Now we consider Kirillov-Kostant Poisson brackets of the regular dual of the semi-direct product of Virasoro Lie algebra with the Affine Kac–Moody Lie algebra. Let 𝓚 be a Lie algebra with a non-degenerated bilinear form 〈., .〉. A function *f* : 𝓚 → ℝ is called regular at *x* ϵ 𝓚 if there exists an element ∇ *f* (*x*) such that

#### Proof

For any regular function *f*(*u*, *a*, *ξ*, *β*) from $\begin{array}{}\stackrel{~}{Vect({S}^{1})}\oplus \stackrel{~}{\mathcal{L}\mathcal{G}}\end{array}$ to ℝ let us define a regular function *f̂* from 𝓢𝓤(𝓖)′ to ℝ by *f̂* (*u*, *a*, *ξ*, *β*) = *f*(*u*′, *a*, *ξ*, *β*). For *f*(*u*, *a*, *ξ*, *β*) a function on 𝓢𝓤(𝓖) or ($\begin{array}{}\stackrel{~}{Vect({S}^{1})}\oplus \stackrel{~}{\mathcal{L}\mathcal{G}}\end{array}$), let us denote by *f*_{u} the function of the variables *a* and *β* that we get when we fix *u* and *ξ*. Let us denote *f*_{a} the function of the variables *u* and *ξ* that we get when we fix *a* and *β*. With the previous notations, one has for *β* ≠ 0 for the bracket {., .}^{U} = {., .}^{𝓢𝓤(𝓖)}

$$\begin{array}{}\{f,\hat{g}{\}}^{U}(u,a,\xi ,\beta )=[\{{f}_{u},{g}_{u}{\}}^{U}+\{{f}_{u},{g}_{a}{\}}^{U}+\{{f}_{a},{g}_{u}{\}}^{U}+\{{f}_{a},{g}_{a}{\}}^{U}](u,a,\xi ,\beta ),\end{array}$$

and for the bracket $\begin{array}{}\{.,.{\}}^{V\mathcal{L}\mathcal{G}}=\{.,.{\}}^{\stackrel{~}{Vect({S}^{1})}\oplus \stackrel{~}{\mathcal{L}\mathcal{G}}}\end{array}$ we have

$$\begin{array}{}\{f,g{\}}^{V\mathcal{L}\mathcal{G}}(u,a,\xi ,\beta )=\{{f}_{u},{g}_{u}{\}}^{V\mathcal{L}\mathcal{G}}+\{{f}_{a},{g}_{a}{\}}^{V\mathcal{L}\mathcal{G}}.\end{array}$$

Then the map *π*_{1} from 𝓢𝓤(𝓖) onto $\begin{array}{}\stackrel{~}{Vect({S}^{1})}\end{array}$ which sends (*u*, *a*, *ξ*, *β*) onto (*u*′, *ξ*) is a Poisson morphism. The map *π*_{2} from 𝓢𝓤(𝓖) onto $\begin{array}{}\stackrel{~}{\mathcal{L}\mathcal{G}}\end{array}$ which sends (*u*, *a*, *ξ*, *β*) to ( *a*, *β*) is a Poisson morphism. For any regular function *f* on $\begin{array}{}\stackrel{~}{Vect({S}^{1})}\end{array}$ and any regular function *G* on $\begin{array}{}\stackrel{~}{\mathcal{L}\mathcal{G}}\end{array}$ we have

$$\begin{array}{}\{{\pi}_{1}^{\ast}f,{\pi}_{2}^{\ast}g{\}}_{U}=0.\end{array}$$

Indeed, for *i* = 1, 2, $\begin{array}{}{\displaystyle ({\delta}_{a}-\frac{a}{\beta}{\delta}_{u}){f}_{i}(\hat{u},\xi )=0.}\end{array}$ We have:

$$\begin{array}{}{\displaystyle \{{f}_{1}(\hat{u},\xi ),{f}_{2}(\hat{u},\xi ){\}}_{\xi ,\beta}^{U}(u,a,\xi ,\beta )}& =\mathcal{J}([\xi (\delta {f}_{1,u}(\hat{u}),\xi {)}_{xxx}\phantom{\rule{thickmathspace}{0ex}}\delta {f}_{2,u}(\hat{u},\xi )+2(\delta {f}_{1,u}(\hat{u},\xi ){)}_{x}\phantom{\rule{thickmathspace}{0ex}}\delta {f}_{2,u}(\hat{u},\xi )\phantom{\rule{thickmathspace}{0ex}}u\\ & +\delta {f}_{1,u}(\hat{u},\xi )\phantom{\rule{thickmathspace}{0ex}}{u}_{x}\phantom{\rule{thickmathspace}{0ex}}\delta {f}_{2,u}(\hat{u},\xi )-{\beta}^{-1}(\delta {f}_{1,u}(\hat{u},\xi {)}_{x})\parallel a{\parallel}^{2}\delta {f}_{2,u}(\hat{u},\xi )\\ & -\u3008(\delta {f}_{1,u}(\hat{u})a/\beta ,\xi {)}_{x},\delta {f}_{2,u}(\hat{u},\xi )\u3009+{\beta}^{-1}\u3008(\delta {f}_{1,u}(\hat{u})a,\xi {)}_{x},\delta {f}_{2,u}(\hat{u},\xi )a\u3009]).\end{array}$$

This gives

$$\begin{array}{}{\displaystyle \{{f}_{1}(\hat{u},\xi ),{f}_{2}(\hat{u},\xi ){\}}_{\xi ,\beta}^{U}(u,a)=\mathcal{J}(\xi [(\delta {f}_{1,u}(\hat{u},\xi ){)}_{xxx}\delta {f}_{2,u}(\hat{u},\xi )}\\ \\ \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{1em}{0ex}}+2(\delta {f}_{1,u}(\hat{u},\xi ){)}_{x}\delta {f}_{2,u}(\hat{u},\xi )(\hat{u},\xi )+\delta {f}_{1,u}(\hat{u},\xi )(\hat{u}{)}_{x}\delta {f}_{2,u}(\hat{u},\xi )]),\end{array}$$

and

$$\begin{array}{}{\displaystyle \{{f}_{1}(\hat{u},\xi ),{f}_{2}(\hat{u},\xi ){\}}_{\xi ,\beta}^{U}(u,a)=\{{f}_{1},{f}_{2}{\}}_{\xi}^{Vir}(\hat{u},\xi ).}\end{array}$$

Let *g*_{i}(*a*, *β*), *i* = 1, 2 be two regular functions on the affine Kac–Moody algebra. One notes that *δ* *g*_{1,u} = *δ* *g*_{2,u} = 0. Therefore,

$$\begin{array}{}{\displaystyle \{{g}_{1},{g}_{2}{\}}_{\xi ,\beta}^{U}(u,a)=\beta \mathcal{J}(\u3008dx(\delta {g}_{1,a}(a,\beta )),\delta {g}_{2,a}(a,\beta )\u3009+\u3008[a,\delta {g}_{1,a}(a,\beta )],\delta {g}_{2,a}(a,\beta )\u3009).}\end{array}$$

Then,

$$\begin{array}{}{\displaystyle \{{g}_{1},{g}_{2}{\}}^{U}(u,a,\xi ,\beta )=\{f,g{\}}^{\stackrel{~}{\mathcal{L}\mathcal{G}}}(a,\beta ).}\end{array}$$

We have:

$$\begin{array}{}{\displaystyle \{f(\hat{u},\xi ),g(a,\beta ){\}}^{U}=\mathcal{J}(\u3008(\delta {f}_{u}(\hat{u}{)}_{x}\phantom{\rule{thickmathspace}{0ex}}a,\xi ),\delta g,a(a,\beta )\u3009}\\ \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}-\beta dx(\delta {f}_{u}(\hat{u},\xi )a),\delta {g}_{a}(a,\beta )\u3009+[a,\delta {f}_{u}a],\delta {g}_{a}\u3009).\end{array}$$

The sum of the first two terms is equal to 0. The last term is 𝓙(*δ**f*_{u}〈[*a*, *a*], *δ**g*_{a}〉), and is equal to zero. One can proceed similarly for 𝓘̃. □

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