1
$\begingroup$

Let $M$ and $N$ be smooth manifolds of dimension $m$ and $n$ respectively and for $p \in M$ denote \begin{equation} \mathcal{T}_p M = \lbrace \gamma \in C^{\infty}(I,M) | 0 \in I \land \gamma(0) = p \rbrace. \end{equation} Also let $\sim$ be a equivalence relation on $\mathcal{T}_p M$ defined by \begin{equation} c_1 \sim c_2 : \iff \exists (U_{\varphi},\varphi) \in \mathcal{A}_M: p \in U_{\varphi} \land \frac{d}{dt}(\varphi \circ c_1)(0) = \frac{d}{dt}(\varphi \circ c_2)(0) \end{equation} where $\mathcal{A}_M$ is a atlas obtained by the smooth structure of $M$ then the tangent space $T_p M$ of $M$ at $p$ is defined by \begin{equation} T_p M := \mathcal{T}_p M / \sim. \end{equation} For $(U_{\varphi}, \varphi) \in \mathcal{A}_M$ define a map $t^M_p \varphi : T_p M \to \mathbb{R}^m$ for $[c]_{\sim} \in T_p M$ by \begin{equation} t^M_p\varphi([c]_{\sim}) := \frac{d}{dt}(\varphi \circ c)(0). \end{equation} Now the product manifold $M \times N$ can be equipped with the product atlas which is contained in a unique maximal atlas hence $M \times N$ is a smooth manifold. For $(p,q) \in M \times N$ the tangent space is then given by $T_{(p,q)}M \times N = \mathcal{T}_{(p,q)}M \times N / \sim$. For $(U_{\varphi_1}, \varphi_1) \in \mathcal{A}_M$ and $(U_{\varphi_2}, \varphi_2) \in \mathcal{A}_N$ define a map $t^{M \times N}_{(p,q)}(\varphi_1 \times \varphi_2) : T_{(p,q)}M \times N \to \mathbb{R}^{m+n} $ by \begin{equation*} t^{M \times N}_{(p,q)}(\varphi_1 \times \varphi_2)(([\gamma_1]_{\sim}, [\gamma_2]_{\sim})) = \left( t^{M}_{p}\varphi_1([\gamma_1]_{\sim}), t^{N}_{q}\varphi_2([\gamma_2]_{\sim})\right) \end{equation*} for two curves $\gamma_1 \in \mathcal{T}_p M$ and $\gamma_2 \in \mathcal{T}_q N$ with domain $I$ and $J$ respectively. Note that \begin{equation*} (t^{M \times N}_{(p,q)}(\varphi_1 \times \varphi_2))^{-1}(\gamma_1(t), \gamma_2(s)) = ((t^{M}_p \varphi_1)^{-1}(\gamma_1(t)), (t^{N}_q \varphi_2)^{-1}(\gamma_2(s)) ) \end{equation*} for some $ t \in I$ and $s \in J$ Also the tangent spaces $T_p M$ and $T_q N$ are identified with vector spaces by \begin{align} [c_1]_{\sim} + [c_2]_{\sim} &:= (t^{M}_p\varphi)^{-1}(t^{M}_p\varphi([c_1]_{\sim}) + t^{M}_p\varphi([c_2]_{\sim})), \\ \lambda[c_1]_{\sim} &:= (t^{M}_p\varphi)^{-1}(\lambda t^{M}_p\varphi([c_1]_{\sim})) \end{align} for $[c_1]_{\sim}, [c_2]_{\sim} \in T_p M$ and $\lambda \in \mathbb{C}$. The same can be done for $T_q N$ and $T_{(p,q)}M \times N$.

Here is the actual problem that I try to solve:

