# Zoran Skoda coproduct for Ugln dual

The realization is

$\hat{x}_{i j} = x_{i j} + t\sum_k x_{i k} \partial_{j k} = \sum_{r s} x_{r s}\phi_{r s,i j}$

where $t$ is a small/formal parameter and

$\phi_{r s,i j} = \phi_{i j,r s}(\{\partial\}) = [\partial_{r s}, x_{i j} + t\sum_k x_{i k}\partial_{j k}] = = \delta^r_i \delta^s_j + t\delta^r_i \partial_{j s}$
$\phi_{i j,r s} = [\partial_{i j}, x_{r s} + t\sum_k x_{r k}\partial_{s k}] = = \delta_i^r \delta_j^s + t\delta^r_i \partial_{s j}$

We calculate

$P_{ij}(\mu k,\partial) = e^{-i\mu k\hat{x}}(-i\partial_{i j})e^{i\mu k\hat{x}}$
$\frac{d P_{i j}}{d\mu} = e^{-i\mu k\hat{x}}[-i k\hat{x},-i\partial_{i j}]e^{i\mu k\hat{x}}$

As we know that $[\partial_{i j},\hat{x}_{r s}] = \phi_{i j,r s}$ (and $k_{r s}$ are central), then

$\frac{d P_{i j}}{d\mu} = \sum_{r s} e^{-i\mu k\hat{x}}k_{r s}\phi_{i j,r s}(\partial)e^{i\mu k\hat{x}}$

Act with $e^{-i\mu k\hat{x}}(equation)e^{i\mu k\hat{x}}$ on $e^{i q x}$ to obtain

$\frac{d P_{i j}}{d\mu} = \sum_{r s} k_{r s} \phi_{i j,r s}(P)$
$\frac{d P_{i j}}{d\mu} = k_{i j} + t \sum_{s} k_{i s} P_{s j})$

(or in the matrix form

$\frac{d P}{d\mu}(k,q) = k + t k P(k,q)$

where the argument of $P$ is $(\mu k, q)$ and the initial condition is

$P_{i j}(\mu = 0)= P_{i j}(0,q) = q_{i j}$

For $t = 0$ we clearly have the solution $P_{i j}(\mu) = k_{i j} + q_{i j}$. For $t\neq 0$, the matrix substitution $R = t P-I$ to get simply the homogeneous matrix equation $d R/d\mu = -t k R$ (here $t$ is a scalar, $k$ and $R$ are $n\times n$-matrices and $k R$ denotes the matrix product) with the initial condition $R(0) = t q -I$ i.e. $R(0)_{i j} = t q_{i j} - \delta_{i j}$. This is now much easier than the one with the transpose I had before.

To relate it to known examples, one sees that the equation in this case, without the transpose, do not mix different $j$-s, so one has $n$ independent systems, one for each $j$ (what is not the case in 1 where we had a transpose).

So for a fixed $j$, call (for emphasis) $R_{i j} =: \tilde{R}_i$ and we have simply

$d \tilde{R}_i/d\mu = -t \sum_s k_{i s}\tilde{R}_s$

what is in a rather standard form for homogeneous systems of first order ODEs with standard exponential matrix solution.

Thus the formal solution is

$R = e^{-\mu t k} R(0)$

i.e. $R_{i j} = \sum_s exp(-\mu t k)_{i s} R(0)_{s j}$. where $exp(-\mu t k)$ is the matrix exponential.

Now this, with $R(0)_{s j} = t q_{s j}-\delta_{s j}$ gives, in matrix form,

$t P - I = e^{-\mu t k} (t q- I)$

hence

$P(\mu) = e^{-\mu t k} q + \frac{I - e^{-\mu t k}}{t}$

Note that $q$ and $k$ are constant matrices and $t$,$\mu$ are scalars. For $\mu = 0$ we have $P(0) = q$ and for small $t$ we have $P = \mu k + q + (\mu^2 k^2/2 - \mu k q) t + O(t^2)$, agreeing with (i.e. with continuous limit to) $t = 0$ result $P = \mu k + q$.

To get the coproduct, one needs to find $P(K^{-1}(k),q)$, where $K^{-1}$ is the inverse of the function $k\mapsto P(k,0)$, and $P(k,q)$ denotes the function $P(\mu = 1)$ calculated above with $k$ and $q$ as above. In our case, we need to invert

$k \mapsto \frac{I - e^{-t k}}{t}$

what gives $k \mapsto -\frac{1}{t} ln(I-t k)$ where we symbolically use the matrix logarithm. Thus we obtain

$P(K^{-1}(k),q) = k + q - t k q$

what gives a simple quadratic coproduct by the correspondence between $P$ and $\Delta$ given in entry coproduct from exponentials. So (up to some rescaling of $t$ by $\pm 1, i$ or so which I am not careful now)

$\Delta(\partial_{i j}) = \partial_{i j}\otimes 1 + 1\otimes \partial_{i j} - i t \partial_{i l}\otimes\partial_{l j}$

with sum over repeated indices and $\partial_{i j}$ are the dual variables to $\hat{x}_{i j}$ and are interpreted as partial derivatives with respect to commutative $x_{i j}$. The Kronecker delta is denoted $\delta_{i j}$. Introduce a new matrix quantity $Z$ by the formula

