Joe Hannon
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Registered User
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Apr 25 |
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de Rham cohomology and flat vector bundles @Aaron: en.wikipedia.org/wiki/Litotes |
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Apr 24 |
revised |
looping and delooping spaces and categories deleted 3 characters in body |
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Apr 23 |
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is there an anyon structure analogous to spin structure for rank 2 bundle? Thank you. The input from physics is that anyons satisfy complicated statistics, due to living in a representation of the braid group rather than the symmetric group. I thought the complicated topological issues with the representation theory of anyons would necessarily imply a more complicated theory of anyon structure (analogue of spin structure). I guess I was wrong! |
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Apr 23 |
revised |
looping and delooping spaces and categories remove spurious tag |
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Apr 23 |
asked | looping and delooping spaces and categories |
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Apr 3 |
asked | is there an anyon structure analogous to spin structure for rank 2 bundle? |
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Mar 21 |
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Exactness of filtered colimits See also Steve Lack's similar question (as yet not answered successfully) mathoverflow.net/questions/57099/… |
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Dec 11 |
revised |
Splitting lemma under assumption of the axiom of choice added 115 characters in body |
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Dec 10 |
awarded | ● Yearling |
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Dec 8 |
awarded | ● Critic |
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Dec 7 |
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What is the free monoidal category generated by a monoid? I found Tom Leinster's post on nCafe on this very question. golem.ph.utexas.edu/category/2010/01/… He says the same thing you've all been saying: it's just a monoidal category with a monoid and no other generators or relations. I know you've all been saying this, thanks for your patience for my stubbornness. I still find it surprising that this concept can't be described as an adjunction, though. |
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Dec 7 |
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What is the free monoidal category generated by a monoid? Yes, my mistake is in #4. In fact the free monoidal category generated by the terminal category 1 is not ∆. The morphisms in $F(M)$, where $M$ is a category with one object, are $k$-tuples of elements of $M$. They are endomorphisms $k\to k$. In particular, there is no morphism $2\to1$ in this category. I thought there was because 1 was terminal. But there is no terminal object in a disjoint category. So it's not $\Delta.$ $F(X)$ contains no extra information above $X$, so it cannot define a monoid structure for $X$, as we expect by the properties of adjunctions. Back to the drawing board... |
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Dec 7 |
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What is the free monoidal category generated by a monoid? 1. if a monoidal category $C$ has countable coprods preserved by tensor then $U\colon Mon(C)\to C$ has an adjoint "free mndl obj functor $F$" $C\to Mon(C)$,2. The category of categories has countable coprods and preserved by cartesian monoidal structure,3. Hence there is an adjoint functor $F\colon\text{Cat}\to\text{MonCat}$ 4. $F(1)=\Delta$,5. There is a 1-1 correspondence between functors $1\to UM$ and $F(1)\to M$ by adjointness,6. $M$ may admit distinct monoid structures, yielding distinct morphisms $F(1)\to M$ for a single $1\to UM$. This is a contradiction, something's wrong. #4? |
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Dec 6 |
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What is the free monoidal category generated by a monoid? @Tom: that's the point. That a free group on one generator is initial in the category of groups equipped with an element = that there is a universal arrow from the singleton to forgetful functor. Universal arrow from all sets too, so there's a left adjoint. I'm looking for same thing to occur for free monoidal category. I'll eventually have to accept a negative answer, but maybe my guess might do: there is a universal arrow from 1 to forgetful functor MonCat->Cat. Then $\Delta=F(1).$ But this gives bjctn between mndl functors $\Delta\to C$ and functors $1\to UC$, all objects not just monoids? |
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Dec 6 |
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What is the free monoidal category generated by a monoid? @Todd: It sounds like you, Andreas, and Adrien are all saying the same thing. It's "free" in name only. It's free because initial objects are like free objects. By analogy, the initial object in the category of monoids with maps from set $S$ is $F(S)$, the free monoid of words in $S$. I find this answer somewhat dissatisfying, it seems somehow circular. How about this: there's a forgetful functor $\mathbf{MonCat}\times Mon(C)\to Mon(C)$... hmm no that doesn't make sense. I guess we have the "evaluation at 1" functor $C^{\Delta}\to C$, for $C$ monoidal... no, that's not going to work. Hmmm |
