User tom lagatta - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T05:35:08Z http://mathoverflow.net/feeds/user/238 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/123493/what-is-a-gaussian-measure What is a Gaussian measure? Tom LaGatta 2013-03-03T21:40:32Z 2013-04-16T20:45:04Z <p>Let $X$ be a topological affine space. A Gaussian measure on $X$ is characterized by the property that its finite-dimensional projections are multivariate Gaussian distributions.</p> <p>Is there a direct characterization of a Gaussian measure which does not rely on finite-dimensional projections? This definition is analogous to describing a duck as the animal whose shadows look like $2$-dimensional ducks. The definition is sufficient for doing analysis, but to me it misses the essence of what a Gaussian measure <i>is</i> as a mathematical object in and of itself. </p> <hr> <p>Here is the precise definition of a Gaussian measure that I usually work with, which relies on the fact that Gaussians are entirely described by their covariance structure.</p> <p>For $X$ a topological affine space as above, let <code>$X^*$</code> denote its dual space of affine functionals. The dual space is a linear space, since there there is a natural zero functional $0 \in X^*$.</p> <p>Let $K : X^* \to X$ be a continuous affine operator which is symmetric and non-negative-definite. i.e., $f'(Kf) = f(Kf')$ and $f(Kf) \ge 0$ for all $f, f' \in X^*$. Let $m_K := K(0)$ denote the image of the zero functional.</p> <p>There is a unique Gaussian measure $P_K$ on $X$ with mean point $m_K \in X$ and covariance operator $K : X^* \to X$. That is, if $\pi : X \to \mathbb R^n$ denotes a finite-dimensional projection, then the push-forward measure $\pi_* P_K := P_K \circ \pi^{-1}$ is an $n$-dimensiona Gaussian distribution with mean vector $\pi(m_K) \in \mathbb R^n$ and covariance matrix <code>$\pi K \pi^*$</code>, where $\pi^* : (\mathbb R^n)^* \to K^*$ denotes the formal adjoint operator.</p> <p>Furthermore, the <a href="http://en.wikipedia.org/wiki/Structure_theorem_for_Gaussian_measures" rel="nofollow">structure theorem for Gaussian measures</a> states that all Gaussian measures arise in this way. Consequently, we may parametrize the space of Gaussian measures by the space $\mathcal K(X)$ of symmetric, non-negative operators from $X^*$ to $X$.</p> <p>This provides a weak answer to the question stated at the top of this post: yes, Gaussian measures can be directly characterized by their covariance structure. Consequently, here is the stronger form of my question:</p> <ul> <li>Is there a geometric description of the space $\mathcal K(X)$ of Gaussian covariance operators?</li> </ul> <p>For example, is the space $\mathcal K(X)$ an infinite-dimensional manifold? What is its symmetry group?</p> <hr> <p><b>Edit:</b> My above post implicitly defines the covariance form incorrectly. In the affine setting, the covariance form is defined by $\langle f', f \rangle_K := f'(Kf) - f'(0)$, and the conditions of symmetry and non-negative-definiteness are $\langle f', f \rangle_K = \langle f, f' \rangle_K$ and $\langle f, f \rangle_K \ge 0$, respectively. It is an easy exercise to verify that this defines a bilinear form on the dual space $X^*$ of affine functionals.</p> http://mathoverflow.net/questions/123067/is-every-bornological-space-measurable Is every bornological space measurable? Tom LaGatta 2013-02-27T04:06:02Z 2013-04-02T04:25:04Z <p>Every topological space is measurable, since we may canonically equip a topological space with its Borel $\sigma$-algebra. A bornological space is like a topological space, except the structure described "bounded" sets instead of "open" sets. Similarly, the morphisms are "bounded" instead of "continuous".</p> <p>Formally, that a bornological space is a pair $(X, \mathbf B)$, where $\mathbf B$ is a set of subsets of $X$ which covers $X$, is downward-closed, and is closed under finite unions [<a href="http://en.wikipedia.org/wiki/Bornological_space" rel="nofollow">wiki</a> | <a href="http://ncatlab.org/nlab/show/bornological+set" rel="nofollow">nLab</a>]. We may trivially assign a measurable structure to every bornological space by defining simply $\mathcal B = \sigma(\mathbf B)$, the minimal $\sigma$-algebra which contains the bornology $\mathbf B$.</p> <p>What kind of measure theory can one do in this setting? For example, are there necessary and sufficient conditions for a bornological space to admit a non-trivial measure? Is there a classification of <a href="http://en.wikipedia.org/wiki/Sigma-ideal" rel="nofollow">$\sigma$-ideals</a> for a bornological space?</p> http://mathoverflow.net/questions/124852/riesz-representation-theorem-for-vector-valued-fields Riesz representation theorem for vector-valued fields Tom LaGatta 2013-03-18T12:42:55Z 2013-03-23T13:52:29Z <p>Let $Q$ be a locally compact Hausdorff space, and let $V$ be a topological vector space. Consider the space $X = C_0(Q, V)$ of $V$-valued fields which vanish at infinity. Let $X^*$ denote the dual space of continuous linear functionals of $X$, and let $M = M(Q,V)$ denote the space of regular $V$-valued Borel measures on $Q$. </p> <p>Under what conditions on the space $V$ is the space of measures $M$ isomorphic to the dual space $X^*$?</p> <p>Is there a good reference for general Riesz representation theorems? Preferably one which takes a category-theoretic approach.</p> http://mathoverflow.net/questions/125049/open-problems-in-pdes-dynamical-systems-mathematical-physics/125054#125054 Answer by Tom LaGatta for Open problems in PDEs, dynamical systems, mathematical physics Tom LaGatta 2013-03-20T10:05:22Z 2013-03-20T10:05:22Z <p>Percolation is a major outstanding research area in both theoretical and applied probability. I recommend reading the recent survey <a href="http://arxiv.org/pdf/1207.0373v1.pdf" rel="nofollow">Percolation Since St. Flour</a> by Geoffrey Grimmett and Harry Kesten (July 2012) which gives an up-to-date list of references.</p> <p>Please feel free to send me an email, AJ, if you've got any specific questions.</p> http://mathoverflow.net/questions/124610/extending-a-hilbert-space-isometrically Extending a Hilbert space isometrically Tom LaGatta 2013-03-15T12:53:24Z 2013-03-15T18:46:50Z <p>Let $H$ be a Hilbert space, and let $X$ be a topological vector space.</p> <ul> <li>Under what conditions on the topologies of $X$ and $H$ does there exist an injective, continuous linear map $f : H \to X$?</li> </ul> <p>Supposing that such an injective embedding $f : H \to X$ exists, consider the topological closure $F := \overline{fH}$ in $X$. </p> <ul> <li>When does the closure $F$ admit the structure of a Fréchet space, with $f : H \to F$ an isometric embedding?</li> </ul> http://mathoverflow.net/questions/124361/how-to-triangulate-a-math-reference/124373#124373 Answer by Tom LaGatta for How to triangulate a math reference? Tom LaGatta 2013-03-13T02:46:31Z 2013-03-13T02:46:31Z <p>MathOverflow itself serves as an excellent reference tool for this sort of triangulation. When a future researcher googles the topic of lextensive categories with strongly orthogonal orbits (perhaps wondering about their quintessential monoidality) she will find a negative answer in Todd Trimble's comment, as well as your reformulation of her question as "Is the category of continuous functors a cartesian closed subcategory of $\operatorname{Cat}$?"