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Consider the Lie algebra $sl_2$ with the standard basis $(e,f,h),$ where \begin{equation*}\label{sl2} [h,e]=2\,e, [h,f]=-2\,f,[e,f]=h. \end{equation*}

Let $V$ be finite-dimensional $sl_2$-module and let we know that element $e$ acts on $V$ as linear operator with a matrix $E$ (in some fixed basis).

Question. How one may find the matrix $F,H$ which correspond to elements $f$ and $h.?$

It is posible to recover $F,H$ from $E?$

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4 Answers 4

up vote 2 down vote accepted

Edited in light of clarifications made by OP:

Given a nilpotent matrix $E$ acting on a finite dimensional vector space $V$, it is always possible to extend it to a representation of $sl_2$ in such a way that it represents $e$. The extension is almost never unique: conjugating the representing matrices $F$ and $H$ by anything in the centralizer of $E$ gives a new extension.

The existence statement is the Jacobson-Morozov lemma (part of whose standard proof is reproduced in another answer) applied to the semisimple Lie algebra $sl(V)$. See Proposition 2 of section 2 of paragraph 11 of Bourbaki's "Lie Groups and Lie Algebras", Chapter VIII (see the Corollary following the Proposition for the extent to which uniqueness is true: basically, up to conjugacy).

On the other hand, if you have some additional rigid structure, there might be a unique extension. For instance, if you know a contravariant form and have an orthonormal basis at your disposal then $F$ is the transpose of $E$ (written in terms of the given orthonormal basis) and $H$ is determined by $H=[E,F]$.

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Thanks for good answer and right direction. Please some clarify the " you need a bit more structure". In other words - Any sl_2 representation define representation of maximal unipotent subalgebra $u_2$(generated by $e$.) Do you mean that the inverse is wrong and a representation of $u_2$ can not lift to representation $sl_2$? –  Melania Oct 29 '10 at 16:22

This is all very classical and dealt with in numerous books on Lie algebras. For instance, analyzing finite dimensional representations depends on complete reducibility and the (easy) explicit construction of all irreducibles. The dependence of h and f on e (up to certain conjugacy conditions) is also standard in treatments of $\mathfrak{sl}_2$ embeddings. The question isn't really about "linear algebra" as tagged, but rather belongs to the general theory of finite dimensional representations of semisimple Lie algebras over fields like $\mathbb{C}$. There is no extra benefit in treating it just as a matrix computation problem, especially because E doesn't strictly determine the choice of F and H in matrix terms.

ADDED: I still don't understand what is actually being asked here, I guess. The theory of nilpotent elements in a (complex) semisimple Lie algebra, including $\mathfrak{sl}_2$ triples and their finite dimensional representations, is well understood.

On the other hand, if one starts with a (nonzero) nilpotent matrix E, there are many possible choices of a matrix F and resulting semisimple matrix H. Concretely, one could transform E by similarity to its rational canonical form (computable if given enough time), then get an F from the transpose by applying the inverse similarity. As noted in some answers, conjugating F by a matrix which commutes with E would give a possibly new choice. Why would one want to do all of this? Note that a resulting $\mathfrak{sl}_2$ representation would typically be reducible. And at the outset, it's hard to recognize that a large matrix E is truly nilpotent, due to possible roundoff errors, etc.

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Thanks. So, the matrix $E$ doest determine $F$ and $H$ in unique way? It's good. But how to find any such $F$ and $H$ in the general case when the $E$ is not its Jordan form? –  Melania Oct 29 '10 at 15:09
    
Is it possible to determine any $F$ from $E$ without similarity $E$ to its canonical form? Is there a simple trick to do it? –  Melania Oct 29 '10 at 18:05
    
Simple trick? Not that I know of, for a randomly chosen nilpotent matrix of arbitrarily large size (and keeping mind my comment about the difficulty of knowing in practice that such a matrix is nilpotent). Your question still seems out of focus to me, given the lack of uniqueness in the choice of $F$. –  Jim Humphreys Nov 9 '10 at 15:35

The element $E$ is nilpotent; if the representation is irreducible, it has the maximal possible rank. Denote the rank $k$ so that the dimension of the underlying vector space $V$ will be $k+1$. There is a basis $(v_0,\ldots,v_k)$ of $V$ such that

  1. $F$ will be a Jordan cell i.e. $Fv_i=v_{i-1}, Fv_0=0$. $v_{k}$ is the highest weight vector.

  2. $Hv_i=(2i-k)v_i$.

  3. $Ev_i=i(k-i+1)v_{i+1},Ev_k=0$.

These formulae in some form or another can be found in many sources (e.g in Serre's Lie algebra book).

[upd: one can reconstruct all this from $F$ given in some basis by setting $v_0$ to be any nonzero vector in $Im F^k=\ker F$ and then defining $v_1,\ldots, v_k$ by formula 3 above. With this definition formulae 1 and 2 will be true as well.]

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Yes, But I mean other problem. Let $e$ acts as the following nilpotent matrix $$ E:= \begin{pmatrix} 1 & 1&-1 \\ -3 & -3 & 3 \\ -2 & -3 &3 \end{pmatrix} $$ What is the corresponding matrix to the actions $f$ and $h.?$ –  Melania Oct 29 '10 at 14:43
    
@Melania: This matrix E isn't nilpotent, which confuses me further. –  Jim Humphreys Oct 29 '10 at 19:15
    
$$ E:= \begin{pmatrix} 1 & 1&-1 \\ -3 & -3 & 3 \\ -2 & -2 &-2 \end{pmatrix} $$ –  Melania Oct 30 '10 at 9:51
    
@Melania: Also obviously not nilpotent. –  Jim Humphreys Oct 30 '10 at 18:58
    
I am sorry $$ E:= \begin{pmatrix} 1 & 1&-1 \\ -3 & -3 & 3 \\ -2 & -2 &2 \end{pmatrix} $$ –  Melania Oct 31 '10 at 7:52

For every irreducible finite-dimensional representation there's a basis in which the matrices take particularly simple form (this should be in most of the introductory representation theory books). You can cook up you matrices F,H from there by an appropriate change of basis.

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Yes, I know a solution for simple suitable basis. I am interested to find solution in most general form. What is the matrix $F$ - for example $F=(-E)^T$ or something like to it. –  Melania Oct 29 '10 at 13:05
    
Then I suggest you try to find such a formula for F in the simple basis at first. –  Vít Tuček Oct 29 '10 at 13:33
    
yes, but I don't want to reduce $E$ to its Jordan form. Please see below my answer to algori. –  Melania Oct 29 '10 at 14:44

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