Contents

Contents

Idea

Where the center of a category is in general just a commutative monoid (the endomorphism monoid of its identity functor formed in the functor category), for additive categories this commutative monoid carries the further structure of a commutative ring: the endomorphism ring of its identity functor. (In fact this is also true for Ab-enriched categories, which are more general.)

Definition

Definition

For $\mathcal{C}$ an Ab-enriched category (e.g. an additive category) its center is

$Z(\mathcal{C}) \;\coloneqq\; [\mathcal{C}, \mathcal{C}] \big(id_{\mathcal{C}}, id_{\mathcal{C}}\big) \;\; \in \; CRing \,,$

where

• $[\mathcal{C}, \mathcal{C}]$ denotes the enriched functor category of enriched endofunctors on $\mathcal{C}$,

• $id_{\mathcal{C}}$ denotes the identity functor on $\mathcal{C}$.

By Ab-enrichment, this means that $Z(\mathcal{C})$ carries the structure of a commutative monoid object internal to Ab, hence: the structure of a commutative ring.

Examples

Proposition

Let $R Mod$ denote the category of left modules of a ring $R$, and let $Z(R)$ be the center of $R$. Then $Z(R Mod) \cong Z(R)$.

Proof

First note that for any $r \in Z(R)$, multiplication by $r$ acts as an endomorphism of each $R$-module, and this endomorphism is natural. This gives a ring homomorphism from $Z(R)$ to $Z(R Mod)$ which is injective because distinct elements of $r$ act differently as multiplication on the $R$-module given by $R$ itself. To see that it is also surjective and hence bijective, suppose $\alpha$ is a natural transformation of the identity functor on $R \Mod$. Then $\alpha_R \colon R \to R$ must be right multiplication by some $r \in R$, since every endomorphism of $R \in R Mod$ is given by right multiplication by some $r \in R$. Because $\alpha_R$ is natural and right multiplication by any $s \in R$ gives an endomorphism of $R \in R Mod$, we have

$(x s) r \, = \, \alpha_R(x s) \, = \, \alpha_R(x)s \, = \, (x r) s$

for all $x, s \in R$, so $r \in Z(R)$. More generally, for any module $M$ and any $m \in M$ there is a module homomorphism $f \colon R \to M$ with $f(1) = m$, which by naturality implies

$\alpha_M (m) \,=\, \alpha_M\big(f(1)\big) \,=\, f\big(\alpha_R(1)\big) \,=\, f(r) \,=\, r m \,,$

which shows that $\alpha_M$ is multiplication by $r$.

Remark

Two rings whose categories of modules are equivalent as Ab-enriched categories are said to be Morita equivalent. As a consequence of Prop. , Morita equivalent commutative rings are already isomorphic.

As a further illustration of these ideas we show how the topology on a compact Hausdorff space is determined by the category of vector bundles over this space. For any compact Hausdorff space $X$ let $Vect(X)$ denote the category of (finite-rank complex) vector bundles over $X$. This category is $\mathbb{C}$-linear, i.e. enriched over the category of complex vector spaces. Thus, the center of $Vect(X)$ is a commutative algebra over $\mathbb{C}$. Moreover:

Proposition

If $X$ is a compact Hausdorff space then the center of $Vect(X)$ is $C(X)$, the function algebra of complex-valued continuous functions on $X$.

Proof

For any field $k$ suppose $A$ is a commutative $k$-algebra. Let $A Proj$ be the category of finitely generated projective $A$-modules. This is a $k$-linear category, and a straightforward extension of the proof of Prop. shows that the center of $A Proj$ is isomorphic to $A$, not merely as a commutative ring, but as a commutative $k$-algebra.

Let $X$ be a compact Hausdorff space. By Swan's theorem, $Vect(X)$ is equivalent, as a $\mathbb{C}$-linear category, to $C(X) Proj$. Thus the center of $Vect(X)$ is isomorphic to $C(X)$.

Corollary

Suppose $X$ and $Y$ are compact Hausdorff spaces such that $Vect(X)$ and $Vect(Y)$ are equivalent as $\mathbb{C}$-linear categories. Then $X$ and $Y$ are homeomorphic.

Proof

By Prop. , if $Vect(X)$ and $Vect(Y)$ are equivalent as $\mathbb{C}$-linear categories then $C(X)$ and $C(Y)$ are isomorphic as complex algebras. The Gelfand-Naimark theorem implies that when $X$ is compact Hausdorff, it is homeomorphic to the set of algebra homomorphisms $C(X) \to \mathbb{C}$, given the topology of pointwise convergence. Thus the isomorphism of algebras $C(X) \cong C(Y)$ implies that $X$ and $Y$ are homeomorphic. (Note that we did not need to show $C(X) \cong C(Y)$ as $C^\ast$-algebras here, but this follows.)

Remark

Note that it is much easier to recover $C(X)$ and thus $X$ starting from $\Vect(X)$ as a symmetric monoidal $\mathbb{C}$-linear category, since then the endomorphism algebra of the unit object, the trivial line bundle over $X$, is $C(X)$.

Remark

An analogue of Prop. also holds for real vector bundles: the center of the $\mathbb{R}$-linear category of real vector bundles over $X$ is the algebra of continuous real-valued functions on $X$, and from this we can recover $X$, either by using the real version of the Gelfand-Naimark theorem, or by complexifying this algebra and using the usual complex version of the Gelfand-Naimark theorem.

Properties

• In general, the construction of centers is not functorial (except with respect to equivalence of categories); but it is functorial in some important special circumstances, such as certain reconstruction theorems.

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References

Early occurrence of the definition of the center of an additive category:

Last revised on May 18, 2023 at 14:42:20. See the history of this page for a list of all contributions to it.