</p> <p>The actual semantic indexing that's going on is a complex mix of expert networking, StackExchange's software, and Google's search index. How these three work together in a formal, structured way is anybody's guess. Start recording some data and see what structures they elucidate.</p> http://mathoverflow.net/questions/124354/symmetry-group-for-the-frame-bundle-of-a-g-space Symmetry group for the frame bundle of a G-space Tom LaGatta 2013-03-12T22:30:12Z 2013-03-12T23:54:39Z <p>Let $Q$ be a smooth manifold, and let $G$ be a Lie group which acts smoothly on $Q$ on the left. </p> <ul> <li><b>Question 1:</b> does the group $G$ act naturally on the tangent bundle $TQ \to Q$?</li> </ul> <p>My motivation here is $Q = \mathbb R^d$ with group of symmetries the Euclidean group $G = \operatorname{Euc}(d) \cong \mathbb R^d \rtimes O(d)$. In this case, $G$ admits an equivariant action on the tangent bundle, where translations act by translating the underlying space and ignoring tangent spaces, and rotations act by rotating both the space and the tangent spaces. I am wondering if this generalizes nicely, or if this lifted action is special to Euclidean space.</p> <p>Next, consider the frame bundle $GL(TQ) \to TQ$ as a principal $GL(d)$-bundle over the tangent bundle. Note that in local coordinates, $GL(TQ)$ looks like $TQ \times GL(d)$, and there is a natural <i>right</i> action of $GL(d)$ on the frame bundle.</p> <ul> <li><b>Question 2:</b> does the group $G$ act naturally on the frame bundle $GL(TQ) \to TQ$?</li> </ul> <p>If the answer to Question 1 is yes, then the answer to Question 2 should also be yes, since we would just act on each of the vectors comprising a frame in the "obvious" way. </p> http://mathoverflow.net/questions/124083/gaussian-measures-on-non-separable-spaces Gaussian measures on non-separable spaces Tom LaGatta 2013-03-09T15:41:14Z 2013-03-11T06:32:12Z <p>Let $X$ be a topological affine space which is neither separable nor metrizable. There are plenty of trivial Gaussian measures: each Dirac point-mass $\delta_x$ are the Gaussian measure with zero covariance and mean $x \in X$</p> <p>When does a topological affine space $X$ admit a non-trivial Gaussian measure? Namely, one whose support is the whole space $X$? One can always construct a Gaussian measure supported on a finite-dimensional subspace of $X$, such as a line.</p> <p>Is local convexity of $X$ a sufficient criterion for there to exist a Gaussian measure supported on all of $X$?</p> http://mathoverflow.net/questions/123665/the-limiting-behavior-of-geometric-random-walk The limiting behavior of geometric random walk Tom LaGatta 2013-03-05T20:52:44Z 2013-03-06T09:32:08Z <p>I would like to know what the asymptotic limiting behavior is for the following random walk on $\mathbb Z^d$. By Donsker's invariance principle, I suspect that its behavior is diffusive, i.e., the distribution of its path converge to $d$-dimensional Brownian motion, but I am not certain.</p> <p>The state space for the walker is the discrete unit tangent bundle <code>$X = \mathbb Z^d \times U^d$</code>, where <code>$U^d = \{ \pm e_1, \cdots, \pm e_d \}$</code>. The walker is a Markov chain, so we need only specify its transition probabilities. </p> <p>Here's the basic idea. From position $x = (q,u)$, the walker samples an independent geometric random variable, moves that distance in direction $u$, then chooses a new orthogonal direction uniformly and independently.</p> <p>Let me make this a little more precise, since there's a free parameter here tuning the step-length distribution, and I want to know how the limiting behavior depends on this parameter.</p> <p>Fix some $p \in (0,1]$, representing the inverse-mean of the step-length distribution. Given that its position is $x = (q,u)$, the walker samples a geometric random variable <code>$S \in \{1, 2, 3, \cdots\}$</code> with parameter $p$, then moves to position $x + Su$. Then, the walker samples a new direction uniformly and independently from the set <code>$u^\perp := U^d - \{ u, -u \}$</code>.</p> <ul> <li><p>Is the scaling limit for this random walker equal to a $d$-dimensional Brownian motion? How does one make this scaling limit precise?</p></li> <li><p>If so, how does the diffusion constant depend on the parameter $p$ and the dimension $d$?</p></li> </ul> http://mathoverflow.net/questions/123136/why-do-we-choose-the-standard-total-order-on-the-integers Why do we choose the standard total order on the integers? Tom LaGatta 2013-02-27T20:10:27Z 2013-02-27T21:21:17Z <p>I understand why the set of natural numbers <code>$\mathbb N = \{ 0, 1, 2, \cdots \}$</code> is equipped with a total order. Indeed, every monoid has a <a href="http://ncatlab.org/nlab/show/preorder" rel="nofollow">pre-order</a>, where $$n' \succeq n \quad \mathrm{if~and~only~if} \quad n' = n + m \quad \mathrm{~for~some~} m.$$ In the case of $\mathbb N$, this pre-order is a total order. </p> <p>However, the same construction does not result in a total order on the set of integers $\mathbb Z$. Indeed, this set is a group, so its canonical monoid pre-order is trivial. i.e., $n' \succeq n$ for all $n, n' \in \mathbb Z$, since $n' = n + (n' - n)$.</p> <p>Nonetheless, $\mathbb N$ is a subset of $\mathbb Z$, so it makes sense to assign an order relation to $\mathbb Z$ which extends the natural order on $\mathbb N$. I see (at least) two natural ways to do this:</p> <ul> <li>The standard total order $\ge$ on $\mathbb Z$.</li> <li>The pre-order on <code>$\mathbb Z$</code> which corresponds to the absolute value norm. i.e., $n' \succeq n$ if and only if $|n'| \succeq |n|$. </li> </ul> <p>There is obvious pragmatic justification for choosing the standard total order; it's utility is not in question. However, there are also pragmatic advantages for the alternate pre-order. For example, it admits $0$ as a minimal element ($n \succeq 0$ for all $n \in \mathbb Z$), and it extends the canonical pre-orders on the monoids $\mathbb N$ and $-\mathbb N$. It also generalizes nicely to higher-dimensional settings such as $\mathbb Z^d$, where no natural total order exists. </p> <p>The integers <i>exist</i> in a universal mathematical sense: they form the Grothendieck group for the natural numbers. However, there seem to be (at least) two order-theoretic models for the integers: the totally ordered set $(\mathbb Z, \ge)$ and the pre-ordered set $(\mathbb Z, \succeq)$. </p> <p>I taught an undergraduate discrete mathematics course last semester and the book never even acknowledged the second model, nor did it provide justification for the first. I suppose this is acceptable for undergraduates, but as a mathematician I am bothered by the implicit choice for an ubiquitous mathematical structure.</p> <p>Ergo my question:</p> <ul> <li>What is a mathematical justification for "always" choosing the "standard order" $(\mathbb Z, \ge)$? </li> </ul> <p>That is, is there a universal characterization of this order structure which can be adapted to the setting of a Grothendieck group over general monoid? When does it it result in a total or partial order on the group instead of just a pre-order?