Blog Post number 1
Published:
N∞-operads and associahedra
June 16, 2023
Defining an N∞-operad using transfer systems
This time, we want a combinatorial understanding of N∞-operads in a suitable G-equivariant setting. The goal is establishing, for G=Cpn, a one-one correspondence between the natural order on the collection of N∞-operads and the poset structure of Stasheff’s (n+1)-associahedron, An+1. This is work by Balchin-Barnes-Roitzheim[1].
Instead of the traditional way, as mentioned right at the article’s outset, we want to understand N∞-operads differently in a combinatorial way. The key player is the following result that was conjectured by Blumberg-Hill [2] and proved by a series of works by Bonventre-Pereira, Gutiérrez-White, and Rubin independently. I just state the result by getting into the details of the proof
Thm 1.1.
Ho(N∞-op)∼→Indwhere N∞-op is the category of N∞-operads, and Ind is the category of indexing systems with poset structure.
Now we’ll see what indexing systems are next.
For that, they’re some machinery that we introduce first.
Def 1.2. A categorical coefficient system is a contravariant functor
C_:OrbG→Catfrom the orbit category of G to the category of small categories.
Typically, there’s an abuse of notation by omitting the word categorical.
We have the notion of a symmetric monoidal coefficient system, which is a contravariant functor from the orbit category of G to the category of symmetric monoidal categories and strong symmetric monoidal functors.
The category of interest for us is the coefficient system of finite G-sets.
Def 1.3. Suppose Set_ is the symmetric monoidal coefficient system of finite sets, with Set_(H)=SetH. The symmetric monoidal operation is the disjoint union.
Now the idea is to associate to every N∞-operad a sub coefficient system of Set_.
Next is the definition of an indexing system using all the machinery we’ve introduced.
Def 1.4. An indexing system is a sub symmetric monoidal coefficient system that’s closed under subobjects and self-induction.
Now the latter condition, for a full sub-symmetric monoidal coefficient system Z, translates to the following.
H/H′∈Z(H)∧K∈Z(H′)⇒H×H′K∈Z(H)We now compare indexing systems to transfer systems [].
Lemma 1.5. An indexing system determines (and is also determined by) a set ZH for H≤G, with subgroups of H satisfying the following axioms:
H/H∈Z(H)
H/H′∈Z(H)⇒gHg−1/gH′g−1∈gZg−1
H/H′∈Z(H) ∧ H′/K∈Z(h′)⇒H/K∈Z(H)
H/H′∈Z(H)⇒K(K∩H′)∈K ,∀ K≤H
All this data is referred to as a transfer system.
Alternatively, we have the following definition.
Def 1.6. For a transfer system Z(H) we can define a set abstractly
NHH′∣H/H′∈Z(H),H≤GWe call these norm maps.
Using all these, now we have an alternate description of an N∞-operad that I state next.
Cor 1.7.
G is a finite group. An N∞ operad for G, upto homotopy is defined as the data of the set X={NKH∣1≤K<H≤G} satisfying the following (and all conjugates thereof)
- NKH∈X ∧M<H⇒NMK∩M∈X (Restriction Axiom aka RA)
- NKL∈X ∧NKH∈X⇒NHL∈X (Composition Axiom aka CA)
It’s important to note that the notation choice makes sense so that the Mackey functor of an algebra A over an N∞-operad O has multiplicative norm maps NHK:πK0(A)→πH0(A).
The case of G=Cpn
We now look at the case of interest, i.e. when G=Cpn where p is a prime.
The goal is to understand the sequence {∣N∞(Cpn)∣} .
The cases of n=0 and n=1 are trivial. We look at the case of n=2.
We start by drawing out all our possibilities and then using the axioms RA and CA to eliminate redundant structures.
The third structure i.e. {N10,N21} isn’t an N∞-structure using CA and the sixth and eighth structures i.e. {N20,N21} , {N20} aren’t N∞-structures by an application of RA.
Thus we can conclude that ∣N∞(Cp2)∣=5
There’s a noticeable pattern that we’ve here.
We now spend some time recalling the Catalan Numbers, which obviously have been popping up so far.
The Catalan sequence is a widespread sequence popping up in various enumerations problems, two of which we’ll use directly. We first define the sequence.
Def 2.1. The Catalan numbers are terms of the sequence {Cn} defined as follows:
C_n = \frac{1}{n+1}\binom{2n}{n} = \frac{(2n)!}{(n+1)!n!}We also recall that they satisfy the following recurrence relation.
C_0 = 1 C_{n+1} = \sum_{l=0}^{n}C_l C_{n-l} \\ , \\ n\geq 0Some examples of enumeration problems where Catalan numbers pop up are the following:
- C_n = # rooted binary trees with n+1 vertices
- C_n = # ways of cutting a convex polygon with n+2 sides into triangles
- C_n= # permutations of \{1, \cdots , n\} with no three terms in increasing order
BBR product
We now define an operation on N_{\infty}-operads that I call the Balchin-Barnes-Roitzheim (BBR) product, also called the o-dot product.
