User chris heunen - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T06:52:06Z http://mathoverflow.net/feeds/user/10368 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/60328/what-kind-of-completion-is-this What kind of completion is this? Chris Heunen 2011-04-01T23:19:09Z 2013-05-10T16:23:41Z <p>Let $X$ be a compact Hausdorff space, and $C(X)$ the unital commutative C*-algebra of continuous functions on it. The double Banach dual <code>$C(X)^{**}$</code> is a commutative von Neumann algebra and hence has a compact Hausdorff space <code>$X^{**}$</code> as Gelfand spectrum again. What is $X^{**}$, in terms of $X$?</p> <p>This gives an (idempotent?) endofunctor (monad?) on the category of compact Hausdorff spaces, that I don't recognize as any of the usual ones like Stone-Cech. What completion is it? Is it related to the functor taking a compact Hausdorff space to the $\sigma$-algebra generated by its opens?</p> <p>Accounts of enveloping von Neumann algebras of (commutative) C*-algebras in terms of double Banach duals seem hard to find in the literature, and any references are welcome. What is the von Neumann algebra $C(X)^{**}$, in the first place? </p> http://mathoverflow.net/questions/60328/what-kind-of-completion-is-this/130275#130275 Answer by Chris Heunen for What kind of completion is this? Chris Heunen 2013-05-10T16:23:41Z 2013-05-10T16:23:41Z <p>For what it's worth, I found a lot of information in [Dales, Lau &amp; Strauss, "Second duals of measure algebras", Dissertationes Mathematicae 481:1-121, 2012]. The assignment $X \mapsto X^{\ast\ast}$ is functorial, and called the <i>hyper-Stonean cover</i>. It loses information: if $X$ is countable, then $X^{\ast\ast} \cong \beta\mathbb{N}$. </p> <p>If $X$ is metrizable and uncountable, a lot of the structure of $X^{\ast\ast}$ is known -- it is characterised as follows:</p> <ul> <li>$X^{\ast\ast}$ is hyper-Stonean;</li> <li>the set $D$ of isolated points of $X^{\ast\ast}$ has cardinality $2^{\aleph_0}$, its closure $Y$ is a clopen subspace homeomorphic to $D_d$;</li> <li>$X\setminus Y$ contains a family of $2^{\alpha_0}$ pairwise disjoint, clopen subspaces, each homeomorphic to $\mathbb{H}$;</li> <li>the union $U$ of the above sets is dense in $X \setminus Y$, and $\beta U = X \setminus Y$. </li> </ul> <p>In general, there exist a continuous projection $p \colon X^{\ast\ast} \to X$ and a (not necessarily injective) injection $i \colon X \to X^{\ast\ast}$ with $i \circ p = 1_{X^{\ast\ast}}$. Moreover, $X$ consists of the isolated points of $X^{\ast\ast}$, and is therefore open.</p> http://mathoverflow.net/questions/127045/fixed-point-theorems/127103#127103 Answer by Chris Heunen for Fixed point theorems Chris Heunen 2013-04-10T15:50:45Z 2013-04-10T15:50:45Z <p><strong>Lawvere's fixed point theorem.</strong> If $f \colon A \to Y^A$ is a surjective morphism in a Cartesian closed category, then any $t \colon Y \to Y$ has a fixed point. </p> <p>(Surjectivity is a technical term, which basically means that any $g \colon A \to Y$ equals $f(a)$ on points for some point $a$ of $A$. See <a href="http://www.tac.mta.ca/tac/reprints/articles/15/tr15abs.html" rel="nofollow">here</a>)</p> <p>Applications: Cantor's diagonal argument, Turing's halting problem, Russell's paradox, G&ouml;del's incompleteness theorem, Tarski's incompleteness theorem, Rice's theorem, and many more, see <a href="http://arxiv.org/abs/math/0305282" rel="nofollow">here</a>.</p> http://mathoverflow.net/questions/120802/are-all-endomorphisms-of-c-just-power-maps/120849#120849 Answer by Chris Heunen for Are all endomorphisms of C^* just power maps? Chris Heunen 2013-02-05T10:17:27Z 2013-02-05T10:17:27Z <p>There are many "wild" automorphisms of the complex numbers, that preserve addition and multiplication, but not much else. These are hard to write down, however, as their existence seems to rely on the axiom of choice. See e.g. the great expository paper "<a href="http://www.jstor.org/stable/2689301" rel="nofollow">Automorphisms of the complex numbers</a>" by P. Yale, Mathematics Magazine 39(3), 1966.</p> http://mathoverflow.net/questions/116150/triangularizing-a-function-matrix-with-smooth-eigenvlaues/116164#116164 Answer by Chris Heunen for Triangularizing a function matrix with smooth eigenvlaues Chris Heunen 2012-12-12T10:55:31Z 2012-12-12T10:55:31Z <p>Not precisely what you are asking, but if you look at continuous functions (instead of smooth and homogeneous ones), Grove and Pedersen ["Diagonalizing Matrices over $C(X)$", Journal of Functional Analysis 59, 65--89, 1984] prove the following. $N \times N$ matrices can be diagonalized for all $N$ if and only if $X$ is a sub-Stonean topological space with $\dim X \leq 2$ and $X$ carries no nontrivial $G$-bundles over any closed subset, for $G$ a symmetric group or the circle group.