Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.