Questions on the proof of Beilinson-Bernstein localization theorem












2














I am trying to understand the Beilinson-Bernstein localization theorem (following the book by Hotta, Takeuchi and Tanisaki). I got stuck at the following two steps. Any help will be greatly appreciated.



For an integral weight $lambda$, I will write $mathcal L(lambda)$ for the sheaf associated to the line bundle $G times_B mathbb C_{lambda}$ on the flag variety $X=G/B$, where $mathbb C_{lambda}$ means the one-dimensional $B$-representation on which the torus $H$ in $B$ acts by $lambda$. I will write $D_{lambda}$ for the sheaf of differential operators acting on sections of the sheaf $mathcal L(lambda+rho)$, where $rho$ is the Weyl vector.




  1. On page 282 (of the book by HTT), they asserted that if $mathcal M$ is a $D_{lambda}$-module, then $mathcal M otimes_{mathcal O_X} mathcal L(mu)$ is a $D_{lambda+mu}$-module. I am confused by how $D_{lambda+mu}$ acts on $mathcal M otimes_{mathcal O_X} mathcal L(mu)$.

  2. On page 281 they introduced a filtration on the trivial bundle $X times L^-(nu)$, where $L^-(nu)$ stands for the lowest weight $G$-representation with lowest weight $nu$, as follows. First filter $L^-(nu)$ by $L^-(nu)=L^1 supset L^2 supset cdots supset L^r =0$, where each quotient $L^i/L^{i+1}$ is a $B$-representation associated to some weight $mu_i$. Then filter $Xtimes L^-(nu)$ by $X times L^-(nu) = U^1 supset U^2 supset cdots supset U^r$, where $U^i = {(gB, l): l in g.L^i}$. This induces a filtration on $mathcal O_X otimes_{mathbb C} L^-(nu)$ and, then, on $mathcal M otimes_{mathbb C} L^-(nu)$. Is this filtration on $mathcal M otimes_{mathbb C} L^-(nu)$ preserved by the action of the center of the universal enveloping algebra of the Lie algebra $mathfrak g$ of $G$?


Thanks a lot!










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  • For part (1), this is just the fact that $D_{lambda + mu}$ can be viewed as $L(mu) otimes D_{lambda} otimes L(mu)^{*}$. So the way it acts on $mathcal{M} otimes L(mu)$ is by letting the $L(mu)^{*}$ part contract the $L(mu)$ part, then acting by $D_{lambda}$ and then recovering the $L(mu)$ part. This is not entirely rigorous so when I have more time, I can try to elaborate.
    – Siddharth Venkatesh
    Feb 10 '16 at 14:57










  • @SiddharthVenkatesh Thanks! I've already figured out that part. I am still struggling on part 2. Do you have any idea on that?
    – user312073
    Feb 12 '16 at 2:52










  • Not off the top of my head, sorry.
    – Siddharth Venkatesh
    Feb 16 '16 at 21:29
















2














I am trying to understand the Beilinson-Bernstein localization theorem (following the book by Hotta, Takeuchi and Tanisaki). I got stuck at the following two steps. Any help will be greatly appreciated.



For an integral weight $lambda$, I will write $mathcal L(lambda)$ for the sheaf associated to the line bundle $G times_B mathbb C_{lambda}$ on the flag variety $X=G/B$, where $mathbb C_{lambda}$ means the one-dimensional $B$-representation on which the torus $H$ in $B$ acts by $lambda$. I will write $D_{lambda}$ for the sheaf of differential operators acting on sections of the sheaf $mathcal L(lambda+rho)$, where $rho$ is the Weyl vector.




  1. On page 282 (of the book by HTT), they asserted that if $mathcal M$ is a $D_{lambda}$-module, then $mathcal M otimes_{mathcal O_X} mathcal L(mu)$ is a $D_{lambda+mu}$-module. I am confused by how $D_{lambda+mu}$ acts on $mathcal M otimes_{mathcal O_X} mathcal L(mu)$.

  2. On page 281 they introduced a filtration on the trivial bundle $X times L^-(nu)$, where $L^-(nu)$ stands for the lowest weight $G$-representation with lowest weight $nu$, as follows. First filter $L^-(nu)$ by $L^-(nu)=L^1 supset L^2 supset cdots supset L^r =0$, where each quotient $L^i/L^{i+1}$ is a $B$-representation associated to some weight $mu_i$. Then filter $Xtimes L^-(nu)$ by $X times L^-(nu) = U^1 supset U^2 supset cdots supset U^r$, where $U^i = {(gB, l): l in g.L^i}$. This induces a filtration on $mathcal O_X otimes_{mathbb C} L^-(nu)$ and, then, on $mathcal M otimes_{mathbb C} L^-(nu)$. Is this filtration on $mathcal M otimes_{mathbb C} L^-(nu)$ preserved by the action of the center of the universal enveloping algebra of the Lie algebra $mathfrak g$ of $G$?


