Do you know how to compute highest weight vectors in practice?

  • 8
    $\begingroup$ You are going to work a bit harder to turn this into a real question. For example: you want to compute it from what data? $\endgroup$ – Mariano Suárez-Álvarez Feb 22 '10 at 3:07
  • 1
    $\begingroup$ It would help if you gave some idea of what you're trying to do and what you have to start with. The tag [ag.algebraic geometry] and absence of [rt.representation-theory] suggests that maybe you mean something totally different by "highest weight vector" than I would have thought. See mathoverflow.net/howtoask for some ideas about how to improve this question. $\endgroup$ – Anton Geraschenko Feb 22 '10 at 3:12

As Mariano says, it will obviously depend what data you are given.

I am going to assume that you are given a representation $\rho: \mathfrak{g} \to \mathrm{Mat}_{n \times n}$ of your Lie algebra $\mathfrak{g}$. Let $\mathfrak{b}$ be a Borel, with $\mathfrak{n}$ the nilpotent part.

Highest weight vectors $v$ satisfy $\rho(n) \cdot v=0$ for all $n$ in $\mathfrak{n}$, and it is enough to test this condition as $n$ runs through a basis of $\mathfrak{n}$. This is a finite list of linear conditions on $v$, so finding the space of vectors that satisfy them is just linear algebra. Let $V$ be the space of such vectors.

Then the action of $\mathfrak{b}$ on $V$ descends to the quotient $\mathfrak{b}/\mathfrak{n} = \mathfrak{t}$. This is an action of an abelian Lie algebra. Diagonalize it. Explicitly, let $t_1$, $t_2$, ... $t_r$ be a basis of $\mathfrak{t}$. Split $V$ into eigenspaces for the action of $\rho(t_1)$; split those spaces further into eigenspaces for the action of $\rho(t_2)$ and so on. Again, finding the eigenvectors of a matrix is a mechanical computation. At the end of the day, you'll have $V$ decomposed into spaces on which $\mathfrak{t}$ acts by scalars. The highest weight vectors are the vectors in these spaces.

If you are given a representation of the Lie group instead of the Lie algebra, you will want to solve $\rho(N) v = v$ instead of $\rho(n) v=0$, and then diagonalize $\rho(T)$ instead of $\rho(\mathfrak{t})$.

| cite | improve this answer | |
  • 1
    $\begingroup$ In other words, "by applying the definition" :) $\endgroup$ – Mariano Suárez-Álvarez Feb 22 '10 at 3:47
  • 3
    $\begingroup$ Well, yeah. But it's surprising how often people don't get that far, or don't realize that a definition is computably effective. I thought it would be silly not to point out this out. $\endgroup$ – David E Speyer Feb 22 '10 at 3:51
  • $\begingroup$ I know. This can be seen in some cases already in the angst caused upon a student by asking him/her to find eigenvectors of an endomorphism of a vector space! $\endgroup$ – Mariano Suárez-Álvarez Feb 22 '10 at 4:00

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