An operad-like structure, is there a name for it?

Here is an example which I'd like to have a name for.

Let $$P$$ be a compact smooth manifold of dimension $$p$$, possibly with non-empty boundary.

Define $$E(k,P)$$ to be the space of smooth (codimension zero) embeddings $$\coprod_{k} D^p \to P \, ,$$ that is the space of embeddings of $$k$$ disjoint $$p$$-disks in $$P$$, where the image of each such embedding lies in the interior.

In particular, when $$P = D^p$$, the spaces $$\{E(k,D^p)\}_{k\ge 0}$$ form an operad.

There is an evident "action" map $$E(\ell,P) \times (E(k_1,D^p) \times \cdots \times E(k_\ell,D^p)) \to E(k_1 + \cdots +k_\ell,P)$$ given by insertion.

Question: What is this action an example of? (Does it have a name?)

• I believe that this is a right action of disc-embedding operad $\{E(k,D^p)\}_{k \geq 0}$ on the symmetric sequence $\{E(k, P)\}_{k \geq 0}$; I think I have seen this in a talk of Turchin's on embedding calculus. – Tyler Lawson Jan 3 '19 at 16:27
• @TylerLawson so, does that mean the symmetric sequence forms a module (or algebra) over the disk-embedding operad? – John Klein Jan 3 '19 at 16:32
• Yes: this is called a right module over an operad. – Dan Petersen Jan 3 '19 at 16:42
• @DanPetersen in the symmetric sequence context (that was what I was missing). I was used to modules over an operad in spaces. Thanks. – John Klein Jan 3 '19 at 16:45

As others have said, this is precisely the structure of a right module over the operad $$E(D^p)$$.

Since it doesn't feel right to give such a short answer that's already in the comments, let me expand a bit; I don't claim I'm saying anything new, but hopefully, maybe some readers will find something interesting.

If $$P = \{P(k)\}_{k \ge 0}$$ is an operad, a right $$P$$-module is given by a symmetric sequence $$M$$ equipped with composition maps $$M(k) \otimes P(r_1) \otimes \dots \otimes P(r_k) \to M(r_1 + \dots + r_k)$$ satisfying the obvious axioms. If you define the plethysm of symmetric sequences $$X \circ Y = \bigoplus_{i \ge 0} X(i) \otimes_{\mathfrak{S}_i} Y^{\otimes i}$$, then an operad is a monoid wrt plethysm, and a right module is a right module over this monoid.

Since operads have units, this is equivalent to giving composition maps $$\circ_i : M(k) \otimes P(l) \to P(k+l-1)$$ for $$1 \le i \le k$$ (simply put, $$m \circ_i p = m(1,\dots,1,p,1,\dots,1)$$ where $$1$$ is the operadic unit and $$p$$ is in position $$i$$).

Plethysm is only linear on the left, so this actually gives two different notions of "left" stuff. On the one hand, a left $$P$$-module is a symmetric sequence $$M$$ equipped with composition maps: $$P(k) \otimes M(r_1) \otimes \dots \otimes M(r_k) \to M(r_1 + \dots r_k).$$ This is exactly a left module over the monoid $$P$$ in the category of symmetric sequences and plethysms. Note that a left $$P$$-module concentrated in arity $$0$$ is the same thing as a $$P$$-algebra. On the other hand, you can define an "infinitesimal" or "abelian" left $$P$$-module, by only requiring composition maps $$\circ_i : P(k) \otimes M(l) \to M(k+l-1)$$. Since you don't have a unit in $$M$$, this is a different notion. (To be entirely honest, I do not know if the terminology "abelian/infinitesimal left module" is used. I am more used to the terminology "abelian/infinitesimal bimodule", where you have a left and a right action defined as above.)

The precise modules in your question are very interesting, too. First, the operad $$E(D^p)$$ is weakly equivalent to the semi-direct product $$E_p \rtimes O(p)$$ of the little $$p$$-disks operad with the orthogonal group $$O(p)$$. If you require your embeddings to be oriented, you get a new operad $$E^{or}(D^p)$$ which is weakly equivalent to the usual framed little $$p$$-disks operad $$E_p \rtimes SO(p)$$. If you require your embeddings to preserve the canonical framing of $$D^p$$, then you get the usual little $$p$$-disks operad $$E_p$$.

If you have a smooth manifold $$P$$, then you can define the right $$E(D^p)$$-module $$E(P)$$ as you did in your question. Suppose you have another manifold $$P'$$ of dimension $$p' \ge p + 3$$. You can still define a right $$E(D^p)$$-module structure on $$E(P')$$, by considering embeddings of $$D^p$$ in $$P'$$.

Then Goodwillie--Weiss manifold calculus (following Arone, Boavida, Turchin, Weiss...) shows that the space of embeddings $$\operatorname{Emb}(P,P')$$ is weakly homotopy equivalent to the derived mapping space: $$\operatorname{hMap}_{E(D^p)\text{-RMod}}(E(P), E(P')).$$ Here, ones takes e.g. a simplicial cofibrant replacement of $$E(P)$$ viewed as a right $$E(D^p)$$-module to obtain a simplicial set, given degree-wise by morphisms of right modules.

If $$P$$ is oriented then you can get away with using $$E^{or}(D^p)$$ instead, and if $$P$$ is parallelized you can even use the little $$p$$-disks operad instead. If the codimension $$\dim P' - \dim P$$ is less than $$3$$, then you obtain the analytic approximation $$T_\infty \operatorname{Emb}(P,P')$$ instead of the space of embeddings.

• I am not aware that Weiss and de Brito claim that the map $E(P,P') \to \operatorname{hMap}_{E(D^p)\text{-RMod}}(E(P), E(P'))$ is an equivalence; they merely assert that it sits in a pullback square. – John Klein Jan 4 '19 at 14:33
• @JohnKlein See the introduction of arxiv.org/abs/1202.1305 (as far as I can tell, this is the oriented case). – Najib Idrissi Jan 4 '19 at 14:44
• Ah...good: you are right. I was looking at a different a paper of theirs which uses the configuration category model. Thanks for the heads up! – John Klein Jan 4 '19 at 14:49
• @JohnKlein I think in their other paper, they are more interested in the $\overline{\operatorname{Emb}}$ space, i.e. the fiber of the map $\operatorname{Emb} \to \operatorname{Imm}$. Which is thus, by their theorem, the fiber of $\mathbb{R}\operatorname{map}_\Gamma(\operatorname{con}(-), \operatorname{con}(-)) \to \Gamma$. – Najib Idrissi Jan 4 '19 at 14:56