Sorry in advance if this question ended up a little bit confused, I just wanted to share my thoughts and make clear where my problems come from. In general the question is to answer the following task, but I think I need some clarification of the definitions, because I am not really satisfied by what I have to work with. So I am thankful if you just answer the original question, which is the given task, or comment my thoughts on it, what is however not necessary, but I would appreciate it more, then just a full answer. Thanks in advance.

Let $M$ and $N$ be differentiable manifolds of dimension $m$ and $n$ with atlases $A_M$, $A_N$. We set $A_M\times A_N:=\{(U\times V, x\times y)~| (U,x)\in A_M, (V,y)\in A_N\}$, where $x\times y: U\times V\to x(U)\times y(U), (p,q)\mapsto (x(p), y(q))$ for chards $(U,x)\in A_M$ and $(V,y)\in A_N$.

I want to show that $A_M\times A_N$ is a differentiable atlas on $M\times N$.

I have already showen that $M\times N$ is a topological manifold.

I have problems with the definition and how to use (formalize) it here correctly:

Definition: A atlas $A=\{h_i: U_i\to U_i'~|i\in J\}$ is 'smooth', if every $h_{ji}:h_i(U_i\cap U_j)\to h_j(U_i\cap U_j)$ is smooth ($C^k$). Where $h_{ji}=h_j\circ h_i^{-1}$ is the transition map.

My first problem is that we only defined the term 'smooth atlas', which I interpret as $C^\infty$ atlas. Next in the definition the function $h_{ji}$ can be smooth or $C^k$. So I think that with differentiable atlas is here meant, that every transition map is $C^1$, and what is really meant is that we call an atlas smooth ($C^\infty$), or more general $C^k$ if the transition maps are $C^k$.

So $A_M$ and $A_N$ are differentiable atlases, which I interpet now as $C^1$ atlases. How ever it would not make a difference.

I now have to show that every transition map is again $C^1$.

My next problem is the notation of the transition maps.

The $(U,x)$ and $(V,y)$ are charts of $A_M$ and $A_N$, so $\bigcup_{i\in I} U_i=M$ and $\bigcup_{j\in J} V_j=N$.

For every pair $(U_i\times V_j, x_i\times y_j)$

We have then $x_i\times y_j: U_i\times V_j\to x_i(U_i)\times y_j(V_j)$, and the indices take over.

How do I note the transistion maps now? If I note $x_i\times y_j$ as $x\times y_{(i,j)}$, I would note a transistion map now as

$x\times y_{(i,j)(i',j')}$ and I feel like this is overly complicated and not correct.

Can you help me. Thank you.

  • 1
    $\begingroup$ No, if your manifold is smooth, then the transition functions $h_{ij}$ have to be smooth as well. If your transition functions are $C^k$, then your manifold is $C^k$. $\endgroup$ Jun 25 '20 at 5:33
  • $\begingroup$ @PankajTiwari Yes, that is what I meant. The problem is that we have a 'differentiable manifold' and not a 'smooth manifold', and we have not defined what a 'differentiable manifold' is. But I am also aware that it does not matter for the proof. Should be the same for C^1, C^2, C^42 and $C^\infty$ manifolds. $\endgroup$
    – Cornman
    Jun 25 '20 at 5:35

I think I've addressed your first question in the comments. For the second question, let $(U_1 \times U_2,\phi_1 \times \phi_2)$ and $(V_1\times V_2, \psi_1\times \psi_2)$, be two charts of the product manifold $M \times N$. Now by definition, $(\phi_1 \times \phi_2) \circ (\psi_1 \times \psi_2)^{-1} = (\phi_1\circ \psi_1^{-1})\times( \phi_2 \circ \psi_2^{-1})$. Since $\phi_1\circ \psi_1^{-1}$ and $\phi_2\circ \psi_2^{-1}$ are individually smooth,$(\phi_1 \times \phi_2) \circ (\psi_1 \times \psi_2)^{-1}$ is also smooth(note that this is a function from $(\psi_1 \times \psi_2)(V_1 \times V_2 \cap U_1 \times U_2) \to (\phi_1 \times \phi_2)(V_1 \times V_2 \cap U_1 \times U_2)$).

  • $\begingroup$ +1. Even if one answers a question in the comments, it's good practice to include it in an answer as well. That way future readers with the same question can find a full answer in an answer, without having to dig through the comments. $\endgroup$ Jun 25 '20 at 17:49

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