# Tag Info

7

what about $\mathbb{Z}_2$ vector spaces over a field $k$? $(k,0)$ is the unit in this category and $(0,k)\otimes (0,k) = (k,0)$.

6

What's the definition of an additive monoidal category? Is it that tensor product distributes over addition of morphisms? If so, use the fact that a functor between additive categories preserves addition of morphisms iff it preserves biproducts (see for example this blog post).

5

"Uniqueness will fail for noncommutative monoids, so here we must use commutativity of our monoids, " This is not correct. You get a morphism in $V$ also for non-commutative monoids, but it won't be a monoid morphism. This is where commutativity is used. Uniqueness holds in general: If $h : M \otimes N \to X$ is a monoid morphism with $h \circ (M \otimes ... 5 The$\lambda$-structure is given by taking exterior powers. This is the main motivation I know for defining$\lambda$-rings in the first place. (You need an action of$S_n$on an$n^{th}$tensor power$V^{\otimes n}$to define the exterior power, which is what being symmetric monoidal gets you; in the braided monoidal case you only get an action of$B_n$.) 5 This is the classical situation covered by the Eckmann–Hilton argument. The point is that the multiplication on$M$defined by$\mu$is a homomorphism for the original multiplication$\cdot$on$M$and that's enough for the argument given on the Wikipedia page to apply. It seems a bit pointless to repeat it here. A number of references are given on ... 5 http://arxiv.org/abs/math/0401347 4 Let$\mathcal{C}$be the following category: The objects are pairs$(I, X, p)$where$I$and$X$are sets and$p : X \to I$is a surjection. The morphisms$(I, X, p) \to (J, Y, q)$are maps$f : X \to Y$such that$q \circ f = p$. (That means we must have$I \subseteq J$to have a morphism.) Composition and identities are inherited from$\mathbf{Set}$. I ... 4 Group objects can actually be defined in any monoidal category$(M, \otimes, I)$; in particular the monoidal product doesn't need to be the categorical product in$M$. For example, a monoid object in$(\text{Ab}, \otimes, 1)$is just a ring (and note that the categorical product in$\text{Ab}$is the direct product, not the tensor product). If$G$is a group ... 4 You are right that the Yoneda lemma plays the key role in the proof. By the definition of convolution: $$(F \otimes G) \otimes H \approx \int^{C, D} \left(\int^{A, B} F(A) \times G(B) \times \hom(C, A \otimes B)\right) \times H(D) \times \hom(-, C \otimes D)$$ Because products in$\mathbf{Set}$preserve coends, the above is isomorphic to: $$\int^{A, B, C, ... 4 A module is an abelian group. (It's more useful to think of a module as the analog of a vector space, but with the set of scalars coming from a ring instead of a field. Usually, one arrives at this notion of a module in terms of "the action of a ring on a set" where the set is a module.) A monoid is a relaxation of the definition of a group. A monoid has ... 4 There is a chain of forgetful functors which progressively forgets the various operations in the structure:$$\mathrm{Mod_R}\to\mathrm{Ab}\to\mathrm{AbMon}\to\mathrm{Set}$$The interesting thing is that you can go in the opposite direction too with free functors$$\mathrm{Set}\to\mathrm{AbMon}\to\mathrm{Ab}\to\mathrm{Mod_R}$$Each forgetful functor U ... 4 Your observation only means that there is no object (X,\phi) in the Drinfeld center which has X simple and non-isomorphic to V_e (e being the identity element of G) But let C\subseteq G be a conjugacy class, and let V_C=\bigoplus_{g\in C}V_g. Then you should be able to find a natural isomorphism ... 4 Use the Yoneda lemma! By adjunction there is a natural isomorphism$$\mathcal{C}(X \otimes I, C) \to \mathcal{C}(X, [I, C])$$for all objects X in \mathcal{C}. Being an isomorphism is preserved by all functors, so the natural transformation$$\mathcal{C}(X, C) \to \mathcal{C}(X \otimes I, C)$$induced by the right unitor \rho_X : X \otimes I \to X is ... 4 The term "linear monoidal category" doesn't mean "linear and monoidal category", but rather "monoidal (linear category)", i.e. a (weak) monoid in the monoidal bicategory of linear categories (over some fixed base ring R). The monoidal product of two linear categories C,D has as objects pairs of objects of C,D with hom-modules \hom((a,b),(c,d)) = ... 4 I believe you'll find the following article helpful: Change of Base, Cauchy Completeness and Reversibility (pdf) (In A. Labella & V. Schmitt, Theory and Applications of Categories, Vol. 10: 10, 2002, pp. 187–219.) This linked article addresses much of what you're looking for, and includes references which may be of help to you. (E.g. G.M. Kelly, ... 