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Dec 5 |
revised |
What is the free monoidal category generated by a monoid? bigoplus -> coprod; added 4 characters in body |
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Dec 5 |
revised |
What is the free monoidal category generated by a monoid? Mon(Cat)=MonCat; added 66 characters in body |
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Dec 5 |
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What is the free monoidal category generated by a monoid? So you've defined "initial object in the category of monoidal categories with monoid object". Do you propose to also take that as the definition of "free monoidal category on a monoid object"? I guessed "free monoidal category" meant something in terms of an adjoint to a forgetful functor, as is usual with free monoids and related. |
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Dec 5 |
asked | What is the free monoidal category generated by a monoid? |
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Dec 5 |
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Why does tensor product in Ab(V) require colimits in V? I guess I already worked this out for myself, with help from Zhen, but chapter 3 of Steve Awodey's category theory [online lecture notes](andrew.cmu.edu/course/80-413-713/notes/chap03.pdf) have a description of how to construct a presentation out of a coequalizer of the free group on relations. |
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Nov 29 |
awarded | ● Commentator |
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Nov 29 |
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Why does tensor product in Ab(V) require colimits in V? Then $A\otimes B=\text{Coeq}(\lambda,\rho)$. So if all the limits and colimits used in this construction exist, the tensor product exists. |
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Nov 29 |
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Why does tensor product in Ab(V) require colimits in V? Now to restate it without "global elements". Set $\lambda=UF(\mu_A\times\text{id}_B\coprod \text{id}_A\times\mu_B),$ where $\mu_A\colon A\times A\to A$ is the multiplication morphism. And $\rho=UF((\text{pr}_{\hat{2}}+\text{pr}_{\hat{1}})\coprod(\text{pr}_{\hat{3}}+\text{pr}_{\hat{2}}))$, where $\text{pr}_{\hat{2}}=\text{pr}_1\times\text{pr}_3\colon X\times Y\times Z\to X\times Y$ is the projection which drops the second component. I've assumed here that $\textbf{Ab}(\mathcal{V})$ is preadditive (so I can take the sums $\text{pr}_{\hat{2}}+\text{pr}_{\hat{1}}$ etc.) (cont) |
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Nov 29 |
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Why does tensor product in Ab(V) require colimits in V? @Zhen: I guess it must be obvious how to make this explicit if you have experience with generators and relations, but I found this situation confusing, maybe because of the large number of generators. After a while of staring at it, I think I got it. It should model this situation: $R=A\times A\times B\coprod A\times B\times B.$ Then we have $\lambda,\rho\colon UF(R)\rightrightarrows UF(A\times B)$ with $\lambda(a_1,a_2,b_1)=(a_1+a_2,b_1)$,$\lambda(a_3,b_2,b_3)=(a_3,b_2+b_3)$,$\rho(a_1,a_2,b_1)=(a_1,b_1)+(a_2,b_1)$, and $\rho(a_3,b_2,b_3)=(a_3,b_2)+(a_3,b_3).$ (cont) |
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Nov 28 |
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Set Theory and V=L So I gotta ask, what's the least real number? |
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Nov 28 |
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Why does tensor product in Ab(V) require colimits in V? When you say "once we know that $\textbf{Ab}(\mathcal{V})$ has coequalisers ... we can construct tensor products in the usual way", I'm imagining you mean something like: set $A\otimes B=\text{Coker}\{I\to F(A\times B)\}$, where $F$ is the free abelian group object functor (which exists because $\textbf{Ab}(\mathcal{V})\to \mathcal{V}$ is monadic?) and I is some subobject of $F(A\times B)$ generated by elements like $(a+a',b)−(a,b)−(a',b)$, and here "element" means "morphism $1\to A$". And we have 1 and $\times$ since $\mathcal{V}$ is Cartesian. Is that right? |
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Nov 28 |
awarded | ● Scholar |
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Nov 28 |
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Why does tensor product in Ab(V) require colimits in V? The explanation for the Hopf algebra axioms that the comonoid and antipode structures are precisely the thing you need to make group objects work is lovely and surprising. Thank you! As for the construction of the tensor product out of reflexive coequalisers, that pretty much went over my head. But the answer to my question is clear enough: tensor objects need not exist, but some extra axioms, including the existence of the right kinds of colimits, will guarantee them. Thank you also for the link to your notes on algebraic theories. They look very useful to me. |
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Nov 28 |
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Examples of common false beliefs in mathematics. Wait, what ascending chain of free groups is $\mathbb{Q}$ the union of? |
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Nov 27 |
asked | Why does tensor product in Ab(V) require colimits in V? |
