Schur-Weyl duality and q-symmetric functions
Clash Royale CLAN TAG#URR8PPP
Disclaimer: I'm far from an expert on any of the topics of this question. I apologize in advance for any horrible mistakes and/or inaccuracies I have made and I hope that the spirit of the question will still be clear despite them.
The (integral) representation rings of the symmetric groups can be packed together into a hopf algebra $H_1 = oplus_n Rep(Sigma_n)$ where the multiplication (resp. comultiplication) comes from induction (resp. restriction) along $Sigma_n times Sigma_k to Sigma_n+k$. In fact there's a further structure one can put on $H$ corresponding to the inner product of characters and a notion of positivity (all together its sometimes called a "positive self adjoint hopf algebra"), but for simplicity I will disregard this structure in what follows (of course if its not important for the answer that would be great to know).
Its well known that sending the irreducible specht modules to their corresponding schur functions induces an isomorphism of hopf algebras to the (integral) hopf algebra of symmetric functions.
Following the "$mathbbF_1$-philosophy" it is tempting to define a ring of "q-symmetric functions" as the hopf algebra $H_q = oplus_n Rep(GL_n(mathbbF_q))$ equipped with the same structures as above.
Question 1: Is there a hopf algebra over $mathbbZ[q]$ which
specializes at a prime power $q=p^n$ to $H_p^n$ and at $q=1$ to
$H_1$ the classical ring of symmetric functions?
By schur weyl duality we also know that $H_1 cong Rep(GL_infty(mathbbC)):= colim_n Rep(GL_n(mathbbC))$ (at least as rings). It seems natural to ask if there's any form of schur duality going in the other direction.
Question 2: Is there any kind of relationship between the rings $Rep(Sigma_infty) := colim_n Rep(Sigma_n)$ and $oplus_n Rep(GL_n(mathbbC))$?
Question 3: Is there a $mathbbZ[q]$-algebra which specializes to $Rep(GL_infty(mathbbF_q))$ at a prime power $q = p^n$ and to $Rep(Sigma_infty)$ at $q=1$?
rt.representation-theory symmetric-groups symmetric-functions f-1 q-analogs
edited Dec 16 at 21:24
Alexander Chervov
11.1k1260139
asked Dec 16 at 11:21
Saal Hardali
63621567
add a comment |
Disclaimer: I'm far from an expert on any of the topics of this question. I apologize in advance for any horrible mistakes and/or inaccuracies I have made and I hope that the spirit of the question will still be clear despite them.
The (integral) representation rings of the symmetric groups can be packed together into a hopf algebra $H_1 = oplus_n Rep(Sigma_n)$ where the multiplication (resp. comultiplication) comes from induction (resp. restriction) along $Sigma_n times Sigma_k to Sigma_n+k$. In fact there's a further structure one can put on $H$ corresponding to the inner product of characters and a notion of positivity (all together its sometimes called a "positive self adjoint hopf algebra"), but for simplicity I will disregard this structure in what follows (of course if its not important for the answer that would be great to know).
Its well known that sending the irreducible specht modules to their corresponding schur functions induces an isomorphism of hopf algebras to the (integral) hopf algebra of symmetric functions.
Following the "$mathbbF_1$-philosophy" it is tempting to define a ring of "q-symmetric functions" as the hopf algebra $H_q = oplus_n Rep(GL_n(mathbbF_q))$ equipped with the same structures as above.
Question 1: Is there a hopf algebra over $mathbbZ[q]$ which
specializes at a prime power $q=p^n$ to $H_p^n$ and at $q=1$ to
$H_1$ the classical ring of symmetric functions?
By schur weyl duality we also know that $H_1 cong Rep(GL_infty(mathbbC)):= colim_n Rep(GL_n(mathbbC))$ (at least as rings). It seems natural to ask if there's any form of schur duality going in the other direction.
Question 2: Is there any kind of relationship between the rings $Rep(Sigma_infty) := colim_n Rep(Sigma_n)$ and $oplus_n Rep(GL_n(mathbbC))$?
