# From algebra of smooth functions to a smooth manifold

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

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

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

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?

This question has boggled me for a while. We are looking for manifolds $$M_i$$ for which $$\mathcal{F}_i \cong C^\infty(M_i)$$. The problem I see here is that $$C^\infty(M_i)$$ is infinite-dimensional while $$\mathcal{F}_i$$'s are both $$2$$-dimensional so how can there exists such manifolds?

The second thing is that the latter one looks like its almost the product we have with complex numbers except for the first $$+$$ sign which should probably be a $$-$$ sign?

How should one go about solving this kinda problem?

• $C^\infty(M)$ is infinite dimensional only when $dim M>0$. We can consider a discrete manifold, for which $M=\{a,b\}$ will be a solution to the first isomorphism. But I don't see an obvious solution to the second one, unless those algebras are actually isomorphic. Yes, it does sound like there's some kind of mistake in the formulation of this problem. Commented Mar 27 at 13:51
• Can I ask how did you obtain the solution for the first problem? I'm not sure if I should consider ideals of $\mathcal{F}_1$ or something of that nature to be able to obtain the correspondence. @freakish Commented Mar 27 at 13:57
• Take $M=\{a,b\}$ to be a zero-dimensional manifold with exactly two points. Then every function $M\to\mathbb{R}$ is smooth, and a function $f$ is mapped to point $(f(a),f(b))$. Commented Mar 27 at 13:59
• What is the origin of your question? Commented Mar 28 at 10:04
• @PaulFrost Smooth manifolds and observables by Jet Neustrev Commented Mar 28 at 10:16

## 1 Answer

I cannot give a satisfying answer, but I had a look in Neustrev's book which convinced me that the author's exposition in not really clear.

Quote from Chapter 3:

Here we give a detailed answer to the following fundamental question: Given an abstract $$\mathbb R$$-algebra $$\mathcal F$$, find a set (smooth manifold) $$M$$ whose $$\mathbb R$$-algebra of (smooth) functions can be identified with $$\mathcal F$$.

I guess he considers two questions:

1. Find a set $$M$$ such that $$\mathcal F(M) =$$ algebra of functions $$f : M \to \mathbb R$$ can be identified with $$\mathcal F$$.

2. Find a smooth manifold $$M$$ such that $$C^\infty(M)$$ can be identified with $$\mathcal F$$.

In the first case one can regard $$M$$ as a discrete topological space so that $$\mathcal F(M) = C^0(M) =$$ algebra of continuous functions $$f : M \to \mathbb R$$. We may also regard $$M$$ as $$0$$-dimensional smooth manifold, at least if $$M$$ is countable.

Neustrev's approach to smooth manifold is a bit unusual. He introduces them by an algebraic definiton (Chapter 4). In Chapter 5 he gives the "standard" definition (5.8. Coordinate definition of manifolds).

Now consider Example 3.2. The author asks

Can this algebra $$\mathcal F$$ be realized as an algebra of nice functions on some set $$M$$?

His answer is

Thus any sequence $$\{a_i\} \in \mathcal F$$ may be viewed as the function on $$M = \mathbb R$$ given by $$x \mapsto \sum_{i=0}^\infty a_ix^i$$.

This is false. It can be regarded as an element of the polynomial algebra $$\mathbb R[x]$$, but in general not as function $$\mathbb R \to \mathbb R$$ because $$\sum_{i=0}^\infty a_ix^i$$ does in general not convergence (only for $$x = 0$$ we can be sure).

But even if it were true, we would only get an isomorphism between $$\mathcal F$$ and a subalgebra of $$C^\infty(\mathbb R)$$. But perhaps one could accept it as a positive answer in a generalized sense to "Can this algebra $$\mathcal F$$ be realized as an algebra of nice functions on some set $$M$$?"

In Exercise 3.3 the answer is clear for $$\mathcal F_1$$. Take a two-point set $$M = \{a, b\}$$. Then $$\mathcal F_1 \approx \mathcal F(M)$$. Concerning $$\mathcal F_2$$ : I did not check whether multiplication is associative and distributive, but let us assume it (you can do the checkings). I think you wanted to have the complex multiplication here, but also other multiplications producing an algebra are legit.

In 3.4 and 3.5 a general method is developped how to find candidates for solving the general problem. In Example 3.6 he comes back to $$\mathcal F_2$$ and shows that there is no solution.