Is a bracket-preserving map between Lie algebras necessarily Linear? Usually, it is assumed to be linear in advance and the second condition of bracket-preservation males it a Lie homomorphism, by definition.
For all x, x', y in L1, twe have [x+x',y]=[x,y]+[x',y], then T([x+x',y])=T([x,y]+[x',y]), in the other hand T([x+x',y])=[T(x+x'),T(y)]. So, if the definition of T doesn't require linearity (i.e. homomorphism ), then we can't have equality T([x,y]+[x',y])=[T(x+x'),T(y)].
True,but that does not force Linearity of T.It only gives a consequence.
absolutely no, it is necessary, not a consequence. the equality above can't hold.
The answer is NO. Take the a trivial Lie algebra L, that is just a vector space L (over an arbitrary field) with the product [x,y]=0 for all x,y in L. Let a f:L-->L a non-linear mapping satisfying f(0)=0.
Then [f(x),f(y)]=0=f(0)=f([x,y]) for all x,y in L. No "aritmetic" helps there.
Dear Hanifa,
Your calculation is valid,but it does not use linearity ( T(x+x')= Tx+Tx') ! We do have your conclusion without assuming linearity of T.That makes oit all the more interesting!
your condition is about multiplication(commutator operation) not necessary Linearirty
Linearity + preserving multiplication together gives a Lie algebra homomorphism.
Dear Rau,
I let small calculations for the reader. If T is not linear, then , we can find x and x' such that T(x+x') not equal T(x)+T(x'), which yields to [T(x+x'),y] not equal to [T(x),T(y)]+[T(x'),T(y)], then how you can have
T([x,y]+[x',y])=[T(x),T(y)]+[T(x'),T(y)], even linearity holds on the left? I TOLD YOU: linearity of T is NECESSARYYY=means without assumption of linearity, we can't do ANYTHINGGGGGG..
CONCLUSION OF THIS QUESTION: what said Cenap iin his second response is exactly the definition of Lie algebra homomorphism which we can't add in it or remove from it.
Remark: In Cenap's first response he said bi linear, I think he meant brackets operation [. , .] not about the mapping T.
Dear Rau
We need to the multiplication properties
especially scalar multiplication
So it doesnt hold
The answer by Peter Plaumann is perfect. One can construct a nontrivial example combining his example and linear map.
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