</p> http://mathoverflow.net/questions/114938/algebras-with-countable-chains-only/114944#114944 Answer by Chris Heunen for Algebras with countable chains only Chris Heunen 2012-11-30T00:01:09Z 2012-11-30T01:19:49Z <p>The second requirement is too strict: it makes the first one impossible.</p> <p>The commutative von Neumann algebra $\ell^\infty(\mathbb{N})$ has as Gelfand spectrum the Stone-Cech compactification of $\mathbb{N}$ (with the discrete topology). This, in turn, is the Stone space of the Boolean algebra $\mathcal{P}(\mathbb{N})$, the powerset of the natural numbers. So $\ell^\infty(\mathbb{N}) \cong C(\mathop{Stone}(\mathcal{P}(\mathbb{N})))$ embeds in $C(\mathop{Stone}(B))$ if and only if $\mathcal{P}(\mathbb{N})$ embeds in $B$. But the former has uncountable chains. So if $B$ satisfies the second requirement, it has uncountable chains, and cannot satisfy the first requirement.</p> http://mathoverflow.net/questions/53330/double-orthogonal-complement/114807#114807 Answer by Chris Heunen for Double Orthogonal Complement Chris Heunen 2012-11-28T19:52:12Z 2012-11-28T19:52:12Z <p>In relation to the question: is it possible to characterize the class of subspaces $W$ satisfying $W=W^{\perp\perp}$ in another natural or revealing way? A <a href="http://dx.doi.org/10.2977/prims/1195195769" rel="nofollow">theorem of Amemiya and Araki</a> shows that the partially ordered set of such subspaces forms an orthomodular lattice if and only if the surrounding inner product space is a Hilbert space.</p> http://mathoverflow.net/questions/114447/idempotent-elements-in-matrix-ring/114488#114488 Answer by Chris Heunen for Idempotent elements in matrix ring Chris Heunen 2012-11-26T05:54:26Z 2012-11-26T05:54:26Z <p>Any idempotent $e$ of $R$ induces an idempotent $E=\mathop{diag}(e,\ldots,e)$ of $M_n(R)$. In fact, if $e_i$ are idempotents in $R$, then $E=\mathop{diag}(e_1,\ldots,e_n)$ is an idempotent of $M_n(R)$.</p> <p>Conversely, if $R$ is nice enough, an idempotent $E$ in $M_n(R)$ can be diagonalized to $E=U^{-1} \cdot \mathop{diag}(e_1,\ldots,e_n) \cdot U$ for some $U \in M_n(R)$ and idempotents $e_i$ in $R$.</p> <p>Of course, this relies crucially on $R$ being nice enough. One sufficient "nicety" condition is that $R$ is an AW*-algebra; see for example <a href="http://arxiv.org/abs/1208.5120" rel="nofollow">this paper</a>. </p> http://mathoverflow.net/questions/114478/categorical-nomenclature/114487#114487 Answer by Chris Heunen for Categorical nomenclature Chris Heunen 2012-11-26T05:25:57Z 2012-11-26T05:25:57Z <p>I'm not sure if this is what you mean, but here are some terms that might come in handy to guide the search. If the collection ${v_{ij}}$ is always of the same shape, say with $i,j$ ranging over a fixed set, and is closed under composition, then what you're describing sounds like a <i>functor category</i>. </p> <p>More precisely, such a category consists of <i>diagram</i>s $D$ of shape $J$. Here, $J$ is a category itself, with objects $i,j,\ldots$, and $D \colon J \to C$ is a functor. Explicitly, you can describe $D$ by giving all the objects $D(i)$ and all the morphisms $D(f)\colon D(i)\to D(j)$ for $f \colon i \to j$ in $J$. Your notation $c_i=D(i)$ suggests there is only one morphism $c_i \to c_j$. In that case $J$ has only one morphism $(i \leq j) \colon i \to j$, i.e. is a preorder $(J,\leq)$ regarded as a category; then $v_{ij} = D(i \leq j)$. Morphisms $D \to E$ are natural transformations. This category is usually denoted $C^J$.</p> <p>The "witness" you mention at the end then sounds like a <i>(co)limit</i> of a diagram.</p> http://mathoverflow.net/questions/113859/categorifications-of-zorns-lemma Categorifications of Zorn's lemma Chris Heunen 2012-11-19T19:21:08Z 2012-11-19T20:20:48Z <p>I'm wondering about categorifications of Zorn's lemma along the following lines.</p> <blockquote> <p><i>Lemma</i>: if $\mathbf{C}$ is a small category in which every directed diagram of monomorphisms has a cocone of monomorphisms, then there is an object $A$ such that any monomorphism $A \rightarrowtail B$ splits.</p> </blockquote> <p><i>Proofsketch</i>: If there were no such object, we could use the axiom of choice to define a function $f$ that assigns to each directed diagram $D$ of monomorphisms, a monomorphism with domain $\mathop{cocone}(D)$ that does not split. Define an ordinal-indexed diagram $D$ by setting $D(\alpha) = f\big( D(\beta) \mid \beta&lt;\alpha \big)$. Because $D$ consists of monomorphisms that don't split, it cannot contain any cycles. But this contradicts smallness of $\mathbf{C}$.