Tangent Bundle topolgy

I am reading Lee's ''Introduction to Smooth Manifolds'' and it states that the tangent bundle TM has a natural topology and smooth structure that makes it a 2n-dimensional manifold.

I get most of the proof, my only problem is to check the Hausdorff condition. And it is because it does not specify which topology they use, so I thought of one that works but I do not know if it is the usual one.

Having the specified topology does not bother most of the proof, since you have a proyection: $\pi: TM \to M$ defined as $\pi(p,X)=p$

And just take open sets as $\pi^{-1}(U)$ where $U$ is open in $M$. This works well for most of the proof, but when you want to check the Hausdorff condition in two points in the same fiber $(p,X),(p,Y)$.

So to avoid this I can define the open sets as the collection of sets:

$(U,\{V_x\}_{x \in U}):=\{ (x,v) : x \in U, v \in V_x \}$ Where $U$ is an open set of $M$ and $\{V_x\}_{x \in U}$ a collection of open sets $V_x \subset T_xM$ where many but not all can be the empty set.

Is this right? I checked and it does not change the smooth structure since the local charts $(U,\phi^{\sim})$ can be seen as $((U, \{T_xM\}_{x\in M}),\phi^{\sim})$ in my topology. Thanks in advanced.

You know that each chart $\phi : U \to V \subset \mathbb{R}^n$ for $M$ induces a canonical bijection $\tilde{\phi} : TM \mid_U = p^{-1}(U) \to V \times \mathbb{R}^n$. We define $W \subset TM$ to be open if $\tilde{\phi}(W \cap p^{-1}(U))$ is open in $V \times \mathbb{R}^n$ for all $\phi$.

It is an easy exercise to show that this is in fact a Hausdorff topology.

• Ok.. I get it, thanks. I think you can also take $\{V_n\}_{n \in \mathbb{N}}$ a countable basis of $R^n$ and $\{U_n\}_{n \in \mathbb{N}}$ a countable basis of $M$, then define a subbasis where the elements are exacly $\tilde{\phi}(V_n \cap p^{-1}(U_m))$ for positive intergers $n,m$ This is mostly to assure asmooth manifold structure. – Bajo Fondo Jul 25 '18 at 2:18

I guess the theorem that you mention is Proposition 3.18 on Lee's book.

For any smooth chart $(U_{\alpha},\varphi_{\alpha})$ of $M$, we have a map $\widetilde{\varphi}_{\alpha} : \pi^{-1}(U_{\alpha}) \to \Bbb{R}^{2n}$ defined as $$\widetilde{\varphi}_{\alpha} (v_p) = (x^1(p),\dots,x^n(p),v^1,\dots,v^n) .$$ This map is a bijection onto its image $\varphi_{\alpha}(U) \times \Bbb{R}^n$. The topology defined on $TM$ is the one that generated by $\{\widetilde{\varphi}_{\alpha}^{-1}(V) : \forall \alpha \in A, V\subseteq \Bbb{R}^{2n} \text{ is open}\}$. You can check directly that the collection above indeed generate a topology.

To show the Hausdorff property, let $p \in U \subseteq M$, where $U$ is the domain of some smooth chart $(U,\varphi)$ of $M$. And $v_p,w_p \in \pi^{-1}(p) \subseteq \pi^{-1}(U)\subseteq TM$, be a pair of points that lie in the same fiber.

With this topology, the map $\widetilde{\varphi} : \pi^{-1}(U) \to \Bbb{R}^{2n}$ is a homeomorphism onto its image. We can write $\widetilde{\varphi}(v_p) = (\hat{p},v)$, with $\hat{p} = \varphi(p)$. So let $\widetilde{\varphi}(v_p)=(\hat{p},v)$ and $\widetilde{\varphi}(w_p) = (\hat{p},w)$ be their images and $v,w \in \Bbb{R}^n$ be two distinct vectors. To obtain disjoint neighbourhoods of $v_p$ and $w_p$, just take disjoint neighbourhoods $B_1,B_2 \subseteq \Bbb{R}^n$ of $v$ and $w$ resp. and maped back the disjoint open sets $\varphi(U)\times B_1$ and $\varphi(U)\times B_2$ to $\pi^{-1}(U)$. So $\widetilde{\varphi}^{-1}(\varphi(U) \times B_1)$ and $\widetilde{\varphi}^{-1}(\varphi(U) \times B_2)$ are the desired neighbourhoods for $v_p$ and $w_p$.

• Thanks, in my book is Lemma 4.1 (There is also, no prop 3.18), there is a little subtletly here since you are asserting an homeomorphism, that might not be true with your defined topology. Maybe define a subbasis with those pre-images, taking a countable basis of $\mathbb{R}^n$. I mixed your answer with Frost's and I think it works, thanks. – Bajo Fondo Jul 25 '18 at 2:20
• @BajoFondo : Maybe you use different edition (mine was 2ed). Why you say that $\widetilde{\varphi}$ might not be a homeomorphism ? I think its clearly stated in the proposition (by exhibit its continous inverse). – Sou Jul 25 '18 at 2:35
• Take $U \times V \subset \mathbb{R}^n \times \mathbb{R}^n$ open set. Then $\widetilde{\varphi}^{-1}(U \times V)$ is not $\pi^{-1}(V_{\alpha})$ for any $V_{\alpha}$. Fix $p \in U$, then $\{p\} \times T_pM \subset \pi^{-1}(U)$ but take a $v_p$ that is not in $V$, then $(p,v_p) \in \{p\} \times T_pM \subset \pi^{-1}(U)$, but it is not in $\widetilde{\varphi}^{-1}(U \times V)$, hence $\widetilde{\varphi}^{-1}(U \times V)$ cannot be of the form $\pi^{-1}(U)$. The idea is that with your topology your open sets are the union of the elements of a class $\{\{p\} \times T_pM; p \in U_{\alpha}\}$. – Bajo Fondo Jul 25 '18 at 19:30
• @BajoFondo You're right i made a mistake about the topology. Let me amend it. – Sou Jul 25 '18 at 20:06