Compatibility of the KZ connection with operadic composition

2012-03-22T15:48:09Z

In what sense is the Knizhnik-Zamolodchikov connection compatible with the operadic composition in the little discs operad and/or the operad of $\overline{M}_{0,n}$'s?

Here are (some) details, motivation, and a more precise question.

The KZ connection $A_n\in\Omega^1(C_n)\otimes\mathfrak{t}^n$ is given by $$A_n =\sum_{i< j} t^{ij}\,d\log(z_i-z_j).$$ Here $C_n$ is the configuration space of $n$ different points in $\mathbb{C}$ and $\mathfrak{t}^n$ is the Lie algebra with generators $t^{ij}$ ($1\leq i,j\leq n$, $i\neq j$, $t^{ij}=t^{ji}$) and relations $[t^{ij},t^{kl}]=0$ if all $i,j,k,l$ are different and $[t^{ij},t^{ik}+t^{jk}]=0$.

The "regularized holonomy" of $A_3$, when $z_1$ stays at $0$, $z_3$ at $1$, and $z_2$ moves from $0$ to $1$, is the KZ Drinfeld associator $\Phi_{KZ}$.

In general, an associator $\Phi$ (together with its coupling constant $\mu$) is equivalent to a morphism of operads of groupoids $F:PaB\to T$. Here $PaB_n$ is the groupoid of parenthesized braids, and $T_n=\exp \mathfrak{t}^n$ ($\mathfrak{t}^n$'s form an operad of Lie algebras). In the case of $\Phi_{KZ}$, $F_{KZ}$ is given by the "regularized holonomy" of $A_n$'s. The natural explanation of the fact that $F_{KZ}$ is a morphism of operads would be that $A$ is compatible with the operad structure. In what sense it is true? (there are two questions - in what sense is $A$ compatible with operad, and how it implies that $F_{KZ}$ is a morphism of operads)

I'll mention some random stuff that might appear in the answer, but you can ignore it.

There is a flat connection on moduli spaces of rational curves with marked points, closely related to $A$. If we define $\hat{\mathfrak{t}}^n$ by imposing one more relation $\sum_i t^{ij}=0$ ($[t^{ij},t^{ik}+t^{jk}]=0$ is then a consequence), the KZ connection descends to $M_{0,n}$; let us call it $\hat{A}_n\in\Omega^1(M_{0,n})\otimes\hat{\mathfrak{t}}^n$. By putting one of the points to $\infty$ we can identify $M_{0,n+1}$ with $C_n/\{az+b\}$ ($a\in\mathbb{C}^*,b\in\mathbb{C}$), and via the isomorphism $\hat{\mathfrak{t}}^{n+1}\cong\mathfrak{t}^n/\text{center}$ we can identify $\hat A_{n+1}$ with $A_{n}/\text{center}$. The center of $\mathfrak{t}$ is lost in this way. The Lie algebras $\hat{\mathfrak{t}}^{n}$ form a cyclic operad. The compactified moduli spaces $\overline{M}_{0,n}$ also form a cyclic operad.

As $F_{KZ}$ is (modulo the center) given by the parallel transport of $\hat{A}$ between certain tangential base points of $M_{0,n+1}\subset\overline{M}_{0,n+1}$, I would imagine the operadic compositions to be (roughly) maps $$M_{0,n+1}\times M_{0,m+1}\times\text{formal punctured disc}\to M_{0,n+m}$$ coming from the operadic composition of $\overline{M}_{0,k}$'s. But I'm not sure in what category it would be an operad (of nice families over the punctured discs?) and how to make it technically work (something should be also said about maps of the trivial $\hat T$-bundles over these spaces).

In fact, to get a parametrization of the punctured disc, it would be better to consider the moduli spaces $M_{0,n}'$ of rational curves with $n$ marked points and non-zero tangent vectors at those points. There is a flat connection on $M_{0,n}'$ which sees also the center. Let us replace the relations $\sum_i t^{ij}=0$ with $s^j:=\sum_i t^{ij}\textit{ is central}$. The Lie algebras $\check{\mathfrak{t}}^{n}$ that we obtain in this way still form a cyclic operad, and the connection $$\check{A}_n= \sum t^{ij}\,d\log(z_i-z_j)+ \sum s^i\,d\log(v_i)$$ is a flat connection on the configuration space of $n$ different points with chosen non-zero tangent vectors; this 1-form is again $SL(2,\mathbb{C})$-basic, and so it descends to the moduli space $M_{0,n}'$. By putting one of the points to $\infty$ and also normalizing its tangent vector, we can identify $M_{0,n+1}'$ with $(C_n/\text{translations})\times (\mathbb{C}^*)^n$. We have an isomorphism $\check{\mathfrak{t}}^{n+1}\cong\mathfrak{t}^n\oplus \mathbb{C}^n$, and so $\check{A}_{n+1}$ gets identified with $A_n$ plus a central part corresponding to the tangent vectors. It is the framed version of the KZ connection (and is my favorite).