Now I want to find a natural isomorphism $(T_p M \oplus T_q N \cong) T_p M \times T_q N \cong T_{(p,q)}M \times N$. I know this can be done by defining $\alpha(v) = (d(\pi_1)_{(p,q)}(v), d(\pi_2)_{(p,q)}(v)) $. But there is also another approach which I find more interesting. So I define $\iota_{(p,q)} : T_p M \times T_q N \to T_{(p,q)}M \times N$ by \begin{equation*} \iota_{(p,q)} : ([\gamma_1]_{\sim}, [\gamma_2]_{\sim}) \mapsto [t \mapsto (\gamma_1(t), \gamma_2(t))]_{\sim} =: [(\gamma_1, \gamma_2)]_{\sim}. \end{equation*} It is easy to show that $\iota_{(p,q)}$ is injective and surjective hence bijective. But for the linearity I struggle. Let $(U_{\varphi_1}, \varphi_1) \in \mathcal{A}_M$ be a chart with $p \in U_{\varphi_1}$ and $(U_{\varphi_2}, \varphi_2) \in \mathcal{A}_N$ a chart with $q \in U_{\varphi_2}$. I got for $([\gamma_1]_{\sim},[\gamma_2]_{\sim}),([\tilde{\gamma_1}]_{\sim},[\tilde{\gamma_2}]_{\sim}) \in T_p M \times T_q N$ the following \begin{align*} &\iota_{(p,q)}(([\gamma_1]_{\sim},[\gamma_2]_{\sim}) + ([\tilde{\gamma_1}]_{\sim},[\tilde{\gamma_2}]_{\sim})) \\ &= (\iota_{(p,q)} \circ (t^{M \times N}_{(p,q)}(\varphi_1 \times \varphi_2))^{-1})(t^{M \times N}_{(p,q)} (\varphi_1 \times \varphi_2)(([\gamma_1]_{\sim},[\gamma_2]_{\sim})) + t^{M \times N}_{(p,q)}(\varphi_1 \times \varphi_2)(([\tilde{\gamma_1}]_{\sim},[\tilde{\gamma_2}]_{\sim}))) ) \\ &= (\iota_{(p,q)} \circ (t^{M \times N}_{(p,q)}(\varphi_1 \times \varphi_2))^{-1}) \left( \left( \frac{d}{dt}(\varphi_1 \circ \gamma_1)(0), \frac{d}{dt}(\varphi_2 \circ \gamma_2)(0)\right) + \left( \frac{d}{dt}(\varphi_1 \circ \tilde{\gamma_1})(0), \frac{d}{dt}(\varphi_2 \circ \tilde{\gamma_2})(0)\right) \right) \\ &= (\iota_{(p,q)} \circ (t^{M \times N}_{(p,q)}(\varphi_1 \times \varphi_2))^{-1}) \left( \left(\frac{d}{dt}(\varphi_1 \circ \gamma_1)(0) + \frac{d}{dt}(\varphi_1 \circ \tilde{\gamma_1})(0), \frac{d}{dt}(\varphi_2 \circ \gamma_2)(0) + \frac{d}{dt}(\varphi_2 \circ \tilde{\gamma_2})(0) \right) \right) \\ &= (\iota_{(p,q)} \circ (t^{M \times N}_{(p,q)}(\varphi_1 \times \varphi_2))^{-1}) \left( \left( t^M_{p}([\gamma_1]_{\sim}) + t^M_{p}([\tilde{\gamma_1}]_{\sim}) , t^N_{q}([\gamma_2]_{\sim}) + t^N_{q}([\tilde{\gamma_2}]_{\sim}) \right) \right) \\ \end{align*} \begin{align*} &= \iota_{(p,q)} \left((t^{M}_{p}\varphi_1)^{-1}(t^M_{p}\varphi_1([\gamma_1]_{\sim}) + t^M_{p}\varphi_1([\tilde{\gamma_1}]_{\sim})), (t^{N}_q \varphi_2)^{-1}(t^{N}_q \varphi_2([\gamma_2]_{\sim}) + t^{N}_q\varphi_2([\tilde{\gamma_2}]_{\sim}) ) \right)\\ &= \iota_{(p,q)} \left([\gamma_1]_{\sim} + [\tilde{\gamma_1}]_{\sim}, [\gamma_2]_{\sim} + [\tilde{\gamma_2}]_{\sim} \right) \\ &= \iota_{(p,q)}\left(([\gamma_1 + \tilde{\gamma_1}]_{\sim},[\gamma_2 + \tilde{\gamma_2}]_{\sim}) \right) \\ &= [(\gamma_1 + \tilde{\gamma_1}, \gamma_2 + \tilde{\gamma_2})]_{\sim} \\ &= [(\gamma_1, \gamma_2) + (\tilde{\gamma_1}, \tilde{\gamma_2})]_{\sim} \\ &= [(\gamma_1, \gamma_2)]_{\sim} + [(\tilde{\gamma_1}, \tilde{\gamma_2})]_{\sim} \\ &= \iota_{(p,q)}([\gamma_1]_{\sim},[\gamma_2]_{\sim}) + \iota_{(p,q)}([\tilde{\gamma_1}]_{\sim},[\tilde{\gamma_2}]_{\sim}) \end{align*} where I used \begin{align*} [\gamma_1]_{\sim} + [\tilde{\gamma_1}]_{\sim} &= [\gamma_1 + \tilde{\gamma_1}]_{\sim},\\ [\gamma_2]_{\sim} + [\tilde{\gamma_2}]_{\sim} &= [\gamma_2 + \tilde{\gamma_2}]_{\sim},\\ [(\gamma_1, \gamma_2) + (\tilde{\gamma_1}, \tilde{\gamma_2})]_{\sim} &= [(\gamma_1, \gamma_2)]_{\sim} + [(\tilde{\gamma_1}, \tilde{\gamma_2})]_{\sim}. \end{align*} I am not sure if this is true so I tried to justify the second equality by the following argument:
\begin{align*} &[(\gamma_1, \gamma_2)]_{\sim} + [(\tilde{\gamma_1}, \tilde{\gamma_2})]_{\sim} = (t^{M \times N}_{(p,q)}(\varphi_1 \times \varphi_2))^{-1}\left(t^{M \times N}_{(p,q)}(\varphi_1 \times \varphi_2)([(\gamma_1, \gamma_2)]_{\sim}) + t^{M \times N}_{(p,q)}(\varphi_1 \times \varphi_2)([(\tilde{\gamma_1}, \tilde{\gamma_2})]_{\sim}) \right) \\ &= (t^{M \times N}_{(p,q)}(\varphi_1 \times \varphi_2))^{-1}(t^{M \times N}_{(p,q)}(\varphi_1 \times \varphi_2)([(\gamma_1, \gamma_2)]_{\sim}) ) + (t^{M \times N}_{(p,q)}(\varphi_1 \times \varphi_2))^{-1}(t^{M \times N}_{(p,q)}(\varphi_1 \times \varphi_2)([(\tilde{\gamma_1}, \tilde{\gamma_2})]_{\sim}) ) \\ \end{align*} since $t^{M \times N}_{(p,q)}(\varphi_1 \times \varphi_2)$ is linear (this comes from the identification of the tangent space with a vector space ) because $t^{M}_{p}\varphi_1$ and $t^{N}_{q}\varphi_2$ are linear. Now \begin{equation*} t^{M \times N}_{(p,q)}(\varphi_1 \times \varphi_2)([(\gamma_1, \gamma_2)]_{\sim}) + t^{M \times N}_{(p,q)}(\varphi_1 \times \varphi_2)([(\tilde{\gamma_1}, \tilde{\gamma_2})]_{\sim}) \end{equation*} is a vector in $\mathbb{R}^{n + m}$ which gets represented by $(\gamma_1, \gamma_2) + (\tilde{\gamma_1}, \tilde{\gamma_2}) = (\gamma_1 + \tilde{\gamma_1}, \gamma_2 + \tilde{\gamma_2})$ hence we can write \begin{equation*} t^{M \times N}_{(p,q)}(\varphi_1 \times \varphi_2)([(\gamma_1, \gamma_2)]_{\sim}) + t^{M \times N}_{(p,q)}(\varphi_1 \times \varphi_2)([(\tilde{\gamma_1}, \tilde{\gamma_2})]_{\sim}) = t^{M \times N}_{(p,q)}(\varphi_1 \times \varphi_2)([(\gamma_1 + \tilde{\gamma_1}, \gamma_2 + \tilde{\gamma_2}) ]_{\sim}) \end{equation*} thus \begin{align*} [(\gamma_1, \gamma_2)]_{\sim} + [(\tilde{\gamma_1}, \tilde{\gamma_2})]_{\sim} &= (t^{M \times N}_{(p,q)}(\varphi_1 \times \varphi_2))^{-1}(t^{M \times N}_{(p,q)}(\varphi_1 \times \varphi_2)([(\gamma_1 + \tilde{\gamma_1}, \gamma_2 + \tilde{\gamma_2}) ]_{\sim})) \\ [(\gamma_1 + \tilde{\gamma_1}, \gamma_2 + \tilde{\gamma_2} )]_{\sim}. \end{align*}

