Is not the left representation a map $\lambda: L^1(G) to B(L^2(G)$ via convlution? As another question is te left and right representation gives the same reduce C^* structure?

BTW i can not open the two links you provided in the post

Is it the standard coalgebra structure on the symmetric algebra?For example is it identical to this one mathoverflow.net/questions/448656/…

What about a possible infinite dimensional analogues of Hartogs extension theorem?(Instead of separable holomorphicity)?

@BranimirĆaćić Yes I see in fact the problem is that the Laplacian is not $C^{\infty}(M) $ linear. However the tensor product in your first comment is taken over scalars.

So $x^6\otimes 1$ is maped to 0 but $x^3\otimes x^3$ is maped to $6x\otimes 6x$ identified to $36x^2$. I think I am doing a stupid mistake but what is my mistake?(I repeat that I do not product M with M. I just consider the tensor trivial bundle and the tensor operrator.

@BranimirĆaćić Sorry if the point is trivial: where is the contraditory point: The trivial line bundle on R is denoted by $\epsilon_1$ then $\epsilon_1\otimes \epsilon_1 \sim \epsilon_1$ then the section space of tensor product is isomorphisc to the tensor product of the corresponding sectioin(Serre Swan theorem). On the other hand the isomorphisms between $\mathbb{R}\otimes \mathbb{R}$ with $\mathbb{R}$ is in the form $a\otimes b \mapsto ab$. Now we wish to look at $\Delta \otimes \Delta$ acted on , say $x^6$... ok..

@BranimirĆaćić Yes but in the post I did not product manifolds. The tensor product of $\epsilon_1$ with itself is the trivial line bundle on $M$(The base space is always $M$. so $f(x)\otimes g(x)$ is identified with $f(x)g(x)$ So I guess I am leading to triviality!! because $f\otimes 1$ maps to 0 !! am Imissing some thing? how can I improve the question in Riemannian metric case?

I mean module constant functions

@DavidGao Very good point. A similar point for codimension?

@BranimirĆaćić But very good point you indicated to

@BranimirĆaćić ..possible other metrics

So in the compact case there is no any harmonic map so there is a chance of ellupticity (since there is no immediate obstruction for Fredholm ness of tensor product of laplacians). Yes the Laplacian of product metric is not the tensor product of corresponding Laplaciqn but what about possible...

@DavidGao Thanks for your comment