</p> http://mathoverflow.net/questions/122503/what-is-quantum-brownian-motion What is quantum Brownian motion? Tom LaGatta 2013-02-21T05:13:15Z 2013-02-21T11:17:30Z <p>It seems that the current state of quantum Brownian motion is ill-defined. The best survey I can find is <a href="http://arxiv.org/pdf/1009.0843v1.pdf" rel="nofollow">this one</a> by László Erdös, but the closest the quantum Brownian motion comes to appearing is in this conjecture (p. 30):</p> <blockquote> <p>[<b>Quantum Brownian Motion Conjecture</b>]: For small [disorder] $\lambda$ and [dimension] $d \ge 3$, the location of the electron is governed by a heat equation in a vague sense: <code>$$\partial_t \big|\psi_t(x)\big|^2 \sim \Delta_x \big|\psi_t(x)\big|^2 \quad \Rightarrow \quad \langle \, x^2 \, \rangle_t \sim t, \quad t \gg 1.$$</code> The precise formulation of the ﬁrst statement requires a scaling limit. The second statement about the diﬀusive mean square displacement is mathematically precise, but what really stands behind it is a diﬀusive equation that on large scales mimics the Schrödinger evolution. Moreover, the dynamics of the quantum particle converges to the Brownian motion as a process as well; this means that the joint distribution of the quantum densities $\big|\psi_t(x)\big|^2$ at diﬀerent times $t_1 &lt; t_2 &lt; \dots &lt; t_n$ converges to the corresponding ﬁnite dimensional marginals of the Wiener process.</p> </blockquote> <p>This is the Anderson model in $\mathbb R^d$ with disordered Hamiltonian $H = -\Delta + \lambda V$. The potential $V$ is disordered, and is generated by i.i.d. random fields; the parameter $\lambda$ controls the scale of the disorder.</p> <hr> <p><a href="http://en.wikipedia.org/wiki/Brownian_motion" rel="nofollow">Classical Brownian motion</a> admits many characterizations and generalizations. For example, <a href="http://en.wikipedia.org/wiki/Wiener_measure" rel="nofollow">Wiener measure</a> leads to the construction of an <a href="http://en.wikipedia.org/wiki/Abstract_Wiener_space" rel="nofollow">abstract Wiener space</a>, which is the appropriate setting for the powerful <a href="http://en.wikipedia.org/wiki/Malliavin_calculus" rel="nofollow">Mallivin calculus</a>. The <a href="http://en.wikipedia.org/wiki/Structure_theorem_for_Gaussian_measures" rel="nofollow">structure theorem of Gaussian measures</a> says that <i>all</i> <a href="http://en.wikipedia.org/wiki/Gaussian_measure#Gaussian_measures_on_infinite-dimensional_spaces" rel="nofollow">Gaussian measures</a> are abstract <a href="http://ncatlab.org/nlab/new/Wiener+measure" rel="nofollow">Wiener measures</a> in this way. I would love to know what all this theory looks like in the language of <a href="http://www.mitchener.staff.shef.ac.uk/free.pdf" rel="nofollow">non-commutative probability theory</a>.</p> <p>The QBM Conjecture states roughly that a quantum particle in a weakly disordered environment should behave like a quantum Brownian motion. This is an important open problem, but it doesn't quite capture what a QBM <i>is</i>, nor what different <i>types</i> of QBM may exist. Thus my question:</p> <blockquote> <p>What kind of precise mathematical object is a quantum Brownian motion? </p> </blockquote> http://mathoverflow.net/questions/122125/math-annotate-platform/122379#122379 Answer by Tom LaGatta for Math Annotate Platform? Tom LaGatta 2013-02-20T06:43:23Z 2013-02-20T06:43:23Z <p>I would like such an annotation platform to have a social component, where users could "opt-in" to follow certain people's annotations. This could tie in to existing social media platforms, like Facebook, Twitter, G+, etc.</p> <p>Prof. Daubechies: please feel free to contact me directly via email if you would like to discuss a social aspect of annotation software more. My address is <i>mylastname</i> @cims.nyu.edu.</p> http://mathoverflow.net/questions/121775/what-is-a-good-example-of-a-hyperspace-where-the-base-space-is-non-hausdorff What is a good example of a hyperspace where the base space is non-Hausdorff? Tom LaGatta 2013-02-14T04:42:41Z 2013-02-14T19:37:12Z <p>Let $X$ be a topological space, and let $\operatorname{CL}(X)$ be its <a href="http://en.wikipedia.org/wiki/Hyperspace_(topology)" rel="nofollow">hyperspace</a>. That is, $\operatorname{CL}(X)$ is the set of closed subsets of $X$, equipped with the minimal topology so that the canonical map <code>$$x \mapsto \overline{\{x\}}$$</code>is a homeomorphism onto its image. </p> <p><a href="http://www.heldermann-verlag.de/jca/jca01/jca01013.pdf" rel="nofollow">Typically</a>, the base space is assumed to be Hausdorff (or at least $T_1$), so that the closure of a singleton is the singleton itself. However, the definition of a hyperspace is perfectly suitable when the space is not Hausdorff, and surely this comes in handy sometimes. </p> <p>What is a good example of a hyperspace $\operatorname{CL}(X)$ where the base space $X$ is non-Hausdorff?</p> http://mathoverflow.net/questions/121638/does-every-commutative-monoid-admit-a-translation-invariant-measure Does every commutative monoid admit a translation-invariant measure? Tom LaGatta 2013-02-12T19:31:48Z 2013-02-13T13:03:45Z <p>Let $T$ be a commutative monoid, written additively. The set $T$ is equipped with a canonical pre-order, defined by $s \le t$ when there exists $s' \in T$ so that $s + s' = t$. Consequently, $T$ may be equipped with the <a href="http://ncatlab.org/nlab/show/specialization+topology" rel="nofollow">specialization topology</a> for this pre-order, where the closed sets are those which are downward-closed. Note that $T$ is typically not Hausdorff, since the closure of a singleton is its down-set: <code>$\overline{\{t\}} = t\!\downarrow\, := \{ s : s \le t \}$</code>. </p> <p>Let $\mathcal B(T)$ denote the Borel $\sigma$-algebra with respect to this topology. In this way, every commutative monoid is canonically a measurable space.</p> <p>Equipped with the $\sigma$-algebra $\mathcal B(T)$, does every commutative monoid $T$ admit a (non-trivial) family of translation-invariant measures?</p> http://mathoverflow.net/questions/121408/what-are-the-symmetries-of-a-principal-homogeneous-bundle What are the symmetries of a principal homogeneous bundle? Tom LaGatta 2013-02-10T17:44:21Z 2013-02-13T08:16:58Z <p>Let $\operatorname{Klein}$ denote the category of principal homogeneous bundles. An object in this category is a tuple $\mathbf Q = (Q, P; G, H; q, a, \tilde a)$, where:</p> <ul> <li><p>$G$ is a Lie group, and $H$ is a closed subgroup;</p></li> <li><p>$P$ is a <a href="http://math.ucr.edu/home/baez/torsors.html" rel="nofollow">$G$-torsor</a> and the manifold $Q$ is diffeomorphic to the quotient $G/H$;</p></li> <li><p>The (left) actions $\tilde a : G \to \operatorname{Aut}(P)$ and $a : G \to \operatorname{Aut}(Q)$ are morphisms of topological groups; and</p></li> <li><p>$q :P \to Q$ is a principal $H$-bundle.</p></li> </ul> <p>A morphism $\Phi : \mathbf Q \to \mathbf Q'$ is described by a tuple $\Phi = (\varphi, \tilde \varphi, f)$, where $\varphi : Q \to Q'$ and $\tilde \varphi : P \to P'$ are diffeomorphisms which commute with the bundle maps $q : P \to Q$ and $q' : P' \to Q'$, and $f : G \to G'$ is a morphism of Lie groups mapping $H$ to $H'$, and which satisfies the identity $$\tilde a'_{f(g)} \circ \tilde \varphi = \tilde \varphi \circ \tilde a_g$$ for all $g \in G$. </p> <hr> <p>I think that this is the correct categorical description of principal homogeneous bundles; please correct me if I'm wrong. I selected the name $\operatorname{Klein}$ in homage to Felix Klein and his Erlangen Program.