</p> <p>Can this be generalised? (E.g. do we need smallness and monics?) Are other versions known? Are there similar categorical existence statements that are provable without the axiom of choice?</p> http://mathoverflow.net/questions/106298/iterating-monoid-categories/106326#106326 Answer by Chris Heunen for Iterating monoid categories Chris Heunen 2012-09-04T09:42:58Z 2012-09-04T09:42:58Z <p>If you take <i>commutative</i> monoids in the first step, there is another natural choice of monoidal product on $\mathrm{CMon}(C)$. Namely, the universal "bilinear" construction, due in this generality to Anders Kock: in case $(C,\otimes) = (\mathrm{Set},\times)$, morphisms $M \otimes N \to K$ correspond to functions $M \times N \to K$ that are monoid morphisms in each argument separately when the other is fixed. In a sense this is more interesting, because no collapse of the kind you describe occurs. In fact, $\mathrm{Mon}(\mathrm{CMon}(C))$ is precisely the category of <i>semirings</i> in $C$. So heightening the "tower" of monoid categories then adds structure that occurs naturally and often in algebra.</p> <p>See also <a href="http://dx.doi.org/10.1016/j.entcs.2008.10.012" rel="nofollow">http://dx.doi.org/10.1016/j.entcs.2008.10.012</a> and references therein. Incidentally, that paper also has a precise proof of the result Mike Shulman's answers refers to.</p> http://mathoverflow.net/questions/102005/what-is-a-de-vries-algebra/102030#102030 Answer by Chris Heunen for What is a de Vries algebra? Chris Heunen 2012-07-12T10:01:01Z 2012-07-12T10:01:01Z <p>See his article "Stone duality and Gleason covers through de Vries duality" in Topology and its Applications 157:1064--1080, 2010, Definition 3.2.</p> http://mathoverflow.net/questions/100576/intersections-of-maximal-abelian-von-neumann-algebras/100583#100583 Answer by Chris Heunen for Intersections of maximal abelian von Neumann algebras Chris Heunen 2012-06-25T10:48:16Z 2012-06-25T10:48:16Z <p>This is only a partial answer, but it didn't fit in the comment box.</p> <p>In finite dimension, say $\dim(H)=n$, a maximal abelian von Neumann algebra $A \subseteq B(H) \cong M_n(\mathbb{C})$ just comes down to (the set of matrices that are diagonal in) a choice of basis for $H$. Similarly, $B$ consists of diagonal matrices in some (second) basis, possibly with repeated eigenvalues. So maximality forces $C$ to consist of all diagonal matrices in some (third) basis that spans the eigenspaces of the second one. The question is whether this third basis can be chosen while respecting $A \cap C=\mathbb{C}I$. If each eigenspace of $B$ has dimension an integer power of a prime number, then <a href="http://en.wikipedia.org/wiki/Mutually_unbiased_bases" rel="nofollow">mutually unbiased bases</a> are known to exist, and the answer is affirmative.</p> http://mathoverflow.net/questions/89877/can-every-nonempty-set-carry-abelian-group-structure Can every nonempty set carry abelian group structure? Chris Heunen 2012-02-29T16:12:41Z 2012-02-29T16:20:23Z <blockquote> <p><strong>Possible Duplicate:</strong><br> <a href="http://mathoverflow.net/questions/12973/does-every-non-empty-set-admit-a-group-structure-in-zf" rel="nofollow">Does every non-empty set admit a group structure (in ZF)?</a> </p> <p>Let $X$ be an arbitrary nonempty set. Can you define a multiplication making it into an abelian group?</p> </blockquote> <p>If $X$ is finite, say $|X|=n$, we can just use $X \cong \mathbb{Z}/n\mathbb{Z}$. What if $X$ is infinite?</p> <p>If I'm not mistaken, the group of permutations of $X$ with finite support has the same cardinality as $X$. So at least any nonempty set carries a group structure. But abelianizing this particular group structure changes the cardinality.</p> <p>Apologies if it is obvious, my group theory knowledge is just insufficient.</p> http://mathoverflow.net/questions/8846/proofs-without-words/58578#58578 Answer by Chris Heunen for Proofs without words Chris Heunen 2011-03-15T22:02:59Z 2012-02-15T04:47:04Z <p>Algebraic manipulations in monoidal categories can also be performed in a graphical calculus. And the best part is that this is completely rigorous: a statement holds in the graphical language if and only if it holds (in the algebraic formulation). See for example Peter Selinger's "<a href="http://dx.doi.org/10.1007/978-3-642-12821-9_4" rel="nofollow">A survey of graphical languages for monoidal categories</a>". There are many instances, for example in knot theory studied via braided categories. The following specific example comes from Joachim Kock's book "<a href="http://books.google.com/books?id=6dZZW08Z04MC" rel="nofollow">Frobenius Algebras and 2D Topological Quantum Field Theories"</a>, and proves that the comultiplication of a Frobenius algebra is cocommutative if and only if the multiplication is commutative.