LieAlgebraMap * LieAlgebraMap -- composition of homomorphisms

Operator: *
Usage:

h = f*g
Inputs:
- f, an instance of the type LieAlgebraMap, a homomorphism from $M$ to $L$
- g, an instance of the type LieAlgebraMap, a homomorphism from $N$ to $M$
Outputs:
- h, an instance of the type LieAlgebraMap, the composition from $N$ to $L$

Description

If $f$ or $g$ is not well defined, then it may happen that $f*g$ is well defined and not equal to the map $x \ \to\ f(g(x))$ but equal to the map induced by the values $f(g(gen))$ where $gen$ is a generator for $M$. Here is an example of this fact for rings.

i1 : R = QQ[x]

o1 = R

o1 : PolynomialRing

i2 : S = R/(x*x)

o2 = S

o2 : QuotientRing

i3 : f = map(R,S)

o3 = map (R, S, {x})

o3 : RingMap R <-- S

i4 : g = map(S,R)

o4 = map (S, R, {x})

o4 : RingMap S <-- R

i5 : h = f*g

o5 = map (R, R, {x})

o5 : RingMap R <-- R

i6 : isWellDefined f

o6 = false

i7 : isWellDefined h

o7 = true

i8 : use R

o8 = R

o8 : PolynomialRing

i9 : h(x*x)

      2
o9 = x

o9 : R

i10 : f(g(x*x))

o10 = 0

o10 : R

i11 : L = lieAlgebra{a,b}

o11 = L

o11 : LieAlgebra

i12 : f = map(L,L,{b,-a})

o12 = f

o12 : LieAlgebraMap

i13 : describe(f*f+id_L)

o13 = a => 0
      b => 0
      source => L
      target => L

Ways to use this method:

LieAlgebraMap * LieAlgebraMap -- composition of homomorphisms

The source of this document is in GradedLieAlgebras/doc.m2:2635:0.