Integrals from a non-analytic point of view - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T08:36:19Z http://mathoverflow.net/feeds/question/38439 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/38439/integrals-from-a-non-analytic-point-of-view Integrals from a non-analytic point of view James D. Taylor 2010-09-12T03:08:53Z 2010-12-15T08:55:39Z <p>I've mentioned before that I'm using this forum to expand my knowledge on things I know very little about. I've learnt integrals like everyone else: there is the Riemann integral, then the Lebesgue integral, and then we switch framework to manifolds, and we have that trick of using partitions of unity to define integrals.</p> <p>This all seems very ad hoc, however. Not natural. I'm aware this is a pretty trivial question for a lot of you (which is why I'm asking it!), but what is the "correct" natural definition we should think of when we think of integrals?</p> <p>I know there's some relation to a perfect pairing of homology and cohomology, somehow relating to Poincare duality (is that right?). And there's also something about chern classes? My geometry, as you can see, is pretty confused (being many years in my past).</p> <p>If you can come up with a natural framework that doesn't have to do with the keywords I mentioned, that would also be very welcome.</p> http://mathoverflow.net/questions/38439/integrals-from-a-non-analytic-point-of-view/38453#38453 Answer by Theo Johnson-Freyd for Integrals from a non-analytic point of view Theo Johnson-Freyd 2010-09-12T07:41:10Z 2010-09-13T04:35:07Z <p>If you are interested in manifolds, then you might be interested in various related notions of <em>measures</em> and <em>distributions</em>. Let $M$ be a smooth manifold with algebra of functions $C^\infty(M)$. There is a very general notion of a <em>distribution</em>, which is that of any linear function $C^\infty(M) \to \mathbb R$. In this framework, a distribution is a <em>measure</em> if it satisfies a positivity condition, namely that it takes everywhere-nonnegative functions to nonnegative numbers.</p> <p>Another, different definition of "distribution" corresponds to the <em>distribution bundle on $M$</em>, which is a canonical trivializable line bundle on $M$. It can be presented by gluing data and transition amplitudes as follows. Let $U,V \subseteq M$ with $\phi: U \to \mathbb R^n$ and $\psi: V \to \mathbb R^n$ be coordinate charts, and consider the trivial bundles one-dimensional bundles over $U,V$. We glue them together by giving transition data: if $f$ is a section over $U$ of the trivial bundle, on $U\cap V$ we identify it with the section $f \cdot \left| \det \frac{\partial \phi}{\partial \psi}\right|$ of the trivial bundle over $V$. (When $M$ is oriented, this bundle is the same as the <em>determinant bundle</em> $\wedge^{\operatorname{top}} {\rm T}^*M$; the determinant bundle is always a line bundle, and so its square is trivializable, and has a trivializable square root, which is the distribution line bundle whether $M$ is oriented or not.) Note that the transition functions preserve positivity of the sections, and so the notion of "positive distribution" and so on are well-defined.</p> <p>Finally, if you are interested in a totally algebraic notion of integration for $\mathbb R^n$, you might be interested in the following observation, which in some form is older but nevertheless deserves to be called an observation of Berezin. Namely, the <em>integral</em>, as a linear map <code>$C^\infty_{\operatorname{compact}}(\mathbb R^n) \to \mathbb R$</code>, is uniquely defined up to scalar multiple by the fact that it vanishes on the images of <code>$\frac{\partial}{\partial x_i} : C^\infty_{\operatorname{compact}}(\mathbb R^n) \to C^\infty_{\operatorname{compact}}(\mathbb R^n)$</code>. Here <code>$C^\infty_{\operatorname{compact}}(\mathbb R^n)$</code> is the algebra of smooth functions with compact support, and $x_1,\dots,x_n$ are the usual coordinate functions on $\mathbb R^n$. There are many situations in which by naming an algebra "of functions" and some "partial derivatives" you can uniquely (up to scalar) determine an "integral". An example is the algebra of de Rham differential forms on an oriented manifolds $M$, and the "partial derivatives" are the de Rham $d$ and the Lie derivatives for all vector fields on $M$. This <strike>uniquely</strike> picks out the integral that is zero on non-top forms and integrates top forms over $M$ as a canonical "measure" on the "space" whose "algebra of functions" is the differential forms. This is an example of a "superintegral", and it was to motivate a definition of superintegrals that Berezin made the above observation.