</p> <p><img src="http://oi55.tinypic.com/5k58uf.jpg" alt="alt text"></p> http://mathoverflow.net/questions/80884/when-are-technical-assumptions-critical When are technical assumptions critical? Chris Heunen 2011-11-14T10:27:27Z 2011-11-14T18:01:04Z <p>Apart from their technical statement and proof, a usual presentation of theorems is by leading up to them with a definite motivation or intuition, for example putting the results in the wider context of a research programme. This focus gives the author the ability to distinguish between "critical" and "technical" assumptions. By critical assumptions I mean the ones that are crucial to making the proof idea work in the first place. Technical assumptions are then the ones that just happen to be needed to straighten the details out. </p> <p>For example, from this point of view, a topological space being compact Hausdorff could be said to be a mere' technical assumption in Gelfand duality, but a critical assumption in Urysohn's lemma.</p> <p>For another example, an irate referee once taught me not to speak about "mild assumptions", because "under mild assumptions every group is a ring".</p> <p>But the line between technical and critical assumptions is vague and flexible at best. This question is about when the line moves.</p> <blockquote> <p>Are there interesting examples of critical assumptions that later (e.g. via a new proof for a generalized setting) turn out to be technical? And, more interestingly, vice versa, are there good examples of technical assumptions that later (e.g. with a different motivation) turn out to be critical?</p> </blockquote> http://mathoverflow.net/questions/23408/reference-for-the-gelfand-neumark-theorem-for-commutative-von-neumann-algebras/78322#78322 Answer by Chris Heunen for Reference for the Gelfand-Neumark theorem for commutative von Neumann algebras Chris Heunen 2011-10-17T10:08:23Z 2011-10-17T10:08:23Z <p>I think we established that the literature is lacking on this question. But I think the "correct" definition of morphisms between hyperstonean spaces can be puzzled together from G. Bezhanishvili's paper "<a href="http://dx.doi.org/10.1016/j.topol.2010.01.007" rel="nofollow">Stone duality and Gleason covers through de Vries duality</a>" (Topology and its Applications 157:1064-1080, 2010), especially section 6. </p> <p>He proves in detail a duality between the category of complete Boolean algebras and complete Boolean algebra homomorphisms, and the category of extremally disconnected compact Hausdorff spaces and continuous open maps. But commutative von Neumann algebras and normal *-homomorphisms form a full subcategory of the former (via taking projections), which corresponds to the full subcategory of the latter consisting of hyperstonean spaces. </p> <p>So Gelfand duality really restricts quite cleanly: commutative von Neumann algebras and normal *-homomorphisms are dual to hyperstonean spaces and <em>open</em> continuous maps.</p> http://mathoverflow.net/questions/75921/what-is-the-tensor-product-for-the-eilenberg-moore-category-of-a-commutative-mona/75929#75929 Answer by Chris Heunen for What is the tensor product for the Eilenberg-Moore category of a commutative monad? Chris Heunen 2011-09-20T09:41:49Z 2011-09-20T09:41:49Z <p>Caveat: this construction only works if your category of algebras has coequalizers of reflexive pairs, i.e. coequalizers of parallel pairs of arrows with a common right inverse.</p> <p>Let $T \colon C \to C$ be your monad. Being commutative, it comes with maps $\mathrm{dst} \colon T(A) \otimes T(B) \to T(A \otimes B)$. Let $\phi \colon TA \to A$ and $\psi \colon TB \to B$ be algebras. Then $\phi \otimes \psi$ is the coequalizer in $\mathrm{Alg}(T)$ of $T(\phi \otimes \psi)$ and $\mu \circ T(\mathrm{dst})$ (which is a reflexive pair of morphisms from the free algebra on $T(A) \otimes T(B)$ to the free algebra on $A \otimes B$). The unit $I$ in $\mathrm{Alg}(T)$ is the free algebra $\mu \colon T^2(I) \to T(I)$. Moreover, the free functor $C \to \mathrm{Alg}(T)$ preserves monoidal structure. </p> <p>A good example to keep in mind is where $T$ is the free vector space monad on the category of sets. The coequalizers then is pretty much directly the usual tensor product construction with bilinear maps.