# Tag Info

11

This is my understanding of this yoga. It may not be exactly what you seek and may differ from another person's point of view. Also I apologize for my bad english. For Grothendieck, many things should have a relative version. So instead of considering just a space $X_0$, consider a morphism $f:X\rightarrow S$ thought as a family of spaces $s\mapsto ... 8 Certainly not, for roughly the same reason that if$X,Y$are monoids, then a bijection between the underlying sets need not respect the monoid operations. If you understand this claim, you can answer your question too. 7 what about$\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)$. 7 In general if you don't want to start with the monoidal structure, you start with a closed category. In a closed category$\mathsf C$, you have a bifunctor $$[-,-] : \mathsf C^{op} \times \mathsf C \to \mathsf C,$$ called the internal hom, and various other data that are somewhat "dual" to the axioms of a monoidal category (I put dual in quotes because this ... 6 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 ... 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). 6 "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

I hope what follows will clear up your confusion. A lax monoidal functor consists of the following data (satisfying some axioms that I won't spell out): One functor $\color{red}{F : \mathsf{C} \to \mathsf{D}}$. This means that for all objects $A \in \mathsf{C}$ you have an object $F(A) \in \mathsf{D}$, and for all morphisms $f : A \to B$ in $\mathsf{C}$ ...

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

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 ...

5

An object $X$ of an abelian category has finite length if it has a (finite) composition series: i.e., a chain of subobjects $$0=X_0<X_1<\dots<X_{n-1}<X_n=X$$ such that $X_i/X_{i-1}$ is simple for $1\leq i\leq n$.

5

http://arxiv.org/abs/math/0401347

5

Yes. This is a corollary of Schur-Weyl duality. You need at least the additional assumption that your symmetric monoidal category is enriched over $k$-vector spaces. In general I don't see any reason to expect that the action of $k[S_n]$ is faithful; consider, for example, the special case where we only look at $1$-dimensional vector spaces. Sometimes. The ...

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

Edit, 3/25/15: In an earlier version of this answer I claimed that group objects can be defined in any monoidal category. In fact this isn't true; one of the axioms requires access to a diagonal map, and so it's standard to only define group objects in cartesian monoidal categories. But since talking about actions only uses the monoid structure this doesn't ...

4

I think you are pointing to two different phenomena, and I don't know which one you're more interested in. Can you clarify? In the second example you're just unwrapping the definition of a functor $BM \to \text{Set}$. You can always describe what a functor $C \to D$ looks like in a manner "internal to $D$" in a sort of trivial way: of course it's the same ...

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

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: $$\int^{A, B, C, ... 4 Use the Yoneda lemma! By adjunction there is a natural isomorphism$$\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)$$induced by the right unitor \rho_X : X \otimes I \to X is ... 4 I think you want to be a little more careful, as a priori there might be multiple ways of writing a given object as F(X). This isn't a really problem though because if you have some object A, you know basically how to write it as F(X): just let X = G(A)! So we just have$$G(A \bullet B) \cong G(F(G(A))\bullet F(G(B))) \cong GF(G(A) \otimes G(B)) \cong G(A) ...

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 $\hom((a,b),(c,d)) = ... 3 It's standard to assume that a "linear monoidal category" means that the linear and monoidal structures are compatible. For example, see Def 0.1.2 of Catègories Tensorielles, Def 1.12.3 of Tensor Categories or Def 3.2.4 of Dualizable Tensor Categories. If I ran across a paper that didn't make this compatibility explicit, I'd assume that the authors were ... 3 The only dualisable object in a cartesian monoidal category is the terminal object. Thus, a cartesian compact closed category must be trivial. Indeed, suppose$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$. This decomposes into components as a ... 3 The additional structure is that there is a natural further quotient of the isomorphism classes you can take where you impose the additional relation$[X] - [Y] + [Z] = 0$for every distinguished triangle$X \to Y \to Z \to \Sigma X$. As Martin says in the comments, any commutative monoid gives a discrete braided monoidal category. (Recall that a discrete ... 3 In general it is very rare for a braided monoidal category to commute "on the nose". For a very concrete example, the usual category of sets and functions is cartesian monoidal, so braided (even symmetric) monoidal, but in general$A \times B \neq B \times A$. As for the braids, there is a PRO (monoidal category with$\mathbb{N}$as the set of objects and ... 3 Since you are interested in higher categories, you should see the Eckmann-HIlton argument as a special case of the use of the interchange law, or exchange law. A general formulation of this is for double categories: this is a set, or class,$C$, with two category structures$\circ_1, \circ_2\$ each of which is a morphism for the other. This amounts to the ...

3

You can think of it as a extranatural transformation or a dinatural transformation. In fact the equation you wrote out is precisely the dinaturality condition in question.

3

So to clarify, this is true in a spherical category (when you assume that left and right traces are equal, this is the definition of a spherical category). The best way (IMO) to prove this is through graphical calculus. Here is a picture (which I hope is not too horrible). The only non-obvious equality is the first: it follows from the monoidality of the ...

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