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Let $G$ be a reductive algebraic group, and let $M$ be a Levi subgroup of $G$. In Urban's paper Eigenvarieties for Reductive Groups, it seems to be assumed that if $G(\mathbb{R})$ and $M(\mathbb{R})$ both have discrete series, then the region of convergence of Eisenstein series for the pair ($G$,$M$) is (up to translation) a union of Weyl chambers. (Otherwise it would not make sense to define the set $W^M_{Eis}$.) This assumption seems to be true in all cases that I have checked but I am not sure how to prove it except by exhaustive case analysis. Is a proof written down anywhere?

EDIT: Maybe this isn't true? If $G=Sp(6)$ with simple roots $(1,-1,0)$, $(0,1,-1)$, $(0,0,2)$ and $M=GL(2) \times GL(1)$ with simple root $(1,-1,0)$, then I think the region of convergence should be the cone whose dual is generated by $(1,-1,0)$, $(1,1,-2)$, $(0,0,1)$ (i. e. the positive root of $M$ and the projections of the other positive roots to the space orthogonal to the roots of $M$). Since $(1,1,-2)$ is not a root, the region is not a union of Weyl chambers. Am I computing this incorrectly?

Let $G$ be a reductive algebraic group, and let $M$ be a Levi subgroup of $G$. In Urban's paper Eigenvarieties for Reductive Groups, it seems to be assumed that if $G(\mathbb{R})$ and $M(\mathbb{R})$ both have discrete series, then the region of convergence of Eisenstein series for the pair ($G$,$M$) is (up to translation) a union of Weyl chambers. (Otherwise it would not make sense to define the set $W^M_{Eis}$.) This assumption seems to be true in all cases that I have checked but I am not sure how to prove it except by exhaustive case analysis. Is a proof written down anywhere?

EDIT: I discussed this with Urban and it seems that the assumption is not true in general. For example, it is not true when $G=Sp(6)$ or $SO(4,3)$ and $M=GL(2) \times GL(1) \subset GL(3)$. However, the assumption does seem to be true when $M$ is maximal (and we still require $G$ and $M$ to have discrete series). I would still be interested in seeing a proof in that case.

Let $G$ be a reductive algebraic group, and let $M$ be a Levi subgroup of $G$. In Urban's paper Eigenvarieties for Reductive Groups, it seems to be assumed that if $G(\mathbb{R})$ and $M(\mathbb{R})$ both have discrete series, then the region of convergence of Eisenstein series for the pair ($G$,$M$) is (up to translation) a union of Weyl chambers. (Otherwise it would not make sense to define the set $W^M_{Eis}$.) This assumption seems to be true in all cases that I have checked but I am not sure how to prove it except by exhaustive case analysis. Is a proof written down anywhere?

EDIT: I think this assumption is actually not true. If $G=Sp(8)$ with simple roots $(1,-1,0,0)$, $(0,1,-1,0)$, $(0,0,1,-1)$, $(0,0,0,2)$ and $M=GL(2) \times GL(2)$ with simple roots $(1,-1,0,0)$ and $(0,0,1,-1)$, then the region of convergence is dual to the cone generated by $(1,-1,0,0)$, $(0,0,1,-1)$, $(0,0,1,1)$, $(1,1,-1,-1)$. The last vector is not a root, so the region is not a union of Weyl chambers. I think the correct statement might be:

If $G$ and $M$ have discrete series, $\lambda$ is a sufficiently regular weight, and $X$ is the region of convergence, then $\sum_{w \in W, w(\lambda) \in X} (-1)^{\ell(w)}$ does not depend on $\lambda$.

Here $W$ is the Weyl group and $\ell(w)$ is the length of the Weyl group element $w$. Is that something that is known?

It is necessary to assume that $G$ has discrete series. If $G=GL(3)$, $M=GL(2) \times GL(1)$, then the region of convergence contains a full Weyl chamber and half of another. I am not sure if it is necessary to assume that M has discrete series.