</p> <p>This goes back to Anders Kock, see his papers "Closed categories generated by commutative monads" (J. Austr. Math. Soc. 12:405-424, 1975), and "Monads on symmetric monoidal closed categories" (Archiv der Mathematik, 21:1-10, 1970).</p> http://mathoverflow.net/questions/73652/adjoint-of-multiplication-operator-in-a-commutative-algebra/73655#73655 Answer by Chris Heunen for adjoint of multiplication operator in a commutative algebra Chris Heunen 2011-08-25T12:23:47Z 2011-08-25T12:23:47Z <p>In case $k=\mathbb{C}$, what you're describing is a finite-dimensional <i>H*-algebra</i>. More generally, these are Banach algebras, whose carrier space is a Hilbert space, satisfying the adjoint property you mention. </p> <p>It is natural to make a nondegeneracy assumption: $A$ is called <i>proper</i> when $\forall a \in A\,.\, a \circ A = 0 \Rightarrow a = 0$. This turns out to be equivalent to the adjoint $L_a^\ast$ being unique, or, in other words, $\ast$ being an involution. Every H*-algebra is a direct sum of a proper one and an algebra in which $a \circ b=0$ for all $a$ and $b$.</p> <p>There is a neat structure theorem by Warren Ambrose (see <a href="http://www.jstor.org/stable/1990182" rel="nofollow">http://www.jstor.org/stable/1990182</a>), showing that proper H*-algebras are always direct sums of full matrix algebras. In particular, commutative H*-algebras are direct sums of 1-dimensional algebras, and hence correspond precisely to orthogonal basis of their carrier space! </p> <p>I don't know about other fields $k$, but Ambrose's proof basically comes down to carefully analyzing idempotents, which is feasible to repeat for other fields $k$.</p> http://mathoverflow.net/questions/67824/why-does-hom-need-an-identity-in-the-definition-of-the-category/67869#67869 Answer by Chris Heunen for Why does Hom need an identity in the definition of the category? Chris Heunen 2011-06-15T15:27:58Z 2011-06-15T15:27:58Z <p>I suppose the extent to which hell breaks loose depends entirely on the purpose you're using categories for. Dropping identities presumably invalidates the Yoneda lemma, and therefore all the results in category theory that depend on it. But if you just want to use (monoidal) functors as bookkeeping devices without expecting "deeper" category theory to predict things for you, nothing much happens. To follow up on Scott's answer, there is for example a perfectly good theory about adjunctions when one ignores identities, and there is in fact a relation to formally adding identities. See Hayashi "Adjunction of semifunctors" in Theoretical Computer Science 41:95--104, 1985, and Hoofman and Moerdijk "A remark on the theory of semi-functors" in Mathematical Structures in Computer Science 5:1--8, 1995.</p> http://mathoverflow.net/questions/64407/when-is-this-map-completely-positive When is this map completely positive? Chris Heunen 2011-05-09T16:33:37Z 2011-05-11T12:06:50Z <p>Consider the complex $n$-by-$n$ matrices $M_n$. Suppose that $A_i$, for $i=1,\ldots,n^2$, satisfy $\mathrm{Tr}(A_i^* A_j)=\delta_{ij}$, so that together they form an orthonormal basis for $M_n$. Define a linear map $T \colon M_n \to M_n \otimes M_n$ by $T(A_i) = A_i \otimes A_i$.</p> <blockquote> <p>Question: when is $T$ completely positive?</p> </blockquote> <p>For example, if $A_i$ are the matrices with a single entry one and the rest zeroes in some fixed basis of $\mathbb{C}^n$, then $T$ is completely positive. In fact, I think these might be the only examples. If $T$ is completely positive, then the following are equivalent to $A_i$ being matrix units as in the above example:</p> <ul> <li>each $A_i$ has rank one;</li> <li>each positive semidefinite $A_i$ has trace one;</li> <li>the set $\{0,A_1,\ldots,A_{n^2}\}$ is closed under multiplication;</li> <li>$T(1)$ is idempotent;</li> <li>$T^*(1) \leq 1$;</li> <li>$T$ preserves trace.</li> </ul> <p>These are sufficient conditions, but proving they are sufficient doesn't use $\mathrm{Tr}(A_i^* A_j)=\delta_{ij}$ at all. Are they necessary?</p> http://mathoverflow.net/questions/63258/seemingly-emergent-structures-in-mathematics/63271#63271 Answer by Chris Heunen for Seemingly emergent structures in mathematics Chris Heunen 2011-04-28T09:01:31Z 2011-04-28T09:01:31Z <p>I'm not exactly sure what you're after, but one could think of <a href="http://en.wikipedia.org/wiki/Ramsey_theory" rel="nofollow">Ramsey theory</a> as saying that any large enough structure will necessarily contain an orderly substructure. Or, even more loosely, that order is unavoidable in a large enough chaos. So I suppose the following would be examples of answers to the question:</p> <ul> <li>Ramsey's theorem;</li> <li>van der Waerden's theorem;</li> <li>the Hales-Jewett theorem;</li> <li>Szemeredi's theorem;</li> <li>the Green-Tao theorem.