It is well-known that when passing to $\infty$-categories the notion of commutativity gets replaced by an infinite array of notions of commutativity: $\mathbb{E}_{1}$, $\mathbb{E}_{2}$, ..., $\mathbb{E}_{\infty}$. This is already apparent when passing from sets to categories and $2$-categories:

  • For sets, we have monoids ($\mathbb{E}_{1}$) and commutative monoids ($\mathbb{E}_2=\cdots=\mathbb{E}_\infty$);
  • For categories, we have monoidal ($\mathbb{E}_{1}$), braided ($\mathbb{E}_{2}$), and symmetric monoidal categories ($\mathbb{E}_{3}=\mathbb{E}_{4}=\cdots=\mathbb{E}_{\infty}$);
  • For $2$-categories, we have monoidal ($\mathbb{E}_{1}$), braided ($\mathbb{E}_{2}$), sylleptic ($\mathbb{E}_{3}$), and symmetric monoidal categories ($\mathbb{E}_{4}=\mathbb{E}_{5}=\cdots=\mathbb{E}_{\infty}$).

A similar phenomenon happens to bilinearity:

  • A morphism $f\colon A\times B\to C$ of commutative monoids is bilinear if, for each $a,a'\in A$ and each $b,b'\in B$, we have \begin{gather*} f(a,b+b') = f(a,b)+f(a,b'),\\ f(a+a',b) = f(a,b)+f(a',b),\\ f(1_A,b) = 1_C,\\ f(a,1_B) = 1_C. \end{gather*}

  • For categories, these relations are replaced by morphisms: we say that a strong bilinear structure on a functor $F\colon\mathcal{C}\times\mathcal{D}\to\mathcal{E}$ of symmetric monoidal categories is a collection of isomorphisms \begin{align*} F^{\mathsf{bil}}_{A,B\otimes_{\mathcal{D}}B'} &\colon F(A,B\otimes_{\mathcal{D}}B') \longrightarrow F(A,B)\otimes_{\mathcal{E}}F(A,B'),\\ F^{\mathsf{bil}}_{A\otimes_{\mathcal{C}}A',B} &\colon F(A\otimes_{\mathcal{C}}A',B) \longrightarrow F(A,B)\otimes_{\mathcal{E}}F(A',B),\\ F^{\mathsf{bil}}_{\mathbf{1}_{\mathcal{C}},B} &\colon F(\mathbf{1}_{\mathcal{C}},B) \longrightarrow \mathbf{1}_{\mathcal{E}},\\ F^{\mathsf{bil}}_{A,\mathbf{1}_{\mathcal{D}}} &\colon F(A,\mathbf{1}_{\mathcal{D}}) \longrightarrow \mathbf{1}_{\mathcal{E}} \end{align*} satisfying coherence conditions.


  1. Is there a similar array of notions of bilinearity in higher algebra?
  2. In particular, can we speak of "$\mathbb{B}_{k}$-morphisms of spectra"?
  3. Tensor products can be characterised/defined as universal bilinear maps; do we also have intermediate tensor products corresponding to "$\mathbb{B}_{k}$"-bilinearity?
  • 2
    $\begingroup$ I don't quite see "A similar phenomenon" happening here: reading you there seems to be only one notion of bilinearity for cartegories. Can you clarify what makes you think there is going to be several different notion when going to higher categories ? Are you refering to the fact that the notion of bilinearity will depends on "how commutative" are A ,B and C ? $\endgroup$ Jul 23 at 18:54
  • $\begingroup$ @SimonHenry For categories there are four: strict, strong, lax, and oplax, depending on whether the morphisms $F^{\mathsf{bil}}_{A,B\otimes_{\mathcal{D}}B'}$, $F^{\mathsf{bil}}_{A\otimes_{\mathcal{C}}A',B}$, $F^{\mathsf{bil}}_{\mathbf{1}_{\mathcal{C}},B}$, and $F^{\mathsf{bil}}_{A,\mathbf{1}_{\mathcal{D}}}$ are taken to be identities, isomorphisms, or morphisms going in one or the other direction. For $2$-categories one would probably replace the coherence conditions satisfied by those by $2$-morphisms, which would again satisfy more coherence conditions. $\endgroup$
    – Théo
    Jul 23 at 19:27
  • $\begingroup$ Once again one could take those to be identities, $2$-isomorphisms, or $2$-morphisms going in some direction; so there would again be more variants. This is similar to monoidal categories themselves, which come also in lax, oplax, ordinary, and strict versions. $\endgroup$
    – Théo
    Jul 23 at 19:30
  • $\begingroup$ I think one can define bilinearity regardless of how commutative $A$, $B$, and $C$ are, but tensor products will need them to be $\mathbb{E}_{\infty}$ because of Eckmann–Hilton: one can only put a monoid structure on the set of monoid maps $\mathrm{Hom}_{\mathsf{Mon}}(B,C)$ from $B$ to $C$, to which $\otimes_{\mathbb{N}}$ is the left adjoint, if $B$ and $C$ are commutative. $\endgroup$
    – Théo
    Jul 23 at 19:32

Let me clarify a bit what I meant in my comment on how the notion of bilinearity will depends on "how commutative" are $A$, $B$ and $C$, and this is one way to define a hierarchy of notion of bilinear maps in higher algebras.

The idea is that up to equivalence of $\infty$-categories, (and let's say in a cartesian monoidal $(\infty,1)$-category for simplicity), an $\mathbb{E}_{n+k}$-algebra is the same as an $\mathbb{E}_n$-algebra in the (cartesian monoidal) category of $\mathbb{E}_{k}$-algebra.

So if $A$ is an $\mathbb{E}_k$-algebra, $B$ is an $\mathbb{E}_n$-algebra and $C$ is an $\mathbb{E}_{n+k}$-algebras then I can define a $(n,k)$-bilinear map $A \times B \to C$ to be a morphism of $\mathbb{E}_k$-algebra from $A$ to the $\mathbb{E}_k$-algebra $\operatorname{Map}_{\mathbb{E}_n}(B,C)$.

Where by $\operatorname{Map}_{\mathbb{E}_n}(B,C)$ I mean the space of morphisms of $\mathbb{E}_n$-algebra morphism from $B$ to $C$, which has an $\mathbb{E}_k$-algebra structure induced by the fact that $C$ is an $\mathbb{E}_k$-algebra when seen as an object of the category of $\mathbb{E}_n$-algebra.

But this is very different from the notion you mention in your coment when instead one goes to "lax/colax" notion of bilinearity.

  • $\begingroup$ I see what you meant now. Thanks! $\endgroup$
    – Théo
    Jul 24 at 18:19

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