Question 3: Is there a $mathbbZ[q]$-algebra which specializes to $Rep(GL_infty(mathbbF_q))$ at a prime power $q = p^n$ and to $Rep(Sigma_infty)$ at $q=1$?
rt.representation-theory symmetric-groups symmetric-functions f-1 q-analogs
edited Dec 16 at 21:24
Alexander Chervov
11.1k1260139
asked Dec 16 at 11:21
Saal Hardali
63621567
1
Ignoring the Hopf algebra aspects of the question, people do study the representation theory of $GL_n(mathbbF_q)$ as a q-analog of the representation theory of $Sigma_n$. A thing to note immediately is that $GL_n(mathbbF_q)$ has many more irreps that $Sigma_n$. But $GL_n(mathbbF_q)$ has a particularly nice family of irreps called unipotent representations $U^lambda(q)$, which are indexed by partitions of $n$. And the degree of $U^lambda(q)$ is a polynomial in $q$ (the ``fake degree polynomial'') which at $q=1$ becomes $f^lambda$, the degree of the $Sigma_n$ irrep.
– Sam Hopkins
Dec 16 at 14:52
This is probably just showing my ignorance, but please could you explain the colimit you have in mind on the right hand side of $mathrmRep(Sigma_infty) := mathrmcolim_n mathrmRep(Sigma_n)$? Since $oplus_n mathrmRep(mathrmGL_n(mathbbC))$ is isomorphic to the ring of symmetric functions, which is an inverse limit (i.e. a limit, not a colimit), do you expect the required relationship to involve some kind of duality?
– Mark Wildon
Dec 16 at 17:46
2
@MarkWildon I think its more a question of convention than anything else. If I write a sum over all representation ring of symmetric group that means that an element is a finite sum while if I define the ring of symmetriic functions as a limit i get series with an infinite number of terms. There are probably two ways to fix this, one is to take the product in my original definition, the other is taking some kind of colimit in the definition of symmetric functions (as is done in the wikipedia article on symmetric functions).
– Saal Hardali
Dec 16 at 17:53
add a comment |
Disclaimer: I'm far from an expert on any of the topics of this question. I apologize in advance for any horrible mistakes and/or inaccuracies I have made and I hope that the spirit of the question will still be clear despite them.
The (integral) representation rings of the symmetric groups can be packed together into a hopf algebra $H_1 = oplus_n Rep(Sigma_n)$ where the multiplication (resp. comultiplication) comes from induction (resp. restriction) along $Sigma_n times Sigma_k to Sigma_n+k$. In fact there's a further structure one can put on $H$ corresponding to the inner product of characters and a notion of positivity (all together its sometimes called a "positive self adjoint hopf algebra"), but for simplicity I will disregard this structure in what follows (of course if its not important for the answer that would be great to know).
Its well known that sending the irreducible specht modules to their corresponding schur functions induces an isomorphism of hopf algebras to the (integral) hopf algebra of symmetric functions.
Following the "$mathbbF_1$-philosophy" it is tempting to define a ring of "q-symmetric functions" as the hopf algebra $H_q = oplus_n Rep(GL_n(mathbbF_q))$ equipped with the same structures as above.
Question 1: Is there a hopf algebra over $mathbbZ[q]$ which
specializes at a prime power $q=p^n$ to $H_p^n$ and at $q=1$ to
$H_1$ the classical ring of symmetric functions?
By schur weyl duality we also know that $H_1 cong Rep(GL_infty(mathbbC)):= colim_n Rep(GL_n(mathbbC))$ (at least as rings). It seems natural to ask if there's any form of schur duality going in the other direction.
Question 2: Is there any kind of relationship between the rings $Rep(Sigma_infty) := colim_n Rep(Sigma_n)$ and $oplus_n Rep(GL_n(mathbbC))$?
