# Regarding diffeomorphism on manifolds

I am trying to show these claims:

1. If $$M,N$$ are smooth manifolds without boundary. Prove that: $$T(M\times N)$$ is diffeomorphic to $$TM\times TN$$.

2. Prove that $$TT^n$$ is diffeomorphic to $$T^n\times R^n$$. ($$T^n=S^1\times \cdots \times S^1$$).

Definition: The tangent bundle of a smooth manifold $$M$$ is a smooth manifold $$TM$$. $$TM$$ as a set:

$$(x,u)\in TM=\cup_{x\in M}$$ {x}$$\times T_xM$$. We define $$\pi: TM\mapsto M$$ by $$\pi(x,u)=x$$ (canonical projection). $$TM$$ as a topological space:

Let $$dim M=n$$. If $$(U,\phi)$$ is a chart on $$M$$. Define:

$$(x,u)\in \hat{U}=TU=\pi^{-1}(U)=\cup {x} \times T_x M$$, where we define $$\hat{\phi}:\hat{U}\to R^n \times R^n$$ by $$\hat{\phi}(x,u)=(\phi(x), \phi_{*,x})(u)$$.

Given a chart $$(U,\phi)$$ on $$M$$ near $$x$$ we define the map: $$\phi_{*,x}:T_xM\to R^n$$ by $$\phi_{*,x}(\gamma)=(\phi\circ\gamma) '(0)$$

For 1: an element of $$TM$$ (the tangent bundle of $$M$$) is of the form $$(x,v)$$ where $$x\in M$$, and same for $$TN$$: $$(y,u)$$ where $$y\in N$$. An element of $$TM\times TN$$ is: $$((x,v),(y,u))$$. And of $$T(M\times N)$$: $$((x,y),(v,u))$$. Now we define a map $$f:TM\times TN\to T(M\times N)$$ by

$$f((x,v),(y,u))=((x,y),(v,u)).$$

We want to show that it is a diffeomorphism, i.e it is smooth, a bijection and its inverse is smooth. I see that it is obvious that f is injective and surjective. I am not sure of how to check smoothness on a chart. Can you please explain how it is applied?

Instead (maybe another approach), I know that $$T_{(x,y)} (M \times N) \simeq T_xM \times T_yN$$ for every $$(x,y)\in M\times N$$. How could this be useful for showing what I want?

For 2, if I look at the universal cover of $$TT^n$$ that is $$TR^n$$ which is diffeomorphic to $$R^{2n}$$ by the identity map, using the coordinates $$(x_i,y_i)$$ where $$x_i$$ span $$R^n$$ and $$y_i$$ span the tangent directions. How I can continue from here?

• the point of the exercise is that, being a beginner, you make the effort to be clear about what the exercise ask and, after, try to reach an answer. The exercise is very simple as far as you know what is the smooth structure of a tangent bundle and know what is the canonical product manifold smooth structure, and what is a smooth map between manifolds (each tangent bundle is a manifold itself with a canonical structure). Try it (I dont downvoted you)
– user173262
Commented Feb 24, 2022 at 10:48
• You know how to construct the natural coordinates of product of manifolds, and that the tangent bundle is a manifold, then since the tangent bundle is locally a product then that should give you an idea of how to describe the the tangent bundle of a product and identifiy it with product of tangent bundles Commented Feb 24, 2022 at 13:29
• For 1, as a warm up, can you prove that for $(p,q)\in M\times N$, that $T_{(p,q)}(M\times N)\cong T_p M\times T_q N$ by coming up with a "natural" isomorphism (as opposed to just noting they are both vector spaces over $\mathbb{R}$ of the same dimension.) For 2, start with $n=1$. Commented Feb 24, 2022 at 14:08
• "I am not sure of how to show formally smoothness." Check it on a chart. Commented Feb 27, 2022 at 6:37

Here are some further details along the lines of the comments above.

For 1., suppose both $$M$$ and $$N$$ are $$C^r$$ manifolds ($$r\in\mathbb{Z}_{\geq2}$$, because we are asked to provide a diffeomorphism), and put $$\dim(M)=m$$ and $$\dim(N)=n$$. Then we have locally

\begin{align*} TM\cong_{\text{loc}} \mathbb{R}^m\times \mathbb{R}^m&\text{ with typical point } (x,\partial_x)=(x_1,x_2,...,x_m,\partial_{x_1},\partial_{x_2},...,\partial_{x_m}), \text{ and }\\ TN\cong_{\text{loc}} \mathbb{R}^n\times \mathbb{R}^n &\text{ with typical point } (y,\partial_y)=(y_1,y_2,...,y_n,\partial_{y_1},\partial_{y_2},...,\partial_{y_n}). \end{align*}

(Since $$M$$ and $$N$$ are $$C^r$$ manifolds, $$TM$$ and $$TN$$ are $$C^{r-1}$$ manifolds, which means that "locally" means "in $$C^{r-1}$$ charts".)

