Embeddings of flag manifolds












7












$begingroup$


Consider the flag manifold $mathbb{F}(a_1,dots,a_k)$ parametrizing flags of type $F^{a_1}subseteqdotssubseteq F^{a_k}subseteq V$ in a vector spaces $V$ of dimension $n+1$, where $F^{a_i}$ is a sub-vector space of dimension $a_i$.



Then $mathbb{F}(a_1,dots,a_k)$ embeds in the product of Grassmannians $G(a_1,V)timesdotstimes G(a_k,V)$ which in turn embeds in $mathbb{P}^{N_1}timesdotstimesmathbb{P}^{N_k}$ via the product of the Plücker embeddings. Now we can embed $mathbb{P}^{N_1}timesdotstimesmathbb{P}^{N_k}$ in a projective space $mathbb{P}^N$ via the Segre embedding.



Finally, we get an embedding $mathbb{F}(a_1,dots,a_k)hookrightarrowmathbb{P}^N$. Is this embedding the minimal rational homogeneous embedding of $mathbb{F}(a_1,dots,a_k)$?










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$endgroup$








  • 2




    $begingroup$
    What @VictorPetrov writes is completely correct. I just want to clarify one point. When you write "minimal", do you mean that the dimension of the projective space is minimal, or do you mean that the ample cone is the translate of the nef cone by the divisor class of this embedding (the embedding is the "vertex" of the ample cone)?
    $endgroup$
    – Jason Starr
    Mar 17 at 11:07
















7












$begingroup$


Consider the flag manifold $mathbb{F}(a_1,dots,a_k)$ parametrizing flags of type $F^{a_1}subseteqdotssubseteq F^{a_k}subseteq V$ in a vector spaces $V$ of dimension $n+1$, where $F^{a_i}$ is a sub-vector space of dimension $a_i$.



Then $mathbb{F}(a_1,dots,a_k)$ embeds in the product of Grassmannians $G(a_1,V)timesdotstimes G(a_k,V)$ which in turn embeds in $mathbb{P}^{N_1}timesdotstimesmathbb{P}^{N_k}$ via the product of the Plücker embeddings. Now we can embed $mathbb{P}^{N_1}timesdotstimesmathbb{P}^{N_k}$ in a projective space $mathbb{P}^N$ via the Segre embedding.



Finally, we get an embedding $mathbb{F}(a_1,dots,a_k)hookrightarrowmathbb{P}^N$. Is this embedding the minimal rational homogeneous embedding of $mathbb{F}(a_1,dots,a_k)$?










share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    What @VictorPetrov writes is completely correct. I just want to clarify one point. When you write "minimal", do you mean that the dimension of the projective space is minimal, or do you mean that the ample cone is the translate of the nef cone by the divisor class of this embedding (the embedding is the "vertex" of the ample cone)?
    $endgroup$
    – Jason Starr
    Mar 17 at 11:07














7












7








7





$begingroup$


Consider the flag manifold $mathbb{F}(a_1,dots,a_k)$ parametrizing flags of type $F^{a_1}subseteqdotssubseteq F^{a_k}subseteq V$ in a vector spaces $V$ of dimension $n+1$, where $F^{a_i}$ is a sub-vector space of dimension $a_i$.



Then $mathbb{F}(a_1,dots,a_k)$ embeds in the product of Grassmannians $G(a_1,V)timesdotstimes G(a_k,V)$ which in turn embeds in $mathbb{P}^{N_1}timesdotstimesmathbb{P}^{N_k}$ via the product of the Plücker embeddings. Now we can embed $mathbb{P}^{N_1}timesdotstimesmathbb{P}^{N_k}$ in a projective space $mathbb{P}^N$ via the Segre embedding.



Finally, we get an embedding $mathbb{F}(a_1,dots,a_k)hookrightarrowmathbb{P}^N$. Is this embedding the minimal rational homogeneous embedding of $mathbb{F}(a_1,dots,a_k)$?










share|cite|improve this question











$endgroup$




Consider the flag manifold $mathbb{F}(a_1,dots,a_k)$ parametrizing flags of type $F^{a_1}subseteqdotssubseteq F^{a_k}subseteq V$ in a vector spaces $V$ of dimension $n+1$, where $F^{a_i}$ is a sub-vector space of dimension $a_i$.