Question 3: Is there a $mathbbZ[q]$-algebra which specializes to $Rep(GL_infty(mathbbF_q))$ at a prime power $q = p^n$ and to $Rep(Sigma_infty)$ at $q=1$?
rt.representation-theory symmetric-groups symmetric-functions f-1 q-analogs
edited Dec 16 at 21:24
Alexander Chervov
11.1k1260139
asked Dec 16 at 11:21
Saal Hardali
63621567
Disclaimer: I'm far from an expert on any of the topics of this question. I apologize in advance for any horrible mistakes and/or inaccuracies I have made and I hope that the spirit of the question will still be clear despite them.
The (integral) representation rings of the symmetric groups can be packed together into a hopf algebra $H_1 = oplus_n Rep(Sigma_n)$ where the multiplication (resp. comultiplication) comes from induction (resp. restriction) along $Sigma_n times Sigma_k to Sigma_n+k$. In fact there's a further structure one can put on $H$ corresponding to the inner product of characters and a notion of positivity (all together its sometimes called a "positive self adjoint hopf algebra"), but for simplicity I will disregard this structure in what follows (of course if its not important for the answer that would be great to know).
Its well known that sending the irreducible specht modules to their corresponding schur functions induces an isomorphism of hopf algebras to the (integral) hopf algebra of symmetric functions.
Following the "$mathbbF_1$-philosophy" it is tempting to define a ring of "q-symmetric functions" as the hopf algebra $H_q = oplus_n Rep(GL_n(mathbbF_q))$ equipped with the same structures as above.
Question 1: Is there a hopf algebra over $mathbbZ[q]$ which
specializes at a prime power $q=p^n$ to $H_p^n$ and at $q=1$ to
$H_1$ the classical ring of symmetric functions?
By schur weyl duality we also know that $H_1 cong Rep(GL_infty(mathbbC)):= colim_n Rep(GL_n(mathbbC))$ (at least as rings). It seems natural to ask if there's any form of schur duality going in the other direction.
Question 2: Is there any kind of relationship between the rings $Rep(Sigma_infty) := colim_n Rep(Sigma_n)$ and $oplus_n Rep(GL_n(mathbbC))$?
Question 3: Is there a $mathbbZ[q]$-algebra which specializes to $Rep(GL_infty(mathbbF_q))$ at a prime power $q = p^n$ and to $Rep(Sigma_infty)$ at $q=1$?
rt.representation-theory symmetric-groups symmetric-functions f-1 q-analogs
rt.representation-theory symmetric-groups symmetric-functions f-1 q-analogs
edited Dec 16 at 21:24
Alexander Chervov
11.1k1260139
asked Dec 16 at 11:21
Saal Hardali
63621567
edited Dec 16 at 21:24
Alexander Chervov
11.1k1260139
asked Dec 16 at 11:21
Saal Hardali
63621567
edited Dec 16 at 21:24
Alexander Chervov
11.1k1260139
edited Dec 16 at 21:24
Alexander Chervov
11.1k1260139
edited Dec 16 at 21:24
Alexander Chervov
11.1k1260139
11.1k1260139
asked Dec 16 at 11:21
Saal Hardali
63621567
asked Dec 16 at 11:21
Saal Hardali
63621567
asked Dec 16 at 11:21
Saal Hardali
63621567
63621567
1
Ignoring the Hopf algebra aspects of the question, people do study the representation theory of $GL_n(mathbbF_q)$ as a q-analog of the representation theory of $Sigma_n$. A thing to note immediately is that $GL_n(mathbbF_q)$ has many more irreps that $Sigma_n$. But $GL_n(mathbbF_q)$ has a particularly nice family of irreps called unipotent representations $U^lambda(q)$, which are indexed by partitions of $n$. And the degree of $U^lambda(q)$ is a polynomial in $q$ (the ``fake degree polynomial'') which at $q=1$ becomes $f^lambda$, the degree of the $Sigma_n$ irrep.
