# Swan like theorem and covering spaces

Let $X$ be a finite CW complex. Swan's theorem provide an equivalence $Vec(X)~\xrightarrow{\sim} ~ProjMod(hom_{Top}(X,\mathbb{R}))$ between the category of finite dimensional vector bundles over $X$ and the category of finitely generated projective modules over the ring of continuous functions from $X$ to the reals. This isomorphism behaves well with the monoidal structures $\oplus$.

There is an intermediate step in this construction: The category $Vec(X)$ of finite dimensional vector bundles over $X$ is equivalent to locally free modules of finite rank over the sheaf $C_X(-)=hom_{Top}(-,\mathbb{R})$ on $X$.

The category $Cov(X)$ of covers of $X$ is equivalent to the category of locally constant sheaves of sets on $X$. Is it possible to formulate this analogously to the above correspondence? So maybe locally constant sheaves are somehow special modules over $C_X(-)$ and this category possibly corresponds to some special modules over $C_X(X)$. Maybe, this is also compatible with disjunct unions of coverings and sums of the corresponding modules. Maybe it is also necessary to require, that the covering is regular.

(The fat things are edits made, partially based on the answers below.)

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It might be worth exploring the connection between vector bundles with flat connections and covering spaces via representations of the fundamental groupoid. It then remains to think about what vector bundles with flat connections are modules for, but my guess is that stacks would come in and/or differential cohomology. –  David Roberts Oct 27 '10 at 22:31

If I can take only the finite covers, then yes, I think. (After all, Swan's theorem is a characterization of finite-dimensional vector bundles, not all vector bundles.) This is easier to do over $\mathbb{C}$ than over $\mathbb{R}$. In addition to the entire sheaf $C_X(-)$, let $C(X) = C_X(X)$ be the algebra of global continuous functions.

Since the module $M$ is locally free, what you want to do is to choose a basis for $C(U) = C_X(U)$ for enough open sets $U$, and such that the bases agree when you restrict to smaller open sets. You could just ask for this directly, but there is an indirect algebraic condition that comes to the same thing. Namely, you can ask for $M$ to not only be finitely generated and projective, but also a semisimple commutative algebra over $C(X)$. This gives you the unordered basis in each fiber.

Over $\mathbb{R}$, it's not quite enough to require that $M$ be a semisimple algebra, because you could end up creating $C(Y,\mathbb{C})$ for a finite cover $Y$ of $X$. So, you could also impose the condition that $f^2 + g^2 = 0$ has no non-trivial solutions in $M$.

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Unfortunately I do not understand your answer. What is $M$ here? If the cover of $X$ is regular, the fibers are discrete groups and in particular normal subgroups of $\pi_1(X)$. I wonder what the action of $C(X)$ is. –  roger123 Oct 28 '10 at 9:32
In the end $M = C(Y)$, where $Y$ is the covering space of $X$. The projection from $Y$ to $X$ induces a ring homomorphism from $C(X)$ to $C(Y)$, and then $C(X)$ acts on $M$ via this ring homomorphism. –  Greg Kuperberg Oct 28 '10 at 10:05
What I'm saying is, if $M$ is a $C(X)$-algebra which (a) satisfies Swan's conditions as a module, (b) is a semisimple algebra, and (c) has no solutions to $f^2 + g^2 = 0$, then, conclusion, it is isomorphic to $C(Y)$ for a finite cover $Y$. –  Greg Kuperberg Oct 28 '10 at 17:40
@roger123 You can construct a locally constant set-theoretic sheaf directly this $C(X)$-algebra $M$. You can first make a sheaf version of $M$, and then you can look at the corresponding sheaf of minimal idempotents. (Recall that an idempotent in a ring is an element $a$ such that $a^2 = 0$. An idempotent is minimal if it is not the sum of two [commuting] idempotents.) Or, if you like, the minimal idempotents are a canonical basis of $M(U)$ for an open set $U$ on which $M$ is trivial. –  Greg Kuperberg Oct 29 '10 at 17:57
@roger123 If $Y$ is a finite-sheeted covering of $X$, then the gluing maps between local trivializations of $Y$ are functions from $U \subset X$ to permutations. You can get a vector bundle if you replace these permutations by the same permutation matrices. Or, in the present context, $C(Y)$ satisfies Swan's theorem as a module over $C(X)$ and thus gives you a vector bundle, exactly the same bundle as you would get from using permutation matrices. –  Greg Kuperberg Oct 30 '10 at 6:03
For every locally connected space $X$ there exists a category equivalence between the category of regular covering spaces of $X$ and the category of Galois extensions of $C(X)$.