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Let $T= (\mathbb{C}^*)^2$ be embedded in $GL_2$ along its diagonal entries, and suppose $T$ acts on $M_2(\mathbb{C})$ via conjugation. Denote $\chi_i(g)=z_i$ where $$ g = \left( \begin{array}{cc} z_1 & 0 \\ 0 & z_2 \\ \end{array}\right). $$

Then we have a weight space decomposition of $M_2(\mathbb{C})$:

$$ M_2(\mathbb{C}) = \underbrace{\mathbb{C}\cdot E_{12}}_{\mbox{weight } \chi_1\chi_2^{-1} } \oplus \underbrace{\mathbb{C}\cdot E_{21}}_{\mbox{weight }\chi_1^{-1}\chi_2} \oplus \underbrace{ \left\{ \left( \begin{array}{cc} a & 0 \\ 0 & b \\ \end{array} \right) : a, b \in \mathbb{C}\right\} }_{\mbox{weight } 0} $$ where $E_{ij}$ is in $M_2(\mathbb{C})$ with a 1 in the $(i,j)$-entry and zeros everywhere else.

How does one determine which one is the highest or positive weight? Or it's possible that I am misunderstanding weights with roots; are they related?

When I mean roots, I am thinking of those that arise in root systems.

Added: please feel free to impose any noncanonical ordering on the above weights if there doesn't exist a canonical choice.

Added question: under what conditions or circumstances would there be a canonical choice on the weights so that the highest weight can be determined? Any simple example would be fine.

Thank you.

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Highest and positive weights are not canonical; you need to make certain choices first. What text are you working from? –  Qiaochu Yuan Jul 11 '12 at 19:47
@QiaochuYuan I was just reading some lecture notes on linear algebraic groups. They refer to books by Humphreys, Borel, and Springer. So there is a choice to be made? –  math-visitor Jul 11 '12 at 19:50
Yes. This should be explained in any reference. –  Qiaochu Yuan Jul 11 '12 at 19:54
@QiaochuYuan Please feel free to impose any noncanonical choice/s. The lecture notes basically summarize some key ideas in linear algebraic groups, and doesn't discuss anything about root systems. I made up this problem, just wanting to relate some simple Dynkin diagrams and the above. –  math-visitor Jul 11 '12 at 19:58
It is not clear to me what your background is here. Could you summarize it? –  Qiaochu Yuan Jul 11 '12 at 20:01

1 Answer 1

up vote 2 down vote accepted

You have a Lie group $G=GL(2,\mathbb C)$ with a chosen maximal torus $T$, acting on its Lie algebra $M_2(\mathbb C)$ by conjugation (the adjoint representation of $G$), so the roots are precisely the non-zero weights $\alpha=\chi_1\chi_2^{-1}$ and $\beta=\chi_1^{-1}\chi_2$ (considered as characters of $T$, that is algebraic group morphisms $T\to\mathbb C^\times$). Note that since the group $\mathrm{Hom}(T,\mathbb C^\times)$ of all characters is isomorphic to $\mathbb Z^n$ (here $n=2$), it is customary to write characters additively rather than multiplicatively: $\beta=-\alpha$ rather than $\beta=\alpha^{-1}$, and indeed $\alpha= \chi_1-\chi_2$ rather than $\alpha=\chi_1\chi_2^{-1}$.

In order to give sense to the notions of "positive root" and that of "highest weight" derived from it, one needs a choice between a finite number (here $2$) of possibilities, conveniently embodied by the choice of a Borel subgroup $B$ of $G$ containing $T$. Here the two possible choices are that of the group of upper triangular matrices in $G$ and the group of lower triangular matrices; it is conventional to choose the upper ones. Once that is done, the roots whose root subspace lies in the Lie algebra of $B$ are called positive, and the others negative. Here the root subspace $E_{1,2}$ lie in that Lie algebra, so $\alpha$ is considered positive, and $\beta=-\alpha$ negative. Thereby $\alpha$ is also the highest weight of the adjoint representation (but the adjoint representation is not always a highest weight representation). In general a weight of a representation is a highest weight if all other weights can be obtained from it by subtracting (a multiset of) positive roots.

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Thanks Marc. This is very helpful! –  math-visitor Jul 12 '12 at 9:37
Just one quick question: so assume the above scenario but suppose now $T$ is the maximal torus embedded naturally in $GL_n(\mathbb{C})$ along its diagonal coordinates. Using the same notation as above ($\chi_i(g)=z_i$), we can write down the weight space decomposition with the root $\chi_i-\chi_j$ corresponding to $E_{ij}$ (and we also have the weight space corresponding to the trivial character). Do we still have only two choices to choose a Borel $B$ of $G$ containing $T$? And which one of these $\chi_i-\chi_j$ is the highest weight if we choose the upper triangular matrices? –  math-visitor Jul 12 '12 at 11:56
For your convenience, I've placed each positive root (assuming we have chosen the upper triangular matrices containing $T$) in the entry of the corresponding weight space in $M_n(\mathbb{C})$: $$\left(\begin{array}{ccccc} \;\;\;\; & \chi_1-\chi_2 & \chi_1-\chi_3 & \ldots &\chi_1-\chi_n \\ & & \ddots & & \vdots \\ & & & & \chi_{n-1}-\chi_n \\ & & & & \\ \end{array}\right).$$ I don't see a natural way to choose which one is the highest weight. Is this an example of when you say "the adjoint representation is not always a highest weight representation"? –  math-visitor Jul 12 '12 at 12:24
In $GL(n,\mathbb C)$ there are $n!$ different Borel subgroups contining a fixed maximal torus (in general it is the order of the Weyl group); these are related to the possible choices of an ordering of the standard basis (which basis defines the chosen $T$). The highest root (for the choice of upper triangular matrices) would be $\chi_1-\chi_n$; this is the only one from which the others can be obtained by subtraction of positive roots. The adjoint not having a highest weight occurs for groups like $GL_n\times GL_m$. –  Marc van Leeuwen Jul 12 '12 at 13:23
Thanks, this is very helpful to me too. With SL $\chi_i-\chi_j$ are a basis of $X(T) = \operatorname{Hom}(T,\mathbb{C}^\times)$, but with GL the $\chi_i$ themselves are a basis. What happens to the other dimension? Does it not get a root? Is there a name for the subgroup of $X(T)$ generated by the roots? –  Jack Schmidt Jul 12 '12 at 14:48

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