In category theory, a dagger category is a precategory $\mathcal{C}$ such that for every pair of objects $A:\mathcal{C}$ and $B:\mathcal{C}$ there is a function $(-)^{\dagger\_{A,B}}:\mathrm{Mor}(A,B) \to \mathrm{Mor}(B,A)$ such that


* for all objects $A:\mathcal{C}$, $\mathrm{id}\_A^{\dagger\_{A,A}} = \mathrm{id}\_A$,
* for all objects $A:\mathcal{C}$, $B:\mathcal{C}$, and $C:\mathcal{C}$, and morphisms $f:\mathrm{Mor}(A,B)$ and $g:\mathrm{Mor}(B,C)$, $(g \circ f)^{\dagger\_{A,C}} = f^{\dagger\_{A,B}} \circ g^{\dagger\_{B,C}}$
* ${(f^{\dagger\_{A,B}})}^{\dagger\_{B,A}} = f$


and which satisfies an equivalent Segal condition using unitary isomorphisms as categories do using isomorphisms.


In the context of higher category theory, does the concept of $(n,1)$-dagger categories make sense, and if so, how does one go about defining them?



$(n,1)$-dagger categories
Well, it is easy to give definitions, the problem is finding the "right" one.


Here "right" can mean that gives the correct notion up to homotopy (many definition will be equivalent) but also it can mean the one that let you the most easily talk about the examples and constructions you care about ( this much more subjective and it depends on why you want this definition)


I'll propose two (edit, three) definitions - there is a good chance that they are equivalent in some appropriate homotopical sense given they are both fairly reasonable, but I don't know if they are.


If you don't mind I'll work with $(\infty,1)$ instead of $(n,1)$ because it is simpler. If you are interested in $n=2$ or $n=3$ more specifically, then that's definitely a different story - and here we might come up with a long but explicit definition.


Before we start, a little bit of introspection on the definition is needed. I think in the case of Dagger category that has been nicely done [here](https://mathoverflow.net/questions/220032/are-dagger-categories-truly-evil). The problem with the definition of Dagger categories is that it break the "equivalence principle" because it impose an equation $x^\*=x$ on objects, or more fundamentally because it is not invariant under equivalence of categories (a category equivalent to a Dagger category is not automatically a dagger category), so even if you phrase the definition in a way that avoid writting $x^\* = x$ something else will break the equivalence principle.


The MO discussion [linked](https://mathoverflow.net/questions/220032/are-dagger-categories-truly-evil) above discuss this point (it's interesting, read it !), and arrive at the conclusion ([this answer](https://mathoverflow.net/a/220111/22131)) that the good way to conceptualize a dagger category is a pair of a category $C$ together with a groupoid $C\_u \to C$ where $C\_u$ is the groupoids of object and dagger-isomorphism between them. The structure of Dagger category can be then added in a "non evil" way on top of this.


This idea of a category $C$ together with a groupoid $C\_u$ and an essentially surjective functor $C\_u \to C$ is conveniently represented in $\infty$-category theory by a [Segal space](https://ncatlab.org/nlab/show/Segal+space) not satisfying the completeness condition.


So it is natural to define Dagger $(\infty,1)$-categories as Segal spaces with structure.


This leads to:


**First definition :**


For $n \in \mathbb{N}$ we denote as usual by $[n]$ the poset (seen as a category)


$$ 0 \to 1 \to \dots \to n $$


and we denote by $[n]\_\dagger$ the free dagger category on $[n]$. Note that $[n]\_\dagger$ is relatively easy to describe explicitely : $[n]$ is free on the graph with only arrow $i \to i+1$, so $[n]\_\dagger$ is free on the graph


$$ 0 \leftrightarrows 1 \leftrightarrows \dots \leftrightarrows n $$


and so arrows in $[n]\_\dagger$ are justs path in this fairly simple graph.


I denote by $\Delta\_\dagger$ the full subcategory of Cat on the objects $[n]\_\dagger$, it admit $\Delta$ as a non-full subcategory.


**Definition :** A Dagger $(\infty,1)$-precategory is a presheaf of spaces on $\Delta\_\dagger$ whose restriction to $\Delta$ is a Segal space.


A Dagger $(\infty,1)$-category should then be a Dagger $(\infty,1)$-precategory that satisfies an appropriate Segal condition, but a version of the Segal condition that uses the Dagger structure: given a Dagger $(\infty,1)$-precategory $X$ one should construct a space of "Dagger isomorphisms" which factor the diagonal $X([0]) \to X\_I \to X([0])^2$ and define a Dagger $(\infty,1)$-category as being a precategory such that the map $X([0]) \to X\_I$ is an equivalence.


