Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Consider the $2$-categories

  • $\mathsf{MonCat}$ of monoidal categories, with strong monoidal functors and monoidal transformations,
  • $\mathsf{SymMonCat}$ of symmetric monoidal categories, with strong symmetric monoidal functors and symmetric monoidal transformations.

There is a forgetful functor $\mathsf{SymMonCat} \to \mathsf{MonCat}$, which probably has a left adjoint. I.e. for a given monoidal category $C$ there should be a free symmetric monoidal category $C_{\text{sym}}$ over $C$. Now my question is: How does this symmetric monoidal category look like explicitly? I would like to know an explicit description of its objects and its morphisms (not just some general construction etc. which produces it somehow).

For example, if $C$ is the free monoidal category on one object, which is discrete and given by $C=\{1,X,X^2,\dotsc\}$, then $C_{\mathrm{sym}}$ is the free symmetric monoidal category on one object, i.e. the permutation groupoid, where we have the same objects, but $$\mathrm{Hom}(X^n,X^m)=\left\{\begin{array}{cc} \Sigma_n & n = m, \\ \emptyset & \text{else}. \end{array}\right.$$

The decategorified analog is the forgetful functor $\mathsf{CMon} \to \mathsf{Mon}$ which has left adjoint $M \mapsto M_{\mathrm{sym}}:=M / \langle xy = yx : x,y \in M\rangle$. Two elements of $M$ become equal in $M_{\mathrm{sym}}$ iff they have the form $x_1 \cdot \dotsc \cdot x_n$ and $x_{\sigma(1)} \cdot \dotsc \cdot x_{\sigma(n)}$ for some $\sigma \in \Sigma_n$. Essentially the same works with strict (symmetric) monoidal categories. As for the general case, I think that one has to "impose a reason" why $X \otimes Y$ and $Y \otimes X$ become isomorphic in $C_{\text{sym}}$, for $X,Y \in C$.

share|improve this question
    
You start talking about $2$-categories, but you suddenly change to consider ordinary functors between them. Is this deliberately ambiguous? –  Fernando Muro Dec 29 '12 at 16:42
2  
@Fernando: in the culture of higher categories, one frequently says "functor" when one really means $n$-functor, and I expect this is the case here. –  Todd Trimble Dec 29 '12 at 17:35
1  
Thank you for the comments. @Fernando: In fact I mean $2$-functors, $2$-adjunctions etc. @Bruce: In fact one of my aims is to understand the left adjoint of $\mathsf{SymMonCat} \to \mathsf{Cat}$. Do you know an explicit description? –  Martin Brandenburg Dec 29 '12 at 18:33
1  
I thought there was a construction where a morphism is the string diagram of a permutation with each string labelled by a morphism in the original category. –  Bruce Westbury Dec 29 '12 at 18:45
1  
@Fernando: didn't mean to imply you were uncultured in any sense! –  Todd Trimble Dec 29 '12 at 23:27

3 Answers 3

up vote 7 down vote accepted

First, having seen the edited version of Martin's question, let's quickly dispose of the construction of the free symmetric monoidal category generated by a category $C$. Objects are tuples $(x_1, \ldots, x_n)$ of objects of $C$. Morphisms are labeled permutations, where permutations are conveniently visualized as string diagrams, each string being labeled by a morphism in $C$. To compose such labeled diagrams, just compose the string diagrams, composing the labels of strings in $C$ along the way.

The free symmetric monoidal category $Sym(M)$ on a monoidal category $M$ is formed from the free symmetric monoidal category $S (U M)$ on the underlying category $U M$ by adjoining isomorphisms $\phi_{x_1, \ldots, x_n}: (x_1, \ldots, x_n) \to x_1 \otimes \ldots \otimes x_n$, where the $x_i$ are objects of $M$, the expression $(x_1, \ldots, x_n)$ is the formal monoidal product in $S(UM)$, and $x_1 \otimes \ldots \otimes x_n$ is the tensor product in $M$.

More precisely, define the objects of $Sym(M)$ to be tuples $(x_1, \ldots, x_n)$ of objects of $M$. Define morphisms $(x_1, \ldots, x_n) \to (y_1, \ldots, y_m)$ to be equivalence classes of pairs $(p, f)$ where $p$ is a permutation on $n$ elements, and $f$ is an $M$-labeled planar forest of $m$ rooted trees. (You should think here of the free multicategory generated by a category.) Formally, a planar forest can be described as a functor $[n]^{op} \to \Delta$ where $n$ is the category $1 \leq \ldots \leq n$ and $\Delta$ is the category of finite (possibly empty) ordinals. Under the obvious way of drawing such forests, the ordered list of leaves of the forest is labeled by $(x_{p(1)}, \ldots, x_{p(n)})$, and the ordered list of roots by $(y_1, \ldots, y_m)$. Edges are labeled by objects of $M$, and each internal node with $k$ inputs labeled (in order) by $m_1, \ldots, m_k$ and output $m$ is labeled by a morphism $f: m_1 \otimes \ldots \otimes m_k \to m$.

