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## Functoriality

I wanted to talk about the geometric interpretation of localization, but before I do so I should talk more generally about the relationship between ring homomorphisms on the one hand and continuous functions between spectra on the other. This relationship is of utmost importance, for example if we want to have any notion of when two varieties are isomorphic, and so it’s worth describing carefully.

The geometric picture is perhaps clearest in the case where $X$ is a compact Hausdorff space and $C(X) = \text{Hom}_{\text{Top}}(X, \mathbb{R})$ is its ring of functions. From this definition it follows that $C$ is a contravariant functor from the category $\text{CHaus}$ of compact Hausdorff spaces to the category $\mathbb{R}\text{-Alg}$ of $\mathbb{R}$-algebras (which we are assuming have identities). Explicitly, a continuous function

$f : X \to Y$

between compact Hausdorff spaces is sent to an $\mathbb{R}$-algebra homomorphism

$C(f) : C(Y) \to C(X)$

in the obvious way: a continuous function $Y \to \mathbb{R}$ is sent to a continuous function $X \xrightarrow{f} Y \to \mathbb{R}$. The contravariance may look weird if you’re not used to it, but it’s perfectly natural in the case that $f$ is an embedding because then one may identify $C(X)$ with the restriction of $C(Y)$ to the image of $f$. This restriction takes the form of a homomorphism $C(Y) \to C(X)$ whose kernel is the set of functions which are zero on $f(X)$, so it exhibits $C(X)$ as a quotient of $C(Y)$.

Question: Does every $\mathbb{R}$-algebra homomorphism $C(Y) \to C(X)$ come from a continuous function $X \to Y$?

A short digression

It’s not completely obvious whether we should talk about $\mathbb{R}$-algebra homomorphisms or ring homomorphisms. Wonderfully, it turns out not to matter.

Proposition: $\mathbb{R}$ has no nontrivial ring automorphisms.

Proof. Any ring automorphism $\phi : \mathbb{R} \to \mathbb{R}$ has a fixed field $F$ which contains the identity, hence contains $\mathbb{Q}$. Since $\phi(x^2) = \phi(x)^2$, it also follows that $\phi$ preserves sign, hence is monotonic. But a monotonic function which fixes $\mathbb{Q}$ must in fact fix $\mathbb{R}$ since we can approximate any real number by rational numbers from above and from below.

Note that the corresponding claim for functions preserving only addition is false in the presence of the axiom of choice.

As a corollary, any ring homomorphism between $\mathbb{R}$-algebras is in fact an $\mathbb{R}$-algebra homomorphism, so we don’t need to worry about distinguishing between the two.

The above claim seems to have been too strong. However, the argument we presented above is enough to show that, if $f : C(Y) \to C(X)$ is a ring homomorphism, any evaluation homomorphism $e_x : C(X) \to \mathbb{R}$ composes to give a homomorphism $e_x f : C(Y) \to \mathbb{R}$. Such a homomorphism restricts to a homomorphism from the copy of $\mathbb{R}$ sitting inside $C(Y)$ to $\mathbb{R}$ which must fix $\mathbb{Q}$, hence $\mathbb{R}$ by the above argument. Since $e_x$ is an isomorphism from the copy of $\mathbb{R}$ sitting inside $C(X)$ to $\mathbb{R}$, it follows that $f$ is $\mathbb{R}$-linear.

Opposite categories

The motivation for our question is this: we know that $C$ is a functor $\text{CHaus} \to \mathbb{R}\text{-Alg}$ and we know that $\text{MaxSpec}$ is the inverse of this functor on objects, at least up to isomorphism. It would be wonderful if $\text{MaxSpec}$ were also a functor (on some subcategory of $\mathbb{R}\text{-Alg}$ given by the image of $C$) and if it were also the inverse of $C$ on morphisms, since we would then know that the category of compact Hausdorff spaces is equivalent to the opposite of some subcategory of the category of $\mathbb{R}$-algebras! If this were true then it would make no difference whether we studied the one or the other, since any statements about one category would translate directly into statements about the other. (This is one motivation behind the modern definition of the category of affine schemes, but for now I am trying to avoid having to define this category.)

In this particular case, all maximal ideals $m$ of $C(X)$ have the property that $C(X)/m \simeq \mathbb{R}$ and the ring homomorphisms $C(X) \to \mathbb{R}$ are actually $\mathbb{R}$-algebra homomorphisms, so we can write $\text{MaxSpec } C(X) = \text{Hom}_{\mathbb{R}\text{-Alg}}(C(X), \mathbb{R})$, which makes its symmetry with $C$ even clearer. Now let’s follow our noses. Given an $\mathbb{R}$-algebra homomorphism $f : C(Y) \to C(X)$ and given a maximal ideal $m_x, x \in X$ of $C(X)$, we consider the quotient $C(X) \to \mathbb{R}$ induced by $m_x$ and compose it with $f$, giving a map

$C(Y) \xrightarrow{f} C(X) \to \mathbb{R}$.