</p> http://mathoverflow.net/questions/38439/integrals-from-a-non-analytic-point-of-view/38479#38479 Answer by Dmitri Pavlov for Integrals from a non-analytic point of view Dmitri Pavlov 2010-09-12T14:58:40Z 2010-09-12T14:58:40Z <p>Here is my own favorite construction of the (Lebesgue) integral.</p> <p>Suppose M is an arbitrary smooth manifold. Denote by Or(M) the orientation line bundle of M. This bundle is equipped with a canonical Riemannian metric. Vectors of length 1 in the fiber of Or(M) over a point p∈M correspond canonically to the two orientations of the tangent space at the point p. The manifold M is orientable if and only if the bundle Or(M) is trivializable. Choosing an orientation of M amounts to choosing an isometric trivialization of Or(M).</p> <p>The bundle Or(M) together with its natural metric is flat. Hence we can twist the de Rham complex Ω^0(M)→⋯→Ω^n(M) by Or(M) and obtain the following <em>twisted de Rham complex</em>: Ω^0(M)⊗Or(M)→⋯→Ω^n(M)⊗Or(M). (Here by a complex I mean a complex of sheaves.) The line bundle Ω^n(M)⊗Or(M) is called the <em>bundle of densities</em> and is denoted by Dens(M). This bundle has a canonical orientation (hence it is trivializable), but does not have a canonical metric or a canonical trivialization.</p> <p>The cohomology of the twisted de Rham complex (with compact support) is called the <em>twisted de Rham cohomology</em> (with compact support). We have a canonical map C^∞_cs(Dens(M))→H^n_cs(M,Or(M)). Here C^∞_cs is the space of global sections of a vector bundle with compact support and H^n_cs denotes the <em>n</em>th cohomology with compact support.</p> <p>The Poincaré duality gives us a canonical isomorphism H^n_cs(M,Or(M))→H_0(M). Finally, the map from M to the point induces a map in homology H_0(M)→H_0(∙)=R.</p> <p>The composition of maps C^∞_cs(Dens(M))→H^n_cs(M,Or(M))→H_0(M)→H_0(∙)=R gives us a map ∫: C^∞_cs(Dens(M))→R, which is the integration map. Note that the actual integration (over each connected component) happens in the first map. The second map is an isomorphism and the third map simply sums integrals over individual connected components.</p> <p>The map f∈C^∞_cs(Dens(M))→∫|f|∈[0,∞) is a norm on C^∞_cs(Dens(M)). Completing C^∞_cs(Dens(M)) in this norm yields L_1(M)(=L^1(M)), which can be identified with the space of finite complex-valued measures on M.</p> <p>The space of bounded measurable functions on M (=L_0(M)=L^∞(M)) can be constructed by completing C^∞(M) in the σ-weak topology induced by L_1(M). Other L_p spaces can be constructed in a similar way to L_1(M) by completing sections of the bundle of p-densities instead of 1-densities (=Dens(M)).</p> <p>The development of the remainder of measure theory in this approach largely parallels the one explained in <a href="http://mathoverflow.net/questions/20740/is-there-an-introduction-to-probability-theory-from-a-structuralist-categorical-p/20820#20820" rel="nofollow">one of my previous answers</a>.</p> <p>I want to stress that these constructions do not rely on any existing integration theory. In fact, they can be used to build integration theory on smooth manifolds from scratch without ever referring to the usual measure theory with its lengthy and technical proofs.</p> http://mathoverflow.net/questions/38439/integrals-from-a-non-analytic-point-of-view/38486#38486 Answer by Paul Siegel for Integrals from a non-analytic point of view Paul Siegel 2010-09-12T16:18:08Z 2010-09-12T16:18:08Z <p>As the others have mentioned, integration over a connected oriented smooth manifold $M$ can be characterized (modulo some technicalities) according to the fact that it fits into an exact sequence:</p> <p>$\Omega_c^{n-1}(M) \to \Omega_c^n(M) \to \mathbb{R} \to 0$ </p> <p>where $\Omega_c^*$ refers to the De Rham complex, the first arrow is the De Rham differential, and the second arrow is integration. In particular, $\int_M$ defines an isomorphism $H^n(M^n) \to \mathbb{R}$. This line of thinking leads to Poincare duality: if one takes a submanifold $V$ representing a dimension $p$ homology class of $M$ (this is always possible <em>ratianally</em>) then the integration isomorphism determines a dimension $p$ cohomology class $[\omega]$, and this can be paired with a cohomology class $[\eta]$ in dimension $n-p$ via the exterior product (cup product):</p> <p>$([\omega],[\eta]) \mapsto \int_M \omega \wedge \eta$</p> <p>Poincare duality is precisely the assertion that this pairing is nondegenerate. I think this point of view on integration - as a pairing between homology and cohomology - leads to many of the genuinely non-analytic formulations of integration.</p> <hr> <p>Still, I would not consider this to have the final say as the correct natural definition of integration. For one thing, manifolds are far from the only spaces that one could want to integrate over, and I highly doubt integration has a cohomological formulation in any serious generality beyond manifolds. Second, the theory of measures truly is fundamental to integration and should be involved in an essential way; dynamical systems people often care as much about the measure as they do the integral.