Given a graded vector space $V$ over a field $k$, consider it's suspension $\Sigma V$ such that $(\Sigma V)^i=V^{i-1}$. For an operad of graded vector spaces over a field $\mathcal{O}$, the operadic suspension $\mathfrak{s}\mathcal{O}$ is defined in several different ways depending on the author. Some standard references might be An Alpine Expedition through Algebraic Topology and Operads in Algebra, Topology and Physics. All the definitions I've seen yield isomorphic graded vector spaces, but the operadic structures differ slightly. In the reference above, the operadic structure is not explicitely defined, it is just said to be induced by the one on $\mathcal{O}$, but it seems to be obvious that $\mathfrak{s}^{-1}\mathfrak{s}\mathcal{O}\cong \mathcal{O}\cong \mathfrak{s}\mathfrak{s}^{-1}\mathcal{O}$ as operads (not only as collections of graded vector spaces).

Here I'm interested in the definition given by Benjamin C. Ward in his Thesis (Section 2.1.2), for which I think that property does not hold.

Background definitions

He defines the operadic suspension as


where $sign_n$ is the sign representation of the symmetric group on $n$ letters. The symmetric group action on the graded vector spaces is the obvious diagonal action, and a diagonal operadic composition is given by the following operadic insertion on $\{\Sigma^{n-1}sign_n\}$. We may identify $\Sigma^{n-1}sign_n$ with the exterior power $\bigwedge^n k$, so it is spanned by the element $e_1\wedge\cdots\wedge e_n$. Therefore, define the $i$-th insertion map

$$\circ_i:\Sigma^{n-1}sign_n\otimes\Sigma^{m-1}sign_m\to \Sigma^{n+m-2}sign_{n+m-1}$$

as the map

$$(e_1\wedge\cdots\wedge e_n)\otimes (e_1\wedge\cdots\wedge e_m)\mapsto (-1)^{(i-1)(m-1)}(e_1\wedge\cdots\wedge e_{n+m-1}).$$

We may identify the elements of $\mathcal{O}$ with elements of its operadic suspension, so for $a,b\in\mathcal{O}$ we may write $a\tilde{\circ}_i b$ for the insertion in the suspension. We can compute it in terms of $a\circ_i b$ (the insertion in $\mathcal{O}$) in the following way:

$$\tilde{\circ}_i=(\mathcal{O}(n)\otimes\Sigma^{n-1}sign_n)\otimes (\mathcal{O}(m)\otimes\Sigma^{m-1}sign_m)\cong (\mathcal{O}(m)\otimes \mathcal{O}(m))\otimes (\Sigma^{n-1}sign_n\otimes \Sigma^{m-1}sign_m)\to \mathcal{O}(n+m-1)\otimes \Sigma^{n+m-2}sign_{n+m-1}$$

The Koszul sign rule on the isomorphism produces a sign with exponent $(n-1)\deg(b)$ and then the insertions are performed diagonaly, so after the identification we get

$$a\tilde{\circ}_i b=(-1)^{(n-1)\deg(b)+(i-1)(m-1)}a\circ_i b.$$

The operadic desuspension $\mathfrak{s}^{-1}\mathcal{O}$ is defined similarly using $\Sigma^{1-n}sign_n$, so the signs are the same.


I expected $\mathfrak{s}^{-1}\mathfrak{s}\mathcal{O}\cong \mathcal{O}$ as operads, but I think that the insertions are different. If I compute the insertion induced on $\mathfrak{s}^{-1}\mathfrak{s}\mathcal{O}$ in a similar way as above using the isomorphism

$$(\mathcal{O}(n)\otimes\Sigma^{n-1}sign_n\otimes \Sigma^{1-n}sign_n)\otimes (\mathcal{O}(m)\otimes\Sigma^{m-1}sign_m\otimes \Sigma^{1-m}sign_m)\cong (\mathcal{O}(m)\otimes \mathcal{O}(m))\otimes (\Sigma^{n-1}sign_n\otimes \Sigma^{m-1}sign_m)\otimes (\Sigma^{1-n}sign_n\otimes \Sigma^{1-m}sign_m)$$

Then, the insertion induced on this product is identified with

$$(-1)^{(1-n)(m-1)}a\circ_i b$$

which is of course not the same as $a\circ_i b$. So, for this new operad created by the suspension and desuspension to be isomorphic to the original one, we must have an automorphism $f$ on $\mathcal{O}$ such that $f(a\circ_i b)=f(a)\circ_i f(b)=(-1)^{(1-n)(m-1)}a\circ_i b$. I think this automorphism must be then of the form $f(a)=(-1)^{\varepsilon(a)}a$, with $\varepsilon(a)=\pm 1$. But this implies that $(-1)^{(n-1)^2}f(a\circ_i a)=(-1)^{2\varepsilon(a)}a\circ_i a=a\circ_i a$, which is not true for all $n$.


Is my conclusion about this suspension true or am I mistaken? I'm not so sure that $f$ really needs to be of that form, but I can't really find an morphism that makes the two structures isomorphic. Is this definition of operadic suspension used by any other author?


1 Answer 1


What you really need to show is that $$f(a\circ_ib)=(-1)^{(n-1)(m-1)}f(a)\circ_if(b).$$ Here, $n$ is the arity of $a$, $m$ is the arity of $b$, and $\circ_i$ is the infinitesimal composition in $\mathcal{O}$ (once you twist the definition of the infinitesimal composition by your sign, you get the usual equation for operad morphisms). You achieve this with $$f(a)=-(-1)^{\frac{n(n+1)}{2}}a.$$

  • $\begingroup$ Thank you, Fernando, it was very silly after all! $\endgroup$
    – Javi
    Commented Jul 9, 2020 at 12:26
  • 1
    $\begingroup$ @javi you're welcome. I've faced this kind of sign confusion several times, I know the formulas by heart. $\endgroup$ Commented Jul 9, 2020 at 12:52
  • $\begingroup$ Me too, and sometimes I get stuck in the combinatorics. Is there any idea of how someone would come up with that sign for $f$? Or any kind of tcommon tricks? $\endgroup$
    – Javi
    Commented Jul 10, 2020 at 12:40
  • 3
    $\begingroup$ You can set $f(a)=(-1)^{\phi(n)}a$ impose the formula you want to get to and solve the recursion to get a definition of $\phi$. I proceeded like this the first time, but now I just try the usual formulas and slightly modify them if necessary. $\endgroup$ Commented Jul 10, 2020 at 21:12

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