TITLE: What properties should $C(M,\mathbb{R})$ have when $M$ is a $n$-dimensional manifold?
QUESTION [6 upvotes]: Let $M$ be a n-dimensional manifold, $C(M,\mathbb{R})$ be the function space of continuous function from $M$ to $\mathbb{R}$. What kind of properties should $C(M,\mathbb{R})$ has, to reflect the manifold structure?
Denote a L-shape line be $X$, we know it is not a manifold, but what difference can we observe from the algebras of continuous function compared to circle? i.e. What properties does $C(M,\mathbb{R})$ has and $C(X,\mathbb{R})$ does not? In this case their topological dimension is the same, so they should have same stable rank in terms of $C^*$-algebras.

REPLY [8 votes]: For compact orientable manifolds, this is accomplished by the notion of a spectral triple.
See the question Commutative spectral triples for additional information, including a precise statement of the theorem.
See also the answer to Noncommutative smooth manifolds for additional remarks about orientability.
As a side remark, if one is willing to use smooth functions instead
of continuous functions, then there is a very satisfactory answer:
the resulting contravariant functor from second countable Hausdorff smooth manifolds to commutative real algebras is a fully faithful functor, i.e., manifolds are (contravariantly) identified with a full subcategory of real algebras.  This is known as Milnor's exercise.
See also the answer How much of differential geometry can be developed entirely without atlases? for a discussion on how to characterize the essential image of the embedding functor.

REPLY [4 votes]: This is not really an answer but rather a long-ish comment.
First of all, if $M$ is not compact, $\mathfrak{A}=C(M,\mathbb{R})$ is not really a C${}^*\!$-algebra but actually only a locally C${}^*\!$-algebra, that is, a locally convex ${}^*\!$-algebra which admits a separating family of submultiplicative C${}^*\!$-seminorms, unless you require e.g. its elements to vanish at infinity. If $M$ is Hausdorff, paracompact and second countable, this family of seminorms can be in addition chosen to be countable (i.e. the underlying locally convex vector space is Fréchet). So, one could see non-compactness represented algebraically as the non-(C${}^*\!$-)normability of $\mathfrak{A}$. The second axiom of countability, on its turn, is characterized by the metrizability of $\mathfrak{A}$. If the elements of $\mathfrak{A}$ are instead required to vanish at infinity, you get the well-known characterization of non-compactness of $M$ by the fact that $\mathfrak{A}$ is then not unital.
About the Hausdorff property, it is related to the capability of $\mathfrak{A}$ to separate the points of $M$. More precisely, $C(X,\mathbb{R})$ only actually sees the Tychonoffication of $X$ for any topological space $X$, which is defined by taking the quotient of $X$ modulo the equivalence relation $\sim$ which states that $x\sim y$ if $f(x)=f(y)$ for all $f\in C(X,\mathbb{R})$. Particularly, since a Tychonoff topological space is always Hausdorff, $\mathfrak{A}$ cannot see at all whether $M$ is Hausdorff or not. As for paracompactness, there is a nice characterization of it in terms of $C(X,\mathbb{R})$ if $X$ is locally compact (which is always the case if $X=M$ is finite dimensional): $X$ is paracompact if and only if the ${}^*\!$-ideal $C_c(X,\mathbb{R})\subset C(X,\mathbb{R})$ of continuous real-valued functions with compact support is a projective $C(X,\mathbb{R})$-module. This, of course, leaves open the question about how to characterize this compactly supported ideal in algebraic terms.
The above discussion shows how the "size" and the "separation" of $M$ are encoded in the structure of $\mathfrak{A}$. Other aspects one may look for an characterization of in terms of properties of $\mathfrak{A}$ are:

The (topological) dimension of $M$ (particularly, whether it is finite or not, which on its turn will determine whether $M$ is locally compact or not);
The local structure of $M$, i.e. the fact that $M$ is locally Euclidean;
The smooth structure of $M$ (if any).

The local structure of $M$ requires one to look at not only $\mathfrak{A}$ alone, but at a whole (pre)sheaf of ${}^*\!$-algebras over $M$. Local algebras over (say) contractible domains should look like $\mathfrak{A}_0=C(\mathbb{R}^n,\mathbb{R})$. Therefore, at the local level, one should ask what is special about $\mathfrak{A}_0$ as a (Fréchet) locally C${}^*\!$-algebra.
As for an occasional smooth structure on $M$, recall that such a structure is not necessarily fixed by the topology of $M$ alone. Typically this is reflected in the structure of (unbounded) derivations on $\mathfrak{A}$ in suitable dense domains, which on their turn should play the role of "smooth" functions on $M$. This structure is, of course, absent for topological manifolds which do not admit a smooth structure or only do so in (open) subsets thereof, which once again displays the need for looking at the local (i.e. presheaf) structure. They can also be used together to encode the dimension of $M$, as in e.g. the top de Rham cohomology of contractible domains. Edit: As pointed in Dmitri Pavlov's answer, a  more precise formulation of this idea is provided by the notion of a spectral triple, which is pretty well developed in the case of a compact $M$.