:H^{i-k}(E)\to H^{i}(E)$$ is $$\alpha\mapsto \alpha\smile \kappa^*u$$ where $$\kappa^*: H^j(E, E_0)\to H^j(E)$$ is the natural map. We then replace $$H^j(E)$$ by $$H^j(M)$$ and $$g$$ is replaced by $$h: \alpha\mapsto \alpha\smile w_{k}(E)$$ where $$w_k(E)$$ is the $$k$$th Stiefel-Whitney class. This gives the long exact sequence $$\cdots\to H^{i-k}(M)\to H^i(M)\to H^i(E_0)\to H^{i-k+1}(M)\to\cdots$$

In our case, we take $$k=2$$ and replace $$E_0$$ by $$T^1K$$ since $$E_0$$ deformation retracts onto the unit sphere bundle. In degree $$3$$ we have the sequence $$\cdots\to H^1(K)\to H^3(K)\to H^3(T^1K)\to H^2(K)\to H^4(K)\to \cdots$$ Since $$K$$ is $$2$$ dimensional $$H^3(K)=0$$ and $$H^4(K)=0$$ so $$H^3(T^1K)\cong H^2(K)\cong \mathbb{Z}/(2)$$. In degree $$2$$ we have the long exact sequence $$\cdots\to H^0(K)\to H^2(K)\to H^2(T^1K)\to H^1(K)\to H^3(K)=0$$ The map $$H^0(K)\to H^2(K)$$ is given by $$\smile w_2(K)$$. Since $$0=\chi(K)=\langle w_2(K), [K]\rangle$$ we see that $$w_2(K)=0$$ and we get a short exact sequence $$0\to H^2(K)\to H^2(T^1K)\to H^1(K)\to 0$$ meaning that $$H^2(T^1K)\cong \mathbb{Z}/(2) ^3$$

In degree $$1$$ we have $$0=H^{-1}(K)\to H^1(K)\to H^1(T^1K)\to H^0(K)\to H^3(K)=0$$ or the short exact sequence $$0\to H^1(K)\to H^1(T^1K)\to H^0(K)\to 0$$ Since every short exact sequence of $$\mathbb{Z}/(2)$$ vector spaces splits, $$H^1(T^1(K))\cong \mathbb{Z}/(2)^3$$.