</p> <p>Here is what I think integration is all about. Let $X$ be a locally compact Hausdorff space (this is enough generality for the vast majority of applications of integration that I know of) and consider the space $C_0(X)$ of continuous functions on $X$ vanishing at infinity (in the sense of the one-point compactification). Whatever intuition you have about integration, it must tell you that the integral should be a way of assigning a real number to a continuous function (maybe other functions too) which depends linearly on the function. In other words, it has to be some sort of linear functional on $C_0(X)$. Of course, not every linear functional deserves to be called an integral - if $x \in X$ then $f \mapsto f(x)$ is a linear functional on $C_0(X)$, but it doesn't make much sense to call it an "integral". </p> <p>So we allow the topology of $X$ to play a greater role. Recall that $C_0(X)$ is a Banach space if it is equipped with the uniform norm, and as such it comes equipped with a preferred collection of linear functionals: the set $C_0(X)^*$ of continuous linear functionals. If we pretend for a moment that we have already worked really hard and built the theory of integration with respect to a Borel measure $\mu$, then assuming the measure is tied closely enough to the topology of $X$ (precisely, it must be a "Radon measure") we would have a continuous linear functional $I_\mu$ on $C_0(X)$ given by $I_\mu(f) = \int f d\mu$.</p> <p>Riesz Representation Theorem: Let $M(X)$ denote the Banach space of Radon measures on $X$. The map $M(X) \mapsto C_0(X)^*$ given by $\mu \mapsto I_\mu$ is an isometric isomorphism.</p> <p>Consequently, if we hadn't already invented a notion of integration it would be perfectly possible to simply define $\int_X f d\mu$ to be $I_\mu(f)$. I personally think this is the right way to think about integrals.</p> http://mathoverflow.net/questions/38439/integrals-from-a-non-analytic-point-of-view/39534#39534 Answer by G. Rodrigues for Integrals from a non-analytic point of view G. Rodrigues 2010-09-21T19:15:15Z 2010-09-21T19:15:15Z <p>My answer here is realy just a footnote to Paul Siegel's excellent answer, but it has become too long to fit in a comment box. Integrals are siamese brothers to measures; leaving them out seems rather perverse to me. Anyway, here is how I think of integrals. The objective here is to tackle the "categorical" part; the analytical viewpoint will necessarily obtrude. But bear with me a little, this is a somewhat long post, with a punchline at the end.</p> <p>Fix a Boolean algebra $\Omega$. A map $\nu: \Omega\to V$ with values on a linear space $V$ is finitely additive if $\nu(E\cup F)= \nu(E) + \nu(F)$ for every disjoint $E, F$. Denote the linear space of such maps by $\mathbf{A}(\Omega, V)$.</p> <p>Theorem 1: There is a linear space $\mathbf{S}(\Omega)$ and a finitely additive map $\chi:\Omega\to \mathbf{S}(\Omega)$ universal among all finitely additive maps. proof: just follow the universal property and do the obvious thing (yeah, I suppose you can use the adjoint functor theorem but why would you?).</p> <p>The universal property recast in terms of representability gives the natural isomorphism ($\mathbf{Vect}$ is the category of linear spaces)</p> <p>$$\mathbf{A}(\Omega, V)\cong \mathbf{Vect}(\mathbf{S}(\Omega), V)$$</p> <p>Before continuing, let me elucidate a little bit of the structure of $\mathbf{S}(\Omega)$.</p> <p>Theorem 2: Let $f$ be a non-zero element of $\mathbf{S}(\Omega)$. Then there are non-zero scalars $k_n$ and non-zero, pairwise disjoint $E_n\in \Omega$ such that $f= \sum_n k_n\chi(E_n)$. Furthermore, if $\nu$ is a finitely additive map and $\widehat{\nu}$ the map induced on $\mathbf{S}(\Omega)$ by universality, then $\widehat{\nu}(f)= \sum_{n}k_n \nu(E_n)$.</p> <p>To put it simply, $\mathbf{S}(\Omega)$ is the linear space of "simple functions on $\Omega$" and the map induced by universality is the integral. Now use theorem 2 to put a norm on $\mathbf{S}(\Omega)$:</p> <p>$$\|\sum_n k_n\chi(E_n)\|= \max\{|k_n|\}$$</p> <p>Denote the completion by $\mathbf{L}_{\infty}(\Omega)$. On the other hand, put on the linear subspace of the <em>bounded</em> finitely additive maps $\Omega\to B$ with $B$ a Banach space, the semivariation norm (which I am not going to define). Denote this space by $\mathbf{BA}(\Omega, B)$. Then:</p> <p>Theorem 3: There is a bounded finitely additive $\chi:\Omega\to \mathbf{L}_{\infty}(\Omega)$ universal among all bounded finitely additive maps.