I am trying to show these claims:
-
If $M,N$ are smooth manifolds without boundary.
Prove that: $T(M\times N)$ is diffeomorphic to $TM\times TN$. -
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?
Best Answer
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.