# Definition of a “root” of a Lie Algebra

I am using the notation that $g$ is the Lie algebra of the Lie group $G$ and $T$ is the maximal torus of $G$ and $t$ is the Lie algebra of $T$ (and hence $t$ is the Cartan subalgebra of $g$). A superscript of $^*$ would indicate a dual vector space.

From the literature I seem to get two "different" ways of thinking about roots of a Lie Algebra,

1. They are the non-trivial weights of the adjoint representation of $G$ on $g$ i.e they are the non-trivial representations of $T$ that occur when the Adjoint representation of $G$ on $g$ is restricted to $T$.

2. They are the elements in $t^*$ which need to act on the elements of $t$ to give the eigen-values of the commutator/adjoint action of the elements of $t$ (complexified) on $g/t$ (complexified).

Some aspects of this which are not clear to me are,

• Its not clear that the above two pictures are "equal" but I guess one can map from one to the other. But is there an explicit bijection between the two descriptions above?

• Given the structure of representations of a torus I guess here it follows that in the second picture the elements of $t^*$ that are talked of are precisely those that map the $\mathbb{Z}$-lattice of $t$ to $\mathbb{Z}$. But I would like to know of a clear proof.

• To do the second picture was it necessary to complexify $t$ and $\cal{g}/t$?

When one has complexified the vector spaces then how does one do the thinking in the language of $\mathbb{Z}$-lattice of $t$ being mapped to $\mathbb{Z}$? Is there a canonical way to embed the lattice?

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A detailed answer to this is probably out of the scope of a math.SE answer and more in the line of a chapter in a book. Have you looked in the standard textbooks on the subject? –  Mariano Suárez-Alvarez Jan 9 '12 at 1:56
@Mariano I have been looking through some lecture notes and books for this topic. May be a brief sketch of the idea that I am missing will also help. –  Anirbit Jan 9 '12 at 18:41

A quick answer: the root in the second picture is the derivative (map induced on tangent spaces) of the first picture. You should think of the root as a map that sends elements of $t$ to the corresponding eigenvalue. And you need to complexify in order to ensure the existence of eigenvalues. I may explain more later, but I hope this helps.
Thanks for the reply. I guess when you say "corresponding eigenvalue" you mean a map which when evaluated on the certain element of $t$ in question gives the eigenvalue of the adjoint action. I can see that derivative picture but is that obviously one-to-one? But there is this issue of some people like to work with "real" roots and some with "complex" roots - depending on whether they are thinking of the root as mapping $t$ to $\mathbb{R}$ or as mapping $t_{\mathbb{C}}$ to $\mathbb{C}$. It would be helpful if you can connect these different things. –  Anirbit Jan 10 '12 at 21:56
Also in which which picture or with/without complexification is it legitimate to talk of the decomposition of $g$ as $g=t\oplus _ \alpha g_\alpha$ ? –  Anirbit Jan 10 '12 at 21:57