Thus locally the map you defined $$\Phi:TM\times TN\to T(M\times N), ((p,v),(q,w))\mapsto ((p,q),(v,w))$$ looks like

$$\Phi_{\text{loc}}: (x,\partial_x,y,\partial_y)\mapsto (x,y,\partial_x,\partial_y).$$

Thus its derivative has the matrix (w/r/t the bases on $$T_{((p,v),(q,w))}(TM\times TN)$$ and $$T_{((p,q),(v,w))} T(M\times N)$$ associated to the local coordinates)

$$T_{((p,v),(q,w))}\Phi = \begin{pmatrix} I_{m} & 0 & 0 & 0 \\ 0 & 0 & I_n & 0 \\ 0 & I_m & 0 & 0 \\ 0 & 0 & 0 & I_n \end{pmatrix},$$

where $$I_k$$ is the $$k\times k$$ identity matrix. Either by the inverse function theorem (see e.g. Diffeomorphism from Inverse function theorem ), or by applying the same reasoning directly to $$\Phi^{-1}$$ (whose derivative has the same matrix w/r/t the same choice of bases).

Note that (the tangent bundle of any manifold is orientable as a manifold (see e.g. Why is the tangent bundle orientable? ) and the product of orientable manifolds is orientable (see e.g. Product of manifolds & orientability ), and) $$\Phi$$ is orientation reversing. In this sense the map $$\Psi: ((p,v),(q,w))\mapsto ((p,q),(w,v))$$ could be considered more natural.

A more highbrow way of arguing for 1. would be to draw diagrams. First note that $$\Phi$$ acts as identity at the base level:

Here $$\pi_K: TK\to K$$ is the natural projection of the manifold $$K$$. In light of this, wlog one could use the products of local trivializing charts $$U,V$$ for $$\pi_M:TM\to M$$, $$\pi_N:TN\to N$$ respectively as local trivializing charts for $$\pi_{M\times N}:T(M\times N)\to M\times N$$. Then the following diagram makes the above argument more rigorous:

(As I'm writing this part somewhat in jest I'll leave it here.)

Part 2. boils down to showing that $$T\mathbb{T}^1\cong \mathbb{T}^1\times \mathbb{R}^1$$ by way of part 1.. It seems to me that any proof of this uses either the group structure of the circle $$\mathbb{T}^1$$ or that it can be considered as an embedded submanifold in $$\mathbb{R}^2$$. See e.g. How to know if a tangent bundle is trivial from its defining equations , How do I see that the tangent bundle of torus is trivial , Show that $T\mathbb S^1$ is diffeomorphic to $\mathbb S^1\times\mathbb R$ , $TS^1$ is Diffeomorphic to $S^1\times \mathbf R$. , or Is this map a diffeomorphism? .

As a final note, for any Lie group $$G$$ with Lie algebra $$\text{lie}(G)$$, $$TG$$ has a unique Lie group structure that makes $$\pi_G:TG\to G$$ a Lie group homomorphism (and coincides with vector addition along fibers) (see e.g. Lie group structure of the tangent bundle ). Further, as Lie groups $$TG\cong \text{lie}(G)\rtimes_{\operatorname{Ad}^G} G$$, where $$\operatorname{Ad}^G_\bullet: G\curvearrowright\text{lie}(G)$$ is the adjoint action. In the case of the torus $$G=\mathbb{T}^d$$, since $$\mathbb{T}^d$$ is abelian $$\operatorname{Ad}^{\mathbb{T}^d}_\bullet$$ is the trivial action, so we have that $$T\mathbb{T}^d\cong \mathbb{T}^d\times \mathbb{R}^d$$ as Lie groups.

• @Maths1999_ If you could disclose what your definition of the tangent bundle is maybe I can be of help. In any event, by definition the vectors in the tangent bundle ought to be related to the directions of the manifold, in particular if there are coordinates $(x_1,...,x_n)$ locally applicable on the manifold then there should be $n$ associated directions, one along each component, which ought to give coordinates locally applicable on the tangent bundle. Commented Apr 19, 2022 at 15:37
• For differentiability the argument I'm proposing above is the standard one: realizing that it has a derivative, which is invertible at any point. Commented Apr 19, 2022 at 15:38
• For part 2. $T (\mathbb{T}^d) \cong T(\mathbb{T}\times \mathbb{T}\times \cdots \times \mathbb{T}) \cong (T\mathbb{T})\times(T\mathbb{T})\times \cdots \times (T\mathbb{T}) \cong (\mathbb{T}\times \mathbb{R})\times(\mathbb{T}\times \mathbb{R})\times \cdots \times (\mathbb{T}\times \mathbb{R})\cong \mathbb{T}^d\times \mathbb{R}^d$. Commented Apr 19, 2022 at 15:41
• The definition you gave is not a complete definition of the tangent bundle, formally speaking. $T_xM$ is intimately related to the vicinity of $x$ in $M$. Yes, $\mathbb{T}=\mathbb{T}^1=S^1$ all denote the circle. Yes, yes and yes. For the question in the paren, I think eventually you are going to need to use the fact that tangent vectors are parameterized by partial derivatives, though all that's needed here is to use partial derivatives as coordinates, not differentiate anything using them. Commented Apr 19, 2022 at 17:42
• @AlpUzman I think your notation is somewhat confusing. $\partial_{x_i}$ may be viewed as a basis vector for the tangent-space but not as the coordinate of a tangent-vector. In the present context you may also avoid using differential operators altogether but that is more a matter of taste. Commented Apr 22, 2022 at 9:17