IsPerfect - Maple Help
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DerivedAlgebra

calculate the derived algebra of a LAVF object.

IsPerfect

check if a LAVF object is perfect.

 Calling Sequence DerivedAlgebra ( obj) IsPerfect( obj)

Parameters

 obj - a LAVF object that is a Lie algebra i.e. IsLieAlgebra(obj) returns true, see IsLieAlgebra.

Description

 • Let L be a LAVF object which is a Lie algebra. Then DerivedAlgebra method returns the derived algebra $\left[L,L\right]$ of L, as a LAVF object.
 • Let L be a LAVF object and is Lie algebra. Then IsPerfect(L) returns true if and only if DerivedAlgebra(L) = L.
 • These methods are associated with the LAVF object. For more detail, see Overview of the LAVF object.

Examples

 > $\mathrm{with}\left(\mathrm{LieAlgebrasOfVectorFields}\right):$
 > $\mathrm{Typesetting}:-\mathrm{Settings}\left(\mathrm{userep}=\mathrm{true}\right):$
 > $\mathrm{Typesetting}:-\mathrm{Suppress}\left(\left[\mathrm{\xi }\left(x,y\right),\mathrm{\eta }\left(x,y\right)\right]\right):$
 > $V≔\mathrm{VectorField}\left(\mathrm{\xi }\left(x,y\right)\mathrm{D}\left[x\right]+\mathrm{\eta }\left(x,y\right)\mathrm{D}\left[y\right],\mathrm{space}=\left[x,y\right]\right)$
 ${V}{≔}{\mathrm{\xi }}{}\left(\frac{{ⅆ}}{{ⅆ}{x}}\right){+}{\mathrm{\eta }}{}\left(\frac{{ⅆ}}{{ⅆ}{y}}\right)$ (1)
 > $\mathrm{E2}≔\mathrm{LHPDE}\left(\left[\mathrm{diff}\left(\mathrm{\xi }\left(x,y\right),y,y\right)=0,\mathrm{diff}\left(\mathrm{\eta }\left(x,y\right),x\right)=-\mathrm{diff}\left(\mathrm{\xi }\left(x,y\right),y\right),\mathrm{diff}\left(\mathrm{\eta }\left(x,y\right),y\right)=0,\mathrm{diff}\left(\mathrm{\xi }\left(x,y\right),x\right)=0\right],\mathrm{indep}=\left[x,y\right],\mathrm{dep}=\left[\mathrm{\xi },\mathrm{\eta }\right]\right)$
 ${\mathrm{E2}}{≔}\left[{{\mathrm{\xi }}}_{{y}{,}{y}}{=}{0}{,}{{\mathrm{\eta }}}_{{x}}{=}{-}{{\mathrm{\xi }}}_{{y}}{,}{{\mathrm{\eta }}}_{{y}}{=}{0}{,}{{\mathrm{\xi }}}_{{x}}{=}{0}\right]{,}{\mathrm{indep}}{=}\left[{x}{,}{y}\right]{,}{\mathrm{dep}}{=}\left[{\mathrm{\xi }}{,}{\mathrm{\eta }}\right]$ (2)

We first construct a LAVF object for E(2).

 > $L≔\mathrm{LAVF}\left(V,\mathrm{E2}\right)$
 ${L}{≔}\left[{\mathrm{\xi }}{}\left(\frac{{ⅆ}}{{ⅆ}{x}}\right){+}{\mathrm{\eta }}{}\left(\frac{{ⅆ}}{{ⅆ}{y}}\right)\right]\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{&where}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}\left\{\left[{{\mathrm{\xi }}}_{{y}{,}{y}}{=}{0}{,}{{\mathrm{\xi }}}_{{x}}{=}{0}{,}{{\mathrm{\eta }}}_{{x}}{=}{-}{{\mathrm{\xi }}}_{{y}}{,}{{\mathrm{\eta }}}_{{y}}{=}{0}\right]\right\}$ (3)
 > $\mathrm{IsLieAlgebra}\left(L\right)$
 ${\mathrm{true}}$ (4)
 > $\mathrm{DL}≔\mathrm{DerivedAlgebra}\left(L\right)$
 ${\mathrm{DL}}{≔}\left[{\mathrm{\xi }}{}\left(\frac{{ⅆ}}{{ⅆ}{x}}\right){+}{\mathrm{\eta }}{}\left(\frac{{ⅆ}}{{ⅆ}{y}}\right)\right]\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{&where}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}\left\{\left[{{\mathrm{\xi }}}_{{x}}{=}{0}{,}{{\mathrm{\eta }}}_{{x}}{=}{0}{,}{{\mathrm{\xi }}}_{{y}}{=}{0}{,}{{\mathrm{\eta }}}_{{y}}{=}{0}\right]\right\}$ (5)

Since L and DL are not the same, therefore L is not perfect.

 > $\mathrm{IsPerfect}\left(L\right)$
 ${\mathrm{false}}$ (6)

Compatibility

 • The DerivedAlgebra and IsPerfect commands were introduced in Maple 2020.
 • For more information on Maple 2020 changes, see Updates in Maple 2020.