# differentiable atlas, differentiable manifold

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.

• 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 $C^k$. Jun 25 '20 at 5:33
• @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. 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 $$(\psi_1 \times \psi_2)(V_1 \times V_2 \cap U_1 \times U_2) \to (\phi_1 \times \phi_2)(V_1 \times V_2 \cap U_1 \times U_2)$$).