I am studying representations of complex semisimple Lie algebras, their root system, weight spaces etc.

I am a very beginner so this question is about relating key definitions to one another.

Question 1: How are the concepts of weights and roots related? Do you have any intuitive way to interpret these? I see that roots "live" in the Lie algebra, while weights "live" in its representation space, but I fail to connect this in a bigger picture.

My understanding:

Generally, roots live in an Euclidean vector space, they are like generalized eigenvalues, they span the vector space. The root system generates hyperplanes in the space and if we make reflections along these hyperplanes, these reflections generate the Weyl group of this root system.

If the Euclidean vector space is a semisimple Lie algebra $\mathfrak{g}$ and $\mathfrak{h}$ its cartan subalgebra, the roots of $\mathfrak{g}$ are nonzero elements $\alpha$ such that we can take a nonzero vector $X$ from $\mathfrak{g}$ and any vector $H$ from $\mathfrak{h}$ makes $[X, H] = \alpha(H) X$ true.

Question 2: What is the intuition behind this equality? Can I visualize $X$ and $H$ as some vectors and $\alpha$ also as some vector? Then how can we put $H$ "into" $\alpha$, what does $\alpha(H)$ mean?

As for weights, they live in the representation space of the $\mathfrak{g}$. Weights are linear functionals such that corresponding weight spaces are nonzero. Again, I am not really sure, how to interpret the weight space definition $V_\lambda = \{v \in V \mid \forall H \in \mathfrak{h}: H * v = \lambda(H) * v\}$ and the expression $H * v = \lambda(H) * v$.

Thank you very much for any intuition to this.

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    $\begingroup$ The Lie algebra $\mathfrak{g}$ is itself also a representation, namely the space of the ad-representation. The roots are just the weights of the ad representation. We can also embed them into a Euclidean space in such a way that the inner products reflect structures within the Lie algebra (look up the Killing form). $\endgroup$ Commented Mar 27, 2022 at 13:24
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    $\begingroup$ I think you may be getting confused by the abstract concept of root systems and the root system of a semisimple Lie algebra. Weights and roots are both a kind of generalised eigenvalue. That is they are elements of $\mathfrak{h}^*$ which sends each $H \in \mathfrak{h}$ to its eigenvalue on their corresponding weight/root space. The fact that the roots form an abstract root system in $\mathfrak{h}^*$ is then a theorem. $\endgroup$
    – Callum
    Commented Mar 27, 2022 at 16:25
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    $\begingroup$ As for the understandable problem that on the one hand, the roots are elements of $\mathfrak h^*$ (so you can plug $H$ into it), and on the other hand, we visualize them as vectors (like arrows) in plain old $\mathbb R^n$, compare math.stackexchange.com/q/3312731/96384. $\endgroup$ Commented Mar 27, 2022 at 16:49

1 Answer 1


Actually, the roots live in the dual space of $\mathfrak h$. In other words, roots are linear maps from $\mathfrak h$ into the field $\Bbbk$ over which you are working. They are the elements $\alpha\in\mathfrak h^*\setminus\{0\}$ such that, for some $X\in\mathfrak g$, $(\forall H\in\mathfrak h):[H,X]=\alpha(H)X$. So, basically the roots are the non-zero eigenvalues of the maps$$\begin{array}{ccc}\mathfrak g&\longrightarrow&\mathfrak g\\X&\mapsto&[H,X]\end{array}$$($H\in\mathfrak g$). And the weights have a similar definition, but this time we are working with an arbitrary representation of $\mathfrak g$ (and we don't have the restriction that they cannot be equal to $0$).

  • $\begingroup$ Thank you, this is helpful! One clarifying question: is the $[H, X]$ some "equivalent" of the $H \star v$ in the definition of weights? $\endgroup$ Commented Apr 9, 2022 at 19:43
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    $\begingroup$ Yes; the natural action of $\mathfrak g$ on itself is the one that I've mentioned in my answer:$$\begin{array}{rccc}\operatorname{ad}(H)\colon&\mathfrak g&\longrightarrow&\mathfrak g\\&X&\mapsto&[H,X].\end{array}$$The weights of this action are the roots of $\mathfrak g$ (together with $0$). $\endgroup$ Commented Apr 9, 2022 at 20:13

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