</li> </ul> <p>And there are many more in this vein. Especially infinite versions of such theorems seem to match nicely with your example of the central limit theorem.</p> http://mathoverflow.net/questions/60833/definition-of-initialterminal-objects-in-an-object-free-category/60838#60838 Answer by Chris Heunen for Definition of Initial&Terminal Objects in an `Object-Free'' Category Chris Heunen 2011-04-06T17:25:27Z 2011-04-06T17:25:27Z <p>In this view, objects are equated with morphisms that are identities, or "units" in their terminology. So a morphism $x$ is initial when it is a unit and for every unit $y$ there is a unique morphism $f$ for which $f \circ x$ and $y \circ f$ are defined. Similarly, a morphism $y$ is terminal when it is a unit and for every unit $x$ there is a unique morphism $f$ for which $f \circ x$ and $y \circ f$ are defined.</p> http://mathoverflow.net/questions/60241/why-dont-ideals-and-quotients-work-well-for-categories Why don't ideals and quotients work well for categories? Chris Heunen 2011-03-31T22:38:20Z 2011-04-03T18:20:28Z <p>Ideals are intimately related to quotients and congruence relations. They clearly play a very important role in ring theory and order theory. So do normal subgroups in group theory. (Enriched) category theory could be regarded as a common generalization of all these settings. Why is it that such important structures don't work well for categories?</p> <p>I am aware that there is a categorical notion of <a href="http://en.wikipedia.org/wiki/Quotient_category" rel="nofollow">congruence relation</a>. However, this doesn't seem to take the spirit of multiple objects to heart: all it does is keep the same objects and relate morphisms within homsets. For one thing, the accompanying notion of quotient category doesn't correspond to coequalizers in $\mathbf{Cat}$ (of which there are many more).</p> <p>It is not even clear how to define an ideal of a category. To allow for proper ideals, it probably shouldn't simply be a subcategory. Naively one thinks of a subset $I(X,Y)$ of each $\mathrm{Hom}(X,Y)$ that is invariant under composition with arbitrary morphisms, or just of subsets $I(X)$ of each $\mathrm{Hom}(X,X)$, or of $I(X)$ just for some objects; but this doesn't really take objects into account. Thinking of an appropriate definition is even more perplexing for higher categories.</p> <blockquote> <p>Question: are there related notions of ideal and quotient for categories that have interesting consequences but are not trivial on the level of objects?</p> </blockquote> <p>It is left open what roles left (postcomposition) or right (precomposition) ideals should play; a related question is if there is a notion of commutativity for categories with interesting consequences.</p> <p>A convincing explanation why one shouldn't consider such questions would also be a good answer.</p> http://mathoverflow.net/questions/58416/mono-and-epi-morphisms-for-c-algebras/58445#58445 Answer by Chris Heunen for Mono- and epi-morphisms for C*-algebras Chris Heunen 2011-03-14T16:49:15Z 2011-03-14T16:49:15Z <p>I'm not sure why <code>*</code>-homomorphisms would be too restrictive a choice, but taking those as morphisms between C*-algebras as objects, the epimorphisms are precisely the surjective <code>*</code>-homomorphisms. This is proposition 2 in G. A. Reid's "Epimorphisms and surjectivity", Inventiones Mathematicae 9:295-307, 1970.</p> http://mathoverflow.net/questions/57388/are-these-ideals-in-rings-of-operators-on-hilbert-space-unique Are these ideals in rings of operators on Hilbert space unique? Chris Heunen 2011-03-04T19:13:19Z 2011-03-04T19:54:57Z <p>Suppose that, for every Hilbert space $H$, we have a subset $I(H) \subseteq B(H)$ of bounded linear operators on $H$, and that together all $I(H)$ form a two-sided ideal, in the sense that whenever $h \in I(H)$, also $f \circ h \circ g \in I(K)$ for any bounded linear maps $f \colon H \to K$ and $g \colon K \to H$. To prevent degeneration, additionally assume $I(\mathbb{C})=B(\mathbb{C})$ and $I(H) \neq B(H)$ for some $H$.</p> <blockquote> <p><strong>Question:</strong> When do such two-sided ideals $I$ satisfy the following: <br> if $f \colon H \to K$ and $g \colon K \to H$ are bounded linear maps, and $g \circ f \in I(H)$, then also $f \circ g \in I(K)$?