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A root datum is given by:

  • A subset $R$ of a free abelian group $M$
  • A subset $C$ of the dual free abelian group Hom$(M,\mathbf{Z})$
  • A bijection between $R$ and $C$

subject to conditions. A root system is given by a

  • A vector space V over the real numbers $\mathbf{R}$
  • A pairing $V \times V \to \mathbf{R}$
  • A subset $R$ of $V$

subject to conditions.

A root datum should determine a pair of root systems, one in $M \otimes \mathbf{R}$ and one in Hom$(M,\mathbf{Z}) \otimes \mathbf{R}$. But what is the scalar product on either of these real vector spaces?

Sweet Potatum
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    Dear Sweet Potatum, Unless I'm blundering, the statement of your last sentence is wrong. If the root datum corresponds to the reductive group $G$, then there will be an associated root system attached to its derived subgroup $\mathcal D G$ (which is semisimple). But the latter will be on the character lattice of the maximal torus in $\mathcal D G$, while the root datum involves the character and cocharacter lattice of the maximal torus in $G$ itself (which is bigger, unless $G$ happened to be semisimple). Regards, – Matt E Apr 29 '14 at 01:59
  • P.S. I recommend working through the examples of $GL_n$ (maybe even for some small explicit values of $n$, such as $n = 2$ and $n = 3$) to see how this works. – Matt E Apr 29 '14 at 02:01
  • The answer lies in"some conditions", which are set up so that the bijection of roots and coroots extends to a linear isomorphism of V and its dual, which then can be promoted to an inner product. – Moishe Kohan Apr 29 '14 at 02:57
  • Matt E: I'm a little bit familiar with these examples. Does it mean that there is no natural pairing at all on the weight space of $\mathrm{GL}_n$, or just no nondegenerate pairing? Studiosus: I think the bijection between roots and coroots usually does not extend to a linear map. – Sweet Potatum Apr 29 '14 at 05:43

1 Answers1

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The conditions defining a root datum $(M,R,M',C)$ (in your notation, $M'=\mathrm{Hom}(M,\mathbb{Z})$, but I prefer to take a perfect pairing $\left<\cdot,\cdot\right> : M'\times M\to\mathbb{Z}$ between the free abelian groups $M'$ and $M$, which is equivalent to an identification of $M'$ with the dual of $M$) immediately give you that $R$ is a (classical) root system in $M_\mathbb{R}:=M\otimes_\mathbb{Z}\mathbb{R}$, and $C$ is a root system (dual to $R$) in $M'_\mathbb{R}:=M'\otimes_\mathbb{Z}\mathbb{R}$. Now, since the group $M$ is assumed to be free, the perfect pairing $\left<\cdot,\cdot\right>$ extends to $\left<\cdot,\cdot\right> : M'_\mathbb{R}\times M_\mathbb{R} \to\mathbb{R}$ and the bilinear form

$$(x,y):=\sum_{\alpha\in R} \left<\alpha^\vee,x\right>\left<\alpha^\vee,y\right>$$

(where of course $\alpha^\vee\in C$ is the image of $\alpha\in R$ under the bijection $R\to C$) certainly defines an inner product on $M_\mathbb{R}$. Moreover, this inner product also makes sense if you just had a Euclidean root system from the beginning, and coincides with the given Euclidean inner product.

Xander Henderson
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Arthur Garnier
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