– Sam Hopkins
Dec 16 at 14:52
This is probably just showing my ignorance, but please could you explain the colimit you have in mind on the right hand side of $mathrmRep(Sigma_infty) := mathrmcolim_n mathrmRep(Sigma_n)$? Since $oplus_n mathrmRep(mathrmGL_n(mathbbC))$ is isomorphic to the ring of symmetric functions, which is an inverse limit (i.e. a limit, not a colimit), do you expect the required relationship to involve some kind of duality?
– Mark Wildon
Dec 16 at 17:46
2
@MarkWildon I think its more a question of convention than anything else. If I write a sum over all representation ring of symmetric group that means that an element is a finite sum while if I define the ring of symmetriic functions as a limit i get series with an infinite number of terms. There are probably two ways to fix this, one is to take the product in my original definition, the other is taking some kind of colimit in the definition of symmetric functions (as is done in the wikipedia article on symmetric functions).
– Saal Hardali
Dec 16 at 17:53
add a comment |
1
Ignoring the Hopf algebra aspects of the question, people do study the representation theory of $GL_n(mathbbF_q)$ as a q-analog of the representation theory of $Sigma_n$. A thing to note immediately is that $GL_n(mathbbF_q)$ has many more irreps that $Sigma_n$. But $GL_n(mathbbF_q)$ has a particularly nice family of irreps called unipotent representations $U^lambda(q)$, which are indexed by partitions of $n$. And the degree of $U^lambda(q)$ is a polynomial in $q$ (the ``fake degree polynomial'') which at $q=1$ becomes $f^lambda$, the degree of the $Sigma_n$ irrep.
– Sam Hopkins
Dec 16 at 14:52
This is probably just showing my ignorance, but please could you explain the colimit you have in mind on the right hand side of $mathrmRep(Sigma_infty) := mathrmcolim_n mathrmRep(Sigma_n)$? Since $oplus_n mathrmRep(mathrmGL_n(mathbbC))$ is isomorphic to the ring of symmetric functions, which is an inverse limit (i.e. a limit, not a colimit), do you expect the required relationship to involve some kind of duality?
– Mark Wildon
Dec 16 at 17:46
2
@MarkWildon I think its more a question of convention than anything else. If I write a sum over all representation ring of symmetric group that means that an element is a finite sum while if I define the ring of symmetriic functions as a limit i get series with an infinite number of terms. There are probably two ways to fix this, one is to take the product in my original definition, the other is taking some kind of colimit in the definition of symmetric functions (as is done in the wikipedia article on symmetric functions).
– Saal Hardali
Dec 16 at 17:53
1
1
Ignoring the Hopf algebra aspects of the question, people do study the representation theory of $GL_n(mathbbF_q)$ as a q-analog of the representation theory of $Sigma_n$. A thing to note immediately is that $GL_n(mathbbF_q)$ has many more irreps that $Sigma_n$. But $GL_n(mathbbF_q)$ has a particularly nice family of irreps called unipotent representations $U^lambda(q)$, which are indexed by partitions of $n$. And the degree of $U^lambda(q)$ is a polynomial in $q$ (the ``fake degree polynomial'') which at $q=1$ becomes $f^lambda$, the degree of the $Sigma_n$ irrep.
– Sam Hopkins
Dec 16 at 14:52
Ignoring the Hopf algebra aspects of the question, people do study the representation theory of $GL_n(mathbbF_q)$ as a q-analog of the representation theory of $Sigma_n$. A thing to note immediately is that $GL_n(mathbbF_q)$ has many more irreps that $Sigma_n$. But $GL_n(mathbbF_q)$ has a particularly nice family of irreps called unipotent representations $U^lambda(q)$, which are indexed by partitions of $n$. And the degree of $U^lambda(q)$ is a polynomial in $q$ (the ``fake degree polynomial'') which at $q=1$ becomes $f^lambda$, the degree of the $Sigma_n$ irrep.
– Sam Hopkins
Dec 16 at 14:52
This is probably just showing my ignorance, but please could you explain the colimit you have in mind on the right hand side of $mathrmRep(Sigma_infty) := mathrmcolim_n mathrmRep(Sigma_n)$? Since $oplus_n mathrmRep(mathrmGL_n(mathbbC))$ is isomorphic to the ring of symmetric functions, which is an inverse limit (i.e. a limit, not a colimit), do you expect the required relationship to involve some kind of duality?
