# Coalgebra differential on reduced symmetric algebra

Definitions/setup

(I am using "Formality and star products" by Cattaneo)

Let $g=\bigoplus_{i\in \mathbb{Z}} g^i$ be a graded vector space over some field. We can then look at the reduced symmetric space $\overline{S}(g)$ (where $g=\bigoplus_{i\in \mathbb{Z}} g^{i+1})$ which is defined by dividing the tensor algebra $\overline{T}(g)=\bigoplus_{n=1}^\infty g^{\otimes n}$ by the ideal generated by elements of the form $x\otimes y-T(x\otimes y)$ where $T$ is the twisting map. This graded vector space is made into an algebra using tensor products as multiplication.

We can make this graded symmetric tensor algebra into a graded coalgebra by defining the comultiplication $\Delta(v)=1\otimes v +v\otimes 1$ on $v \in g$ and extend this using tensor products.

My question

Suppose now that we have a degree 1 coalgebra differential on $\overline{S}(g)$ say $Q$. In the document "Formality and starproducts" it is stated after definiton 3.6 that it is enough to know $Q^1$ which is the composition of $Q$ and the projection onto the first component. I dont understand this, because since $Q$ is a degree 1 differential, it means that it sends pure tensors of degree $n$ to tensors of degree $n+1$, this means that for any element in $\overline{S}(g)$ the first component of the image under $Q$ will be $0$.

Clearly, I must be missing something here. I hope someone can help me.

There are two gradings on $\bar{S}(g)$. The first is the usual degree grading, given by $\deg(x_1 \otimes \dots \otimes x_n) = \deg(x_1) + \dots + \deg(x_n)$ where $x_i \in g$. The second, also commonly called weight, is given by $w(x_1 \otimes \dots \otimes x_n) = n$ (the number of "letters" in the "word", if you want).

The $Q^1$ they mention is the projection on the first weight component $\bar{S}^1(g) = g$. More generally define $\bar{S}^n(g)$ to be the image of $(g)^{\otimes n}$ in the quotient, then $$\bar{S}(g) = \bigoplus_{n=1}^\infty \bar{S}^n(g),$$ and $Q^1 : \bar{S}(V) \to V$ is the composite of $Q$ with the projection $\operatorname{pr}_1 : \bar{S}(V) \to \bar{S}^1(V) = V$

It's true that $Q$ is a degree $1$ differential, so $\deg(Q(x)) = \deg(x) + 1$. But even if $x$ is a "word with $n$ letters" (i.e. has weight $n$) then $Q(x)$ will typically be a sum of monomials of varying lengths, not necessarily just $n$, or $n+1$ letters.

Now why is $Q$ uniquely determined by $Q^1$? $\bar{S}(g)$ is what is called a cofree cocommutative coalgebra (more precisely, it's a cofree conilpotent cocommutative coalgebra). This is a notion formally dual to the notion of a free commutative algebra, that is, a polynomial algebra $S(V)$.

For ease of notation I'll consider some graded vector space $V$ (that you can specialize as $g$ later). It is well known that a given derivation $d : S(V) \to S(V)$ (i.e. a map satisfying $d(ab) = d(a)b \pm a d(b)$) is uniquely determined by its restriction $f$ on $V = S^1(V) \subset S(V)$, because if you have a monomial $v_1 \dots v_n \in S(V)$, you can use the derivation relation to compute $$d(v_1 \cdot \dots \cdot v_n) = \sum_{i=1}^n \pm v_1 \dots f(v_i) \dots v_n.$$ What's more, given any linear map $f : V \to S(V)$, you get a uniquely defined derivation $d_f : S(V) \to$ with the above formula.

Well, since the notion of cofree coalgebra is formally dual to the notion of free algebra, it's not surprising that a coderivation $Q : \bar{S}(V) \to \bar{S}(V)$ is uniquely determined by its "corestriction" to $\bar{S}^1(V)$, i.e. the composite $\operatorname{pr}_1 \circ Q : \bar{S}(V) \to \bar{S}(V) \to \bar{S}^1(V) = V$. If you are not used to working with coalgebras, it might be harder to see, but I'll try to explain.