Then $mathbb{F}(a_1,dots,a_k)$ embeds in the product of Grassmannians $G(a_1,V)timesdotstimes G(a_k,V)$ which in turn embeds in $mathbb{P}^{N_1}timesdotstimesmathbb{P}^{N_k}$ via the product of the Plücker embeddings. Now we can embed $mathbb{P}^{N_1}timesdotstimesmathbb{P}^{N_k}$ in a projective space $mathbb{P}^N$ via the Segre embedding.



Finally, we get an embedding $mathbb{F}(a_1,dots,a_k)hookrightarrowmathbb{P}^N$. Is this embedding the minimal rational homogeneous embedding of $mathbb{F}(a_1,dots,a_k)$?







ag.algebraic-geometry projective-geometry homogeneous-spaces flag-varieties embeddings






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edited Mar 17 at 14:56









Michael Albanese

7,77655293




7,77655293










asked Mar 17 at 9:35









gxggxg

1538




1538








  • 2




    $begingroup$
    What @VictorPetrov writes is completely correct. I just want to clarify one point. When you write "minimal", do you mean that the dimension of the projective space is minimal, or do you mean that the ample cone is the translate of the nef cone by the divisor class of this embedding (the embedding is the "vertex" of the ample cone)?
    $endgroup$
    – Jason Starr
    Mar 17 at 11:07














  • 2




    $begingroup$
    What @VictorPetrov writes is completely correct. I just want to clarify one point. When you write "minimal", do you mean that the dimension of the projective space is minimal, or do you mean that the ample cone is the translate of the nef cone by the divisor class of this embedding (the embedding is the "vertex" of the ample cone)?
    $endgroup$
    – Jason Starr
    Mar 17 at 11:07








2




2




$begingroup$
What @VictorPetrov writes is completely correct. I just want to clarify one point. When you write "minimal", do you mean that the dimension of the projective space is minimal, or do you mean that the ample cone is the translate of the nef cone by the divisor class of this embedding (the embedding is the "vertex" of the ample cone)?
$endgroup$
– Jason Starr
Mar 17 at 11:07




$begingroup$
What @VictorPetrov writes is completely correct. I just want to clarify one point. When you write "minimal", do you mean that the dimension of the projective space is minimal, or do you mean that the ample cone is the translate of the nef cone by the divisor class of this embedding (the embedding is the "vertex" of the ample cone)?
$endgroup$
– Jason Starr
Mar 17 at 11:07










2 Answers
2






active

oldest

votes


















9












$begingroup$

In general there is a more efficient way: $a_1,ldots,a_k$ determines a Young diagram, and you can realize the flag variety as the stabilizer of a point in the unique closed orbit of ${mathbb P}(U)$, where $U$ is the representation of $GL(V)$ corresponding to this diagram. Its dimension is given by the "hook formula".






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Am I missing something or should the words "the stabilizer of a point in" be omitted? The flag variety will be the orbit itself while the stabilizer of a point will be the parabolic subgroup.
    $endgroup$
    – imakhlin
    Mar 20 at 1:34










  • $begingroup$
    You are right, I have had in mind the respective parabolic subgroup. The variety itself is just the closed orbit.
    $endgroup$
    – Victor Petrov
    Mar 20 at 6:48



















2












$begingroup$

Victor Petrov essentially answered your question showing that this projective embedding is, in general, not minimal. I'll just try to explain why this other embedding is, in fact, minimal by dimension. (I'm assuming everything is complex.)



First, the embedding. Let $n$ and $a_1,ldots,a_k$ be as in your question, $F=mathbb F(a_1,ldots,a_k)$ and $G=SL_{n+1}$. Consider the dominant $G$-weight $lambda=omega_{a_1}+ldots+omega_{a_k}$ where the $omega_i$ are the fundamental weights. Let $L_lambda$ be the corresponding irreducible representation with highest weight vector $v_lambda$. (Your $F$ is $G/P$ where $P$ is the parabolic subgroup preserving the line $mathbb Cv_lambda$.) Consider the projectivization $mathbb P(L_lambda)$ and the point $mathrm v_lambda$ therein corresponding to the line $mathbb Cv_lambda$. Then, $F$ can be realized as the (closed) orbit $Gmathrm v_lambdasubset mathbb P(L_lambda)$.



