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Suppose I have an irreducible representation $\phi:\mathfrak{sl}_3 \to \mathfrak{gl}(V)$ of the Lie algebra $\mathfrak{sl}_3$. Now I have been asked to express the heighest weight of the corresponding dual representation on $\mathfrak{gl}(V^\ast)$ in terms of those for that on $V$. The definition I have of a weight is: A linear function $\mu : \mathfrak{h} \to \Bbb{C}$ is said to be a weight for $\phi$ if there is $v \in V$ such that

$$\phi(H)v = \mu(H)v$$

for all $H \in \mathfrak{h}$. $\mathfrak{h}$ is the usual Cartan subalgebra of $\mathfrak{sl}_3$. Now I seem to only be able to calculate explicitly the highest weight of $V^\ast$ in the case that I have a concrete representation, such as the standard representation. Furthermore, what does one mean by "express the highest weight of $V^\ast$" in terms of that for $V$?

For example the highest weight of the standard representation of $\mathfrak{sl}_3$ is the linear functional $L_1$ defined by

$$L_i \left( diag(a_1,a_2,a_3) \right) = a_i \hspace{2mm} \text{for $i=1,2,3$}.$$

Here $diag(a_1,a_2,a_3)$ is a matrix in the Cartan subalgebra $\mathfrak{h}$. The highest weight of $V^\ast$ here is now $-L_3$. How do I translate this into "expressing" $-L_3$ in terms of $L_1$? I am quite confused as to what I need to show.

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Do you know about the Weyl group? –  Eric O. Korman Oct 13 '12 at 22:25
    
@Eric No we have not learned about the Weyl group yet. –  fpqc Oct 13 '12 at 22:25
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1 Answer

up vote 2 down vote accepted

Your description of the weights of the standard rep is correct, you might be confused by the fact that $L_1+L_2+L_3=0$ on $\mathfrak{sl}_3$, so $L_1 = -L_2 - L_3$.

One way to approach this would be to write everything in terms of the two fundamental weights $\varpi_i$ (that is, the basis of $\mathfrak{h}^*$ dual to $h_1=\operatorname{diag}(1,-1,0)$ and $h_2=\operatorname{diag}(0,1,-1)$). What they are probably looking for is some statement of the form "if $V$ has highest weight $a\varpi_1 + b \varpi_2$ then $V^*$ has highest weight $f(a,b)\varpi_1 + g(a,b)\varpi_2$".

$v$ in your first displayed equation is called "a weight vector of $V$ with weight $\mu$". You might try proving a result that says that if you take a basis of $V$ consisting of weight vectors and $v$ is an element of this basis with weight $\mu$, then the element $v^*$ of the dual basis of $V^*$ has weight XXX....

Do you know how to draw the irreducible modules on the weight/root lattice? That might help you, since then you can see what the highest/lowest weights look like.

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Thanks for your answer. In your third paragraph, I do know that the weight of $v^\ast$ is the negative of that for $v$. However, it is not automatic that because $v$ is the highest weight vector for $V$ that $v^\ast$ is that for $V^\ast$. As for your third paragraph, I do know how to draw the weight lattice for $V,V^\ast$ and even things like $(\textrm{Sym}^2 V)^\ast$. However, I do not know what a "lowest weight" is. –  fpqc Oct 13 '12 at 22:11
    
@BenjaLim It's not true in general that if $v$ is a highest weight vector for $V$ then $v^*$ is a hwv for $V^*$: indeed the dual of a lowest weight vector will be a hwv and vice versa. A lowest weight vector is one killed by all the $F$s, just as a highest weight vector is killed by the $E$s. If you know that $V$ lives in a hexagon on the weight lattice, you can use that to find what the lowest weight is in terms of the highest. –  mt_ Oct 15 '12 at 7:52
    
I submitted my assignment today, feel free to edit your answer/ post a complete solution. –  fpqc Oct 15 '12 at 8:07
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