$\endgroup$

1 Answer 1

0
$\begingroup$

I would rather avoid proving linearity directly from the definitions because as you've noted, the vector space operations on $T_pM$ are defined via those on $\Bbb{R}^n$ and using charts via "transport of structure". You've correctly noted that the map you defined (which is indeed the obvious map) is bijective. To prove linearity, one way is to verify (in your mind by unwinding the definitions) that the map you provided equals the map $\alpha$, which you know is linear.

Alternatively, you can verify that the map $[(\gamma_1,\gamma_2)]\mapsto ([\gamma_1],[\gamma_2])$ you defined equals the following composition: \begin{align} T_{(p,q)}(M\times N)\to \Bbb{R}^{m+n}\to \Bbb{R}^m\times \Bbb{R}^n\to T_pM \times T_qN \end{align} The first map being $t^{M\times N}_{(p,q)}(\varphi_1\times\varphi_2)$ in your notation (which is a linear isomorphism, because we defined the operations on the tangent space to ensure precisely this). The second map is the obvious $(\xi_1,\dots, \xi_{m+n})\mapsto ((\xi_1,\dots, \xi_m), (\xi_{m+1},\dots, \xi_{m+n}))$, and the final one is the map $(v,w)\mapsto ((t_p^M\varphi_1)^{-1}(v), (t_q^N\varphi_2)^{-1}(w))$, which is linear since $t_p^M\varphi_1,t_q^N\varphi_2$ are linear isomorphisms (i.e you're looking at the 'product' of linear maps $t_p^M\varphi_1 \times t_q^N\varphi_2:T_pM\times T_qN\to \Bbb{R}^m\times\Bbb{R}^n$ and inverting it).

In short, the isomorphism $T_{(p,q)}(M\times N)\cong T_pM\times T_qN$ corresponds in coordinates to the obvious $\Bbb{R}^{m+n}\cong \Bbb{R}^m\times\Bbb{R}^n$ (such an 'obvious' isomorphism that we may even treat it as if the two sets were equal and the isomorphism is just the identity).

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .