Does the Gray tensor product exhibit Gray as a monoidal Gray-category?

Question

Crans constructs a lax Gray tensor product for strict omega-categories that defines a monoidal structure on the category of strict omega-categories $\omega Cat$. We write Gray for this monoidal category $(\omega Cat, \otimes)$. It is easy to see that Gray does not refine to a monoidal strict omega-category, i.e. a weak monoid for the cartesian product on the category of strict omega-categories.

The lax Gray tensor product is bi-closed and so equips the category of strict omega-categories with an enrichment in Gray. Is the Gray tensor product a Gray-enriched functor? Can Gray be enhanced to a weak monoid in Gray-enriched categories? If not, does Gray refine to a weak monoid for some other "weaker" enrichment?

Let $C$ be a strict omega-category and $Fun(C,\omega Cat)$ be the strict $\omega$-category of $\omega$-functors $C \to \omega Cat$ and strict transformations and let $Funlax(C,\omega Cat)$ be the strict $\omega$-category of $\omega$-functors $C \to \omega Cat$ and lax transformations. Are there degree-wise Gray tensor products on the underlying categories $Fun(C,\omega Cat), Funlax(C,\omega Cat)$ that makes these categories to monoidal categories?

Answer 1

If $V$ is a symmetric (or at least braided) monoidal closed category then there is a natural tensor product on the category of $V$-enriched categories. A $V$-enriched monoidal category is a (weak) monoid with respect to this tensor product.

But when $V$ is a nonbraided monoidal category, there is no such tensor product of $V$-enriched categories. So there is no obvious notion of "$V$-enriched monoidal category".

This is analogous to how a commutative ring has a tensor product on its category of modules but a noncommutative ring does not have a canonical tensor product on its category of (left, say) modules.

You can think of a tensored $V$-enriched category as a (left, say) $V$-module, so you could simply define a tensored $V$-enriched monoidal category to be a category with a monoidal functor from $V$, just as you can define an algebra over a noncommutative ring to be a ring with a homomorphism from your base ring. In this sense every mooidal category is an algebra over itself via the identity functor. Some care will be warranted though.