Let $G$ be a reductive algebraic group, and let $M$ be a Levi subgroup of $G$. In Urban's paper Eigenvarieties for Reductive Groups, it seems to be assumed that if $G(\mathbb{R})$ and $M(\mathbb{R})$ both have discrete series, then the region of convergence of Eisenstein series for the pair ($G$,$M$) is (up to translation) a union of Weyl chambers. (Otherwise it would not make sense to define the set $W^M_{Eis}$.) This assumption seems to be true in all cases that I have checked but I am not sure how to prove it except by exhaustive case analysis. Is a proof written down anywhere?

EDIT: Maybe this isn't true? If $G=Sp(6)$ with simple roots $(1,-1,0)$, $(0,1,-1)$, $(0,0,2)$ and $M=GL(2) \times GL(1)$ with simple root $(1,-1,0)$, then I think the region of convergence should be the cone whose dual is generated by $(1,-1,0)$, $(1,1,-2)$, $(0,0,1)$ (i. e. the positive root of $M$ and the projections of the other positive roots to the space orthogonal to the roots of $M$). Since $(1,1,-2)$ is not a root, the region is not a union of Weyl chambers. Am I computing this incorrectly?

Let $G$ be a reductive algebraic group, and let $M$ be a Levi subgroup of $G$. In Urban's paper Eigenvarieties for Reductive Groups, it seems to be assumed that if $G(\mathbb{R})$ and $M(\mathbb{R})$ both have discrete series, then the region of convergence of Eisenstein series for the pair ($G$,$M$) is (up to translation) a union of Weyl chambers. (Otherwise it would not make sense to define the set $W^M_{Eis}$.) This assumption seems to be true in all cases that I have checked but I am not sure how to prove it except by exhaustive case analysis. Is a proof written down anywhere?

It is necessary to assume that both $G$ and $M$ have discrete series. If $G=GL(3)$, $M=GL(2) \times GL(1)$, then the region of convergence contains a full Weyl chamber and half of another. If $G=Sp(6)$ or $SO(4,3)$ and $M=GL(3)$, then the region of convergence contains three Weyl chambers and parts of two others.

Let $G$ be a reductive algebraic group, and let $M$ be a Levi subgroup of $G$. In Urban's paper Eigenvarieties for Reductive Groups, it seems to be assumed that if $G(\mathbb{R})$ and $M(\mathbb{R})$ both have discrete series, then the region of convergence of Eisenstein series for the pair ($G$,$M$) is (up to translation) a union of Weyl chambers. (Otherwise it would not make sense to define the set $W^M_{Eis}$.) This assumption seems to be true in all cases that I have checked but I am not sure how to prove it except by exhaustive case analysis. Is a proof written down anywhere?

EDIT: I think this assumption is actually not true. If $G=Sp(8)$ with simple roots $(1,-1,0,0)$, $(0,1,-1,0)$, $(0,0,1,-1)$, $(0,0,0,2)$ and $M=GL(2) \times GL(2)$ with simple roots $(1,-1,0,0)$ and $(0,0,1,-1)$, then the region of convergence is dual to the cone generated by $(1,-1,0,0)$, $(0,0,1,-1)$, $(0,0,1,1)$, $(1,1,-1,-1)$. The last vector is not a root, so the region is not a union of Weyl chambers. I think the correct statement might be:

If $G$ and $M$ have discrete series, $\lambda$ is a sufficiently regular weight, and $X$ is the region of convergence, then $\sum_{w \in W, w(\lambda) \in X} (-1)^{\ell(w)}$ does not depend on $\lambda$.

Here $W$ is the Weyl group and $\ell(w)$ is the length of the Weyl group element $w$. Is that something that is known?

It is necessary to assume that $G$ has discrete series. If $G=GL(3)$, $M=GL(2) \times GL(1)$, then the region of convergence contains a full Weyl chamber and half of another. I am not sure if it is necessary to assume that M has discrete series.