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In the paper B. Bakalov, A.D'Andrea and V.G.Kac, Theory of finite pseudoalgebras, section 3, one finds the following definition of pseudotensor category:

A pseudotensor category is a class of objects $\mathcal {M}$ together with vector spaces Lin$(\{L_i, M\}),$ equipped with actions of symmetric groups $S_I$ among them and composition maps, satisfying Associativity, Unit, and Equivariance axioms.

My question is:

  1. Why does the definition require the action of symmetric groups $S_I$ ?

  2. How does the action of $S_I$ work? Is there an example?

  3. A Lie $H$-pseudo algebra is a Lie algebra in the pseudotensor category $\mathcal{M}^*(H)$, but how does $S_I$ act on the composition of a pseudotensor category $\mathcal{M}^*(H)$?

  4. The definition of pseudotensor category given by the book by A. Beilinson and V. Drinfeld, "Chiral algebras," didn't mention the equivariance axiom. Does the equivariance follow from the other properties they give? How?

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The third tag does not fit. – Sönke Hansen Jun 30 '13 at 10:03

That's the same as a symmetric multicategory (a.k.a. coloured operad) enriched in vector spaces, I believe, see Wikipedia. The action of symmetric groups is not necessary, you also have a notion of non-symmetric multicategory, but you often have it, e.g. in the canonical example which is in the reference you give: the category of vector spaces itself, where $\operatorname{Lin}(\{L_i\}_{i\in I},M)$ is the vector space of multilinear maps. The action of symmetric groups in an abstract example is required to satisfy the same properties as in this concrete example, where it is given by permutation of coordinates in the source. I think this answers 1 and 2. As for 3, you have a description of that multicategory in the paper you link to. Question 3 is better answered by Pavel Safronov.

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4) The definition in Beilinson-Drinfeld does contain $S_n$-equivariance, although it is hidden in the composition map.

Let me explain where the action of $\mathrm{Aut}(I)$ comes from. For $\pi:I_2\stackrel{\sim}\rightarrow I_1$ you have a map (in the BD notaion $K_i=L_{\pi(i)}$) $$P_{I_1}(\{L_i\}, M)\rightarrow P_{I_1}(\{L_i\}, M)\times \prod_{I_1} P_1(L_i, L_i)\rightarrow P_{I_2}(\{L_{\pi(i)}\}, M),$$ where the first map is the identity operation in $P_1(L_i, L_i)$ and the second map is the composition.

The equivariance of the composition under this $\mathrm{Aut}(I)$ action follows by combining associativity and the identity axiom.

2) Let $\phi:J\twoheadrightarrow I_2$. Then you want the composition morphism $$P_{I_1}(\{L_i\}, M)\times \prod_{I_1} P_{J_i}(\{K_j\}, L_i)\rightarrow P_J(\{K_j\}, M)$$ obtained from $\pi\circ\phi:J\twoheadrightarrow I_1$ to coincide with the composite $$P_{I_1}(\{L_i\}, M)\times \prod_{I_1} P_{J_i}(\{K_j\}, L_i)\rightarrow P_{I_2}(\{L_{\pi(i)}\}, M)\times \prod_{I_2} P_{J_i}(\{K_j\}, L_i)\rightarrow P_J(\{K_j\}, M),$$ where the first map is the action of $\pi$ and the second map is the composition morphism associated to $\phi$.

For example, let $J=I_1=I_2$ be the set $\{1, 2\}$, $\phi$ is the identity and $\pi$ is the flip. Then the equivariance implies that $$P_2(\{L_1, L_2\}, M)\times \mathrm{End}(L_1)\times \mathrm{End}(L_2)\rightarrow P_2(\{L_2, L_1\}, M)$$ coincides with $$P_2(\{L_1, L_2\}, M)\times \mathrm{End}(L_1)\times \mathrm{End}(L_2)\rightarrow P_2(\{L_2, L_1\}, M)\times \mathrm{End}(L_2)\times \mathrm{End}(L_1)\rightarrow P_2(\{L_2, L_1\}, M).$$

In other words, if your operations are representable (e.g. you are in an honest tensor category), equivariance expresses the fact that the braiding $L_1\otimes L_2\rightarrow L_2\otimes L_1$ is a natural transformation.

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Sorry , I really can not understand the Equivariance. Could you please give me more details? – Ruby Jun 27 '13 at 5:08
Ones know the definition of a $G$-equivariant homomorphism, where $G$ is a group.Let $G$ be a group and let $X$ and $Y$ be two associated $G$-sets. A function $f : X\longrightarrow Y$ is said to be equivariant if $$f(g\cdot x) = g\cdot f(x) \leqno(*)$$ for all $g \in G$ and all $x \in X$. But about the pseudotensor category, how does $S_I$ acts on a composition?What is the meaning the equivariace? – Ruby Jun 27 '13 at 5:16
In the definition of symmetric multicategory, link page 2, it said:`` Explicitly, for $\sigma \in \Sigma_n,$ and each sequence of objects $x_1, \cdots x_n; x$ we have a right action of $\Sigma_n$, i.e., a morphism $$\sigma^* : \mathcal{P}(x_1, \cdots x_n; x)\longrightarrow \mathcal{P}(x_{\sigma (1)}, \cdots x_{\sigma (n)}; x) $$ The action maps are well behaved, in the sense that all composition operations are invariant under the $\Sigma n$-actions, and$(\sigma \tau)^*=\sigma^* \tau^*.$ – Ruby Jun 27 '13 at 5:18
@FernandoMuro @PavelSafronov I really can not image what does the all composition operations are invariant under the $\Sigma n$-actions", mean, can you give me a formula like (*)? @PavelSafronov Sorry, I think your explanation can tell that $(\sigma \tau)^*=\sigma^* \tau^*,$ but how could you explain the equivariace"? I am really sorry , please hele me. – Ruby Jun 27 '13 at 5:19
@Ruby, it's hard to understand, I know, we all had to understand it for the first time at some point in our lives. Unfortunately, it is something you'll only learn by looking at the definitions and working out examples on your own. My suggestion: start with symmetric operads and then go for coloured operads. – Fernando Muro Jun 27 '13 at 16:52

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