Thanks a lot!










share|cite|improve this question






















  • For part (1), this is just the fact that $D_{lambda + mu}$ can be viewed as $L(mu) otimes D_{lambda} otimes L(mu)^{*}$. So the way it acts on $mathcal{M} otimes L(mu)$ is by letting the $L(mu)^{*}$ part contract the $L(mu)$ part, then acting by $D_{lambda}$ and then recovering the $L(mu)$ part. This is not entirely rigorous so when I have more time, I can try to elaborate.
    – Siddharth Venkatesh
    Feb 10 '16 at 14:57










  • @SiddharthVenkatesh Thanks! I've already figured out that part. I am still struggling on part 2. Do you have any idea on that?
    – user312073
    Feb 12 '16 at 2:52










  • Not off the top of my head, sorry.
    – Siddharth Venkatesh
    Feb 16 '16 at 21:29














2












2








2







I am trying to understand the Beilinson-Bernstein localization theorem (following the book by Hotta, Takeuchi and Tanisaki). I got stuck at the following two steps. Any help will be greatly appreciated.



For an integral weight $lambda$, I will write $mathcal L(lambda)$ for the sheaf associated to the line bundle $G times_B mathbb C_{lambda}$ on the flag variety $X=G/B$, where $mathbb C_{lambda}$ means the one-dimensional $B$-representation on which the torus $H$ in $B$ acts by $lambda$. I will write $D_{lambda}$ for the sheaf of differential operators acting on sections of the sheaf $mathcal L(lambda+rho)$, where $rho$ is the Weyl vector.




  1. On page 282 (of the book by HTT), they asserted that if $mathcal M$ is a $D_{lambda}$-module, then $mathcal M otimes_{mathcal O_X} mathcal L(mu)$ is a $D_{lambda+mu}$-module. I am confused by how $D_{lambda+mu}$ acts on $mathcal M otimes_{mathcal O_X} mathcal L(mu)$.

  2. On page 281 they introduced a filtration on the trivial bundle $X times L^-(nu)$, where $L^-(nu)$ stands for the lowest weight $G$-representation with lowest weight $nu$, as follows. First filter $L^-(nu)$ by $L^-(nu)=L^1 supset L^2 supset cdots supset L^r =0$, where each quotient $L^i/L^{i+1}$ is a $B$-representation associated to some weight $mu_i$. Then filter $Xtimes L^-(nu)$ by $X times L^-(nu) = U^1 supset U^2 supset cdots supset U^r$, where $U^i = {(gB, l): l in g.L^i}$. This induces a filtration on $mathcal O_X otimes_{mathbb C} L^-(nu)$ and, then, on $mathcal M otimes_{mathbb C} L^-(nu)$. Is this filtration on $mathcal M otimes_{mathbb C} L^-(nu)$ preserved by the action of the center of the universal enveloping algebra of the Lie algebra $mathfrak g$ of $G$?


Thanks a lot!










share|cite|improve this question













I am trying to understand the Beilinson-Bernstein localization theorem (following the book by Hotta, Takeuchi and Tanisaki). I got stuck at the following two steps. Any help will be greatly appreciated.



For an integral weight $lambda$, I will write $mathcal L(lambda)$ for the sheaf associated to the line bundle $G times_B mathbb C_{lambda}$ on the flag variety $X=G/B$, where $mathbb C_{lambda}$ means the one-dimensional $B$-representation on which the torus $H$ in $B$ acts by $lambda$. I will write $D_{lambda}$ for the sheaf of differential operators acting on sections of the sheaf $mathcal L(lambda+rho)$, where $rho$ is the Weyl vector.




  1. On page 282 (of the book by HTT), they asserted that if $mathcal M$ is a $D_{lambda}$-module, then $mathcal M otimes_{mathcal O_X} mathcal L(mu)$ is a $D_{lambda+mu}$-module. I am confused by how $D_{lambda+mu}$ acts on $mathcal M otimes_{mathcal O_X} mathcal L(mu)$.

