Applications of Zariski topology outside alg. geometry

Question

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 (What is the Zariski topology good/bad for? ) but the answers given there are mainly concerned with the geometry stuff.

What does this question mean? You might as well ask whether algebraic geometry is useful (answer: yes) — Igor Rivin, Mar 10 '12 at 01:03
I'm not asking whether Zariski topology or algebraic geometry is useful but if there are examples for the usage of the Zariski topology in other areas of mathematics than the geometric ones. — Todd Leason, Mar 10 '12 at 01:17
Well, Zariski topology is defined in the framework of algebraic geometry, and is basic to the subject, so it would seem close to inconceivable that the question you ask is different from the broader question (with the obvious answer). — Igor Rivin, Mar 10 '12 at 01:45
This is almost an dublicate of http://mathoverflow.net/questions/29271/algebraic-geometry-used-externally-in-problems-without-obvious-algebraic-struc — Martin Brandenburg, Mar 12 '12 at 10:05

score 23 · Accepted Answer · answered Mar 10 '12 at 09:08

23

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 $M_n(k)$

answered Mar 10 '12 at 09:08

Georges Elencwajg

46,833

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+1. To be honest though, one can eliminate Zariski from this exact reasoning completely, keeping its logic unaffected: this is a statement in the ring $R=\mathbb{Z}[a_{ij},b_{ij}]$ generated by matrix elements of $A$. The field of fractions of $R$ can be embedded into $\mathbb{C}$, so it is enough to work over complex numbers, where on the complement to $\det(A)=0$ our identity obviously holds (as you indeed explain), so by continuity we have it everywhere. – Vladimir Dotsenko Mar 10 '12 at 10:29
matrix elements of $A$ and $B$, of course!

Vladimir Dotsenko

Mar 10 '12 at 10:30

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Dear @Vladimir: of course, you can eliminate the Zariski topology from all of linear algebra (where it is in fact generally not invoked) and even (with more difficulty) from huge parts of algebraic geometry: after all there were great algebraic geometers before Zariski or even topology were born!
But I think that if the Zariski topology comes naturally to the minds of mathematicians with certain mathematical habits , there is no need for them to try to hide it. Other mathematicians, like you, are welcome to express the proof in their favourite formulation. – Georges Elencwajg Mar 10 '12 at 12:48

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But of course! I am just still excited after many years since learning that trick I described in my comment... – Vladimir Dotsenko Mar 10 '12 at 13:31

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Very nice! This may be a dense question: Are the char. polynomials equal over any commutative ring or only over a field ? – Todd Leason Mar 11 '12 at 00:46

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Dear @Todd, they are equal over any commutative ring. Indeed they are equal for the generic matrices $A=(X_{ij}), B= (Y_{ij})$ with indeterminates as entries because this happens over the field $\mathbb Q((X_{ij});(Y_{ij}))$ and then you can specialize to elements $a_{ij},b_{ij}\in A$ of an arbitrary commutative ring $A$ (This is the trick that Vladimir mentions). And, by the way, your question is quite judicious : +1. – Georges Elencwajg Mar 11 '12 at 01:49

See also my answer here: http://mathoverflow.net/questions/29271 – Martin Brandenburg Mar 12 '12 at 10:06

score 8 · Answer 2 · answered Mar 10 '12 at 02:20

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.

score 4 · Answer 3 · answered Mar 10 '12 at 02:47

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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.

answered Mar 10 '12 at 02:47

Hailong Dao

30,261

Marc Palm · Answer 4 · 2012-03-12T10:09:47.383

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(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.

edited Mar 12 '12 at 10:09

answered Mar 10 '12 at 12:08

Marc Palm

11,097

Surely Pontryagin duality requires more than just Gelfand-Naimark? If not, would you care to expand on this comment? – Yemon Choi Mar 10 '12 at 23:23
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Regarding sober spaces, what are the algebraic objects dual to these? Certainly not only the commutative $C^\ast$-algebras. (I guess you are referring to https://en.wikipedia.org/wiki/Sober_space ) – Yemon Choi Mar 10 '12 at 23:26
No, there is a comment on the wikipedia side (Stone Duality). I pretty much only know of the existence of this duality, and do not dare more than referring you to it. so perhaps you better check the wikipedia side. Of course, the Pontryagin duality is more special, and more an enhancement of the Gelfand Naimark theorem, since you have additional (group) structure. – Marc Palm Mar 12 '12 at 09:53

Applications of Zariski topology outside alg. geometry

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