Now, the minimality. Suppose we have a minimal projective embedding $iota:Fhookrightarrowmathbb P(U)$. Consider the pullback $mathcal L=iota^*(mathcal O_{mathbb P(U)}(1))$. The minimality implies that $Gamma(F,mathcal L)=U^*$. However, every line bundle on $F$ is $G$-equivariant (see Theorem 1 in http://www.math.harvard.edu/~lurie/papers/bwb.pdf) and every equivariant line bundle with global sections on $F$ is $mathcal L_mu$ for some $mu$ which is a $mathbb Z_{>0}$-linear combination of the $omega_{a_i}$ (by Borel-Weil-Bott, again, see Lurie's text). So $mathcal L=mathcal L_mu$ for some $mu$ but $Gamma(F,mathcal L_mu)=L_mu^*$ and $dim L_mugedim L_lambda$, i.e. $dim Uge dim L_lambda$.






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    9












    $begingroup$

    In general there is a more efficient way: $a_1,ldots,a_k$ determines a Young diagram, and you can realize the flag variety as the stabilizer of a point in the unique closed orbit of ${mathbb P}(U)$, where $U$ is the representation of $GL(V)$ corresponding to this diagram. Its dimension is given by the "hook formula".






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Am I missing something or should the words "the stabilizer of a point in" be omitted? The flag variety will be the orbit itself while the stabilizer of a point will be the parabolic subgroup.
      $endgroup$
      – imakhlin
      Mar 20 at 1:34










    • $begingroup$
      You are right, I have had in mind the respective parabolic subgroup. The variety itself is just the closed orbit.
      $endgroup$
      – Victor Petrov
      Mar 20 at 6:48
















    9












    $begingroup$

    In general there is a more efficient way: $a_1,ldots,a_k$ determines a Young diagram, and you can realize the flag variety as the stabilizer of a point in the unique closed orbit of ${mathbb P}(U)$, where $U$ is the representation of $GL(V)$ corresponding to this diagram. Its dimension is given by the "hook formula".






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Am I missing something or should the words "the stabilizer of a point in" be omitted? The flag variety will be the orbit itself while the stabilizer of a point will be the parabolic subgroup.
      $endgroup$
      – imakhlin
      Mar 20 at 1:34










    • $begingroup$
      You are right, I have had in mind the respective parabolic subgroup. The variety itself is just the closed orbit.
      $endgroup$
      – Victor Petrov
      Mar 20 at 6:48














    9












    9








    9





    $begingroup$

    In general there is a more efficient way: $a_1,ldots,a_k$ determines a Young diagram, and you can realize the flag variety as the stabilizer of a point in the unique closed orbit of ${mathbb P}(U)$, where $U$ is the representation of $GL(V)$ corresponding to this diagram. Its dimension is given by the "hook formula".






    share|cite|improve this answer









    $endgroup$



    In general there is a more efficient way: $a_1,ldots,a_k$ determines a Young diagram, and you can realize the flag variety as the stabilizer of a point in the unique closed orbit of ${mathbb P}(U)$, where $U$ is the representation of $GL(V)$ corresponding to this diagram. Its dimension is given by the "hook formula".







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Mar 17 at 10:08









    Victor PetrovVictor Petrov

    1,24968




    1,24968












    • $begingroup$
      Am I missing something or should the words "the stabilizer of a point in" be omitted? The flag variety will be the orbit itself while the stabilizer of a point will be the parabolic subgroup.
      $endgroup$
      – imakhlin
      Mar 20 at 1:34










    • $begingroup$
      You are right, I have had in mind the respective parabolic subgroup. The variety itself is just the closed orbit.
      $endgroup$
      – Victor Petrov
      Mar 20 at 6:48


















    • $begingroup$
      Am I missing something or should the words "the stabilizer of a point in" be omitted? The flag variety will be the orbit itself while the stabilizer of a point will be the parabolic subgroup.
      $endgroup$
      – imakhlin
      Mar 20 at 1:34










    • $begingroup$
      You are right, I have had in mind the respective parabolic subgroup. The variety itself is just the closed orbit.
      $endgroup$
      – Victor Petrov
      Mar 20 at 6:48
















    $begingroup$
    Am I missing something or should the words "the stabilizer of a point in" be omitted? The flag variety will be the orbit itself while the stabilizer of a point will be the parabolic subgroup.
    $endgroup$
    – imakhlin
    Mar 20 at 1:34