– Mark Wildon
Dec 16 at 17:46
This is probably just showing my ignorance, but please could you explain the colimit you have in mind on the right hand side of $mathrmRep(Sigma_infty) := mathrmcolim_n mathrmRep(Sigma_n)$? Since $oplus_n mathrmRep(mathrmGL_n(mathbbC))$ is isomorphic to the ring of symmetric functions, which is an inverse limit (i.e. a limit, not a colimit), do you expect the required relationship to involve some kind of duality?
– Mark Wildon
Dec 16 at 17:46
2
2
@MarkWildon I think its more a question of convention than anything else. If I write a sum over all representation ring of symmetric group that means that an element is a finite sum while if I define the ring of symmetriic functions as a limit i get series with an infinite number of terms. There are probably two ways to fix this, one is to take the product in my original definition, the other is taking some kind of colimit in the definition of symmetric functions (as is done in the wikipedia article on symmetric functions).
– Saal Hardali
Dec 16 at 17:53
@MarkWildon I think its more a question of convention than anything else. If I write a sum over all representation ring of symmetric group that means that an element is a finite sum while if I define the ring of symmetriic functions as a limit i get series with an infinite number of terms. There are probably two ways to fix this, one is to take the product in my original definition, the other is taking some kind of colimit in the definition of symmetric functions (as is done in the wikipedia article on symmetric functions).
– Saal Hardali
Dec 16 at 17:53
add a comment |
1 Answer
1
active
oldest
votes
As Sam Hopkins says, the category of all representations of $GL_n(mathbb F_q)$ is too large to give what you want. Instead, let's consider the category of unipotent representations, i.e. those appearing in the irreducible decomposition of $mathbb Q [GL_n(mathbb F_q)/B_n(mathbb F_q)]$.
Unipotent representations are not closed under the naive induction product, but they are closed under parabolic induction $V*W = rm Ind_P(n,m)^GL_n+m V otimes W$. This gives $oplus_n rm Rep^un(GL_n(mathbb F_q))$ the structure of a monoidal category. Instead of being symmetric monoidal, it is now braided monoidal! The Grothendieck ring is a $q$ deformation of the ring of symmetric functions.
Finally, by Morita theory, unipotent representations are equivalent to representations of $mathcal H_n(q) = rm End_GL_n(mathbb Q GL_n/B_n )$, here $mathcal H_n(q)$ is the Iwahori-Hecke algebra which $q$-deforms the group ring of $S_n$. It is Schur-Weyl dual to representations of the quantum group $U_q(GL_infty)$.
answered Dec 16 at 16:09
Phil Tosteson
853158
Does the category being braided monoidal mean that the product "q-commutes"?
– Sam Hopkins
Dec 16 at 16:41
The product is still commutative, because the braiding still gives an isomorphism between the two products. One difference is that the Grothendieck ring won't naturally be a lambda ring anymore.
– Phil Tosteson
Dec 16 at 16:52
Could you elaborate please on the last point. What does it mean that they are dual to the representations of the quantum group? What kind of object is it? Is it a hopf algebra over $mathbbZ[q]$ specializing at $q=1$ to the hopf algebra of the general linear group?
– Saal Hardali
Dec 16 at 17:06
1
Yes, there are different versions: the one I'm most familiar with is a $q$ deformation of the universal envoloping algebra as a hopf algebra. There is a standard rep $V$, and endomorphisms of $n$th tensor power of $V$ is the $n$th Hecke algebra-- giving a Morita equivalence like in Schur-Weyl duality.
– Phil Tosteson
Dec 16 at 21:14
add a comment |
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As Sam Hopkins says, the category of all representations of $GL_n(mathbb F_q)$ is too large to give what you want. Instead, let's consider the category of unipotent representations, i.e. those appearing in the irreducible decomposition of $mathbb Q [GL_n(mathbb F_q)/B_n(mathbb F_q)]$.