</p> </blockquote> <p>Taking $I(H)$ to be the trace class operators gives one example. Is this the unique one?</p> <p>I know that $I(H)$ at least has to contain the finite rank operators, and has to be contained in the compact operators. Finite rank operators also form a two-sided ideal, but do they satisfy the requirement, i.e. if $g \circ f$ is of finite rank, is $f \circ g$, too?</p> http://mathoverflow.net/questions/54923/is-there-a-disjoint-union-sigma-algebra/54953#54953 Answer by Chris Heunen for Is there a "disjoint union" sigma algebra? Chris Heunen 2011-02-09T23:12:09Z 2011-02-09T23:12:09Z <p>David H. Fremlin's "Measure Theory", vol 2, 214K, gives this construction explicitly. He also proves some elementary properties, but unfortunately stops short of universal properties such as in Peter's insightful answer. </p> http://mathoverflow.net/questions/53767/constructions-unique-up-to-non-unique-isomorphism/53771#53771 Answer by Chris Heunen for Constructions unique up to non-unique isomorphism Chris Heunen 2011-01-30T10:17:54Z 2011-01-30T10:17:54Z <p>Vector spaces have a basis that is unique up to a non-unique isomorphism.<br> Hilbert spaces have an orthonormal basis that is unique up to a non-unique unitary.<br> (At least, if you accept Zorn's lemma, i.e. the axiom of choice)</p> http://mathoverflow.net/questions/53188/decent-texts-on-categorical-logic/53218#53218 Answer by Chris Heunen for Decent Texts on Categorical Logic Chris Heunen 2011-01-25T10:54:32Z 2011-01-25T10:54:32Z <p>I'd like to add Carsten Butz' "<a href="http://www.brics.dk/LS/98/2/" rel="nofollow">Regular categories and regular logic</a>". It is available online in the BRICS lecture series, and is very accessible. Though perhaps a bit too basic at times for readers with some background knowledge, it is very suitable for a first introduction before jumping into texts on toposes.</p> http://mathoverflow.net/questions/49426/is-there-a-category-structure-one-can-place-on-measure-spaces-so-that-category-th/49431#49431 Answer by Chris Heunen for Is there a category structure one can place on measure spaces so that category-theoretic products exist? Chris Heunen 2010-12-14T20:33:57Z 2010-12-15T06:39:19Z <p>If preserving measure is too strong a choice of morphisms to guarantee products, one could weaken that a bit. Of course, taking measurable functions as the morphisms would give a category equivalent to that of measurable spaces, but that completely ignores the measures. How about taking as morphisms measurable functions $\phi \colon X \to Y$ such that $\phi_* \mu_X \ll \mu_Y$? The latter means that $\phi_*\mu_X$ is absolutely continuous with respect to $\mu_Y$, i.e. that $\mu_X(\phi^{-1}(E))=0$ whenever $\mu_Y(E)=0$. Then certainly the projection $\pi \colon X \times Y \to X$ is a well-defined morphism, and if I did my doodles right, so is the tuple inherited from the product structure on the category of measurable spaces.</p> <p><strong>UPDATE:</strong> The diagonal map $X \to X \times X$ is not necessarily a morphism as defined above, and hence the category of measure spaces with these morphisms does not have products. Do we really need an even weaker choice of morphisms?</p> http://mathoverflow.net/questions/130788/anomalies-in-the-definition-of-turaevs-tqft Comment by Chris Heunen Chris Heunen 2013-05-16T09:46:56Z 2013-05-16T09:46:56Z Could you clarify a bit further? Are $\tau(M_i)$ Hilbert spaces? Of what algebra is $k$ an invertible element? http://mathoverflow.net/questions/118246/characterising-categories-of-vector-spaces/118269#118269 Comment by Chris Heunen Chris Heunen 2013-01-07T13:18:41Z 2013-01-07T13:18:41Z If you add a contravariant identity-on-objects involution to the requirements (complete, compact, semisimple, monoidal with simple unit), the scalars will be an involutive field $k$, and the category will be (equivalent to) that of finite-dimensional $k$-Hilbert spaces. See <a href="http://www.tac.mta.ca/tac/volumes/22/13/22-13abs.html" rel="nofollow">tac.mta.ca/tac/volumes/22/13/22-13abs.html</a>. http://mathoverflow.net/questions/114938/algebras-with-countable-chains-only/114944#114944 Comment by Chris Heunen Chris Heunen 2012-11-30T01:14:56Z 2012-11-30T01:14:56Z Good point, thanks! I'll edit the answer. http://mathoverflow.net/questions/114703/why-does-tensor-product-in-abv-require-colimits-in-v/114708#114708 Comment by Chris Heunen Chris Heunen 2012-11-27T23:07:04Z 2012-11-27T23:07:04Z Re point 2. This in fact