Stephen Paul King (stephenk1@home.com)
Thu, 25 Mar 1999 18:37:44 GMT
On Thu, 25 Mar 1999 01:36:18 GMT, John Baez wrote:
>In article <36F2740B.96314AA3@uiuc.edu>, Eric Forgy <forgy@uiuc.edu> wrote:
>
>>It looks like maybe I was blurring the notion of duality in my last post. What
>>I described (I thought was Poincare duality) appears to be Fourier duality,
>>i.e. the functions f in L* map vectors A in L to a real number.
>>
>>f(A) = f_1 A^1 + f_2 A^2 + ... + f_n A^n
>>
>>But actually, I thought that notion was defining the space of differential
>>forms and that differential forms were related to Poincare duality. Hmm...
>
>Ah, you're sensing a relationship between Poincare duality and the duality
>for vector spaces!
>
>Like I said, it's good to spend a couple of hours per month trying to
>relate different notions of duality. If you spend less than this you'll
>never see the grander patterns lurking beneath the surface, but if you
>spend much more, you run the danger of going insane.
>
>In my previous post I described Poincare duality as an operation on cell
>decompositions of a fixed n-dimensional manifold M. The Poincare dual of
>a p-dimensional cell is an (n-p)-dimensional cell. If M is compact and
>oriented, we can use this idea to prove that the pth real homology
>of M has as its dual vector space the (n-p)th real homology. In other
>words, H_p(M,R)* is isomorphic to H_{n-p}(M,R). So Poincare duality
>is related to vector space duality.
>
>We can also prove a similar thing for cohomology. Here it's convenient
>to use differential forms. There's an operation called the "Hodge dual"
>that turns a p-form on an oriented Riemannian manifold into an (n-p) form.
>Using this we can prove that that if M is compact and oriented, H^p(M,R)*
>is isomorphic to H^{n-p}(M,R). Sometimes people call this idea "Hodge
>duality".
>
>>The lattice version of Poincare duality that you mentioned is also called
>>barycentric subdivision.
>
>Actually the Poincare dual is not the same as the barycentric subdivision.
>
>If you reread my description you'll see the differences. For example, the
>barycentric subdivision of a cell decomposition X has one vertex for each
>cell of X. The Poincare dual of X only has one vertex for each n-cell of X.
>The barycentric subdivision of a triangulation is always a triangulation,
>while the Poincare dual is usually not. The Poincare dual of the Poincare
>dual of X is X, while the barycentric subdivision of the barycentric
>subdivision of X is not the same as X.
>
>>There is also another way to form a lattice dual via a
>>Voronoi subdivision.
>
>I don't know what this is. Does it really deserve the name "dual"?
>I.e., is the Voronoi subdivision of the Voronoi subdivision of a
>lattice the original lattice? For some reason I doubt it - probably
>because you're calling it a "subdivision".
>
>
>
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