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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.

Thank you in advance!

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    $\begingroup$ 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. $\endgroup$
    – J.V.Gaiter
    Dec 18, 2022 at 23:02

2 Answers 2

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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 $\mathbb{Z}$. $\square$.

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  • $\begingroup$ 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 $H_1$. $\endgroup$ Dec 19, 2022 at 13:18
  • $\begingroup$ 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? $\endgroup$
    – Turtle5
    Dec 19, 2022 at 14:27
  • $\begingroup$ 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. $\endgroup$ Dec 19, 2022 at 15:27
  • $\begingroup$ Yes, I see it now, thank you for your solution! $\endgroup$
    – Turtle5
    Dec 20, 2022 at 10:35
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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, $H^1(T^1(K))\cong \mathbb{Z}/(2)^3$.

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