</p> <p>It seems that such a bundle $\mathbf Q$ contains all the data on its symmetries. Namely, I think that its automorphism group $\operatorname{Aut}(\mathbf Q)$ is isomorphic to its Lie group Lie group $G = G(\mathbf Q)$?</p> <p>It is easy to see that there is a natural map $K : G \hookrightarrow \operatorname{Aut}(\mathbf Q)$, in that each $u \in G$ corresponds to a unique automorphism $K_u \in \operatorname{Aut}(\mathbf Q)$. The morphism $K_u = (k_u, \tilde k_u, c_u)$ is defined by $$k_u(q) = a_u(q), \quad \tilde k_u(p) = \tilde a_u(p), \quad \mathrm{and} \quad c_u(g) = ugu^{-1}.$$ That is, the morphism $K_u$ acts by left-multiplication on both $Q$ and $P$, but by left-conjugation on $G$.</p> <ul> <li>Is this map $K$ surjective? i.e., is $\operatorname{Aut}(\mathbf Q)$ isomorphic to $G = G(\mathbf Q)$?</li> </ul> <p>If the answer is yes, then I think that this captures the notion of "internal symmetries" of the bundle, since these are the transformations which preserve the bundle structure. </p> <p>However, I know that groupoids also show up to describe symmetries in a categorical setting, and I would be interested to hear more on that point of view.</p> <ul> <li>How can groupoids be used to describe symmetries in this category?</li> </ul> http://mathoverflow.net/questions/121495/do-there-exist-almost-surely-c-infty-smooth-gaussian-random-fields Do there exist (almost surely) $C^{\infty}$-smooth Gaussian random fields? Tom LaGatta 2013-02-11T16:14:04Z 2013-02-11T21:31:46Z <p>Let $d \ge 1$. Do there exist Gaussian random fields on $\mathbb R^d$ which are (almost surely) $C^{\infty}$-smooth, but which are not analytic? </p> <p>If so, what are necessary and sufficient conditions on the covariance function to assure that the field is (a.s.) $C^{\infty}$-smooth?</p> http://mathoverflow.net/questions/121379/what-is-an-infty-1-topos-and-why-is-this-a-good-setting-for-doing-different What is an $(\infty,1)$-topos, and why is this a good setting for doing differential geometry? Tom LaGatta 2013-02-10T07:15:35Z 2013-02-11T08:23:51Z <p>In <a href="http://golem.ph.utexas.edu/category/2012/04/principal_bundles.html" rel="nofollow">this post</a> on the <i>n</i>-Category Café, Urs Schreiber says that, "The theory of <a href="http://ncatlab.org/nlab/show/principal+bundle" rel="nofollow">G-principal bundles</a> makes sense in any <a href="http://ncatlab.org/nlab/show/%28infinity%2C1%29-topos" rel="nofollow">$(\infty,1)$-topos</a>." I followed the link to the nLab and tried to chase definitions, but I found too quickly my head spinning.</p> <p>What is an <a href="http://ncatlab.org/nlab/show/%28infinity%2C1%29-topos" rel="nofollow">$(\infty,1)$-topos</a>, and why is this <b>an</b> appropriate setting for the study of principal bundles, i.e., doing differential geometry?</p> http://mathoverflow.net/questions/120942/is-there-a-natural-measurable-structure-on-the-sigma-algebra-of-a-measurable-s Is there a natural measurable structure on the $\sigma$-algebra of a measurable space? Tom LaGatta 2013-02-06T06:03:08Z 2013-02-10T09:25:04Z <p>Let $(X, \Sigma)$ denote a measurable space. Is there a non-trivial $\sigma$-algebra $\Sigma^1$ of subsets of $\Sigma$ so that $(\Sigma, \Sigma^1)$ is also a measurable space? </p> <hr> <p>Here is one natural candidate. I'm not certain, but based on answers to related questions, I think this might be the Effros Borel structure that Gerald Edgar has mentioned <a href="http://mathoverflow.net/questions/1388/is-there-a-natural-measures-on-the-space-of-measurable-functions/1427#1427" rel="nofollow"><b>here</b></a> and <a href="http://mathoverflow.net/questions/104305/nice-sigma-algebra-on-set-of-measurable-functions/104315#104315" rel="nofollow"><b>here</b></a>.</p> <p>The $\sigma$-algebra $\Sigma$ is an ordered set under the canonical relation given by subset inclusion $\subseteq$, and is therefore naturally equipped with a specialization topology. The closed sets are generated by downward-closed sets, and the closure of a singleton is its down-set:<code>$$\overline{\{A\}} = \{ B \in \Sigma : B \subseteq A \}.$$</code> Even though this topology is highly non-Hausdorff, it's still pretty nice. For example, it's an Alexandroff space: arbitrary unions of closed sets are closed. </p> <p>Being a topological space, $\Sigma$ now has a natural measurable structure, namely, the one generated by the Borel $\sigma$-algebra $\Sigma^1 := \mathcal B_{\subseteq}(\Sigma)$. </p> <ul> <li>Is this space $(\Sigma, \Sigma^1)$ a reasonable one on which to do measure theory and probability? </li> </ul> <p>Whether it is or not, there's some non-trivial structure present. For example, we can iterate this procedure. Set $\Sigma^0 = \Sigma$, and define $\Sigma^n := \mathcal B_{\subseteq}(\Sigma^{n-1}).$ Then each one of these spaces $\Sigma^n(X) := (\Sigma^{n}, \Sigma^{n+1})$ is measurable.</p> <ul> <li><p>Is $\Sigma : \mathrm{Meas} \to \mathrm{Meas}$ an endofunctor on the category of measurable spaces?</p></li> <li><p>Under what conditions does the sequence of measurable spaces $\Sigma^n(X)$ have a limit $\Sigma^{\infty}(X)$?</p></li> </ul> http://mathoverflow.net/questions/120738/is-a-measurable-homomorphism-on-a-lie-group-smooth Is a measurable homomorphism on a Lie group smooth? Tom LaGatta 2013-02-04T08:54:37Z 2013-02-07T17:46:22Z <p>Let $G$ be a Lie group, and let $\mathcal B(G)$ its Borel $\sigma$-algebra. Suppose that $f : G \to G$ is a Borel-measurable homomorphism. Is $f$ smooth?</p> <hr> <p><b>Edit:</b> My original question said "measurable function" instead of the more accurate "measurable homomorphism." Marc Palm and other people answered both questions very nicely: </p> <ul> <li>there are obviously non-smooth measurable functions on Lie groups, and</li> <li>all measurable homomorphisms on Lie groups are smooth.</li> </ul> http://mathoverflow.net/questions/120909/can-one-view-the-independent-product-in-probability-categorially/120949#120949 Answer by Tom LaGatta for Can one view the Independent Product in Probability categorially? Tom LaGatta 2013-02-06T07:21:48Z 2013-02-06T07:21:48Z <p>One resource you may like is this recent paper by Culbertson and Sturtz on <a href="http://arxiv.org/abs/1205.1488" rel="nofollow">A Categorical Foundation for Bayesian Probability</a>.</p> <hr> <p>Here are some thoughts on the category $\mathrm{Meas}$ of measurable spaces, where the objects are sets equipped with $\sigma$-algebras, and morphisms are measurable functions (i.e., random variables). This seems like the most natural place to play with foundations of probability, since the category $\mathrm{Meas}$ has a natural concept of a tensor product, where the corresponding $\sigma$-algebra is generated by product sets.</p> <ul> <li>Question: Is $\mathrm{Meas}$ a Cartesian closed category under the tensor product?