# Finding an algebra of smooth functions on a manifold with a given product.

I am having trouble with the following exercise from Jet Nestruev's Smooth Manifolds and Observables. It is one of the first exercises in the book and I have no idea how to approach this type of problem. I have had some exposure to elementary differential geometry (especially of curves and surfaces) but I do not know how I should think about this exercise: is there a standard method to approach it or should I just know the answer from other sources (e.g classical examples of differential geometry)?

Could you please point me in the right direction?

Suppose that the $\mathbb{R}$-algebras $\mathcal{F}_1$ and $\mathcal{F}_2$, as vector spaces, are isomorphic to the plane $\mathbb{R}^2$. Let the multiplication in $\mathcal{F}_1$ and $\mathcal{F}_2$ be respectively given by the relations

$(x_1, y_1) \cdot (x_2, y_2) = (x_1x_2, y_1y_2) \quad (1)$

$(x_1, y_1) \cdot (x_2, y_2) = (x_1x_2 + y_1y_2, x_1y_2 + x_2y_1) \quad (2)$

Find the manifold $M_i$ for which the algebra $\mathcal{F}_i$, $i =1, 2$, is the algebra of smooth functions, explicitly indicating what function on $M_i$ corresponds to the element $(x, y) \in \mathcal{F}_i$. Are the algebras $\mathcal{F}_1$ and $\mathcal{F}_2$ isomorphic?

• I haven't worked through the details at all, but here are two observations: 1. If $M$ contains infinitely many points (in particular, if it's of positive dimension), then the collection of smooth functions is an infinite dimensional vector space. Hence, your $M_i$ must be discrete finite sets of points (so are compact!). 2. If $M$ is compact, then the set of points of $M$ is in natural bijective correspondance with maximal ideals in its ring of smooth functions, given by sending a point $p\in M$ to the maximal ideal given by the kernel of the map which evaluates smooth functions at $p$. Commented Mar 12, 2013 at 19:53
• This exercise has really nothing to do with manifolds. It contains them somehow artificially ... Commented Mar 12, 2013 at 20:18
• I was confused by this question when I could only come up with Jason's answer. Then I arrived at Martin's conclusion and was (dis)satisfied. Commented Mar 12, 2013 at 20:24
• @JasonDeVito Could you explain more about your second observation? I wonder how "maximal ideal" shows up in this context. Thank you! Commented Apr 7, 2013 at 23:06
• @vladimir: It wouldn't be appropriate to post as an answer on this question but if you asked about it on the main site, I (or someone else) could (eventually) answer it there. Commented Apr 8, 2013 at 0:18

It is immediate that $F_1 = \mathbb{R} \times \mathbb{R} \cong C^{\infty}(1+1)$, where $1+1$ is the discrete manifold with two elements. The second algebra has the unit $(1,0)$, and it is generated by $u=(0,1)$, satisfying $u^2=1$. Hence, $F_2 \cong \mathbb{R}[x]/(u^2-1) \cong \mathbb{R}[u]/(u+1) \times \mathbb{R}[u]/(u-1) \cong \mathbb{R} \times \mathbb{R} = F_1$. It maps

$$F_2 \ni (x,y) \mapsto \overline{x+uy} \mapsto (\overline{x+uy},\overline{x+uy}) \mapsto (x-y,x+y) \in F_1.$$