What is the relationship between universal enveloping algebra $U(g)$ and the QUE algebras $U_{q}(g)$











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I just know $U_{q}(sl_2)$ is universal enveloping algebra $U(sl_2)$ when $q$ tends to $1$. Due to Qiaochu Yuan's comment, this is true in general. Conversely, Suppose $sl_2$ has three basises $e, h, f$, then we let $E=e, F=f, K=q^{h}, K^{-1} = q^{-h}$, we can get $U_{q}(sl_2)$. What is the general case?






abstract-algebra quantum-groups







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  • 1




    That's true in general.
    – Qiaochu Yuan
    Nov 25 at 19:05










  • I see, thank you very much.
    – Daisy
    Nov 26 at 2:26















up vote
2
down vote

favorite












I just know $U_{q}(sl_2)$ is universal enveloping algebra $U(sl_2)$ when $q$ tends to $1$. Due to Qiaochu Yuan's comment, this is true in general. Conversely, Suppose $sl_2$ has three basises $e, h, f$, then we let $E=e, F=f, K=q^{h}, K^{-1} = q^{-h}$, we can get $U_{q}(sl_2)$. What is the general case?






abstract-algebra quantum-groups







share|cite|improve this question




















  • 1




    That's true in general.
    – Qiaochu Yuan
    Nov 25 at 19:05










  • I see, thank you very much.
    – Daisy
    Nov 26 at 2:26













up vote
2
down vote

favorite









up vote
2
down vote

favorite











I just know $U_{q}(sl_2)$ is universal enveloping algebra $U(sl_2)$ when $q$ tends to $1$. Due to Qiaochu Yuan's comment, this is true in general. Conversely, Suppose $sl_2$ has three basises $e, h, f$, then we let $E=e, F=f, K=q^{h}, K^{-1} = q^{-h}$, we can get $U_{q}(sl_2)$. What is the general case?






abstract-algebra quantum-groups







share|cite|improve this question















I just know $U_{q}(sl_2)$ is universal enveloping algebra $U(sl_2)$ when $q$ tends to $1$. Due to Qiaochu Yuan's comment, this is true in general. Conversely, Suppose $sl_2$ has three basises $e, h, f$, then we let $E=e, F=f, K=q^{h}, K^{-1} = q^{-h}$, we can get $U_{q}(sl_2)$. What is the general case?






abstract-algebra quantum-groups




abstract-algebra quantum-groups






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 6 at 2:20

























asked Nov 25 at 13:04









Daisy

420313




420313








  • 1




    That's true in general.
    – Qiaochu Yuan
    Nov 25 at 19:05










  • I see, thank you very much.
    – Daisy
    Nov 26 at 2:26














  • 1




    That's true in general.
    – Qiaochu Yuan
    Nov 25 at 19:05










  • I see, thank you very much.
    – Daisy
    Nov 26 at 2:26








1




1




That's true in general.
– Qiaochu Yuan
Nov 25 at 19:05




That's true in general.
– Qiaochu Yuan
Nov 25 at 19:05












I see, thank you very much.
– Daisy
Nov 26 at 2:26




I see, thank you very much.
– Daisy
Nov 26 at 2:26










1 Answer
1






active

oldest

votes

















up vote
2
down vote



accepted










Let me assume $mathfrak{g}$ semisimple for simplicity (I think that also works in general).



The answer to your question is yes, we always have $limlimits_{q to 1} U_q = U$.



This is what Qiaochu said in the comments section. The structure of $U_q$ is richer mainly because of the interesting coproduct, leading to the notion of $R$-matrix.



To check it, $U_q$ has generators $E_i, F_i, K_i^{pm1}$ and different scalars $q_i$ so that each $(E_i,F_i,K_i^{pm 1},q_i)$ is isomorphic to $U_q(mathfrak{sl}_2)$. You also want the $K_i$ to commute and same with the $q_i$. Finally there is a last type of relations, the quantum Serre relation and this is basically the only thing you need to check.