</li> </ul> <p>A measure is a countably-additive function defined on $\sigma$-algebras. Typically we learn that measures take values in the real numbers (a Borel-measurable space), but there is no reason that they cannot take values in more general structures, e.g., vector spaces or topological groups. To the best of my knowledge, the most general formulation is due to Tarski, whose monoids are valued in commutative monoids. To Tarski, a "measure" is really a functor from a subcategory of $\mathrm{Meas}$ to the category $\mathrm{ComMon}$ of commutative monoids. This is some universal object, and all other measures can be derived from it. I'm still iffy on this approach, but understanding the Tarski functor is the current thesis project of Tyler Bryson, a masters student at the Courant Institute. He should have more details in a few months.</p> <p>To define independence, we need to multiply probabilities, so measures should take values in a ring. This multiplicative structure is also important for conditional probabilities, more generally, as well as integration theories. If the space admits symmetries (i.e., a group action), then Tarski's approach spits out a natural multiplicative structure, so I think we're safe in general.</p> <p>Next,, you can normalize all the measures to have size $1$, turning them into probability measures, but I recommend against it from the categorical point of view. In statistical physics (and more generally, statistics), these normalization constants are the hardest things to compute. I recommend keeping track of them combinatorially, then reducing them at the end. Moreover, it may be useful to keep track of a few different "scales" of measurements, where it is not clear which one to normalize to $1$. This is even seen in the case of vector-valued measures. In the quantum setting of nonnegative-operator-valued measures, a uniform scaling can be chosen, but again the normalization constants are difficult spectral integrals to compute. </p> <p>No big deal if you want probability measures in the end, just make sure that $1$ makes sense to use in the context you're studying.</p> <p>Anyway, let $R$ be a topological commutative ring and let $M(X) := M(X,R)$ denote the space of $R$-valued measures on $X$. This is a measurable space, when equipped with the minimal $\sigma$-algebra so that the evaluation map is measurable. This $M$ is an endofunctor on $\mathrm{Meas}$, and closely related to the Giry monad. </p> <p>Consider the tensor product $XY := X \otimes Y$ of measurable spaces. There are natural projection maps $\pi_X : XY \to X$ and $\pi_Y : X \otimes Y \to Y$. Measures push-forward, so there are natural maps $(\pi_X)_* : M(XY) \to M(X)$ and $(\pi_Y)_* : M(XY) \to M(Y)$. Note that $M(XY)$ corresponds to the joint distributions over $X$ and $Y$, and the resulting projections the marginal distributions.</p> <p>Note that $M(XY)$ corresponds to the joint distributions over $X$ and $Y$, and the resulting projections the marginal distributions.</p> <p>There is also the natural space $M(X)M(Y) := M(X) \otimes M(Y)$, and its corresponding projection maps $\pi_{MX} : M(X)M(Y) \to M(X)$ and $\pi_{MY} : M(X)M(Y) \to M(Y)$.</p> <p>As long as there is some Fubini-type theorem present, there should be a natural map $\varphi : M(X)M(Y) \to M(XY)$ sending a pair of measures to their product measure, corresponding to independent random variables. This should correspond to a commutative diagram, but I don't see it. </p> <p>Again, watch for the normalization constants, since there are two $M$'s in the source of this map but only one $M$ in the target. Dependent random variables then should be quantified by a failure of some diagram to commute.</p> <hr> <p>Even though I didn't answer the question, hopefully you will find some of this content useful. </p> <p>If you can improve this community-wiki post, please go for it.</p> http://mathoverflow.net/questions/1388/is-there-a-natural-measures-on-the-space-of-measurable-functions/120736#120736 Answer by Tom LaGatta for Is there a natural measures on the space of measurable functions? Tom LaGatta 2013-02-04T07:12:08Z 2013-02-06T05:54:04Z <p>Sorry for the necromancy. Here's an attempt at constructing a $\sigma$-algebra using the tensor product of $\sigma$-algebras. This should likely <i>not</i> result in a Borel structure (i.e., a $\sigma$-algebra generated as the Borel $\sigma$-algebra of a topological space), so I don't think it contradicts Aumann's work.</p> <p>I made this answer community wiki, so feel free to edit it. If it's wrong, please correct it. </p> <p>I figured I'd answer the question to provide a quick reference for <a href="http://ncatlab.org/nlab/show/future" rel="nofollow">the future</a>.</p> <hr> <p>Let $(X,\Sigma_X)$ and $(Y, \Sigma_Y)$ be two measurable spaces, and let $H = \operatorname{Hom}(X,Y)$ be the set of measurable functions from $X$ to $Y$. Define <a href="http://ncatlab.org/nlab/show/evaluation+map" rel="nofollow">the evaluation map</a> $\operatorname{eval} : H \times X \to Y$ by $$\operatorname{eval}(h,x) = h(x).$$</p> <p>Now, simply define $\Sigma_{H}$ to be the minimal $\sigma$-algebra on $H$ so that the evaluation map $\operatorname{eval} : H \times X \to Y$ is measurable, where $H \times X$ is equipped with the tensor product $\sigma$-algebra $\Sigma_H \otimes \Sigma_X$.</p> <p>I think that $\Sigma_H$ should be well-defined, even though it's unlikely to be Borel in most interesting situations. There should always be <i>some</i> minimal solution, even if it's the whole power set $2^H$.</p> <hr> <p>Here are some general thoughts on why it is important that the evaluation function is measurable, and why this is good enough for most interesting applications, e.g., applied analysis, physics or computation. This means that f $B \in \Sigma_Y$ is any measurable event in $Y$, then <code>$$\operatorname{eval}^{-1}(B) = \big\{ (h,x) : h(x) \in B \big\} \in \Sigma_H \otimes \Sigma_X.$$</code></p> <p>For example, this always describes solution-sets to equations, since <code>$$\{ h(x) = y \} = \operatorname{eval}^{-1}({y}).$$</code></p> <p>When $Y$ is a <a href="http://en.wikipedia.org/wiki/Hierarchy_(mathematics)" rel="nofollow">measurable hierarchy</a> (i.e., a pre-ordered measurable space), then this also includes inequalities, e.g., <code>$$\{ h(x) \le y \} = \operatorname{eval}^{-1}(\downarrow{y}),$$</code> where <code>$\downarrow{y} = \{ y' \le y \}$</code> denotes the down-set of $y \in Y$. Basically, <code>$$\mbox{if you can write it down, it's probably measurable.}$$</code></p> <p>This is very useful computationally, since the hom-set $\operatorname{Hom}(H \times X, Y)$ is adjoint to $\operatorname{Hom}(H,Y^X)$ via the process of <a href="http://en.wikipedia.org/wiki/Currying#Mathematical_view" rel="nofollow">currying</a>. The adjoint to the evaluation map is called <a href="http://ncatlab.org/nlab/show/function+application" rel="nofollow">function application</a>, and in computer science is known as <a href="http://en.wikipedia.org/wiki/Apply" rel="nofollow">Apply</a>. Ultimately, this means that anything you work out computationally is measurable, which means no more appendices full of nasty measurability proofs by hand.</p> <p>Note that $Y^X$ is a measurable space when equipped with the tensor-product $\sigma$-algebra, and in most cases of interest its $\sigma$-algebra is not generated by a topology (<a href="http://mathoverflow.net/questions/87838/is-every-sigma-algebra-the-borel-algebra-of-a-topology/87888#87888" rel="nofollow">reference Jochen Wengenroth's answer to this question</a>).</p> <p>Furthermore, this should be useful in measure theory, and may lead toward an answer to Kenny Easwaran's question. If you can see a way to answer it, go ahead and edit this answer.