\odot: N_{\infty}(C_{p^k}) \times N_{\infty}(C_{p^{l}}) \rightarrow N_{\infty}(C_{p^{k+l+2}})For X \in N_{\infty}(C_{p^k}) and X’ \in N_{\infty}(C_{p^l}), X \odot X’ \in N_{\infty}(C_{p^{k+l+2}}) is defined as follows:
X \odot X' = X \coprod \Sigma^{k+2}X' \coprod \\{N_{k+1}^{m} \\}_{k+1 < m \leq k+k+2}where \Sigma^{j}N_{a}^{b} = N_{a+j}^{b+j}
Now there’s a pictorial way to work and see things that we see next.
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Pictorial illustration of the BBR product from the BBR paper[] |
We now see an example:
Let X,Y \in N_{\infty}(C_{p^2})
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It’s easy to see how the operation works and that it’s not commutative |
We now see the result that lets us use the BBR product for our purposes.
\mathbf{Prop} \ 3.1. \ X \in N_{\infty}(C_{p^k}) and Y \in N_{\infty}(C_{p^l}). X \odot Y is an object in N_{\infty}(C_{p^{k+l+2}}) for k,l \geq -1. Also conversely, if we’ve X \odot Y \in N_{\infty}(C_{p^{k+l+2}}) then X and Y are N_{\infty} structures.
Now we make our next step towards computing \mid N_{\infty}(C_{p^{n}}) \mid
\mathbf{Prop} \ 3.2. \ We have the following recurrence relation
|N_{\infty}(C_{p^{-1}}) | = 1 |N_{\infty}(C_{p^n})| = \sum_{q=0}^{n}|N_{\infty}(C_{p^{q-1}})| |N_{\infty}(C_{p^{n-q-1}})|, \\ n \geq 0The idea of the proof is to use the last proposition after showing that every X \in N_{\infty}(C_{p^n}) can be written as X = Y \cdot Z where Y \in N_{\infty}(C_{p^{q-1}}) and Z \in N_{\infty}(C_{p^{n-q-1}})
Now we begin with j \in \mathbb{Z} such that it’s the minimum integer such that N_{n}^{j} is in X. The cases of j=0 and j=n are straightforward, with Y = \emptyset and Z = \emptyset in both cases.
Now if O < j < n, then clearly we can split off C_{p^i} with 0 \leq i < j. Let’s call this part Y \in N_{\infty}(C_{p^{j-1}}) and the remaining part X’ Now it’s easy to see that this part’s like \emptyset \cdot Z for some Z \in N_{\infty}(C_{p^{n-j-1}}).
We can thus conclude that X = Y \cdot Z
Now a consequence of the previous result is the following corollary.
\mathbf{Cor} \ 3.3. \
For G= C_{p^n}, and X \in N_{\infty}(C_{p^n}) we can write X = Y \cdot Z for some suitable N_{\infty}-operads Y and Z.
Now as a corollary to this, we have the first part of the main theorem.
\mathbf{Cor} \ 3.4. \ We have a bijection
\\{\text{rooted binary trees with} \\ n+2 \\ \text{vertices} \\} \Leftrightarrow \\{N_{C_{p^n}} \\}The recursive construction works as follows:
Then
Relation to Stasheff's Associahedra ------ We now define the associahedron. For binary trees, let's define the following relation: For binary trees X and Y X < Y \Leftrightarrow \\{ \text{Y can be obtained from}\\ X \\ \text{by means of a finite sequence of rotations} \\} \mathbf{Def} \\ 4.1. \\ The poset structure on binary trees with n+1 vertices induced by the above relation is referred to as Stasheff's n-associahedron \mathcal{A}_{n}. We now equip N_{\infty}-operads for G= C_{p^n} with a poset structure by defining the following order relation: For X, Y \in N_{\infty}(C_{p^n}) X < Y \Leftrightarrow \\{ Y \\ \text{can be obtained via}\\ X \\ \text{by means of a sequence of norm map additions} \\} What follows next is the main result in all its glory and completeness, which basically tells us about the agreement of the poset structures we just saw. \mathbf{Thm} \\ 4.2. \\ *We have an order-reflecting and order-preserving bijection* \\{\text{rooted binary trees with} \\ n+2 \\ \text{vertices} \\} \Leftrightarrow \\{N_{C_{p^n}} \\} \Leftrightarrow \mathcal{A}_{n+1} The remaining part of this result is the second bijection. The idea is to prove the correspondence between adding a norm map to the N_{\infty}-operad and a clockwise rotation operation to a tree. Here's a pictorial description of the \Leftarrow part of the equivalence:
\Updownarrow
The \Rightarrow part's not as straightforward as this. I don't get into all the details; it just boils down to the following 3 cases:
from which the first and second cases are almost immediate, while the third needs a bit of work for whose details the reader's redirected to [1]. References ------ 1. Roitzheim, Constanze, Barnes, David, Balchin, Scott (2022) N-infinity operads and associahedra. Pacific Journal of Mathematics, 315 (2). pp. 285-304. E-ISSN 0030-8730. (doi:10.2140/pjm.2021.315.285) (KAR id:77048) [link](https://kar.kent.ac.uk/77048/11/Roitzheim%20published%20version.pdf) 2. Blumberg, Andrew J., and Michael A. Hill. "Operadic multiplications in equivariant spectra, norms, and transfers." Advances in Mathematics 285 (2015): 658-708. [link](https://pdf.sciencedirectassets.com/272585/1-s2.0-S0001870815X00144/1-s2.0-S000187081500256X/main.pdf?X-Amz-Security-Token=IQoJb3JpZ2luX2VjEIL%2F%2F%2F%2F%2F%2F%2F%2F%2F%2FwEaCXVzLWVhc3QtMSJHMEUCIQCO6vfKKcndrwlTgCHNSbiJ2G8VCUH%2FilkcfjXDXcyxTgIgZtDHloTThyJm1sH04X9w7nLaUBSaNozzN2P5W8cezasquwUI2%2F%2F%2F%2F%2F%2F%2F%2F%2F%2F%2FARAFGgwwNTkwMDM1NDY4NjUiDCT15dfFZPANtwdJuiqPBTyJy7Ib9n2n6I1jiycZo62hL01bwagymcPU1U5wh4Fsp6qRYZQewL3jJCPhUwiqbHKpvQ7deLPLDZuMsIfHPbegRL%2F1IOh7egNBX3lwbO8bDBkMdbgjWRecwQMpKIQV3XXZPe4nnK5BD50BkRpw2opoE3GNN%2FelGr6OYde4HT4P%2F730lgzfQyf8JlyPow8HEU4vvoph1T9iw6c346bwIdWZ%2FT%2Fcj8DEGfAXTdHI1c%2Bkd80q6Dx4i1wDWblQFRZrRU921OgPvLUF3uTONERuN85maW%2F7DdgQ3Zip3l3CXYf9cCCKNW%2B7fvHionqKcCmYGbx8p%2Ff2PzKGVssrfqZJ7MX9OZlOWpbJxrXLRNXq%2BrOZAES%2F1SafzrSagVXiI3zYMqoLTDcTjkQi6cacwGNjErZNocPbpSViguC%2FwXJAXMnOILw1BWsvTHsUgYrhspqapslU9yVhCU57dTgOHYeuS4%2BQAgc0qq%2BBAZbKrxTKP%2B08z6GT2CdAU4aJrtH6SVAvwAtqgMNoSn7MCx%2B1mQZ1wjMTMXwG3h5HfQcUT3%2BdsK6UHfiTKKKrHZrwRyUrUTN3V8ocGybVFA3iOAoMi8lWAzpYWbOXdl3ys7nDdw%2BK%2BuISeivXEzSQiQ09u7emrDA0G%2BFaAI9cOfy8vdW6QUn0jbNHrYjDpc%2BTbUoRsXjQPfs33%2FHoBNwDK03tyUkiNbYTQyskE%2FyABbmkRaSKyNx12BX8EnJic1lk6FClJy0XtBluKnqEUsF4QxSbHoyYmxL9IsYCPjwtATsL7M4eERN8XMDCF4SoAxQHCgDcKNrUnnQ3mlug5tH6jC2%2BuZzRA%2Flf8Oy6rnooeLgkCMalB%2FZYRY%2FpNV1uQfamWXuvndfbQn4wqJnSpAY6sQFMkNuESlJQl3AjJPupNwtns3qtMYKW96tb%2FWNJ4GT5RAPrH1qSVgO%2BtI%2Bvz5TmwcBPYqtOCJLMpa0cqjtbeGaTTRDnwPui7DBwWCJGJ5Qor%2ByIVbEhsaahQdGtAq6rWdQnT1Tlt6X51984kibwuU98VVPsCip9l1JNgAcDR7eM8KNcdikU3Xw9h3tVdu3J3OlXdlPE3rl%2B34WDdrtWicA%2FcKqOl2ySqo0PtMrVvtMjQRI%3D&X-Amz-Algorithm=AWS4-HMAC-SHA256&X-Amz-Date=20230622T185116Z&X-Amz-SignedHeaders=host&X-Amz-Expires=299&X-Amz-Credential=ASIAQ3PHCVTYR47UNKXY%2F20230622%2Fus-east-1%2Fs3%2Faws4_request&X-Amz-Signature=c18e3dea7a3496c3ff156ac0be1a78f3c0444ce307595dbf4d1e548b14d38e1e&hash=d24d35890bdfe424228276c45f5dc1c2c3b21a3f7ce2c71da852cab1813cc034&host=68042c943591013ac2b2430a89b270f6af2c76d8dfd086a07176afe7c76c2c61&pii=S000187081500256X&tid=spdf-400ca13a-e793-461d-b061-5aac41cb1648&sid=92a107798e6f5344926abbd44f81c8e93754gxrqb&type=client&tsoh=d3d3LnNjaWVuY2VkaXJlY3QuY29t&ua=0c08520b065400005950&rr=7db6ae823d979a72&cc=in)