To conclude that the kernel of this map is a maximal ideal of $C(Y)$ we need to know that the map $C(Y) \to \mathbb{R}$ is surjective. The important point here is that all of the maps involved send identities to identities and also respect scalar multiplication by $\mathbb{R}$. So $f$ sends $m_x$ to some maximal ideal $m_y = f^{-1}(m_x)$, the contracted ideal $(m_x)^c$, hence it sends $x \in X$ to $y \in Y$. In other words, $\text{MaxSpec}$ sends a function $f : C(Y) \to C(X)$ to a function

$M(f) : X \simeq \text{MaxSpec } C(X) \to \text{MaxSpec } C(Y) \simeq Y$.

Proposition: $M(f)$ is continuous.

Proof. It suffices to show that the preimage of a basis element is open. Recall that the topology on $Y$ has basis the sets $U_r = \{ y | r(y) \neq 0 \}, r \in C(Y)$. To say that $M(f)(x) = y$ is precisely to say that $e_x(f) = e_y$. It follows that $e_x(f(r)) = e_y(r)$, and the RHS is nonzero if and only if the LHS is, so it follows that the preimage of $U_r$ is precisely $U_{f(r)}$. Note that the only topological fact used here is that the topology on $X$ (resp. $Y$) coincides with the Zariski topology.

It’s not hard to see that this implies that $C, \text{MaxSpec}$ are inverses on the level of morphisms, hence define a contravariant equivalence of categories as desired.

Functoriality lets you do a lot of things. Morally, it suggests than any interesting definitions or theorems in the topological setting can be translated to the algebra setting. For example, if $X$ is a compact topological manifold, that means it’s locally isomorphic to Euclidean space. Can that condition be translated to a condition on the algebra $C(X)$? The answer is yes; there is a way to translate the notion of localization to the algebra setting and a way to make precise the condition that $C(X)$ should be locally isomorphic to $C(\mathbb{R}^n)$ as algebras, and this improves our understanding both of the topological and the algebraic situation.

If $X$ is a differentiable manifold, it makes more sense to consider the algebra $C^{\infty}(X)$ of $C^{\infty}$ functions on $X$, and again one can translate constructions on $X$ to constructions on the corresponding algebra. For example, a vector field on $X$ turns out to be the same thing as a derivation on $C^{\infty}(X)$. So one can think of derivations on a ring $R$ as “generalized vector fields” on $\text{MaxSpec } R$. There is a lot more one can do with this perspective, but I’m not very familiar with it.

Contravariance has the important property that it exchanges subobjects and quotients, and looking at how this works is one way to see how nice $\text{CHaus}$ behaves as opposed to the full category $\text{Top}$. As mentioned above, if $X \to Y$ is an injective continuous function of CH spaces, then because it is also a closed map it is an embedding, and it gives rise to a quotient $C(Y) \to C(X)$ of rings. In other words, $X$ being a subobject of $Y$ is equivalent to $C(X)$ being a quotient object of $C(Y)$. In the other direction, if $X \to Y$ is a surjective continuous function of CH spaces, then because it is also a closed map it is a quotient map, and it gives rise to an injection $C(Y) \to C(X)$ of rings. In other words, $Y$ being a quotient object of $X$ is equivalent to $C(Y)$ being a subobject of $C(X)$.

Note that just as ring homomorphisms can be factored into the composition of a quotient and an injection, continuous maps between compact Hausdorff spaces can be factored into the composition of a quotient map and an embedding, something that is not true for general continuous functions between topological spaces. What this suggests is that the power of the algebraic perspective is limited to the case where the spaces and/or continuous functions involved are sufficiently nice.

C*-algebras

What we still don’t have is a good description of the algebras of the form $C(X)$. One clue is that when $X$ is compact, real-valued functions $X \to \mathbb{R}$ have the special property that they attain minima and maxima. In other words, it is possible to define the uniform or supremum norm $||f|| = \sup_{x \in X} |f(x)|$ on $C(X)$. In addition, since the uniform limit of continuous functions is continuous, $C(X)$ is complete with respect to this norm, so it is a Banach space, in fact a Banach algebra. This setup should sound familiar from the setup of the Stone-Weierstrass theorem.

The special thing about the supremum norm is that it satisfies a stronger condition, namely the C*-identity

$|| f^2 || = || f ||^2$

(at least, this is what it looks like over $\mathbb{R}$), turning it into a commutative unital C*-algebra over $\mathbb{R}$ with trivial involution. But now the commutative Gelfand-Naimark theorem states that the maximal spectrum of any such algebra must be compact Hausdorff!

Corollary: $\text{CHaus}$ is contravariantly equivalent to the category of commutative unital C*-algebras over $\mathbb{R}$.