</p> http://mathoverflow.net/questions/120561/given-that-a-conditional-measure-is-gaussian-how-bad-can-the-original-measure-be Given that a conditional measure is Gaussian, how bad can the original measure be? Tom LaGatta 2013-02-01T23:47:38Z 2013-02-01T23:47:38Z <p>Let $X$ and $Y$ be Banach spaces, and let $\varphi : X \to Y$ be a continuous linear map. Suppose that $\mathbb P$ is a probability measure on $X$ which satisfies the <a href="http://arxiv.org/abs/1003.0975" rel="nofollow">continuous disintegration property</a> with respect to $Y$. This means that there exists a measure-valued function $y \mapsto \mathbb P^y(d x)$ which is (1) a <a href="http://en.wikipedia.org/wiki/Regular_conditional_probability" rel="nofollow">regular conditional probability</a> (RCP), and (2) is continuous, where the space of probability distributions $\Delta(X)$ is equipped with the topology of weak-* convergence of measures. </p> <p>Suppose that, for all $y \in Y$, the probability measure $\mathbb P^y$ is Gaussian with mean $m(y) \in X$ and covariance operator $\hat K : X^* \to X$. It is a consequence of the linear structure that the covariance depends only on the function $\varphi$ and not the actual observed value $y$.</p> <p><b>Question:</b> How bad can the original measure $\mathbb P$ be? More precisely, what is the set $G \subseteq \Delta(X)$ of probability measures $\mathbb P$ which result in conditional Gaussian distributions $\mathbb P^y$ of Gaussian type?</p> <p><b>Followup question:</b> Does the answer change if the map $y \mapsto \mathbb P^y(dx)$ is simply measurable, and not continuous?</p> <p>Being a RCP means that $\mathbb P$ satisfies the disintegration equation: for all continuous and bounded functions $f : X \to \mathbb R$, $$\int_X f(x) ~ \mathbb P(dx) = \int_Y \int_X f(x) ~\mathbb P^y(dx) \mathbb P_Y(dy),$$ where $\mathbb P_Y := \mathbb P \circ \varphi^{-1}$ is the push-forward probability measure on $Y$.</p> <p>Even though the conditional measure is Gaussian, there is no guarantee that the original measure $\mathbb P$ or the push-forward measure $\mathbb P_Y$ are Gaussian. However, it is conceivable that this "non-Gaussianness" cancels out in the conditioning. I am not comfortable enough with the structure of these spaces to know what conditions are necessary for a Gaussian to manifest in the conditioning.</p> <p>I am asking this question on behalf of an oceanographer friend of mine, who says, "This is, IMO, one of the most important theoretical questions in data assimilation."</p> http://mathoverflow.net/questions/120363/is-there-a-good-concept-of-a-measurable-fibration Is there a good concept of a measurable fibration? Tom LaGatta 2013-01-30T22:40:19Z 2013-02-01T18:34:57Z <p>In probability theory, there are many results which are valid in purely measurable settings, usually beginning with the assumption, "let $(\Omega, \mathcal F, \mathbb P)$ be an abstract probability space." On the other hand, many results require a more "locally measurable" structure, often taking the form of a Borel $\sigma$-algebra on a topological space. In an effort to compromise between the two concepts, I am wondering if there is a good notion of a <i>measurable fibration</i>. </p> <p>Let $E$ be a measurable space, let $B$ be a topological space, and let $\pi : E \to B$ be a Borel-measurable function. I would like to define a "measurable homotopy lifting property" with respect to an abstract probability space $(\Omega, \mathcal F, \mathbb P)$. </p> <p>Let $I = [0,1]$ be the unit interval, equipped with its Borel $\sigma$-algebra <code>$\mathcal B(I)$</code> and Lebesgue measure $\lambda$. A "measurable homotopy" should be a measurable map $f : \Omega \times I \to B$, where the product $\Omega \times I$ is equipped with the tensor product $\sigma$-algebra $\mathcal F \otimes \mathcal B(I)$. Measures push forward, so this naturally endows $B$ with a probability measure $P_f = f_*(\mathbb P \otimes \lambda)$.</p> <p><a href="http://en.wikipedia.org/wiki/Homotopy_lifting_property" rel="nofollow">Normally</a>, one next considers a lift $\tilde f_0 : \Omega \to E$ of the map <code>$f_0 = f|_{\Omega \times \{0\}}$</code>. This makes sense in topology, because by lifting at a point one can "tug" the rest of the homotopy up to $E$. However, this doesn't make sense in this context: the set <code>$\Omega \times \{0\}$</code> has measure zero, hence is meaningless from the point of view of measure theory. Hence:</p> <p><b>Question:</b> is there a generalization of the homotopy lifting property to this measurable setting?</p> <p>Even thought $0 \in I$ has no special meaning probabilistically, the concept of a random number $\iota \in I$ with distribution $\lambda$ <i>does</i> make sense. In fact, the product measure $\mathbb P \otimes \lambda$ represents choosing a random $\omega \in \Omega$ and random $\iota \in I$ independently from one another. Consequently, a random point $(\omega, \iota)$ picks out a particular function <code>$f_{\iota} := f|_{\Omega \times \{\iota\}}$</code>. Diagrammatically this results in a mess of arrows, so I'll stop the speculation and leave the question to the community.</p> http://mathoverflow.net/questions/119901/kernel-elements-for-the-grothendieck-group-map-of-a-commutative-monoid Kernel elements for the Grothendieck group map of a commutative monoid Tom LaGatta 2013-01-26T01:15:59Z 2013-01-27T04:18:53Z <p>This is just a nomenclature question. Let $T$ be a commutative monoid, and let <code>$T^*$</code> be its Grothendieck group. That is, $T^* \cong T \times T \ / \sim$, where $(s,s') \sim (t, t')$ if $s+t'+e = s'+t+e$ for some $e \in T$. </p> <p>Let $i : T \to T^*$ be the natural inclusion map, where $i(t) = [(t,0)]$, and let $K$ denote the kernel of this map. </p> <p>Does the kernel $K \subseteq T$ have a common name in the literature? What are elements $k \in K$ called?</p> http://mathoverflow.net/questions/48770/controlling-the-lipschitz-norm-of-the-limit-of-a-sequence-of-functions Controlling the Lipschitz norm of the limit of a sequence of functions Tom LaGatta 2010-12-09T15:08:15Z 2012-12-30T14:52:28Z <p>Consider the Fréchet space $\Omega = C(\mathbb R^d)$ of real-valued continuous functions equipped with the seminorms <code>$$\|f\|_D := \sup_{x,y \in D} \left\{ |f(x)|, \tfrac{|f(x)-f(y)|}{|x-y|} \right\}, \qquad \mathrm{for~compact~} D \subseteq \mathbb R^d.$$</code> That is, $\|f\|_D$ is the larger of the supremum norm and the Lipschitz constant of $f$ over $D$.</p> <p>Let $D \subseteq \mathbb R^d$ be compact, and let $f \in \Omega$. Consider a sequence $D_n$ of compact sets and a sequence of functions $f_n$ so that:<ul> <li>$D_n \to D$ in the <a href="http://en.wikipedia.org/wiki/Hausdorff_distance" rel="nofollow">Hausdorff topology</a>,</li> <li>$f_n \to f$ in $\Omega$, and</li> <li>There exists a value $h$ so that $\|f_n\|_{D_n} \le h$ for all $n$. </ul></p> <p>Is it the case that $\|f\|_D \le h$?