edit: there is another (Alekseev-Torossian) connection $A^{AT}_n$ on $C_n$ with values is $\mathfrak{t}^n$. $A^{AT}_n$ is in fact on $FM_2(n)$ ($FM_2(n)$ is a compactification of $C_n/\{az+b\}$, where this time $a\in\mathbb{R}_+$). $FM_2(n)$'s form an operad (a version of the little discs operad) and $A^{AT}_n$'s are compatible with the operad structure in the obvious way: if $o_i:FM_2(m)\times FM_2(n)\to FM_2(m+n-1)$ is one of the compositions then $o_i^*A^{AT}_{m+n-1}$ is equal to the connection $A^{AT}_m\oplus A^{AT}_n$ on $FM_2(m)\times FM_2(n)$, after we apply the corresponding $o_i:\mathfrak{t}^m\oplus\mathfrak{t}^n\to\mathfrak{t}^{m+n-1}$. The operad $PaB$ is a sub-operad of the fundamental groupoid of $FM_2$. $\Phi_{AT}$ and $F_{AT}$ are defined as (ordinary, not regularized) parallel transport of $A^{AT}$, and $F_{AT}$ is obviously a morphism of operads. I would like to understand the corresponding picture for $A_{KZ}$.

Answer by Adrien for Compatibility of the KZ connection with operadic composition

2012-03-23T00:52:35Z

Edit: Answer reformulated

Hi Pavol,

While it's not true that you can extend the KZ connection on the compactified space, there is a well defined procedure to "specialize" it to get a connection on each boundary component. My claim is that these specializations are precisely obtained through application of operadic composition.

The precise story is as follows: De Concini--Procesi introduced a compactification $Y_n$ of the configuration space satisfying the following properties:

The complementary $K=Y_n\backslash C_n$ is a divisor with normal crossings
The KZ connection induces a meromorphic connection $\nabla_n$ on $Y_n$ with logarithmic singularities on $K$.

These properties means the following: let $D$ be a smooth irreducible divisor. Then there exists some local coordinates $z_1,\dots,z_{n-1},t$ such that $D$ is defined by $t=0$, and such that $$\nabla_n=d-\omega_D(z_1,\dots,z_{n-1},t)+A \frac{dt}t$$ where $\omega_D$ is holomorphic at $t=0$ and $A$ is a constant (the residue of the defining 1-form of $\nabla_n$). Now for any choice of a tangential basepoint inside $D$ the monodromy of $\nabla_n$ around $D$ is then given by $e^A$.

Now you can specialize $\nabla_n$ to a flat, meromorphic connection on $D$ by setting $$\nabla_D:=d-\omega_D(z_1,\dots,z_{n-1},0)$$

It turns out that there is a natural identification $D=Y_{n-1}$, and that $D$ is the image of some operadic composition $$\mu:Y_{n-1}\times Y_2 \rightarrow Y_n$$ Let $$\mu:\mathfrak t_{n-1}\times \mathfrak t_{2}\rightarrow \mathfrak t_{n}$$ be the "same" operadic composition.

Now my precise attempt to answer your question is the following

Claim: $$\nabla_D=\mu(\nabla_{n-1})$$

Now you can repeat this process on $D=Y_{n-1}$ and get the compatibility with other operadic maps.

Example

Consider the KZ connection on $Y_4$ (obtained from the original KZ connection by setting $z_1=0,z_2=x,z_3=y,z_4=1$):

$$ \nabla_4=t_{12}d\log x+t_{23}d\log(x-y)+t_{24}d\log(x-1)+t_{13}d\log y+t_{34}d\log(y-1)$$ Let $D \subset Y_4$ be the divisor associated to the hyperplane $x=y$ (that is, to the hyperplane $z_2=z_3$ in $C_n$). Setting $t=x-y$ one gets: $$\nabla_4=t_{12}d\log x+t_{13}d\log(x-t)+t_{24}d\log(x-1)+t_{34}d\log(x-t-1)+t_{23}d \log t$$

Then we can define a connection on $D$ it by setting $t=0$ in the holomorphic part of $\nabla_4$: $$\nabla_D= t_{12}d\log x+t_{13}d\log x+t_{24}d\log(x-1)+t_{34}d\log(x-1)$$

i.e. $$\nabla_D= (t_{12}+t_{13})d\log x+(t_{24}+t_{34})d\log(x-1)=\nabla_3^{1,23,4}$$

Compatibility of the KZ connection with operadic composition - MathOverflow

Compatibility of the KZ connection with operadic composition

Answer by Adrien for Compatibility of the KZ connection with operadic composition

Example