    $begingroup$
    Am I missing something or should the words "the stabilizer of a point in" be omitted? The flag variety will be the orbit itself while the stabilizer of a point will be the parabolic subgroup.
    $endgroup$
    – imakhlin
    Mar 20 at 1:34












    $begingroup$
    You are right, I have had in mind the respective parabolic subgroup. The variety itself is just the closed orbit.
    $endgroup$
    – Victor Petrov
    Mar 20 at 6:48




    $begingroup$
    You are right, I have had in mind the respective parabolic subgroup. The variety itself is just the closed orbit.
    $endgroup$
    – Victor Petrov
    Mar 20 at 6:48











    2












    $begingroup$

    Victor Petrov essentially answered your question showing that this projective embedding is, in general, not minimal. I'll just try to explain why this other embedding is, in fact, minimal by dimension. (I'm assuming everything is complex.)



    First, the embedding. Let $n$ and $a_1,ldots,a_k$ be as in your question, $F=mathbb F(a_1,ldots,a_k)$ and $G=SL_{n+1}$. Consider the dominant $G$-weight $lambda=omega_{a_1}+ldots+omega_{a_k}$ where the $omega_i$ are the fundamental weights. Let $L_lambda$ be the corresponding irreducible representation with highest weight vector $v_lambda$. (Your $F$ is $G/P$ where $P$ is the parabolic subgroup preserving the line $mathbb Cv_lambda$.) Consider the projectivization $mathbb P(L_lambda)$ and the point $mathrm v_lambda$ therein corresponding to the line $mathbb Cv_lambda$. Then, $F$ can be realized as the (closed) orbit $Gmathrm v_lambdasubset mathbb P(L_lambda)$.



    Now, the minimality. Suppose we have a minimal projective embedding $iota:Fhookrightarrowmathbb P(U)$. Consider the pullback $mathcal L=iota^*(mathcal O_{mathbb P(U)}(1))$. The minimality implies that $Gamma(F,mathcal L)=U^*$. However, every line bundle on $F$ is $G$-equivariant (see Theorem 1 in http://www.math.harvard.edu/~lurie/papers/bwb.pdf) and every equivariant line bundle with global sections on $F$ is $mathcal L_mu$ for some $mu$ which is a $mathbb Z_{>0}$-linear combination of the $omega_{a_i}$ (by Borel-Weil-Bott, again, see Lurie's text). So $mathcal L=mathcal L_mu$ for some $mu$ but $Gamma(F,mathcal L_mu)=L_mu^*$ and $dim L_mugedim L_lambda$, i.e. $dim Uge dim L_lambda$.






    share|cite|improve this answer











    $endgroup$


















      2












      $begingroup$

      Victor Petrov essentially answered your question showing that this projective embedding is, in general, not minimal. I'll just try to explain why this other embedding is, in fact, minimal by dimension. (I'm assuming everything is complex.)



      First, the embedding. Let $n$ and $a_1,ldots,a_k$ be as in your question, $F=mathbb F(a_1,ldots,a_k)$ and $G=SL_{n+1}$. Consider the dominant $G$-weight $lambda=omega_{a_1}+ldots+omega_{a_k}$ where the $omega_i$ are the fundamental weights. Let $L_lambda$ be the corresponding irreducible representation with highest weight vector $v_lambda$. (Your $F$ is $G/P$ where $P$ is the parabolic subgroup preserving the line $mathbb Cv_lambda$.) Consider the projectivization $mathbb P(L_lambda)$ and the point $mathrm v_lambda$ therein corresponding to the line $mathbb Cv_lambda$. Then, $F$ can be realized as the (closed) orbit $Gmathrm v_lambdasubset mathbb P(L_lambda)$.



      Now, the minimality. Suppose we have a minimal projective embedding $iota:Fhookrightarrowmathbb P(U)$. Consider the pullback $mathcal L=iota^*(mathcal O_{mathbb P(U)}(1))$. The minimality implies that $Gamma(F,mathcal L)=U^*$. However, every line bundle on $F$ is $G$-equivariant (see Theorem 1 in http://www.math.harvard.edu/~lurie/papers/bwb.pdf) and every equivariant line bundle with global sections on $F$ is $mathcal L_mu$ for some $mu$ which is a $mathbb Z_{>0}$-linear combination of the $omega_{a_i}$ (by Borel-Weil-Bott, again, see Lurie's text). So $mathcal L=mathcal L_mu$ for some $mu$ but $Gamma(F,mathcal L_mu)=L_mu^*$ and $dim L_mugedim L_lambda$, i.e. $dim Uge dim L_lambda$.






      share|cite|improve this answer











      $endgroup$
















        2












        2








        2





        $begingroup$

        Victor Petrov essentially answered your question showing that this projective embedding is, in general, not minimal. I'll just try to explain why this other embedding is, in fact, minimal by dimension. (I'm assuming everything is complex.)