Unipotent representations are not closed under the naive induction product, but they are closed under parabolic induction $V*W = rm Ind_P(n,m)^GL_n+m V otimes W$. This gives $oplus_n rm Rep^un(GL_n(mathbb F_q))$ the structure of a monoidal category. Instead of being symmetric monoidal, it is now braided monoidal! The Grothendieck ring is a $q$ deformation of the ring of symmetric functions.
Finally, by Morita theory, unipotent representations are equivalent to representations of $mathcal H_n(q) = rm End_GL_n(mathbb Q GL_n/B_n )$, here $mathcal H_n(q)$ is the Iwahori-Hecke algebra which $q$-deforms the group ring of $S_n$. It is Schur-Weyl dual to representations of the quantum group $U_q(GL_infty)$.
answered Dec 16 at 16:09
Phil Tosteson
853158
Does the category being braided monoidal mean that the product "q-commutes"?
– Sam Hopkins
Dec 16 at 16:41
The product is still commutative, because the braiding still gives an isomorphism between the two products. One difference is that the Grothendieck ring won't naturally be a lambda ring anymore.
– Phil Tosteson
Dec 16 at 16:52
Could you elaborate please on the last point. What does it mean that they are dual to the representations of the quantum group? What kind of object is it? Is it a hopf algebra over $mathbbZ[q]$ specializing at $q=1$ to the hopf algebra of the general linear group?
– Saal Hardali
Dec 16 at 17:06
1
Yes, there are different versions: the one I'm most familiar with is a $q$ deformation of the universal envoloping algebra as a hopf algebra. There is a standard rep $V$, and endomorphisms of $n$th tensor power of $V$ is the $n$th Hecke algebra-- giving a Morita equivalence like in Schur-Weyl duality.
– Phil Tosteson
Dec 16 at 21:14
add a comment |
As Sam Hopkins says, the category of all representations of $GL_n(mathbb F_q)$ is too large to give what you want. Instead, let's consider the category of unipotent representations, i.e. those appearing in the irreducible decomposition of $mathbb Q [GL_n(mathbb F_q)/B_n(mathbb F_q)]$.
Unipotent representations are not closed under the naive induction product, but they are closed under parabolic induction $V*W = rm Ind_P(n,m)^GL_n+m V otimes W$. This gives $oplus_n rm Rep^un(GL_n(mathbb F_q))$ the structure of a monoidal category. Instead of being symmetric monoidal, it is now braided monoidal! The Grothendieck ring is a $q$ deformation of the ring of symmetric functions.
Finally, by Morita theory, unipotent representations are equivalent to representations of $mathcal H_n(q) = rm End_GL_n(mathbb Q GL_n/B_n )$, here $mathcal H_n(q)$ is the Iwahori-Hecke algebra which $q$-deforms the group ring of $S_n$. It is Schur-Weyl dual to representations of the quantum group $U_q(GL_infty)$.
answered Dec 16 at 16:09
Phil Tosteson
853158
Does the category being braided monoidal mean that the product "q-commutes"?
– Sam Hopkins
Dec 16 at 16:41
The product is still commutative, because the braiding still gives an isomorphism between the two products. One difference is that the Grothendieck ring won't naturally be a lambda ring anymore.
– Phil Tosteson
Dec 16 at 16:52
Could you elaborate please on the last point. What does it mean that they are dual to the representations of the quantum group? What kind of object is it? Is it a hopf algebra over $mathbbZ[q]$ specializing at $q=1$ to the hopf algebra of the general linear group?
– Saal Hardali
Dec 16 at 17:06
1
Yes, there are different versions: the one I'm most familiar with is a $q$ deformation of the universal envoloping algebra as a hopf algebra. There is a standard rep $V$, and endomorphisms of $n$th tensor power of $V$ is the $n$th Hecke algebra-- giving a Morita equivalence like in Schur-Weyl duality.