This theorem leads in all sorts of interesting directions. For example, starting from an algebra $C(X)$ consider the subalgebra of all functions which are zero at a particular point $p \in X$. This algebra is not unital, and this is reflected in the fact that its spectrum is not compact Hausdorff but locally compact Hausdorff – in fact, it’s precisely equal to $X - p$. So we can use the algebraic method to study locally compact Hausdorff spaces, not just compact ones; instead of taking the algebra of all continuous functions, we take the algebra $C_0(X - p)$ of all functions “vanishing at infinity,” since these are precisely the functions that arise from the above construction. (Intrinsically, this means that for every $\epsilon > 0$ there exists a compact set such that the function is bounded by $\epsilon$ outside of that set.)

A basic reason to be interested in locally compact Hausdorff spaces is that, if we in addition allow the algebras to be over $\mathbb{C}$ instead of over $\mathbb{R}$, we recover Pontryagin duality. One constructs a C*-algebra which generalizes the group algebra of a locally compact Hausdorff abelian group $G$, and the spectrum of this algebra is the dual group $\hat{G}$. However, I’m not sure of the technical details, so I’ll just refer to this post of Terence Tao.

The discrete case is simple enough, however. For example, when $G = \mathbb{Z}/n\mathbb{Z}$ then $\mathbb{C}[G] \simeq \mathbb{C}[x]/(x^n - 1)$. The maximal ideals – “points” – of this algebra are precisely the roots of $x^n - 1$, which in turn define homomorphisms $G \to \mathbb{C}$. Since $\mathbb{C}[x]/(x^n - 1)$ is isomorphic to the space of complex-valued functions on $\hat{G}$, we have a way to turn convolution into pointwise multiplication, one of the hallmarks of the Fourier transform. An equivalent way to get the discrete Fourier transform is to observe that $\mathbb{C}[x]/(x^n - 1)$ is isomorphic to the direct sum of the algebras $\mathbb{C}[x]/(x - \zeta_n^i), 0 \le i < n$ by the Chinese Remainder Theorem.

The upshot of all of this is that, on the algebra side, there’s no reason to restrict to commutative algebras. Thus we can instead consider general C*-algebras and think of their spectra as “noncommutative topological spaces.” This is one of the basic motivations behind the noncommutative geometry program, about which I know next to nothing, so we won’t pursue this line of inquiry further.

### 10 Responses

1. […] For compact Hausdorff, the Banach algebras will be particularly important to keep in mind as examples in this post. Keep in mind the following, which was discussed previously here and here. […]

2. Dear Qiaochu,

As you pointed out, consider R as a group, we have $|Aut(\mathbb{R})|=1$. May I know as to why we look at Automorphism groups of different groups. Are they useful at all?

Next, are there other groups which have trivial automorphisms?

3. My preceding post timed 1:34 pm was actually started earlier and was a reply to your comment about ring morphisms being R-algebra morphisms at 10:59 am. But I was interrupted and when I actually submitted my counter-example I hadn’t seen your modified text. I completely agree with it now.

4. Oh, you’re absolutely right in saying that the image of R must be isomorphic to R, but this doesn’t imply that f is a morphism of R-algebras. Here is an example.

Take the R-algebras A and B both to have underlying ring S=R@R ( I am using @ for the tensor product over the rationals).
The R-algebra A is the ring S endowed with the product r.(x@y)=rx@y (r in R).
Similarly, the R-algebra B is the ring S endowed with the product r.(x@y)=x@ry (r in R).
Then the identity f=id: A–>B is of course a ring morphism but not a morphism of R-algebras because
f(r.(1@1))=f(r@1)=r@1 is unequal to r.f(1@1)=r.1@1=1@r
whenever r is an irrational real number.

5. $\mathbb{C}$ is a $\mathbb{R}$ algebra that has a lot of field automorphisms that are not $\mathbb{R}$-linear.

• Here is an exposition about them: http://www.jstor.org/stable/2689301

• Whoops. Hmm. It still seems plausible to me that ring homomorphisms between algebras all of whose residue fields are $\mathbb{R}$ are $\mathbb{R}$-linear.

6. Your statement that the only ring automorphism of R is the identity and its proof are perfectly correct.
However it’s not clear why it implies that a ring morphism from C(Y) to C(X)is an R-algebra morphism.
The problem is that I don’t see why this ring morphism should send R to R.
Thanks for this interesting and well-written post,
Attila

• I am reasonably certain that the following argument works: given a ring homomorphism $f : A \to B$ between two $\mathbb{R}$-algebras, the image of $\mathbb{R}$ must be isomorphic to $\mathbb{R}$ and it must contain the prime subfield $\mathbb{Q}$ of $B$. The above proof then applies.

• That argument seems to not work. However, I think the new argument in the post is correct.