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I am trying to understand the notation $\rho^{\vee}(-1)$. Let $T$ be a maximal torus of a semi-simple algebraic group $G$ and $\mathbb{G}_m$ the multiplicative group. I think that $\rho^{\vee}$ is a map $\mathbb{G}_m \to T$ which satisfies $\alpha_i(\rho^{\vee}(s))=s$ for all $s \in \mathbb{G}_m$, where $\alpha_i: T \to \mathbb{G}_m$ are simple roots.

Suppose that $G=SL_n$. Then elements in $T$ are $n \times n$ matrices. The element $\rho^{\vee}(-1) \in T$ is an $n \times n$ matrix. How to write this matrix explicitly? Thank you very much.

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In general $\rho^\vee$ is an element of the co-weight lattice. The notation $\rho^\vee(-1)$ only makes sense if $\rho^\vee$ is in fact in the co-character lattice, i.e. the lattice $\text{Hom}(\mathbb G_m,T)$. This holds, for example, if $G$ is adjoint.

If this condition holds, $\rho^\vee(s)$ is an element of $T$ satisyfing $\alpha(\rho^\vee(s))=s^{\langle \alpha,\rho^\vee\rangle}$ as you said (this holds for all roots). However this condition only determines an element of $T$ up to the center (the center of $G$ is precisely the kernel of all of the roots).

The precise value of $\rho^\vee(-1)$ as a matrix requires some case-by-case analysis, although for most simple groups the center is very small and it isn't hard.

For $SL(n)$ with the usual choice of positive roots $\rho^\vee=\frac12(n-1,n-2,\dots, -n+1)$. This is in the co-character lattice if and only if $n$ is odd. In this case:

$$ \begin{aligned} \rho^\vee(-1)&=\exp(\pi i\rho^\vee)\\ &=(i^{n-1},i^{n-3},\dots, i^{-n+1})\\ &=(-1,1,\dots,1,-1) \end{aligned} $$ If $n$ is even $\rho^\vee(-1)$ simply doesn't make sense.

Addendum: For a complex group you can use the exponential map, which satisfies $\gamma^\vee(e^z)=\exp(z\gamma^\vee)$. An adjoint group is one with trivial center. Equivalently the co-weight lattice is equal to the co-character lattice.

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  • $\begingroup$ thank you very much. Why $\rho^{\vee}(-1) = \exp(\pi i \rho^{\vee})$? $\endgroup$
    – Jianrong Li
    Commented May 9, 2018 at 8:48
  • $\begingroup$ thank you very much. What is the definition of "G is adjoint"? $\endgroup$
    – Jianrong Li
    Commented May 9, 2018 at 13:45
  • $\begingroup$ Adjoint means that $\Delta$ is a basis for $X = \textrm{Hom}_{\textrm{alg-grp}}(T,\mathbf G_m)$. $\endgroup$
    – D_S
    Commented May 9, 2018 at 22:02
  • $\begingroup$ It may be worth being even more explicit about what you said: that, although $\rho^\vee(-1)$ doesn't make sense in $\mathrm{SL}(n)$ for $n$ even, it still makes perfectly good sense as an element of the adjoint quotient $\mathrm{PGL}(n)$, in which it is the image there of the matrix $\operatorname{diag}((-1)^{n/2}, (-1)^{n/2 - 1}, \dotsc, (-1)^{1 - n/2}) \in \mathrm{GL}(n)$. $\endgroup$
    – LSpice
    Commented Nov 30, 2018 at 23:28

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