Can you provide a link to the lecture notes? On first sight I agree that this seems to be true only for some form of completed version of the tensor product.

@Hanno$:$ faculty.math.illinois.edu/~ruiloja/Poisson2014/…

this does not include a Chapter 8?

Sorry for the confusion, too, but the doc you ref'd stops at p.37 for me.

@Hanno$:$ Oops... See here then acrobat.adobe.com/link/…

I can't open that, either :-(

@Hanno$:$ I have a pdf in my device. I don't know how to share it. Could you please help me with it? Extremely sorry for the inconvenience caused.

I'm sorry, I cannot help you share the document, if only because I don't know if it is meant to be shared. I'm afraid I have to stop at my initial comment that I suspect that some technical details are missing here.

@Hanno$:$ I have added a screenshot of the relevant page. Please have a look at it now.

Oh, you see, the authors assume $G$ to be finite here! So then there's no issue with completion of the tensor product.

@Hanno$:$ Could you please elaborate? You mean completion with respect to which topology? How finiteness of $G$ helps in this completion?

@Hanno$:$ Hopf algebra is supposed to be an algebraic object. Then how does this completion work in this algebraic setting?

If $G$ is finite, then ${\mathcal F}_k(G)\otimes_k {\mathcal F}_k(G)\to {\mathcal F}_k(G\times G)$ is indeed an isomorphism, for the naive algebraic tensor product. Only if you want to carry over this idea to other contexts, e.g. $G$ being a Lie group, then you'd surely need a different version of tensor product. I don't know if the book elaborates on that later on.

@Hanno$:$ Can a finite group have a manifold structure?

Yes, the discrete structure. What the authors are ultimately after is a Hopf-algebra(-like) structure on the smooth functions on a Lie group, but as a motivation, they consider the much simpler special case of a finite group.

@Hanno$:$ Any smooth manifold has to contain infinitely many elements as it is locally homemorphic to $\mathbb R^n$ for some $n.$

A $0$-dimensional manifold is just a discrete space.

Oh! I see @Hanno. But how to prove that for finite groups the algebraic tensor product $C^{\infty} (G) \otimes C^{\infty} (G)$ is isomorphic to $C^{\infty} (G \times G)\ $?

You can just write down bases for both sides and see that the canonical map is an isomorphism.

Do you mean that if $G$ is finite then $C^{\infty} (G)$ is finite dimensional?

Yes -- it's just $k$-valued functions on a finite set. I suggest you sit down and work it out for a bit yourself?

@Hanno$:$ They are generated by the indicators, I guess and the dimension of both the spaces is $\left \lvert G \right \rvert^2.$ If I can prove that the map is injective then we are through.

@Hanno$:$ Got it. Actually basis is going to basis under this embedding and hence it has to be an isomorphism. Many many thanks for your kind help.