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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?)

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    $\begingroup$ 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. $\endgroup$ – Tyler Lawson Jan 3 '19 at 16:27
  • $\begingroup$ @TylerLawson so, does that mean the symmetric sequence forms a module (or algebra) over the disk-embedding operad? $\endgroup$ – John Klein Jan 3 '19 at 16:32
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    $\begingroup$ Yes: this is called a right module over an operad. $\endgroup$ – Dan Petersen Jan 3 '19 at 16:42
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    $\begingroup$ @DanPetersen in the symmetric sequence context (that was what I was missing). I was used to modules over an operad in spaces. Thanks. $\endgroup$ – John Klein Jan 3 '19 at 16:45
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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.

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  • $\begingroup$ 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. $\endgroup$ – John Klein Jan 4 '19 at 14:33
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    $\begingroup$ @JohnKlein See the introduction of arxiv.org/abs/1202.1305 (as far as I can tell, this is the oriented case). $\endgroup$ – Najib Idrissi Jan 4 '19 at 14:44
  • $\begingroup$ 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! $\endgroup$ – John Klein Jan 4 '19 at 14:49
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    $\begingroup$ @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 $\mathbb{R}\operatorname{map}_\Gamma(\operatorname{con}(-), \operatorname{con}(-)) \to \Gamma$. $\endgroup$ – Najib Idrissi Jan 4 '19 at 14:56

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