Applications of Zariski topology outside alg. geometry - MathOverflow

Applications of Zariski topology outside alg. geometry

2012-03-10T00:57:14Z

Are there applications of the Zariski topology in mathematics that are not within the scope of algebraic geometry (including schemes and algebraic groups) ?

There is an older question with a similar title (http://mathoverflow.net/questions/21502/what-is-the-zariski-topology-good-bad-for) but the answers given there are mainly concerned with the geometry stuff.

Answer by Ralph for Applications of Zariski topology outside alg. geometry

2012-03-10T02:20:04Z

The Zariski topology can be used to construct the Stone-Čech compactification $\beta X$ and the real compactification $\nu X$ of a topological space $X$: The Stone-Čech compactification is just the maximal ideal spectrum of $C_b(X)$ (the ring of bounded continuous functions $X \to \mathbb R$) endowed with the Zariski topology. Since there are other ways to construct these compactifications (see the wiki article), the appearance of the Zariski topology is not that obvious.

The Stone-Čech compactification is used intensively in functional analysis and allows to boil down questions concerning rings of continuous functions to the case of a compact space $X$. This is used for instance in Riesz representations of linear functionals of $C_b(X)$ and related spaces.

Answer by Hailong Dao for Applications of Zariski topology outside alg. geometry

2012-03-10T02:47:52Z

In modular representation theory there has been quite a bit of work on classifying certain kind of subcategories using subsets of the spectrum of the cohomology ring. For example see this recent paper and survey as well as the references.

Answer by Georges Elencwajg for Applications of Zariski topology outside alg. geometry

2012-03-10T09:08:49Z

Given two $n\times n$ matrices $A,B$ over a field $k$ let's prove that the characteristic polynomials of $AB$ and $BA$ are equal: $\chi(AB)=\chi(BA)$.
Since the characteristic polynomial of a matrix obviously doesn't change under field extension , we may and do assume $k$ algebraically closed

If $A$ is invertible, the result is clear because $\chi(BA)=\chi (A(BA)A^{-1})=\chi (AB) $.
Now fix $B$ and consider the set $F\subset M_n(k)$ of all $A$ for which $\chi(AB)=\chi(BA)$.
It is closed in the Zariski topology of $M_n(k)\cong \mathbb A^{n^2}(k)=k^{n^2}$ (because the characteristic polynomial of a matrix $M$ has as coefficients polynomials in the entries of $M$).
Since, as we have just seen,it contains the open non-empty set of invertible matrices $A$, it is dense by irreducibility of $\mathbb A^{n^2}(k)$ (which requires that $k$ be algebraically closed).
Since $F$ is closed and dense, we have $F=\mathbb A^{n^2}(k)$: all matrices $A$ satisfy $\chi(AB)=\chi(BA)$

Many theorems in elementary linear algebra can similarly be proved by using the Zariski topology on $M_n(k)$

Answer by Marc Palm for Applications of Zariski topology outside alg. geometry

2012-03-10T12:08:40Z

(In some sense, there is some overlap with Ralph's answer)

Gelfand Naimark theorem.

For a commutative $C^\star$ algebra $A$, the spectrum of $A$ is the set of primitive ideals (=kernel of functionals). With the Zariski topology ($C^\star$ algebraist prefer the notion Jacobson topology/hull-kernel topology), they become a topological space $X$ and we have $C_0(X) \cong A$. This yields an anti equivalence between locally compact Hausdorff spaces with commutative $C^\star$ algebras. This equivalence generalizes to so called to sober spaces, where the dual objects are complete Heyting algebras. So from this experience, it seems natural to topologize the dual of an algebra and see how much is encoded.

Pontryagin duality:

The Gelfand Naimark theorem can be enhanced to the Pontryagin duality of locally compact abelian groups.

Note that the Gelfand Naimark theorem was first, and probably inspired some of the constructions in algebraic geometry. Similar things are happening with spectral triples in Arakelov theory now, I guess.