  2. On page 281 they introduced a filtration on the trivial bundle $X times L^-(nu)$, where $L^-(nu)$ stands for the lowest weight $G$-representation with lowest weight $nu$, as follows. First filter $L^-(nu)$ by $L^-(nu)=L^1 supset L^2 supset cdots supset L^r =0$, where each quotient $L^i/L^{i+1}$ is a $B$-representation associated to some weight $mu_i$. Then filter $Xtimes L^-(nu)$ by $X times L^-(nu) = U^1 supset U^2 supset cdots supset U^r$, where $U^i = {(gB, l): l in g.L^i}$. This induces a filtration on $mathcal O_X otimes_{mathbb C} L^-(nu)$ and, then, on $mathcal M otimes_{mathbb C} L^-(nu)$. Is this filtration on $mathcal M otimes_{mathbb C} L^-(nu)$ preserved by the action of the center of the universal enveloping algebra of the Lie algebra $mathfrak g$ of $G$?


Thanks a lot!







group-theory representation-theory






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asked Feb 9 '16 at 17:32









user312073

111




111












  • For part (1), this is just the fact that $D_{lambda + mu}$ can be viewed as $L(mu) otimes D_{lambda} otimes L(mu)^{*}$. So the way it acts on $mathcal{M} otimes L(mu)$ is by letting the $L(mu)^{*}$ part contract the $L(mu)$ part, then acting by $D_{lambda}$ and then recovering the $L(mu)$ part. This is not entirely rigorous so when I have more time, I can try to elaborate.
    – Siddharth Venkatesh
    Feb 10 '16 at 14:57










  • @SiddharthVenkatesh Thanks! I've already figured out that part. I am still struggling on part 2. Do you have any idea on that?
    – user312073
    Feb 12 '16 at 2:52










  • Not off the top of my head, sorry.
    – Siddharth Venkatesh
    Feb 16 '16 at 21:29


















  • For part (1), this is just the fact that $D_{lambda + mu}$ can be viewed as $L(mu) otimes D_{lambda} otimes L(mu)^{*}$. So the way it acts on $mathcal{M} otimes L(mu)$ is by letting the $L(mu)^{*}$ part contract the $L(mu)$ part, then acting by $D_{lambda}$ and then recovering the $L(mu)$ part. This is not entirely rigorous so when I have more time, I can try to elaborate.
    – Siddharth Venkatesh
    Feb 10 '16 at 14:57










  • @SiddharthVenkatesh Thanks! I've already figured out that part. I am still struggling on part 2. Do you have any idea on that?
    – user312073
    Feb 12 '16 at 2:52










  • Not off the top of my head, sorry.
    – Siddharth Venkatesh
    Feb 16 '16 at 21:29
















For part (1), this is just the fact that $D_{lambda + mu}$ can be viewed as $L(mu) otimes D_{lambda} otimes L(mu)^{*}$. So the way it acts on $mathcal{M} otimes L(mu)$ is by letting the $L(mu)^{*}$ part contract the $L(mu)$ part, then acting by $D_{lambda}$ and then recovering the $L(mu)$ part. This is not entirely rigorous so when I have more time, I can try to elaborate.
– Siddharth Venkatesh
Feb 10 '16 at 14:57




For part (1), this is just the fact that $D_{lambda + mu}$ can be viewed as $L(mu) otimes D_{lambda} otimes L(mu)^{*}$. So the way it acts on $mathcal{M} otimes L(mu)$ is by letting the $L(mu)^{*}$ part contract the $L(mu)$ part, then acting by $D_{lambda}$ and then recovering the $L(mu)$ part. This is not entirely rigorous so when I have more time, I can try to elaborate.
– Siddharth Venkatesh
Feb 10 '16 at 14:57












@SiddharthVenkatesh Thanks! I've already figured out that part. I am still struggling on part 2. Do you have any idea on that?
– user312073
Feb 12 '16 at 2:52




@SiddharthVenkatesh Thanks! I've already figured out that part. I am still struggling on part 2. Do you have any idea on that?
– user312073
Feb 12 '16 at 2:52












Not off the top of my head, sorry.
– Siddharth Venkatesh
Feb 16 '16 at 21:29




Not off the top of my head, sorry.
– Siddharth Venkatesh
Feb 16 '16 at 21:29










1 Answer
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For (1), first note that we have an isomorphism $G/B times L^-(nu) to G times^B L^-(nu)$ given by $(gB,v) to (g, g^{-1}v)$. Under this isomorphism, we can identify $mathcal{U}^i$ with $G times^B L^i$ (following the notation in the proof).