Recall that a coderivation $Q$ is characterized by the relation $\Delta Q = (Q \otimes 1 + 1 \otimes Q) \Delta$. Suppose that you have some coderivation $Q : \bar{S}(V) \to \bar{S}(V)$ and that you know how to compute the corestriction $Q^1$. Write $Q = Q^1 + Q^2 + \dots$, where $Q^n$ is the corestriction to $\bar{S}^n(V)$. How do you compute $Q^2$? Take some $x \in \bar{S}(V)$ and write $\Delta(x) = \sum_{(x)} x_{(1)} \otimes x_{(2)}$ (I'm using Sweedler's notation). Then the coderivation relation tells you: $$\Delta(Q(x)) = \sum_{(x)} \bigl( Q(x_{(1)}) \otimes x_{(2)} \pm x_{(1)} \otimes Q(x_{(1)}) \bigr) \in \bar{S}(V) \otimes \bar{S}(V).$$

Now corestrict (project) both parts of the tensor product onto $\bar{S}^1(V)$. What does it tell you?

• On the one hand, because of how the coproduct of $\bar{S}(V)$ is defined, the projection of the LHS is precisely $Q^2(x)$. Indeed, if $n \ge 3$, then at least one of the factors of $\Delta(Q^n(x)) \in \bar{S}(V) \otimes \bar{S}(V)$ will have at least weight $2$, and so $(\operatorname{pr}_1 \otimes \operatorname{pr}_1) \Delta(Q^n(x)) = 0$. And $\Delta(Q^1(x)) = 0$ anyway, so all you're left is $Q^2$, and by definition on a polynomial of weight $2$ you just get $$(\operatorname{pr}_1 \otimes \operatorname{pr}_1) \Delta(v \cdot w) = v \otimes w.$$
• On the other hand, the projection of the RHS can be expressed in terms of $Q^1$, and so you get: $$Q^2(x) = \sum_{(x)} \bigl( Q^1(x_{(1)}) \otimes x_{(2)} \pm x_{(1)} \otimes Q^1(x_{(1)}) \bigr).$$

(More precisely, $Q^2$ will be the projection of this in the quotient by the transpose map $T$.) A similar reasoning gives: $$Q^n(x) = \sum_{(x)} \sum_{i=1}^n \pm x_{(1)} \otimes \dots \otimes Q^1(x_{(i)}) \otimes \dots \otimes x_{(n)},$$ and so you can recover the whole coderivation just from $Q^1$. And as before, any linear map $f : \bar{S}(V) \to V$ defines a coderivation with the above formula (where $Q^1 = f$).

• You say that even if $x$ is a word in $n$ letters, i.e. $x\in \overline{S}^n(g)$, $Q(x)$ will be a sum if monomials of varying lenghts. If I look at the definition of a coderivation of degree $1$, it says that $Q(\overline{S}^n(g)\subset \overline{S}^{n+1}(g)$. So how can $Q(x)$ be a sum of monomials of varying lengts? – Badshah Oct 13 '15 at 12:39
• @Badshah You're looking at the definition in Cattaneo's lecture notes, right? He says that a coderivation of degree $k$ on $\mathfrak{h}$ satisfies $\delta(\mathfrak{h}^i) = \mathfrak{h}^{i+k}$. The degree here is $\deg$, not the weight. Look at Example 3.8: $Q^1_1(a) = \pm da$ has the same weight as $a$ (but degree $\deg(a)+1$), and $Q^1_2(bc) = \pm [b,c]$ has weight $1$ while $bc$ has weight $2$. – Najib Idrissi Oct 13 '15 at 12:53
• Ah I see it now! Thank you very much – Badshah Oct 13 '15 at 13:00
• I have one more question: How is $(\text{pr}_1\otimes \text{pr}_1)\Delta(Q(x))=Q^2(x)$? – Badshah Oct 13 '15 at 15:07
• @Badshah It comes from the definition of $\Delta$. If $n \ge 3$, then at least one of the factors of $\Delta(Q^n(x)) \in \bar{S}(V) \otimes \bar{S}(V)$ will have at least weight $2$, and so $(\operatorname{pr}_1 \otimes \operatorname{pr}_1) \Delta(Q^n(x)) = 0$. And $\Delta(Q^1(x)) = 0$ anyway. So all you're left is $Q^2$, and by definition on a polynomial of weight $2$ you just get $(\operatorname{pr}_1 \otimes \operatorname{pr}_1) \Delta(v \cdot w) = v \otimes w$. (I edited the question to include this information) – Najib Idrissi Oct 13 '15 at 15:10