</p> <p>Once again, recasting the universal property in terms of representability, we have a natural isometric isomorphism ($\mathbf{Ban}$ is the category of Banach spaces and bounded linear maps).</p> <p>$$\mathbf{BA}(\Omega, B)\cong \mathbf{Ban}(\mathbf{L}_{\infty}(\Omega), B)$$</p> <p>It is illuminating to write down what does the naturality of the isomorphism implies: I will leave that as an exercise to the reader.</p> <p>Note that $\mathbf{L}_{\infty}(\Omega)$ is a Banach algebra in a natural way (use theorem 2 or juggle the universal property around. Or "cheat" all the way up and use Stone duality) and that $\chi$ is <em>spectral</em> or <em>multiplicative</em>, that is, $\chi(E\cap F)= \chi(E)\chi(F)$. Theorem 3 can now be extended by saying that $\chi$ is universal among all spectral measures (with values in Banach algebras). This extension is trivial given theorem 3.</p> <p>The case of <code>$\mathbf{L}_{\infty}(\Omega)$</code> does not need the introduction of measures but of course, this is not so with <code>$\mathbf{L}_{1}$</code>. So fix a finitely additive, positive $\mu:\Omega\to \mathbb{R}$. For the sake of simplification I will assume $\mu$ non-degenerate, that is, $\mu(E)= 0$ implies $E= 0$ (otherwise, you will have to take some quotient along the way). A finitely additive $\nu:\Omega\to B$ with $B$ a Banach space is $\mu$-Lipschitz if there is a constant $C$ such that $\|\nu(E)\|\leq C\mu(E)$ for all $E$. The infimum of all the constants $C$ in the conditions of the inequality gives a norm and a normed space I will denote by $\mathbf{LA}(\Omega, \mu, B)$. On the other hand, endow $\mathbf{S}(\Omega)$ with the norm</p> <p>$$\|\sum_n k_n\chi(E_n)\|= \sum_n |k_n|\mu(E_n)$$</p> <p>and denote the completion by $\mathbf{L}_{1}(\Omega, \mu)$.</p> <p>Theorem 4: There is a finitely additive, $\mu$-Lipschitz $\chi:\Omega\to \mathbf{L}_{1}(\Omega, \mu)$ universal among all such maps.</p> <p>Before the conclusion let me address a few points.</p> <ol> <li><p>Measurable spaces are not needed. If you really want them, use Stone duality (that is, points count for nothing in measure theory so why not leave them out, heh?).</p></li> <li><p>Finitely additive measures are really not that much more general than $\sigma$-additive ones. I will leave this cryptic comment as is, and just note that once again, Stone duality is the key here.</p></li> <li><p>I am <em>not</em> advocating this approach to be used in teaching (unless your goal is to flunk and befuddle as many undergrads as humanly possible). For one, you need some functional analysis under the belt (Banach spaces, completions, semivariation, etc.). Intuition is very hard to come by as I have thrown away the measurable spaces without which THE most important example, Lebesgue measure (arguably, the core of a first measure theory course) cannot be constructed. The whole logic of the approach only makes sense after you have seen other instances of categorical thinking at work. I am sure you can think of other objections.</p></li> </ol> <p>How categorical is this approach? Certainly, the universal properties of the respective spaces are central to the whole business and at least, they make clear that some results are really just a consequence of abstract nonsense. In the words of P. Freyd, category theory is doing what it was invented for: to make the easy things really easy (or some such, my memory is lousy). For example, the Bochner vector integral is obtained simply by taking the projective tensor product. Fubini and Fubini-Tonelli on the equality of iterated integrals are other notable cases of categorical thinking at work. Now pepper with Stone duality and a few more tools (e.g. Hahn-Banach and the compact-Hausdorff monad) and you can get (a slight variation of) the Riesz representation theorem for compact Hausdorff spaces. Use the proper compactifications and generalize to wider classes of topological spaces. Or use Loomis-Sikorski to get Vitali-Hahn-Saks in one line (but this is really "cheating" as the crucial step in establishing Loomis-Sikorski is essentially the same as the one to establish Vitali-Hahn-Saks: a Baire category-theorem application). And a few more.</p> <p>But once again, how categorical is this approach? Well, the argument is categorical enough to be generalized to symmetric monoidal closed categories. See R. Borger -- A categorical approach to integration, in the 23rd volume of TAC available online. For the modifications needed to internalize the arguments to a topos (and much more) see the delightful Phd thesis of Mathew Jackson "A Sheaf theoretic approach to measure theory" -- this is available online, just google for it. Oh, by the way, you can see (almost) everything I have explained above in volume 3 of D. Fremlin's measure theory 5-volume series, also available online.</p>