works for any commutative monad whose algebra category has coequalisers of reflexive pairs. This is due to Anders Kock in the 1970s, see <a href="http://dx.doi.org/10.1007/BF01304852" rel="nofollow">dx.doi.org/10.1007/BF01304852</a> and <a href="http://dx.doi.org/10.1007/BF01220868" rel="nofollow">dx.doi.org/10.1007/BF01220868</a>. http://mathoverflow.net/questions/114478/categorical-nomenclature/114487#114487 Comment by Chris Heunen Chris Heunen 2012-11-26T06:16:34Z 2012-11-26T06:16:34Z @Johan: I'm not sure I follow what you mean. A preorder is a category that has at most one morphism between any two objects. There can also be no arrow. In other words, a disjoint union of preorders is again a preorder. http://mathoverflow.net/questions/113859/categorifications-of-zorns-lemma Comment by Chris Heunen Chris Heunen 2012-11-19T23:04:38Z 2012-11-19T23:04:38Z David, left-cancellative categories (i.e. categories in which every morphism is monic) are not necessarily preorders; think of any groupoid. Having said that, it would definitely be interesting to &quot;derive&quot; a lemma such as that in the question from Zorn's actual lemma applied on a &quot;higher&quot; categorical level; could you make that more precise? http://mathoverflow.net/questions/113859/categorifications-of-zorns-lemma Comment by Chris Heunen Chris Heunen 2012-11-19T20:21:21Z 2012-11-19T20:21:21Z Good spot. It needs that every directed diagram of monomorphisms has a cocone of monomorphisms. I've edited the question, thanks. http://mathoverflow.net/questions/112177/a-generalization-of-a-group-isomorphism Comment by Chris Heunen Chris Heunen 2012-11-12T18:30:16Z 2012-11-12T18:30:16Z For abelian categories, this is known, right? (See e.g. 2.67 in Freyd's book &quot;Abelian categories&quot;.) Are you asking if it still holds without assuming that every monomorphism is a kernel? http://mathoverflow.net/questions/104494/product-operation-name-and-notation/104509#104509 Comment by Chris Heunen Chris Heunen 2012-08-13T09:27:41Z 2012-08-13T09:27:41Z @Wolfgang: &quot;cotupling&quot; has been used. I suppose in Colin's scheme that would become &quot;copairing&quot;? http://mathoverflow.net/questions/60328/what-kind-of-completion-is-this/63393#63393 Comment by Chris Heunen Chris Heunen 2012-07-03T09:53:41Z 2012-07-03T09:53:41Z Thanks for your satisfyingly categorical answer. However, I'm having trouble locating this material in Conway's book. Could I press you for a more precise reference? http://mathoverflow.net/questions/98777/moore-penrose-inverse-as-an-adjoint Comment by Chris Heunen Chris Heunen 2012-06-05T08:50:39Z 2012-06-05T08:50:39Z If you drop the latter two conditions, pseudo-inverses need not be unique anymore. That might get around the obstruction Michal raises. http://mathoverflow.net/questions/15696/category-of-nonempty-metric-spaces-and-contractive-maps Comment by Chris Heunen Chris Heunen 2012-05-21T09:15:37Z 2012-05-21T09:15:37Z Such notions of ideals have been worked out in <a href="http://arxiv.org/abs/math/9805102" rel="nofollow">arxiv.org/abs/math/9805102</a>, for example. They cover Hilbert-Schmidt maps, such as in Andrew's answer, and trace class operators, as in Kevin's comment. One would think the ideal of contractions could be axiomatized similarly. http://mathoverflow.net/questions/88301/constructing-the-stone-space-of-a-distributive-lattice Comment by Chris Heunen Chris Heunen 2012-02-12T21:05:32Z 2012-02-12T21:05:32Z &quot;Stone spaces&quot; by P. T. Johnstone http://mathoverflow.net/questions/82090/when-do-tensor-products-of-c-algebras-commute-with-colimits Comment by Chris Heunen Chris Heunen 2011-11-28T16:43:47Z 2011-11-28T16:43:47Z By the way, preserving filtered colimits is the same as preserving colimits of chains [Adamek&amp;Rosicky, Cambridge Univ Press, 1994, Cors 1.5 &amp; 1.7]. So you may assume that I is a total order. If all the maps $A_i \to A_j$ are inclusions, doesn't that make the colimit just the closure of the union of all the $A_i$, leading to a trivial proof? http://mathoverflow.net/questions/82090/when-do-tensor-products-of-c-algebras-commute-with-colimits Comment by Chris Heunen Chris Heunen 2011-11-28T16:15:17Z 2011-11-28T16:15:17Z In the category of von Neumann algebras, the functor $- \otimes N$ preserves coequalizers [Guichardet, Bull Sci Math 90:41-64, 1966, Prop 8.3]. I would hope the same holds for C*-algebras, so that it would suffice to concentrate on coproducts. But also, in the category of von Neumann algebras, colimits do not preserve flatness [Guichardet, Remark 8.2], which is a bad sign.