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Funny definition of dimension of a topological space

  1. Jul 10, 2005 #1
    I'm reading this book (Hartshorne), and it uses a funny definition of topological dimension, which I'm having a hard time convincing myself is the usual one. The definition is as follows:

    dim X is the supremum of natural numbers such that there exists a chain [itex]Z_0\subset Z_1\subset \dotsb \subset Z_n[/itex] of distinct closed irreducible subsets of X.

    And a subset Y of X is defined to be irreducible if it can expressed as the union of two proper subsets which are closed in Y.

    Compare this with the normal definition that you learn in Munkres:

    The dimension of a space X is one less than the infimum of the orders of refinements of open covers.

    OK, so Hartshorne has some examples using his definitions, but only using the Zariski topology. That's fine, and the results agree with what you expect. But I wanted to prove that the Hartshorne definition will give the same result as the Munkres definition for, eg, the real line with its standard topology.

    It seems to me that there simply are no closed irreducible sets in the real line with its standard topology (while there are closed irreducible sets in the Zariski topology). For take any closed interval, it can be written as the union of two closed intervals. Can I conclude then that the Hartshorne dimension of the standard real line is zero?
     
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  3. Jul 10, 2005 #2

    mathwonk

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    the hartshorne definition is the same as the one on maximal chains of subvarioeties in ym post on dimesnion in the general math forum.

    i would guess it applies in general only to noetherian topological spaces, ones in which there is only a finite descendiong chain of closed sets beginnig from any given closed set.

    look later in hartshorne however, and see it equates to sheaf theoretic dimension by a theorem of grothendieck on vanishing of cohomology. then you can relate it to the munkres definition, which is lebesgue covering dimension and hence directly related to vanishing of cech cohomology as i discussed in the thread mentioned above.
     
  4. Jul 10, 2005 #3

    Hurkyl

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    Don't forget {P} is an irreducible, closed set of R, for any P in R!

    I thought about adding generic points to R, one corresponding to each closed set, but then the dimension would be uncountably infinite.
     
  5. Jul 10, 2005 #4

    mathwonk

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    generic points usually correspond only to irreducible closed sets, but hurkyl is trying to extend this somehow.
     
  6. Jul 10, 2005 #5

    mathwonk

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    we had a discussion last eyar in august 2004 about prime ideals of the rign of continuous functions no the real line and you produced a family of exampels which were noit maximal. so in p[articular prime ideals do not correspond to irreducible subsets, and mroeover are not dwtermiend by their set of common zerowes, since all prime ideals have ezxactly one common zero. so what is the geometric object corresponding to a prime ideal? it is some "scheme" like structure on a point presumably.

    maybe c infinity functions would be ambitious enough at first. but it seems there are arbitrarily large chains of prime ideals in that ring right? i forget a bit of the details from alst year.


    so the krull dimension seems to be infinite, and if we allow irreducible schemes, also the zariski or hartshorne dimension.
     
  7. Jul 11, 2005 #6
    um... what's a "ym post"?

    *Edit: I guess that's "my post", I think I've found it*

    I'm not too familiar with the term "noetherian space", but I guess the real line with the standard topology is not noetherian? I can easily make an infinite descending chain of closed sets. So then I can't use Hartshorne's dimension for this space?

    I'll take a book. I have to admit, though, it's slow going for me with this guy.
     
    Last edited: Jul 11, 2005
  8. Jul 11, 2005 #7
    Oh, of course! For some reason, I was only considering closed intervals. So this gives me a chain of length 1, right? The dimensions agree then.

    I don't know what you're talking about here.
     
  9. Jul 11, 2005 #8

    Hurkyl

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    Yah, but mere coincidence, since they won't for R^2.


    Then ignore it. It was a silly idea anyways!
     
  10. Jul 11, 2005 #9
    Hm, yes, so if the points are the only irreducible sets, then we'll get that R^n is dimension 1, for any n. That's no good.
     
  11. Jul 12, 2005 #10

    mathwonk

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    if you want to actually understand some algebraic geometry, it is better to begin with shafarevich, or even with miles reid, rather than with hartshorne.

    algebraic geometry is a subject in which the evolution of the ideas has been so long, that their intuitive meanings is in the distant past. so most current versions of each idea are very hard to understand.

    people thought riemann's original paper on algebraic plane curves, used some analysis that was not fully justified, so already before 1900, there were attempts to completely algebraicize his analytic and topological concepts, to render them not more understandable, but more provable.

    it is very hard to appreciate even the 1890's version of algebraic geometry fully, much less the 1980's version, without knowing riemann's original version.

    for example, riemann himself would not understand the definition of the genus of a curve in hartshorne, as the dimension of the sheaf cohomology group h^1(O), on page 294, unless he saw the comment on line three of page 414, that it equals the dimension of the space H^0(K).

    Hartshorne apparently does not even mention the basic exact differentiation sheaf sequence 0-->C-->O-->K-->0, for complex curves, yielding 0-->H^0(K)-->H^1(C)-->H^1(O)-->0, where C is the complex numbers. This relates the topology H^1(C) and the analysis H^0(K) (both due to riemann), to the sheaf cohomology group H^1(O), which illustrates the relation between hartshorne's definition and riemann's definition.

    hartshorne has tried to explain the very most modern and abstract version of algebraic geometry, and has done a very good, job, but to understand the intuition takes a lot more preliminary study at a much more elementary level.

    how many people would realize that the coboundary homomorphism

    H^0(O(D)|D)-->H^1(O), induced from the sheaf sequence

    0-->O-->O(D)-->O(D)|D-->0,

    similar to the one on page 296 of hartshorne, line 10,

    is a version of riemann's period mappings for integrals (in roch's normalization), from meromorphic differentials to complex cohomology? (page 244, bottom, the map "phi", in griffiths and harris, principles of algebraic geometry).

    you probably are not going to learn it in either of those excellent books.

    notice that both these arguments for rrt, classical and sheaf theoretic, appear in grifftihs - harris, on pages 244-245, and 246. but they use on p. 246 the "one point at a time" version of the sheaf sequence as in hartshorne, which proves the result inductively but avoids relating the two proofs. the proof on page 244 of GH is however exactly analogous to the version of the sheaf proof given in this post.

    this kind of connection drawn between classical and modern work tends to be found mainly in works by george kempf, in my experience.

    i would however be happy to learn of any other references which contain this sort of connection, in particular any published source for the remark above about the sheaf theoretic version of the classical rrt proof by roch (that the map H^0(O(D)|D)-->H^1(O) is the sheaf version of riemann period map).
     
    Last edited: Jul 13, 2005
  12. Jul 15, 2005 #11
    How do you feel about Joe Harris's book?
     
  13. Jul 15, 2005 #12

    mathwonk

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    i love joe harris'; book. i admire him a great deal. he has a terrific grasp of the geometry of the subject and writes in a very intuitive appealing way. it is completely unlike kempf's writing but it is very instructive.

    joe gives you a great feel for the subject, as much information as possible with as few prerequisites as possible.

    joe is very intuitive and geometric whereas kempf is very precise and more algebraic. one should probably begin with joe's book.
     
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