</p> http://mathoverflow.net/questions/117294/applications-of-the-giry-monad-in-probability-and-statistics Applications of the Giry monad in probability and statistics Tom LaGatta 2012-12-27T01:05:59Z 2012-12-27T11:14:14Z <p>In <a href="http://mathoverflow.net/questions/117274/fixed-objects-of-the-m-endofunctor-on-category-meas" rel="nofollow">another thread</a>, I asked about the $M$ endofunctor on the category $\operatorname{Meas}$ of measurable spaces, which sends a space $X$ to its space of measures $M(X)$. </p> <p>Will Sawin described the monad structure of this endofunctor <a href="http://mathoverflow.net/questions/117274/fixed-objects-of-the-m-endofunctor-on-category-meas/117283#117283" rel="nofollow">in an answer</a>, and in the comments thread Tom Leinster brought up the Giry monad, saying, "it's almost inconceivable that there could be two different reasonable monad structures on this endofunctor, so I bet it's the same." </p> <p>Here's the citation to Giry's original paper:</p> <blockquote> <p><a href="http://link.springer.com/chapter/10.1007%2FBFb0092872?LI=true" rel="nofollow">Michèle Giry</a>, A categorical approach to probability theory. In B. Banaschewski, editor, Categorical Aspects of Topology and Analysis, Springer LNM 915, 1982. </p> </blockquote> <p>It is clear that <a href="http://mathoverflow.net/questions/84309/why-are-monads-useful" rel="nofollow">monads are very useful</a> in general, but I am not familiar enough with the concept to see how they are useful in this context. Fortunately, as of December 2012, <a href="http://scholar.google.com/scholar?hl=en&amp;lr=&amp;cites=9453957818222470478" rel="nofollow">there are 167 citations to Giry's paper on Google Scholar</a>, so clearly many researchers have already recognized her work. There is also a discussion on the nLab page on <a href="http://ncatlab.org/nlab/show/probability+theory#probability_theory_from_the_npov_15" rel="nofollow">probability theory</a>.</p> <p>For the benefit of future researchers, I've created this community wiki thread to aggregate possible applications of the Giry monad in probability and statistics. My hope is that this thread might be a place for the <a href="http://mathoverflow.net/questions/20740/is-there-an-introduction-to-probability-theory-from-a-structuralist-categorical-p/" rel="nofollow">structuralist</a> and <a href="http://mathoverflow.net/questions/20740/is-there-an-introduction-to-probability-theory-from-a-structuralist-categorical-p/20828#20828" rel="nofollow">probability</a> communities to come together and learn from each other.</p> <p>If you see any interesting applications of the Giry monad, please post them here.</p> http://mathoverflow.net/questions/117274/fixed-objects-of-the-m-endofunctor-on-category-meas Fixed objects of the M endofunctor on category Meas Tom LaGatta 2012-12-26T19:59:14Z 2012-12-27T04:22:27Z <p>Consider the category $\operatorname{Meas}$ of measurable spaces: its objects are sets equipped with $\sigma$-algebras, and its morphisms are measurable functions between spaces. </p> <p>As Gerald Edgar &amp; Michael Greinecker pointed out in <a href="http://mathoverflow.net/questions/117118/when-is-a-space-of-measures-a-measurable-space/117145#117145" rel="nofollow">this thread</a>, there is the natural endofunctor $M : \operatorname{Meas} \to \operatorname{Meas}$ which sends a space $X$ to the collection $M(X)$ of (extended-real-valued) measures on $X$. This collection $M(X)$ is a measurable space, equipped with the minimal $\sigma$-algebra so that the evaluation functions $\mu \mapsto \mu(A)$ are measurable. A morphism $f : X \to Y$ is mapped to the push-forwarding map $f_* : \mu \mapsto \mu \circ f^{-1}$.</p> <p>We may naturally iterate this endofunctor. Thinking of a measure on $X$ as a statistical ensemble, the space $M^2(X) = M(M(X))$ consists of ensembles-of-ensembles. Such hierarchical spaces are important in probability theory and dynamical systems. We may go further, defining $M^3(X) = M(M(M(X)))$ and so forth.</p> <p>This simply generates a dynamical system on the category of measurable spaces, where the initial condition $X \in \operatorname{Meas}^{\operatorname{ob}}$ gets mapped to its successors $M(X)$, $M^2(X)$, $M^3(X)$, etc. Understanding these categorical dynamics is a hard problem, to say the least. Understanding the fixed points'', on the other hand, might actually be tractable. Hence the question:</p> <blockquote> <p>What are the fixed objects of the endofunctor $M :\operatorname{Meas} \to \operatorname{Meas}$ ?</p> </blockquote> http://mathoverflow.net/questions/117274/fixed-objects-of-the-m-endofunctor-on-category-meas/117300#117300 Answer by Tom LaGatta for Fixed objects of the M endofunctor on category Meas Tom LaGatta 2012-12-27T04:16:54Z 2012-12-27T04:22:27Z <p>Here is a candidate class of examples. I have made this <b>community wiki</b> so please feel free to edit it if you can answer it. Or, copy the text and make a new answer so we can give you reputation points.</p> <hr> <p>Let $X_0 := X$ be any non-empty measurable space, and for each $n \in \mathbb N$, define the product $X_{n+1} := M(X_n) \times X_n$. This is a measurable space when equipped with the tensor product of $\sigma$-algebras. Define $X_{\infty}$ to be the projective limit of the sequence $X_0 \leftarrow X_1 \leftarrow X_2 \leftarrow \cdots$, where the arrows denote the projections onto the second components of the products. </p> <p>The existence of the limit gives a canonical section $X_{\infty} \to M(X_{\infty}) \times X_{\infty}$. By iterating this map with the projection onto the first component, we have defined a natural measurable function $\varphi : X_{\infty} \to M(X_{\infty})$.</p> <p>Consequently, a point $x$ in $X_{\infty}$ contains the data of a measure $\varphi(x)$ on the space. It may be the case that this is <i>all</i> the data possessed by the point, in which case $X_{\infty} \cong M(X_{\infty})$. Is this the case? That is, </p> <blockquote> <p>is the function $\varphi : X_{\infty} \to M(X_{\infty})$ one-to-one and onto?</p> </blockquote> <p>Note that the requirement that $X_0$ be non-empty is necessary. If $X_0 = \varnothing$, then <code>$M(X_0) = \{0\}$</code> consists of the zero measure, but <code>$X_1 = \{0\} \times \varnothing = \varnothing$</code>. Consequently $X_{\infty} = \varnothing$ and <code>$M(X_{\infty}) = \{0\} \ne X_{\infty}$</code>.</p> <hr> <p>This construction is based on the concept of an <i>epistemic type space</i>, which encodes the belief hierarchies of players in <a href="http://faculty.wcas.northwestern.edu/~msi661/Hierarchies-Palgrave-070328.pdf" rel="nofollow">epistemic game theory</a>.</p> http://mathoverflow.net/questions/117118/when-is-a-space-of-measures-a-measurable-space When is a space of measures a measurable space? Tom LaGatta 2012-12-24T02:56:40Z 2012-12-24T14:46:18Z <p>Let $X$ denote a <a href="http://ncatlab.org/nlab/show/measurable+space" rel="nofollow">measurable space</a>, that is, a set equipped with a $\sigma$-algebra $\Sigma(X)$. Let $M(X)$ denote the space of real-valued measures over $X$. This is a vector space over the real numbers, since the sum of two measures is again a measure, as is a scalar multiple of a measure. I would like to know the most general setting for which $M(X)$ is a measurable vector space.</p> <p>Does $M(X)$ admit a canonical choice of $\sigma$-algebra, turning it into a measurable space? </p> <p>If the answer is "no", then what about the setting where $X$ is a <a href="http://ncatlab.org/nlab/show/measurable+space#relation_to_von_neumann_algebras_14" rel="nofollow">localizable measurable space</a>?</p> <p>If the answer is again "no", then what is the most general setting so that $M(X)$ is admits a canonical measurable structure?