$Z_{i j} := \delta_{i j} - i t \partial_{i j}$

Then

$[Z_{j k},\hat{x}_{m n}] = [Z_{j k},x_{m n}+ t x_{m l} \partial_{n l}] = [-i t \partial_{j k}, x_{m n}+ t x_{m l} \partial_{n l}] = -i t \delta_{j m} Z_{n k}$

and for the matrix $Z^{-1}$ one has

$[\hat{x}_{m n}, Z^{-1}_{i b}] = Z_{i j} [Z_{j k},\hat{x}_{m n}] Z^{-1}_{k b} = -i t Z^{-1}_{i m} \delta_{n b}$

Now one defines $\mathcal{F}$ by

$\mathcal{F} = exp(x_{k i}\partial_{k j}\otimes (ln Z^{-1})_{j i})$

Then if $\Delta_0$ is the undeformed coproduct on $\mathbf{C}[\{\partial_{i j}\}]$ (primitive on all partial derivatives) then we claim that the coproduct written above is

$\Delta(\partial_{m n}) = \mathcal{F}\Delta_0(\partial_{m n}) \mathcal{F}^{-1}$

To see this one expands the exponents by the Hadamard’s formula for ad

$exp(X) Y exp(-X) = \sum_{n= 0}^\infty \frac{(ad X)^n Y}{n!}$

what gives for $\mathcal{F}(\partial_{m n}\otimes 1)\mathcal{F}^{-1}$ (for the other summand just observe that $1\otimes \partial_{m n}$ commutes with $\mathcal{F}$ as $Z$ has only derivatives so it commutes with $\partial_{m n}$)

$exp(x_{k i}\partial_{k j}\otimes (ln Z^{-1})_{j i})(\partial_{m n}\otimes 1)exp(x_{k' i'}\partial_{k' j'}\otimes (ln Z^{-1})_{j' i'}) = \sum_{p = 0}^\infty \frac{(-1)^p}{p!}\partial_{m l} (ln Z^{-1})^p_{l n}$

because, inductively on $p$, one sees that

$[x_{k i}\partial_{k j}\otimes (ln Z^{-1})_{j i},\partial_{m l}\otimes (ln Z^{-1})^p_{l n}] = [x_{k i}\partial_{k j},\partial_{m l}]\otimes (ln Z^{-1})^p_{l n}(ln Z^{-1})_{j i}$

what equals

$-\delta_{k m}\delta_{i l}\partial_{k j} \otimes (ln Z^{-1})^p_{l n}(ln Z^{-1})_{j i} = -\partial_{m j}\otimes (ln Z^{-1})^{p+1}_{j n}$

Now sum up over $p$ and notice that $-ln Z^{-1} = ln Z$, hence

$\sum_{p=0}^\infty \frac{(-1)^p}{p!} (ln Z^{-1})^p = Z$

and thus the final result is

$\array{ \mathcal{F}\Delta_0(\partial_{m n})\mathcal{F}^{-1} &=& \partial_{m j}\otimes Z_{j n} + 1\otimes \partial_{m n} \\&=& \partial_{m j}\otimes (\delta_{j n} - i t \partial_{m j}\otimes \partial_{j n}) + 1\otimes \partial_{m n} \\&=& \partial_{m n}\otimes 1 - i t \partial_{m j}\otimes \partial_{j n} + 1\otimes \partial_{m n} }$

as claimed.

It follows also that

$\Delta(Z_{i j}) = \delta_{i j} 1\otimes 1 - i t (\partial_{i j}\otimes 1 +1\otimes \partial_{i j} - i t \partial_{i l}\otimes\partial_{l j}) = Z_{i l}\otimes Z_{l j}$

and, hence, for the $Z_{i j}^{-1}$ and $ln(Z^{-1})_{i j}$ (understood as formal power series in $\partial_{i j}$), also $\Delta(Z^{-1}_{i j}) = Z_{i l}^{-1}\otimes Z_{l j}^{-1}$ and $ln(Z^{-1})_{i j}$ is primitive

$\Delta(ln(Z^{-1})_{i j}) = ln(Z^{-1})_{i j}\otimes 1 + 1\otimes ln(Z^{-1})_{i j}.$

We will write below the proof that also $\mathcal{F}$ satisfies the equation for the Drinfel’d cocycle

$(1\otimes\mathcal{F})(id\otimes\Delta_0)\mathcal{F} = (\mathcal{F}\otimes 1)(\Delta_0\otimes id)\mathcal{F},$

where $\Delta_0$ is in the sense of the enveloping algebra of the Lie algebra of vector fields on the dual of the Lie algebra $gl_n$ i. e. primitive on vector fields of the form $P(\{x_{i j}\})\partial_{m n}$.

The key is in the factorization property

$(\Delta_0\otimes 1)\mathcal{F} = \mathcal{F}_{1 3}\mathcal{F}_{2 3}$
$(1\otimes\Delta)\mathcal{F} = \mathcal{F}_{1 2}\mathcal{F}_{1 3}$

which can be directly checked using the fact that $ln(Z^{-1})_{i j}$ is primitive for $\Delta$ and $x_{k i} \partial_{k j}$ primitive for $\Delta_0$, and expanding the exponential series.

Using the factorization property, both sides of the cocycle equation become

$\mathcal{F}_{1 2}\mathcal{F}_{1 3}\mathcal{F}_{2 3}$

Cf. also gl2 linear realization.

category: valenciennes

Last revised on July 17, 2014 at 08:36:53. See the history of this page for a list of all contributions to it.