Some additional thought is needed to determine what exactly should be the correct definition of $X\_I$, but I would guess something like the space of $f$ such that $f^\* f \sim Id$ and $f$ is an equivalence. Or maybe more conceptually, take $I$ to be the Dagger version of the walking isomorphism and define $X\_I$ as the space of maps $I \to X$ in the natural model structure for Dagger $(\infty,1)$-precategories (the Bousfiled localization of the projective/injective model structure).


*Note :* the intuition behind that definition is that Dagger categories are monadic over categories (in the 1-categorical sense) and that monads is "Nervous" in the sense of [Bourke and Garner](https://arxiv.org/abs/1805.04346) for the category of arities $\Delta$, so following the idea suggested by Meadow myself [here](https://arxiv.org/abs/2106.02706) this provides a natural way to extend the definition to an $\infty$-categorical one. That's exactly what the above definition of Dagger $(\infty,1)$-precategory above is. To say this more naively, $X([n])$ represents the space of string of n composable arrows in $X$, and the category $\Delta\_\dagger$ encodes all the way you can transform such string in a dagger categories.


**Second definition:** One can also follow a much more naive route. One takes $\infty$-categories to mean quasi-categories, and then it actually make sense to ask for a quasi-category with an identity on object involution.


**Definition 2 :** A Dagger simplicial set is a simplicial set $X$ with for each $n$ an involutive map $\dagger: X(n) \to X(n)$ such $\dagger$ is the identity on $X(0)$ and collectively they form a morphism of simplicial sets $X \to X^{op}$ (where $X^{op}$ is defined by taking $X^{op}(n) = X([n]^{op})$ )


Then I can for exemple define a Dagger Quasi-category as a Dagger simplicial set which is a quasi-category.


**Are they equivalent ?**


Well, the question don't quite make sense yet, because the second definition is missing a key ingredient: *what are the equivalences of Dagger quasi-categories ?* It is not clear it is reasonable to define the equivalence as the morphism of dagger-simplicial sets that are equivalence of quasi-categories, nor to use an explicit definition with morphism having a "dagger inverse".


For the first definition, the notion of equivalence is easy as we have a nice Quillen Model structures on the category of presheaves of spaces on $\Delta\_\dagger$ by starting with either the projective or the injective model structure and then localising so that the fibrant objects are the Dagger $(\infty,1)$-categories.


So I'm proposing:


**Conjecture :** There exists a model structure on the category of Dagger simplicial sets whose fibrant objects are (certain) Dagger quasi-categories and which is Quillen equivalent to the model structure for Dagger $(\infty,1)$-categories.


I'm phrasing it as a conjecture, because I tend to think this is true - and probably within reach with a bit of work, but there are also definitely a few things that could go wrong making it false.




---


**Edit: A third definition**


So the suggestion by Achim Krause doesn't quite work, but one can adjust it so that it does work.


Taking the "opposite category" produce an action of the two elements group $C\_2$ on $Cat\_\infty$. Now, as a generalization of the fact that every groupoid is equivalent to its opposite through the functor which is the identity on object and send every arrow to its inverse, the restriction of this action to $Gpd\_\infty$ should admit a trivialization (I don't know if this has been explicitely constructed anywhere though). In particular, we have functor $Gpd\_\infty \to Cat\_\infty^{C\_2}$.


**Definition:** The $(\infty,1)$-category of Dagger $(\infty,1)$-precategories is defined as the pullback $Gpd\_{\infty} \times\_{Cat\_\infty^{C\_2}} E^{C\_2}$ where $E$ is the full subcategory of $(Cat\_\infty)^{[1]}$ of essentially surjective functor and its map to $Cat\_\infty$ is the source functor.


(the $C\_2$ exponent means the space of homotopy fix point)


Informally, we are saying that a Dagger precategory is an $\infty$-precategory (in the sense of an essentially surjective functor $G \to C$ where $G$ is an $\infty$-groupoid) endowed with an involutive equivalence with $G^{op} \to C^{op}$ (that is what an element of $E^{C\_2}$ is ) such that the component of this equivalence on $G$ identifies (as an involution) the "obvious" equivalence $G \simeq G^{op}$ coming fron the fact that $G$ is an $\infty$-groupoid.


This last condition implies both that the dagger is the identity on object and that arrows in $G$ get sent to dagger-isomorphism.


Now this is only a Dagger precategory, one further needs to impose a Segal type condition that will enforce that $G$ is the $\infty$-groupoid of dagger-isomorphism. So, one needs to give a definition of dagger isomorphism in this structure, build a natural map from the Hom space of $G$ to the spaces of dagger isomorphisms, and ask this map to be an equivalence.


It also seems reasonable to conjecture that this $\infty$-category is equivalent to the other two definitions.