There is an evident way, using the monoidal structure of $M$, of evaluating such a labeled forest $f$ as a morphism $ev(f): x_{p(1)} \otimes \ldots \otimes x_{p(n)} \to y_1 \otimes \ldots \otimes y_m$ in $M$. We consider two arrows $(p, f)$ and $(p', f')$ to be equivalent if $p = p'$ as permutations and $ev(f) = ev(f')$.

Now we define composition of pairs. The main idea is to rewrite the composition of a forest followed by a permutation,

$$(x_1, \ldots, x_n) \stackrel{(1, f)}{\to} (y_1, \ldots, y_m) \stackrel{(q, 1)}{\to} (y_{q(1)}, \ldots, y_{q(m)}),$$

into a form $(p, f')$ forced by the naturality requirement of the symmetry isomorphism. Namely, if $\bar{x}_i$ is the tuple of leaves of the tree whose root is $y_i$, then we have a block permutation $bl(q)$ taking $(\bar{x}_1, \ldots, \bar{x}_m)$ to $(\bar{x}_{q(1)}, \ldots, \bar{x}_{q(m)})$. The permutation $q$ can also be applied to the $m$ trees of the forest by reordering the trees, yielding a new forest $\mathrm{perm}_q(f)$, and the composition $(q, 1) \circ (1, f)$ is defined to be $(bl(q), \mathrm{perm}_q(f))$. Then, if we have $(p, f): (x_1, \ldots, x_n) \to (y_1, \ldots, y_m)$ and $(q, g): (y_1, \ldots, y_m) \to (z_1, \ldots, z_p)$, we define their composite to be

$$(bl(q) \circ p, g \circ \mathrm{perm}_q(f))$$

where the first component is by composing permutations and the second is by the usual way of composing forests by plugging in roots of one planar forest of trees into leaves of another.

Remaining details that all this works will be left to the reader. I will remark that the key rewriting procedure above is an instance of a kind of distributive law; there are some more details to this effect in some notes on my nLab web; see here.

share|improve this answer
    
It seems to me that labelling objects in the free symmetric monoidal category by strings of objects means that this is the free strict symmetric monoidal category. No? –  Zhen Lin Dec 30 '12 at 0:44
1  
Zhen, do you mean strict symmetric monoidal, or symmetric strict monoidal? In any case, I am trying to describe the free symmetric strict monoidal category generated by a strict monoidal category, which seems to me perfectly fine. (Notice that Martin wrote that the free monoidal category on one object is discrete, which is not literally true but is true up to equivalence; I am taking a similar tack here and working with strict monoidal categories throughout.) –  Todd Trimble Dec 30 '12 at 1:25
    
I mean symmetric strict-monoidal. I wasn't aware that there were such things as strict-symmetric monoidal categories! –  Zhen Lin Dec 30 '12 at 15:18
2  
If you have $(f, 1): A\to (B, C)$ with $f: A\to B\otimes C$ and $(1, \sigma): (A, B)\cong (B, A)$ how do write the composition $(1, \sigma)\circ (f, 1)$ in the form $(\tau, g)$? –  Buschi Sergio Jan 21 '13 at 20:44
1  
It's a pertinent question, and I've long been meaning to address this (but it would entail some rewriting that I didn't feel up to). Stay tuned however. –  Todd Trimble Jan 21 '13 at 21:49

I hope to have done well:

Let $\mathcal{C}$ a monoidal category. Define a category $ \mathcal{C}_S$ with objets the class defined for as follow: any object of $\mathcal{C}$ is also a object of $\mathcal{C}_S$ (we call it elementary), give $X, Y \in \mathcal{C}$ then $X\cdot Y$ belong to $\mathcal{C}_S$ (we call it composite object). And for induction: If $X $ is elementary objects and $A, B$ are composite objects then $X\cdot(A)$, $ (A)\cdot X$, $ (A)\cdot (B)$ are composite objets. These are all the objets of $\mathcal{C}_S$. In few word these object are finite string of objects of $\mathcal{C}$ with a (well defined) disposition of parenthesis, on these objects is well defined the length function (the length of the underling string) by $|X|:=1$ if $X$ is elementary and with $|A\cdot B|:= |A| +|B|$, in similar way we have the function "underling string" defined as $stg(X)= X$ if $X$ is elementary and with $ stg(A\cdot B):= stg(A) \ast stg(B)$ (concatenation of string), for example if $A= (X\cdot Y)\cdot Z$ then $stg(A)=(X, Y, Z)$ and $|A|=3$. We have a map $[-]$ from the objet of $\mathcal{C}_S$ to the object of $\mathcal{C}$ that is the identity on elementary objects and with $[A\cdot B]:= [A]\otimes [B]$. The morphisms of $\mathcal{C}_S$ are of type $f: A\to B$ where is (identified to) a morphism $f: [A] \to [B]$ in $\mathcal{C}$, with identities and compositions of $\mathcal{C}$.In this way the monoidal structure $\mathcal{C}$ is translate to a monoidal structure of $\mathcal{C}_S$, and write $A \otimes B:=A\cdot B= [A]\otimes[B]$ for $A, B\in \mathcal{C}_S$.

Then we add to $\mathcal{C}_S$ the permutations (iso)morphisms $(A, \sigma): A^\sigma\to A$ where if $stg(A)=(A_1,\ldots, A_n)$ then $stg(A^\sigma)=(A_{\sigma(1)},\ldots, A_{\sigma(n)})$ and $A^\sigma$ as the some disposition of parenthesis of $A$, with composition $(A, \sigma)\circ (A^\sigma, \tau)=(A, \sigma\circ\tau)$.

And the composition of a morphisms of $\mathcal{C}_S $ with a permutation is the formal concatenation, then we get the category:

$\mathcal{C}_{sy}$ where morphisms are coherent (domain-codomain correspondence) sequences of morphism of $\mathcal{C}_S$ and permutations, the composition is by concatenation where we compose all successive morphisms of $\mathcal{C}_S$ or successive permutations everywhere is possible.

On $\mathcal{C}_{sy}$

considering the congruence $f\otimes 1_{B^\sigma}\circ 1_{A'}\otimes \sim f\otimes1_{B}\circ 1_A\otimes \sigma: A\otimes B^\sigma\to A'\otimes B$ for $f: A\to A'$ in $\mathcal{C}$ and $\sigma: B^\sigma\to B$ permutation and where $1_A\otimes \sigma: A\otimes B^\sigma\to A\otimes B$ is the permutation that in detail is $(A_1,\ldots A_n, B_1,\ldots B_m)\to (A_1,\ldots A_n, B_{\sigma(1)},\ldots B_{\sigma(m)})$ analogously for $1_{A'}\otimes \sigma$ or more un general for two permutations $\tau\otimes\sigma$. In this way we get the category $\mathcal{C}_{Sym}$.

In $\mathcal{C}_{sym}$ we define tensor product of morphisms: given $f_n\circ \sigma_n\ldots f_1\circ \sigma_1$ and $g_n\circ \tau_n\ldots g_1\circ \tau_1$ their tensorial product is $g_n\otimes f_n\circ \tau_n\otimes \sigma_n\ldots g_1\otimes f_1\circ \tau_1\otimes \sigma_1$, if these have different length we can insert identities for have morphism like above, this operation is well defined for the congruence above (and for the bifunctorial property of $\otimes$ in $\mathcal{C}$). The symmetry on $\mathcal{C}_{sym}$ is the obvious permutation $\sigma: A\otimes B \cong B\otimes A$ in detail $(A_1,\ldots A_n, B_1,\ldots B_m)\to (B_1,\ldots B_m, A_1,\ldots A_n)$ . It seems that symmetric monoidal axioms work well .

We have the (strict monoidal) functor $J: \mathcal{C}\to \mathcal{C}_{sym}$ the maps any element in its singleton, that as the universal request property for strict (resp. strong, lax) monoidal functors in symmetrical monoidal categories.

share|improve this answer

More a speculation:

If we think of Symmetric Monoidal Categories as $\Gamma$ objects in $\mathsf{Cat}$ and Monoidal as $\Delta$ objects in $\mathsf{Cat}$, then the restriction from $\mathsf{SymMon}$ to $\mathsf{Mon}$ just comes from pulling back along a certain functor $\Delta^{op}\rightarrow\Gamma^{op}$. Probably the adjoint should be given by pushing forward along this map.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.