Zoran Skoda
coproduct for Ugln dual

The realization is

x^ ij=x ij+t kx ik jk= rsx rsϕ rs,ij \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 tt is a small/formal parameter and

ϕ rs,ij=ϕ ij,rs({})=[ rs,x ij+t kx ik jk]==δ i rδ j s+tδ i r js \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}
ϕ ij,rs=[ ij,x rs+t kx rk sk]==δ i rδ j s+tδ i r sj \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(μk,)=e iμkx^(i ij)e iμkx^ P_{ij}(\mu k,\partial) = e^{-i\mu k\hat{x}}(-i\partial_{i j})e^{i\mu k\hat{x}}
dP ijdμ=e iμkx^[ikx^,i ij]e iμkx^ \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 [ ij,x^ rs]=ϕ ij,rs[\partial_{i j},\hat{x}_{r s}] = \phi_{i j,r s} (and k rsk_{r s} are central), then

dP ijdμ= rse iμkx^k rsϕ ij,rs()e iμkx^ \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 iμkx^(equation)e iμkx^e^{-i\mu k\hat{x}}(equation)e^{i\mu k\hat{x}} on e iqxe^{i q x} to obtain

dP ijdμ= rsk rsϕ ij,rs(P) \frac{d P_{i j}}{d\mu} = \sum_{r s} k_{r s} \phi_{i j,r s}(P)
dP ijdμ=k ij+t sk isP sj) \frac{d P_{i j}}{d\mu} = k_{i j} + t \sum_{s} k_{i s} P_{s j})

(or in the matrix form

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

where the argument of PP is (μk,q)(\mu k, q) and the initial condition is

P ij(μ=0)=P ij(0,q)=q ij P_{i j}(\mu = 0)= P_{i j}(0,q) = q_{i j}

For t=0t = 0 we clearly have the solution P ij(μ)=k ij+q ijP_{i j}(\mu) = k_{i j} + q_{i j}. For t0t\neq 0, the matrix substitution R=tPIR = t P-I to get simply the homogeneous matrix equation dR/dμ=tkRd R/d\mu = -t k R (here tt is a scalar, kk and RR are n×nn\times n-matrices and kRk R denotes the matrix product) with the initial condition R(0)=tqIR(0) = t q -I i.e. R(0) ij=tq ijδ ijR(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 jj-s, so one has nn independent systems, one for each jj (what is not the case in 1 where we had a transpose).

So for a fixed jj, call (for emphasis) R ij=:R˜ iR_{i j} =: \tilde{R}_i and we have simply

dR˜ i/dμ=t sk isR˜ s 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 μtkR(0) R = e^{-\mu t k} R(0)

i.e. R ij= sexp(μtk) isR(0) sjR_{i j} = \sum_s exp(-\mu t k)_{i s} R(0)_{s j}. where exp(μtk)exp(-\mu t k) is the matrix exponential.

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

tPI=e μtk(tqI) t P - I = e^{-\mu t k} (t q- I)

hence

P(μ)=e μtkq+Ie μtkt P(\mu) = e^{-\mu t k} q + \frac{I - e^{-\mu t k}}{t}

Note that qq and kk are constant matrices and tt,μ\mu are scalars. For μ=0\mu = 0 we have P(0)=qP(0) = q and for small tt we have P=μk+q+(μ 2k 2/2μkq)t+O(t 2)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=0t = 0 result P=μk+qP = \mu k + q.

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

kIe tkt k \mapsto \frac{I - e^{-t k}}{t}

what gives k1tln(Itk)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+qtkq P(K^{-1}(k),q) = k + q - t k q

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

Δ( ij)= ij1+1 ijit il lj \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 ij\partial_{i j} are the dual variables to x^ ij\hat{x}_{i j} and are interpreted as partial derivatives with respect to commutative x ijx_{i j}. The Kronecker delta is denoted δ ij\delta_{i j}. Introduce a new matrix quantity ZZ by the formula

Z ij:=δ ijit ij Z_{i j} := \delta_{i j} - i t \partial_{i j}

Then

[Z jk,x^ mn]=[Z jk,x mn+tx ml nl]=[it jk,x mn+tx ml nl]=itδ jmZ nk [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 1Z^{-1} one has

[x^ mn,Z ib 1]=Z ij[Z jk,x^ mn]Z kb 1=itZ im 1δ nb [\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

=exp(x ki kj(lnZ 1) ji) \mathcal{F} = exp(x_{k i}\partial_{k j}\otimes (ln Z^{-1})_{j i})

Then if Δ 0\Delta_0 is the undeformed coproduct on C[{ ij}]\mathbf{C}[\{\partial_{i j}\}] (primitive on all partial derivatives) then we claim that the coproduct written above is

Δ( mn)=Δ 0( mn) 1 \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)Yexp(X)= n=0 (adX) nYn! exp(X) Y exp(-X) = \sum_{n= 0}^\infty \frac{(ad X)^n Y}{n!}

what gives for ( mn1) 1\mathcal{F}(\partial_{m n}\otimes 1)\mathcal{F}^{-1} (for the other summand just observe that 1 mn1\otimes \partial_{m n} commutes with \mathcal{F} as ZZ has only derivatives so it commutes with mn\partial_{m n})

exp(x ki kj(lnZ 1) ji)( mn1)exp(x ki kj(lnZ 1) ji)= p=0 (1) pp! ml(lnZ 1) ln p 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 pp, one sees that

[x ki kj(lnZ 1) ji, ml(lnZ 1) ln p]=[x ki kj, ml](lnZ 1) ln p(lnZ 1) ji [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

δ kmδ il kj(lnZ 1) ln p(lnZ 1) ji= mj(lnZ 1) jn p+1 -\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 pp and notice that lnZ 1=lnZ-ln Z^{-1} = ln Z, hence

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

and thus the final result is

Δ 0( mn) 1 = mjZ jn+1 mn = mj(δ jnit mj jn)+1 mn = mn1it mj jn+1 mn\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

Δ(Z ij)=δ ij11it( ij1+1 ijit il lj)=Z ilZ lj \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 ij 1Z_{i j}^{-1} and ln(Z 1) ijln(Z^{-1})_{i j} (understood as formal power series in ij\partial_{i j}), also Δ(Z ij 1)=Z il 1Z lj 1\Delta(Z^{-1}_{i j}) = Z_{i l}^{-1}\otimes Z_{l j}^{-1} and ln(Z 1) ijln(Z^{-1})_{i j} is primitive

Δ(ln(Z 1) ij)=ln(Z 1) ij1+1ln(Z 1) ij. \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)(idΔ 0)=(1)(Δ 0id), (1\otimes\mathcal{F})(id\otimes\Delta_0)\mathcal{F} = (\mathcal{F}\otimes 1)(\Delta_0\otimes id)\mathcal{F},

where Δ 0\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 ngl_n i. e. primitive on vector fields of the form P({x ij}) mnP(\{x_{i j}\})\partial_{m n}.

The key is in the factorization property

(Δ 01)= 13 23 (\Delta_0\otimes 1)\mathcal{F} = \mathcal{F}_{1 3}\mathcal{F}_{2 3}
(1Δ)= 12 13 (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) ijln(Z^{-1})_{i j} is primitive for Δ\Delta and x ki kjx_{k i} \partial_{k j} primitive for Δ 0\Delta_0, and expanding the exponential series.

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

12 13 23 \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.