To show that each $mathcal{M} otimes_{mathcal{O}_X}mathcal{V}^i$ is closed under the action of $mathfrak{g}$, consider the map of sheaves $mathcal{M} otimes_{mathcal{O}_X} mathcal{V}^i to mathcal{M} otimes_{mathcal{O}_X}mathcal{L}(mu_i)$ induced by the identity on $mathcal{M}$ and the map $mathcal{V}^i to mathcal{L}(mu_i)$ of sections induced by maps on total spaces $G times^B L^i to G times^B L^i/L^{i + 1}$. This map and the identity map are both maps of $mathfrak{g}$-modules, the latter because the maps $G/B times L^-(nu) to G times^B L^-(nu)$ and $G times^B L^i to G times^B L^i/L^{i + 1}$ are $G$-equivariant.



Furthermore, the kernel of the map $mathcal{M} otimes_{mathcal{O}_X} mathcal{V}^i to mathcal{M} otimes_{mathcal{O}_X}mathcal{L}(mu_i)$ is $mathcal{M} otimes_{mathcal{O}_X} mathcal{V}^{i+1}$. Thus $mathcal{M} otimes_{mathcal{O}_X} mathcal{V}^{i+1}$ is the kernel of a map of $mathfrak{g}$-modules, and therefore is closed under the action of $mathfrak{g}$ (meaning for every open $W subset G/B, (mathcal{M} otimes_{mathcal{O}_X} mathcal{V}^{i+1})(W)$ is closed under the $mathfrak{g}$ action.)






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    For (1), first note that we have an isomorphism $G/B times L^-(nu) to G times^B L^-(nu)$ given by $(gB,v) to (g, g^{-1}v)$. Under this isomorphism, we can identify $mathcal{U}^i$ with $G times^B L^i$ (following the notation in the proof).



    To show that each $mathcal{M} otimes_{mathcal{O}_X}mathcal{V}^i$ is closed under the action of $mathfrak{g}$, consider the map of sheaves $mathcal{M} otimes_{mathcal{O}_X} mathcal{V}^i to mathcal{M} otimes_{mathcal{O}_X}mathcal{L}(mu_i)$ induced by the identity on $mathcal{M}$ and the map $mathcal{V}^i to mathcal{L}(mu_i)$ of sections induced by maps on total spaces $G times^B L^i to G times^B L^i/L^{i + 1}$. This map and the identity map are both maps of $mathfrak{g}$-modules, the latter because the maps $G/B times L^-(nu) to G times^B L^-(nu)$ and $G times^B L^i to G times^B L^i/L^{i + 1}$ are $G$-equivariant.



    Furthermore, the kernel of the map $mathcal{M} otimes_{mathcal{O}_X} mathcal{V}^i to mathcal{M} otimes_{mathcal{O}_X}mathcal{L}(mu_i)$ is $mathcal{M} otimes_{mathcal{O}_X} mathcal{V}^{i+1}$. Thus $mathcal{M} otimes_{mathcal{O}_X} mathcal{V}^{i+1}$ is the kernel of a map of $mathfrak{g}$-modules, and therefore is closed under the action of $mathfrak{g}$ (meaning for every open $W subset G/B, (mathcal{M} otimes_{mathcal{O}_X} mathcal{V}^{i+1})(W)$ is closed under the $mathfrak{g}$ action.)






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      For (1), first note that we have an isomorphism $G/B times L^-(nu) to G times^B L^-(nu)$ given by $(gB,v) to (g, g^{-1}v)$. Under this isomorphism, we can identify $mathcal{U}^i$ with $G times^B L^i$ (following the notation in the proof).



      To show that each $mathcal{M} otimes_{mathcal{O}_X}mathcal{V}^i$ is closed under the action of $mathfrak{g}$, consider the map of sheaves $mathcal{M} otimes_{mathcal{O}_X} mathcal{V}^i to mathcal{M} otimes_{mathcal{O}_X}mathcal{L}(mu_i)$ induced by the identity on $mathcal{M}$ and the map $mathcal{V}^i to mathcal{L}(mu_i)$ of sections induced by maps on total spaces $G times^B L^i to G times^B L^i/L^{i + 1}$. This map and the identity map are both maps of $mathfrak{g}$-modules, the latter because the maps $G/B times L^-(nu) to G times^B L^-(nu)$ and $G times^B L^i to G times^B L^i/L^{i + 1}$ are $G$-equivariant.



