A triple system is said to be a Lie triple system if the trilinear map, denoted [.,.,.], satisfies the following identities: {displaystyle [u,v,w]=-[v,u,w]}[u,v,w]=-[v,u,w] {displaystyle [u,v,w]+[w,u,v]+[v,w,u]=0}[u,v,w]+[w,u,v]+[v,w,u]= {displaystyle [u,v,[w,x,y]]=[[u,v,w],x,y]+[w,[u,v,x],y]+[w,x,[u,v,y]].}[u,v,[w,x,y]]=[[u,v,w],x,y]+[w,[u,v,x],y]+[w,x,[u,v,y]]. The first two identities abstract the skew symmetry and Jacobi identity for the triple commutator, while the third identity means that the linear map Lu,v: V → V, defined by Lu,v(w) = [u, v, w], is a derivation of the triple product. The identity also shows that the space k = span {Lu,v : u, v ∈ V} is closed under commutator bracket, hence a Lie algebra. Writing m in place of V, it follows that {displaystyle {mathfrak {g}}:=koplus {mathfrak {m}}}{displaystyle {mathfrak {g}}:=koplus {mathfrak {m}}} can be made into a {displaystyle mathbb {Z} _{2}}mathbb {Z} _{2}-graded Lie algebra, the standard embedding of m, with bracket {displaystyle [(L,u),(M,v)]=([L,M]+L_{u,v},L(v)-M(u)).}[(L,u),(M,v)]=([L,M]+L_{{u,v}},L(v)-M(u)). The decomposition of g is clearly a symmetric decomposition for this Lie bracket, and hence if G is a connected Lie group with Lie algebra g and K is a subgroup with Lie algebra k, then G/K is a symmetric space.