3 The main point is that composition of arrows in bicategories (tensor of objects in monoidal categories) wants to be associative by nature, but it is associative only up to isomorphism. However, we should really speak about multiple composition such as$$e\otimes f\otimes g\otimes h $$in a sense, without parenthesis! This can be made precise in more ways. ... 3 I'm a PhD student in operator algebraic quantum groups, which is a generalisation of topological groups effectively. The other side of quantum groups are the purely algebraic versions, that is to say Hopf algebras. So quantum groups begin with the study of functions on a group (as it's a generalisation), in the Hopf algebra case we have polynomials, for ... 3 Cartesian closure applies to cartesian categories, i.e. categories which are (symmetric) monoidal with respect to the (binary) product bifunctor (basically any finitely complete category is cartesian). Cartesian closed categories are those categories where each functor A\times- has a right adjoint (-)^A realizing the binatural bijection$$ {\cal ... 3 It's standard to assume that a "linear monoidal category" means that the linear and monoidal structures are compatible. For example, see Def 0.1.2 of Catègories Tensorielles, Def 1.12.3 of Tensor Categories or Def 3.2.4 of Dualizable Tensor Categories. If I ran across a paper that didn't make this compatibility explicit, I'd assume that the authors were ... 3 The only dualisable object in a cartesian monoidal category is the terminal object. Thus, a cartesian compact closed category must be trivial. Indeed, suppose$A$has a dual$A^*$. Since the unit object is terminal, the counit$\epsilon : A \times A^* \to 1$is forced. Consider the unit$\eta : 1 \to A^* \times A$. This decomposes into components as a ... 3 In general it is very rare for a braided monoidal category to commute "on the nose". For a very concrete example, the usual category of sets and functions is cartesian monoidal, so braided (even symmetric) monoidal, but in general$A \times B \neq B \times A$. As for the braids, there is a PRO (monoidal category with$\mathbb{N}$as the set of objects and ... 3 I think you want to be a little more careful, as a priori there might be multiple ways of writing a given object as F(X). This isn't a really problem though because if you have some object A, you know basically how to write it as F(X): just let X = G(A)! So we just have $$G(A \bullet B) \cong G(F(G(A))\bullet F(G(B))) \cong GF(G(A) \otimes G(B)) \cong G(A) ... 3 The pentagon identity is a commutative diagram of 2-morphisms, each one induced by \alpha_M. It looks as follows. Because I'm lazy, I will omit the associators from \mathcal{C}.$$\begin{array}{cc} \mu \circ ((\mu \circ (\mu \otimes\mathrm{id}_M)) \otimes \mathrm{id}_M) & \rightarrow & \mu \circ ((\mu \circ (\mathrm{id}_M \otimes \mu)) ... 3 The term you want is "distributes over," not "commutes with." (Whatever "$A$commutes with$B$" means it should be symmetric in$A$and$B$, which the condition you want is not.) Such monoidal categories are called distributive. Examples include any closed monoidal category with coproducts because in this case$A \otimes (-)$has a right adjoint and hence ... 3 The question is not really precise. And you certainly don't look for an equivalent concept. Perhaps you are looking for the notion of a rig object internal to a rig category? This is an object$x$equipped with morphisms$z : 0 \to x$(zero) ,$a : x \oplus x \to x$(addition),$u : 1 \to x$(unit) and$m : x \otimes x \to x$(multiplication) such that the ... 3 [This answer is copied with very minor modifications from a preprint of mine "The Balanced Tensor Product of Module Categories" joint with Chris Douglas and Chris Schommer-Pries which will appear on the arxiv in the next few weeks.] Let$\mathcal{R} \cong \mathrm{Vec} \oplus \mathrm{Vec} \cdot X$be the symmetric monoidal category consisting of pairs of ... 3 The first functor here is actually a composition $$(\mathcal C\times\mathcal C)\times\mathcal C\xrightarrow{(-\otimes-)\times\text{Id}}\mathcal C\times\mathcal C\xrightarrow{\otimes}\mathcal C\\ ((a,b),c)\mapsto (a\otimes b,\ c)\mapsto(a\otimes b)\otimes c$$ and the second functor is the composition$$\mathcal C\times(\mathcal C\times\mathcal ... 3 This condition follows from the other axioms. I suggest you take a look at a beautiful J. Kock's paper "Elementary remarks on units in monoidal categories", which describes monoidal categories from a bit different (and more coherent) perspective. However, proving the condition "by hand" may be a good exercise for you. 3 They are always isomorphic, and by a unique isomorphism that preserves the structure of the duality (see this question for the unicity). Suppose$X^*$and$\hat{X}^*$are two different left duals (with corresponding evaluations and coevaluations). Define$f : X^* \to \hat{X}^*$as the composite$f = (\hat{\epsilon} \otimes 1) \circ (1 \otimes \eta)\$. ...

Only top voted, non community-wiki answers of a minimum length are eligible