        First, the embedding. Let $n$ and $a_1,ldots,a_k$ be as in your question, $F=mathbb F(a_1,ldots,a_k)$ and $G=SL_{n+1}$. Consider the dominant $G$-weight $lambda=omega_{a_1}+ldots+omega_{a_k}$ where the $omega_i$ are the fundamental weights. Let $L_lambda$ be the corresponding irreducible representation with highest weight vector $v_lambda$. (Your $F$ is $G/P$ where $P$ is the parabolic subgroup preserving the line $mathbb Cv_lambda$.) Consider the projectivization $mathbb P(L_lambda)$ and the point $mathrm v_lambda$ therein corresponding to the line $mathbb Cv_lambda$. Then, $F$ can be realized as the (closed) orbit $Gmathrm v_lambdasubset mathbb P(L_lambda)$.



        Now, the minimality. Suppose we have a minimal projective embedding $iota:Fhookrightarrowmathbb P(U)$. Consider the pullback $mathcal L=iota^*(mathcal O_{mathbb P(U)}(1))$. The minimality implies that $Gamma(F,mathcal L)=U^*$. However, every line bundle on $F$ is $G$-equivariant (see Theorem 1 in http://www.math.harvard.edu/~lurie/papers/bwb.pdf) and every equivariant line bundle with global sections on $F$ is $mathcal L_mu$ for some $mu$ which is a $mathbb Z_{>0}$-linear combination of the $omega_{a_i}$ (by Borel-Weil-Bott, again, see Lurie's text). So $mathcal L=mathcal L_mu$ for some $mu$ but $Gamma(F,mathcal L_mu)=L_mu^*$ and $dim L_mugedim L_lambda$, i.e. $dim Uge dim L_lambda$.






        share|cite|improve this answer











        $endgroup$



        Victor Petrov essentially answered your question showing that this projective embedding is, in general, not minimal. I'll just try to explain why this other embedding is, in fact, minimal by dimension. (I'm assuming everything is complex.)



        First, the embedding. Let $n$ and $a_1,ldots,a_k$ be as in your question, $F=mathbb F(a_1,ldots,a_k)$ and $G=SL_{n+1}$. Consider the dominant $G$-weight $lambda=omega_{a_1}+ldots+omega_{a_k}$ where the $omega_i$ are the fundamental weights. Let $L_lambda$ be the corresponding irreducible representation with highest weight vector $v_lambda$. (Your $F$ is $G/P$ where $P$ is the parabolic subgroup preserving the line $mathbb Cv_lambda$.) Consider the projectivization $mathbb P(L_lambda)$ and the point $mathrm v_lambda$ therein corresponding to the line $mathbb Cv_lambda$. Then, $F$ can be realized as the (closed) orbit $Gmathrm v_lambdasubset mathbb P(L_lambda)$.



        Now, the minimality. Suppose we have a minimal projective embedding $iota:Fhookrightarrowmathbb P(U)$. Consider the pullback $mathcal L=iota^*(mathcal O_{mathbb P(U)}(1))$. The minimality implies that $Gamma(F,mathcal L)=U^*$. However, every line bundle on $F$ is $G$-equivariant (see Theorem 1 in http://www.math.harvard.edu/~lurie/papers/bwb.pdf) and every equivariant line bundle with global sections on $F$ is $mathcal L_mu$ for some $mu$ which is a $mathbb Z_{>0}$-linear combination of the $omega_{a_i}$ (by Borel-Weil-Bott, again, see Lurie's text). So $mathcal L=mathcal L_mu$ for some $mu$ but $Gamma(F,mathcal L_mu)=L_mu^*$ and $dim L_mugedim L_lambda$, i.e. $dim Uge dim L_lambda$.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited 2 days ago

























        answered Mar 20 at 1:23









        imakhlinimakhlin

        1,24311020




        1,24311020






























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