</p> <p>Most generally, what is the largest subcategory $\mathcal C$ of $\mathbf{Meas}$ so that $M : \mathcal C \to \mathcal C$ is an endofunctor? </p> http://mathoverflow.net/questions/109549/the-symmetry-group-of-mathbb-zd The symmetry group of $\mathbb Z^d$ Tom LaGatta 2012-10-13T18:23:43Z 2012-10-13T21:32:18Z <p>Let $d \ge 1$, and consider the integer lattice $\mathbb Z^d$. This is a homogeneous space, in the manner of the Erlangan Programm. </p> <p>I would like to write $\mathbb Z^d = G / H$, where $G$ is the symmetry group of the lattice and $H$ is the stabilizer of a point, but I do not see readily how to do so. This should be an easy question, so maybe one of you can answer it quickly.</p> <p>Let <code>$V = \{\pm \mathrm e_1, \cdots, \pm \mathrm e_d\}$</code> denote the $2d$ standard basis vectors in $\mathbb Z^d$, and let $H$ be the group consisting of lattice orthogonal matrices from $V$. i.e., an element $\beta \in H$ describes an orthogonal basis of $\mathbb Z^d$, in the sense that it is a matrix whose columns are an independent set from $V$.</p> <p>The group $G$ should then consist of translations and rotations. Does that mean that it is a semidirect product of $\mathbb Z^d$ and $H$?</p> http://mathoverflow.net/questions/123493/what-is-a-gaussian-measure/127752#127752 Comment by Tom LaGatta Tom LaGatta 2013-04-18T04:30:55Z 2013-04-18T04:30:55Z Thanks, George. This pretty well answers my question, and at a deeper level of generality than I was asking at originally. http://mathoverflow.net/questions/124852/riesz-representation-theorem-for-vector-valued-fields Comment by Tom LaGatta Tom LaGatta 2013-03-24T21:52:51Z 2013-03-24T21:52:51Z Thanks everybody for your helpful comments. http://mathoverflow.net/questions/108749/generalized-skorokhod-spaces Comment by Tom LaGatta Tom LaGatta 2013-03-24T12:54:16Z 2013-03-24T12:54:16Z Here is a good survey on Skorokhod space and generalizations: <a href="http://kpbc.umk.pl/Content/39953/kievtopologies.pdf" rel="nofollow">kpbc.umk.pl/Content/39953/kievtopologies.pdf</a> http://mathoverflow.net/questions/15292/why-cant-there-be-a-general-theory-of-nonlinear-pde/15307#15307 Comment by Tom LaGatta Tom LaGatta 2013-03-20T10:44:29Z 2013-03-20T10:44:29Z Here is a 7-page review of Partial Differential Relations by Dusa McDuff: <a href="http://projecteuclid.org/DPubS/Repository/1.0/Disseminate?view=body&amp;id=pdf_1&amp;handle=euclid.bams/1183554541" rel="nofollow">projecteuclid.org/DPubS/Repository/1.0/&hellip;</a> http://mathoverflow.net/questions/125049/open-problems-in-pdes-dynamical-systems-mathematical-physics Comment by Tom LaGatta Tom LaGatta 2013-03-20T10:01:46Z 2013-03-20T10:01:46Z This is a good question, but should be made Community Wiki. I see that there is already 1 vote to close. To users with closing power: I ask that you keep the question open for at least a few days to collect a few good answers for @AJGibson. If having a big list is annoying at that point, then we can close it. http://mathoverflow.net/questions/124852/riesz-representation-theorem-for-vector-valued-fields Comment by Tom LaGatta Tom LaGatta 2013-03-18T18:11:35Z 2013-03-18T18:11:35Z @Willie Wong: I don't know! I usually work with the real- or complex-valued case, in which case $V$ and $V^*$ are isomorphic. Thanks for raising the issue. @jbc: thanks for the reference, that will definitely be a good starting place for me. http://mathoverflow.net/questions/124610/extending-a-hilbert-space-isometrically Comment by Tom LaGatta Tom LaGatta 2013-03-15T18:49:06Z 2013-03-15T18:49:06Z @Jochen Wengenroth: I specifically made no additional assumptions on the Hilbert space nor the larger topological space $X$, such as separability or local convexity. Your remark on the separable case is interesting, and I thank you for making the point. For the second question, I mean to say, &quot;when does F admit the structure of a Fr&#233;chet space?&quot; Certainly, if its topology is completely metrizable, then the metric on $F$ will be an extension of the metric from $H$. http://mathoverflow.net/questions/124473/ck-topology-of-metrics Comment by Tom LaGatta Tom LaGatta 2013-03-14T11:08:59Z 2013-03-14T11:08:59Z Here is a reference on Cheeger-Gromov theory: <a href="http://arxiv.org/pdf/gr-qc/0208079v2.pdf" rel="nofollow">arxiv.org/pdf/gr-qc/0208079v2.pdf</a> http://mathoverflow.net/questions/124267/origin-of-the-notion-of-network-in-graph-theory Comment by Tom LaGatta Tom LaGatta 2013-03-13T00:07:07Z 2013-03-13T00:07:07Z David Aldous defines a network to be &quot;a graph with context-dependent extra structure.&quot; <a href="http://www.stat.berkeley.edu/~aldous/Talks/sparse_entropy.pdf" rel="nofollow">stat.berkeley.edu/~aldous/Talks/&hellip;</a> http://mathoverflow.net/questions/124354/symmetry-group-for-the-frame-bundle-of-a-g-space/124359#124359 Comment by Tom LaGatta Tom LaGatta 2013-03-12T23:57:06Z 2013-03-12T23:57:06Z Thank you, @Ryan! http://mathoverflow.net/questions/124354/symmetry-group-for-the-frame-bundle-of-a-g-space Comment by Tom LaGatta Tom LaGatta 2013-03-12T23:19:17Z 2013-03-12T23:19:17Z @Ryan Budney, smooth actions are all I am concerned with; I edited the post. Thanks for the quick reply. Could you add a few more details and post that as an answer? http://mathoverflow.net/questions/124083/gaussian-measures-on-non-separable-spaces/124205#124205 Comment by Tom LaGatta Tom LaGatta 2013-03-11T11:25:34Z 2013-03-11T11:25:34Z Indeed, this does answer the question. The Corollary to Theorem 2 states that there is no admissible norm on a non-separable Hilbert space. Since the support of the measure in affine space $X$ is the closure of the Cameron-Martin space corresponding to the covariance operator, the support of the measure must be separable. Consequently, there can be no Gaussian measure with full support in a non-separable affine space. Cheers. http://mathoverflow.net/questions/124083/gaussian-measures-on-non-separable-spaces/124205#124205 Comment by Tom LaGatta Tom LaGatta 2013-03-11T11:10:11Z 2013-03-11T11:10:11Z Thanks, @Anatoly Kochubei! Is is the case that in the non-separable setting, Gaussian measures are always concentrated on separable subspaces? It would seem reasonable that the answer is yes, which provides a negative answer to my original question. I was mulling over your second fact last night (the canonical Gaussian cylinder-set measure cannot be extended to a genuine measure), and am glad to see the reference. I'll take a look at Sat&#244;'s paper and let you know if I've got any questions. http://mathoverflow.net/questions/124153/can-random-elements-be-defined-in-terms-of-a-measure-algebra Comment by Tom LaGatta Tom LaGatta 2013-03-11T00:34:20Z 2013-03-11T00:34:20Z @Gerald: could you please summarize the image measure catastrophe in a comment? http://mathoverflow.net/questions/124083/gaussian-measures-on-non-separable-spaces Comment by Tom LaGatta Tom LaGatta 2013-03-11T00:29:16Z 2013-03-11T00:29:16Z @Gerald (my apologies for misspelling your name before), a topological affine space is a topological vector space with the origin forgotten. Thankfully, the situation you hypothesize never occurs: there is always the zero functional. In that case, the only centered Gaussian measure is the Dirac point-mass concentrated on the origin.