– Phil Tosteson
Dec 16 at 21:14
add a comment |
As Sam Hopkins says, the category of all representations of $GL_n(mathbb F_q)$ is too large to give what you want. Instead, let's consider the category of unipotent representations, i.e. those appearing in the irreducible decomposition of $mathbb Q [GL_n(mathbb F_q)/B_n(mathbb F_q)]$.
Unipotent representations are not closed under the naive induction product, but they are closed under parabolic induction $V*W = rm Ind_P(n,m)^GL_n+m V otimes W$. This gives $oplus_n rm Rep^un(GL_n(mathbb F_q))$ the structure of a monoidal category. Instead of being symmetric monoidal, it is now braided monoidal! The Grothendieck ring is a $q$ deformation of the ring of symmetric functions.
Finally, by Morita theory, unipotent representations are equivalent to representations of $mathcal H_n(q) = rm End_GL_n(mathbb Q GL_n/B_n )$, here $mathcal H_n(q)$ is the Iwahori-Hecke algebra which $q$-deforms the group ring of $S_n$. It is Schur-Weyl dual to representations of the quantum group $U_q(GL_infty)$.
answered Dec 16 at 16:09
Phil Tosteson
853158
As Sam Hopkins says, the category of all representations of $GL_n(mathbb F_q)$ is too large to give what you want. Instead, let's consider the category of unipotent representations, i.e. those appearing in the irreducible decomposition of $mathbb Q [GL_n(mathbb F_q)/B_n(mathbb F_q)]$.
Unipotent representations are not closed under the naive induction product, but they are closed under parabolic induction $V*W = rm Ind_P(n,m)^GL_n+m V otimes W$. This gives $oplus_n rm Rep^un(GL_n(mathbb F_q))$ the structure of a monoidal category. Instead of being symmetric monoidal, it is now braided monoidal! The Grothendieck ring is a $q$ deformation of the ring of symmetric functions.
Finally, by Morita theory, unipotent representations are equivalent to representations of $mathcal H_n(q) = rm End_GL_n(mathbb Q GL_n/B_n )$, here $mathcal H_n(q)$ is the Iwahori-Hecke algebra which $q$-deforms the group ring of $S_n$. It is Schur-Weyl dual to representations of the quantum group $U_q(GL_infty)$.
answered Dec 16 at 16:09
Phil Tosteson
853158
edited Dec 16 at 16:19
answered Dec 16 at 16:09
Phil Tosteson
853158
answered Dec 16 at 16:09
Phil Tosteson
853158
answered Dec 16 at 16:09
Phil Tosteson
853158
853158
Does the category being braided monoidal mean that the product "q-commutes"?
– Sam Hopkins
Dec 16 at 16:41
The product is still commutative, because the braiding still gives an isomorphism between the two products. One difference is that the Grothendieck ring won't naturally be a lambda ring anymore.
– Phil Tosteson
Dec 16 at 16:52
Could you elaborate please on the last point. What does it mean that they are dual to the representations of the quantum group? What kind of object is it? Is it a hopf algebra over $mathbbZ[q]$ specializing at $q=1$ to the hopf algebra of the general linear group?
– Saal Hardali
Dec 16 at 17:06
1
Yes, there are different versions: the one I'm most familiar with is a $q$ deformation of the universal envoloping algebra as a hopf algebra. There is a standard rep $V$, and endomorphisms of $n$th tensor power of $V$ is the $n$th Hecke algebra-- giving a Morita equivalence like in Schur-Weyl duality.
– Phil Tosteson
Dec 16 at 21:14
add a comment |
Does the category being braided monoidal mean that the product "q-commutes"?
– Sam Hopkins
Dec 16 at 16:41
The product is still commutative, because the braiding still gives an isomorphism between the two products. One difference is that the Grothendieck ring won't naturally be a lambda ring anymore.
– Phil Tosteson
Dec 16 at 16:52
Could you elaborate please on the last point. What does it mean that they are dual to the representations of the quantum group? What kind of object is it? Is it a hopf algebra over $mathbbZ[q]$ specializing at $q=1$ to the hopf algebra of the general linear group?