      Furthermore, the kernel of the map $mathcal{M} otimes_{mathcal{O}_X} mathcal{V}^i to mathcal{M} otimes_{mathcal{O}_X}mathcal{L}(mu_i)$ is $mathcal{M} otimes_{mathcal{O}_X} mathcal{V}^{i+1}$. Thus $mathcal{M} otimes_{mathcal{O}_X} mathcal{V}^{i+1}$ is the kernel of a map of $mathfrak{g}$-modules, and therefore is closed under the action of $mathfrak{g}$ (meaning for every open $W subset G/B, (mathcal{M} otimes_{mathcal{O}_X} mathcal{V}^{i+1})(W)$ is closed under the $mathfrak{g}$ action.)






      share|cite|improve this answer
























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        For (1), first note that we have an isomorphism $G/B times L^-(nu) to G times^B L^-(nu)$ given by $(gB,v) to (g, g^{-1}v)$. Under this isomorphism, we can identify $mathcal{U}^i$ with $G times^B L^i$ (following the notation in the proof).



        To show that each $mathcal{M} otimes_{mathcal{O}_X}mathcal{V}^i$ is closed under the action of $mathfrak{g}$, consider the map of sheaves $mathcal{M} otimes_{mathcal{O}_X} mathcal{V}^i to mathcal{M} otimes_{mathcal{O}_X}mathcal{L}(mu_i)$ induced by the identity on $mathcal{M}$ and the map $mathcal{V}^i to mathcal{L}(mu_i)$ of sections induced by maps on total spaces $G times^B L^i to G times^B L^i/L^{i + 1}$. This map and the identity map are both maps of $mathfrak{g}$-modules, the latter because the maps $G/B times L^-(nu) to G times^B L^-(nu)$ and $G times^B L^i to G times^B L^i/L^{i + 1}$ are $G$-equivariant.



        Furthermore, the kernel of the map $mathcal{M} otimes_{mathcal{O}_X} mathcal{V}^i to mathcal{M} otimes_{mathcal{O}_X}mathcal{L}(mu_i)$ is $mathcal{M} otimes_{mathcal{O}_X} mathcal{V}^{i+1}$. Thus $mathcal{M} otimes_{mathcal{O}_X} mathcal{V}^{i+1}$ is the kernel of a map of $mathfrak{g}$-modules, and therefore is closed under the action of $mathfrak{g}$ (meaning for every open $W subset G/B, (mathcal{M} otimes_{mathcal{O}_X} mathcal{V}^{i+1})(W)$ is closed under the $mathfrak{g}$ action.)






        share|cite|improve this answer












        For (1), first note that we have an isomorphism $G/B times L^-(nu) to G times^B L^-(nu)$ given by $(gB,v) to (g, g^{-1}v)$. Under this isomorphism, we can identify $mathcal{U}^i$ with $G times^B L^i$ (following the notation in the proof).



        To show that each $mathcal{M} otimes_{mathcal{O}_X}mathcal{V}^i$ is closed under the action of $mathfrak{g}$, consider the map of sheaves $mathcal{M} otimes_{mathcal{O}_X} mathcal{V}^i to mathcal{M} otimes_{mathcal{O}_X}mathcal{L}(mu_i)$ induced by the identity on $mathcal{M}$ and the map $mathcal{V}^i to mathcal{L}(mu_i)$ of sections induced by maps on total spaces $G times^B L^i to G times^B L^i/L^{i + 1}$. This map and the identity map are both maps of $mathfrak{g}$-modules, the latter because the maps $G/B times L^-(nu) to G times^B L^-(nu)$ and $G times^B L^i to G times^B L^i/L^{i + 1}$ are $G$-equivariant.



        Furthermore, the kernel of the map $mathcal{M} otimes_{mathcal{O}_X} mathcal{V}^i to mathcal{M} otimes_{mathcal{O}_X}mathcal{L}(mu_i)$ is $mathcal{M} otimes_{mathcal{O}_X} mathcal{V}^{i+1}$. Thus $mathcal{M} otimes_{mathcal{O}_X} mathcal{V}^{i+1}$ is the kernel of a map of $mathfrak{g}$-modules, and therefore is closed under the action of $mathfrak{g}$ (meaning for every open $W subset G/B, (mathcal{M} otimes_{mathcal{O}_X} mathcal{V}^{i+1})(W)$ is closed under the $mathfrak{g}$ action.)







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        answered 2 days ago









        Tom Gannon

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