TITLE: Is Joyal's category $\Theta_n$ the "only" Reedy category which is dense in $n$-categories?
QUESTION [8 upvotes]: Let $Cat_n$ denote the category of $n$-categories. Then Joyal's category $\Theta_n$ is (1) a full dense subcategory of $Cat_n$ which (2) is also a Reedy category.
Question: Is Joyal's category $\Theta_n$ somehow uniquely characterized by (1) and (2)?
Notes:

I've been ambiguous about what "category" and "$n$-category" mean -- the above statements are true whether "category" means "1-category" or "$(\infty,1)$-category". In the latter case, "$n$-category" can mean either "strict $n$-category" or "weak $(\infty,n)$-category" or various things in between.

I'm interested in understanding the story for any $n \in \mathbb N \cup \{\infty\}$.

One nice story which makes Joyal's category $\Theta_n$ look very "canonical" is the theory of generalized nerves. I'd be interested, for example, if there were something to say in that framework about getting one's nerves to be indexed by Reedy categories.

I'd also be interested if $\Theta_n$ didn't turn out to be literally "unique" with respect to (1) and (2), but at least "minimal" with respect to (1) and (2), or something like that.

When $n=1$, the question asks whether the simplex category $\Delta = \Theta_1$ is the "unique" full subcategory of $Cat$ which is a Reedy category. It sounds like the answer is probably "no" in this case, but I don't know a good counterexample.

REPLY [2 votes]: EDIT: As $\Theta_n \subseteq Cat_n$ is not "irredunant" in the sense below ((3) fails), the following doesn't really answer the question. If the Claim below could be upgraded to work with "isomorphism" in (3) rather than "split epimorphism", then we'd be in business.

 Here's an answer which really surprises me: I claim that $\Theta_n$ is the unique idempotent-complete Reedy category $\mathcal R$ appearing as a dense full subcategory of $Cat_n$ (where $Cat_n$ means the weak $(\infty,1)$-category of weak $(\infty,n)$-categories) which is irredundant in the following sense: 
Definition: Let $\mathcal C$ be an $\infty$-category. Say that $\mathcal R \subseteq \mathcal C$ is a basis if it is a dense, idempotent-complete, full subcategory. Say that a basis $\mathcal R \subseteq \mathcal C$ is irredundant if the following conditions are satisfied:

$\mathcal R$ is a Reedy category

Every morphism of $\mathcal R^-$ is a split epimorphism.

For every $R \in \mathcal R$, the reflection in $\mathcal C \subseteq Psh(\mathcal R)$ of the inclusion $\partial R \to R$ is not a split epimorphism.


Claim: Irredundant bases are unique and minimal. In other words, let $\mathcal C$ be an $\infty$-category with an irredundant basis $\mathcal R$. Then for any other basis $\mathcal R' \subseteq \mathcal C$, we have $\mathcal R \subseteq \mathcal R'$ up to repletion in $\mathcal C$.
Proof: Let $R \in \mathcal R$, and let us show that $R \in \mathcal R'$. Consider the map of $\mathcal R$-presheaves
$$(\ast) \qquad \varinjlim_{\mathcal R' \ni R' \to R} R' \to R$$
The reflection of $(\ast)$ in $\mathcal C$ is an isomorphism. If $(\ast)$ is not split epi, then by the Lemma below, it factors through $\partial R \to R$. Upon reflection in $\mathcal C$, we obtain a splitting of the $\mathcal C$-reflection of $\partial R \to R$, contradicting irredundancy of $\mathcal C$. So $(\ast)$ is split epi. This means that there exists a split epimorphism $R' \to R$ with $R' \in \mathcal R'$. Since $\mathcal R'$ is closed under idempotent splitting, this implies that $R \in \mathcal R'$.


**UPSHOT:** The main question is given an affirmative answer of some form by the following:
Corollary: Since $\Theta_n$ is an irredundant basis of $Cat_n$, it follows that any other basis of $Cat_n$ contains $\Theta_n$, and no other basis of $Cat_n$ is irredundant.


We have used the following Lemma. Thanks to Chris Schommer-Pries for pointing out the necessity of something like this in the comments below.
Lemma: Let $\mathcal C$ be an $\infty$-category with a bases $\mathcal R, \mathcal R'$. Suppose that $\mathcal R$ satisfies (1) and (2) above. Let $R \in \mathcal R$. If the map $(\ast)$ is not a split epimorphism of $\mathcal R$-presheaves, then it factors through the map $\partial R \to R$.
Proof:
By Yoneda, $(\ast)$ is split epi iff $(\ast)(R)$ (that's the map of spaces obtained by evaluating the map of presheaves $(\ast)$ at the object $R$) is split epi, if and only if $id_R$ is in the image of $(\ast)(R)$. Note that the codomain of $(\ast)(R)$ is the discrete space $R(R)$. Moreover, $\partial R(R) \subset R(R)$ is the complement of $\{id_R\} \subseteq R(R)$. So if $(\ast)$ is not split epi, then the image of $(\ast)(R)$ is contained in $\partial R(R)$.
Now for any $T \in R$, the inclusion of discrete spaces $\partial R(T) \subseteq R(T)$ has complement given by the maps $f : T \twoheadrightarrow R$ which lie in $\mathcal R^-$. Suppose for contradiction that such an $f$ is in the image of $(\ast)(T)$ evaluated at $T$. Then $f$ factors through some $R' \in \mathcal R'$. By condition (2) in the definition of irredundancy, $f$ has a section $g$, and now $id_R = fg$ factors through $R' \in \mathcal R$, which means that $id_R$ is in the image of $(\ast)(R)$, contrary to what we concluded in the previous paragraph. Therefore $(\ast)(T)$ factors through $\partial R(T)$ for each $T \in \mathcal R$. Since $\partial R(T) \to R(T)$ is always a monomorphism of discrete spaces, this means that $\ast$ factors through $\partial R \to R$.
Notes:

Conditions (1) and (2) in the definition of irredundancy are implied by the stronger assumption that $\mathcal R$ is an elegant Reedy category, which is indeed the case when $\mathcal R = \Theta_n$.

$\Theta_n$ is not an irredundant basis of the category of strict $n$-categories (cf. Zhen Lin's comment above), because condition (3) fails in this case. For irredundancy, it's important that we put ourselves in the weak setting.

Another way of stating condition (3) in the definition of irredundancy is this: for every $R \in \mathcal R$, it is not the case that every $C \in \mathcal C$ is weakly right orthogonal to the inclusion $\partial R \to R$.

I suspect that the Claim will remain true if condition (3) in the definition of irredundancy is weakened to stipulate only that $\partial R \to R$ does not reflect to an isomorphism (this corresponds, in the previous bullet, to replacing weak orthogonality with strong orthogonality). I was almost able to modify the proof of the above Claim to verify this, at the cost of introducing into the structure of the proof an induction on the degree of $R$, but there were some coherence issues I was unsure of (I didn't push terribly hard, though).