One universal enveloping algebra
In my search for less abstract or trivial examples of a universal enveloping algebra, I have come upon this one.
Let be a
vector space and
a free associative algebra on
. The mapping
is a vector space morphism
.
can also be viewed as a Lie algebra.
is a part of
; let
be the sub-Lie algebra of
generated by
.
(It appears that, except in trivial cases, is strictly smaller than
; for instance, if we take
as the free associative algebra on
constructed in chapter 18,
, in
it appears that all the
components are antisymmetric.)
We have a natural Lie algebra morphism . That is,
is the canonical injection from
to
.
Proposition: The associative algebra is the universal enveloping algebra of the Lie algebra
.
Proof:
Let us first note that since is the Lie subalgebra of
generated by
, we have
. Let us define
, that is,
is
with its codomain restricted to
. We can then write
.
We must prove that given any associative algebra and any Lie algebra morphism
, there exists one and only one associative algebra morphism
such that
.
So let be an associative algebra and
a Lie algebra morphism
; that is, a linear mapping
such that
.
Then we have vector space morphism
. Since
is a free associative algebra on the vector space
, there exists a unique associative algebra morphism
such that
, that is,
.
Thus and
agree on all elements of
, which is also
, which generates the sub Lie algebra
of
. We wish to show that they actually agree on all elements of
.
Let be the set of all elements of
on which
and
agree. Since
and
are both linear mappings,
is a linear mapping too, even if it may not be an associative algebra or Lie algebra morphism.
is its kernel, and is a sub vector space of
. Furthermore, if
and
agree on vectors
and
of
, then
. Since both
and
are Lie algebra morphisms, this is equal to
, and, since
and
agree on
and
, this expression reduces to zero. Thus
too is an element of
. This completes the proof that
is a sub Lie algebra of
. Since
includes
, and since
is the smallest sub Lie algebra of itself including
, it follows that
, that is, that
and
agree on all
. Since
is their common domain, they are equal:
.
Hence there exists at least one associative algebra , namely
, such that
.
Let us show that such a is unique.
Let be an associative algebra morphism
such that
.
Then , that is,
.
But, because is a free associative algebra on the vector space
, we know that there is a unique associative algebra morphism
having such a property, namely
.
Hence there is one and only one associative algebra morphism such that
.
This shows that is the free associative algebra on the Lie algebra
.