– Saal Hardali
Dec 16 at 17:06
1
Yes, there are different versions: the one I'm most familiar with is a $q$ deformation of the universal envoloping algebra as a hopf algebra. There is a standard rep $V$, and endomorphisms of $n$th tensor power of $V$ is the $n$th Hecke algebra-- giving a Morita equivalence like in Schur-Weyl duality.
– Phil Tosteson
Dec 16 at 21:14
Does the category being braided monoidal mean that the product "q-commutes"?
– Sam Hopkins
Dec 16 at 16:41
Does the category being braided monoidal mean that the product "q-commutes"?
– Sam Hopkins
Dec 16 at 16:41
The product is still commutative, because the braiding still gives an isomorphism between the two products. One difference is that the Grothendieck ring won't naturally be a lambda ring anymore.
– Phil Tosteson
Dec 16 at 16:52
The product is still commutative, because the braiding still gives an isomorphism between the two products. One difference is that the Grothendieck ring won't naturally be a lambda ring anymore.
– Phil Tosteson
Dec 16 at 16:52
Could you elaborate please on the last point. What does it mean that they are dual to the representations of the quantum group? What kind of object is it? Is it a hopf algebra over $mathbbZ[q]$ specializing at $q=1$ to the hopf algebra of the general linear group?
– Saal Hardali
Dec 16 at 17:06
Could you elaborate please on the last point. What does it mean that they are dual to the representations of the quantum group? What kind of object is it? Is it a hopf algebra over $mathbbZ[q]$ specializing at $q=1$ to the hopf algebra of the general linear group?
– Saal Hardali
Dec 16 at 17:06
1
1
Yes, there are different versions: the one I'm most familiar with is a $q$ deformation of the universal envoloping algebra as a hopf algebra. There is a standard rep $V$, and endomorphisms of $n$th tensor power of $V$ is the $n$th Hecke algebra-- giving a Morita equivalence like in Schur-Weyl duality.
– Phil Tosteson
Dec 16 at 21:14
Yes, there are different versions: the one I'm most familiar with is a $q$ deformation of the universal envoloping algebra as a hopf algebra. There is a standard rep $V$, and endomorphisms of $n$th tensor power of $V$ is the $n$th Hecke algebra-- giving a Morita equivalence like in Schur-Weyl duality.
– Phil Tosteson
Dec 16 at 21:14
add a comment |
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1
Ignoring the Hopf algebra aspects of the question, people do study the representation theory of $GL_n(mathbbF_q)$ as a q-analog of the representation theory of $Sigma_n$. A thing to note immediately is that $GL_n(mathbbF_q)$ has many more irreps that $Sigma_n$. But $GL_n(mathbbF_q)$ has a particularly nice family of irreps called unipotent representations $U^lambda(q)$, which are indexed by partitions of $n$. And the degree of $U^lambda(q)$ is a polynomial in $q$ (the ``fake degree polynomial'') which at $q=1$ becomes $f^lambda$, the degree of the $Sigma_n$ irrep.
– Sam Hopkins
Dec 16 at 14:52
This is probably just showing my ignorance, but please could you explain the colimit you have in mind on the right hand side of $mathrmRep(Sigma_infty) := mathrmcolim_n mathrmRep(Sigma_n)$? Since $oplus_n mathrmRep(mathrmGL_n(mathbbC))$ is isomorphic to the ring of symmetric functions, which is an inverse limit (i.e. a limit, not a colimit), do you expect the required relationship to involve some kind of duality?
– Mark Wildon
Dec 16 at 17:46
2
@MarkWildon I think its more a question of convention than anything else. If I write a sum over all representation ring of symmetric group that means that an element is a finite sum while if I define the ring of symmetriic functions as a limit i get series with an infinite number of terms. There are probably two ways to fix this, one is to take the product in my original definition, the other is taking some kind of colimit in the definition of symmetric functions (as is done in the wikipedia article on symmetric functions).
– Saal Hardali
Dec 16 at 17:53