# The Unapologetic Mathematician

## The Category of Lie Algebras is (not quite) Abelian

We’d like to see that the category of Lie algebras is Abelian. Unfortunately, it isn’t, but we can come close. It should be clear that it’s an $\mathbf{Ab}$-category, since the homomorphisms between any two Lie algebras form a vector space. Direct sums are also straightforward: the Lie algebra $L\oplus L'$ is the direct sum as vector spaces, with $[l,l']=0$ for $l\in L$ and $l'\in L'$ and the regular brackets on $L$ and $L'$ otherwise.

We’ve seen that the category of Lie algebras has a zero object and kernels; now we need cokernels. It would be nice to just say that if $\phi:L\to L'$ is a homomorphism then $\mathrm{Cok}(\phi)$ is the quotient of $L'$ by the image of $\phi$, but this image may not be an ideal. Luckily, ideals have a few nice closure properties.

First off, if $I$ and $J$ are ideals of $L$, then $[I,J]$ — the subspace spanned by brackets of elements of $I$ and $J$ — is also an ideal. Indeed, we can check that $[[i,j],x]=[[i,x],j]+[i,[j,x]]$ which is back in $[I,J]$. Similarly, the subspace sum $I+J$ is an ideal. And, most importantly for us now, the intersection $I\cap J$ is an ideal, since if $i\in I\cap J$ then both $[i,x]\in I$ and $[i,x]\in J$, so $[i,x]\in I\cap J$ as well. In fact, this is true of arbitrary intersections.

This is important, because it means we can always expand any subset $X\subseteq L$ to an ideal. We take all the ideals of $L$ that contain $X$ and intersect them. This will then be another ideal of $L$ containing $X$, and it is contained in all the others. And we know that this intersection is nonempty, since there’s always at least the ideal $L$.

So while $\mathrm{Im}(\phi)$ may not be an ideal of $L'$, we can expand it to an ideal and take the quotient. The projection onto this quotient will be the largest epimorphism of $L'$ that sends everything in $\mathrm{Im}(\phi)$ to zero, so it will be the cokernel of $\phi$.

Where everything falls apart is normality. The very fact that we have ideals as a separate concept from subalgebras is the problem. Any subalgebra is the image of a monomorphism — the inclusion, if nothing else. But not all these subalgebras are themselves kernels of other morphisms; only those that are ideals have this property.

Still, the category is very nice, and these properties will help us greatly in what follows.

August 14, 2012 - Posted by | Algebra, Lie Algebras

## 3 Comments »

1. […] category of Lie algebras may not be Abelian, but it has a zero object, kernels, and cokernels, which is enough to get the first isomorphism […]

Pingback by Isomorphism Theorems for Lie Algebras « The Unapologetic Mathematician | August 15, 2012 | Reply

2. […] algebras that we want to take care of right up front, and both of them are defined similarly. We remember that if and are ideals of a Lie algebra , then — the collection spanned by brackets of […]

Pingback by Nilpotent and Solvable Lie Algebras « The Unapologetic Mathematician | August 20, 2012 | Reply

3. The category of Lie algebras is very far from abelian. First, homomorphisms between two Lie algebras just do not form a vector space under pointwise addition. Second, the category doesn’t have biproducts; what most people call the direct sum of two Lie algebras is their categorical product, but the categorical coproduct is more complicated (e.g. the coproduct of $n$ copies of the $1$-dimensional Lie algebra is the free Lie algebra on $n$ elements, not the $n$-dimensional abelian Lie algebra). Comment by Qiaochu Yuan | November 7, 2012 | Reply