# The Unapologetic Mathematician

## Isomorphism Theorems for Lie Algebras

The 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 theorem, just like for rings. Specifically, if $\phi:L\to L'$ is any homomorphism of Lie algebras then we can factor it as follows: $\displaystyle L\twoheadrightarrow L/\mathrm{Ker}(\phi)\cong\mathrm{Im}(\phi)\hookrightarrow L'$

That is, first we project down to the quotient of $L$ by the kernel of $\phi$, then we have an isomorphism from this quotient to the image of $\phi$, followed by the inclusion of the image as a subalgebra of $L'$.

There are actually two more isomorphism theorems which I haven’t made much mention of, though they hold in other categories as well. Since we’ll have use of them in our study of Lie algebras, we may as well get them down now.

The second isomorphism theorem says that if $I\subseteq J$ are both ideals of $L$, then $J/I$ is an ideal of $L/I$. Further, there is a natural isomorphism $(L/I)/(J/I)\cong L/J$. Indeed, if $x+I\in L/I$ and $j+I\in J/I$, then we can check that $\displaystyle[x+I,j+I]=[x,j]+I\in J/I$

so $J/I$ is an ideal of $L/I$. As for the isomorphism, it’s straightforward from considering $I$ and $J$ as vector subspaces of $L$. Indeed, saying $x+I$ and $y+I$ are equivalent modulo $J/I$ in $L/I$ is to say that $(x-y)+I\in J/I$. But this means that $x-y=j$ for some $j\in J$, so $x$ and $y$ are equivalent modulo $J$ in $L$.

The third isomorphism theorem states that if $I$ and $J$ are any two ideals of $L$, then there is a natural isomorphism between $(I+J)/J$ and $I/(I\cap J)$ — we showed last time that both $I+J$ and $I\cap J$ are ideals. To see this, take $i_1+j_1$ and $i_2+j_2$ in $I+J$ and consider how they can be equivalent modulo $J$. First off, $j_1$ and $j_2$ are immediately irrelevant, so we may as well just ask how $i_1$ and $i_2$ can be equivalent modulo $J$. Well, this will happen if $i_1-i_2\in J$, but we know that their difference is also in $I$, so $i_1-i_2\in I\cap J$.

August 15, 2012 Posted by | Algebra, Lie Algebras | 1 Comment