For example, in type $A_2$ you need to check that $$ limlimits_{q to 1} (E_1^2E_2 - (q+q^{-1})E_1E_2E_1 + E_2E_1^2) = e_1^2e_2 - 2e_1e_2e_1 + e_2e_1^2 $$ which is clear.



Here I took the Lusztig presentation indexed by a set of simple roots but of course it also holds with different generators.



Just a remark about representation theory, if $q$ is not a root of unity $U_q$ and $U$ basically have the same representation theory. When $q$ is a root of unity, this is more complicated and look like representation of $U$ but in positive characteristic. This is well explained in Jantzen's book on quantum groups.






share|cite|improve this answer























  • Thank you for your answer, do you know the converse case?
    – Daisy
    Dec 6 at 2:22










  • @Daisy : What do you mean about the converse case ?
    – Nicolas Hemelsoet
    Dec 6 at 11:52










  • ， How to get $U_{q}(g)$ from $U(g)$? I just know the case of $sl_2$, the other cases is similar?
    – Daisy
    Dec 7 at 2:05










  • @Daisy : I edited my answer, I hope this is more clear.
    – Nicolas Hemelsoet
    Dec 7 at 7:57











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
2
down vote



accepted










Let me assume $mathfrak{g}$ semisimple for simplicity (I think that also works in general).



The answer to your question is yes, we always have $limlimits_{q to 1} U_q = U$.



This is what Qiaochu said in the comments section. The structure of $U_q$ is richer mainly because of the interesting coproduct, leading to the notion of $R$-matrix.



To check it, $U_q$ has generators $E_i, F_i, K_i^{pm1}$ and different scalars $q_i$ so that each $(E_i,F_i,K_i^{pm 1},q_i)$ is isomorphic to $U_q(mathfrak{sl}_2)$. You also want the $K_i$ to commute and same with the $q_i$. Finally there is a last type of relations, the quantum Serre relation and this is basically the only thing you need to check.



For example, in type $A_2$ you need to check that $$ limlimits_{q to 1} (E_1^2E_2 - (q+q^{-1})E_1E_2E_1 + E_2E_1^2) = e_1^2e_2 - 2e_1e_2e_1 + e_2e_1^2 $$ which is clear.



Here I took the Lusztig presentation indexed by a set of simple roots but of course it also holds with different generators.



Just a remark about representation theory, if $q$ is not a root of unity $U_q$ and $U$ basically have the same representation theory. When $q$ is a root of unity, this is more complicated and look like representation of $U$ but in positive characteristic. This is well explained in Jantzen's book on quantum groups.






share|cite|improve this answer























  • Thank you for your answer, do you know the converse case?
    – Daisy
    Dec 6 at 2:22










  • @Daisy : What do you mean about the converse case ?
    – Nicolas Hemelsoet
    Dec 6 at 11:52










  • ， How to get $U_{q}(g)$ from $U(g)$? I just know the case of $sl_2$, the other cases is similar?
    – Daisy
    Dec 7 at 2:05










  • @Daisy : I edited my answer, I hope this is more clear.
    – Nicolas Hemelsoet
    Dec 7 at 7:57















up vote
2
down vote



accepted










Let me assume $mathfrak{g}$ semisimple for simplicity (I think that also works in general).



The answer to your question is yes, we always have $limlimits_{q to 1} U_q = U$.



This is what Qiaochu said in the comments section. The structure of $U_q$ is richer mainly because of the interesting coproduct, leading to the notion of $R$-matrix.



To check it, $U_q$ has generators $E_i, F_i, K_i^{pm1}$ and different scalars $q_i$ so that each $(E_i,F_i,K_i^{pm 1},q_i)$ is isomorphic to $U_q(mathfrak{sl}_2)$. You also want the $K_i$ to commute and same with the $q_i$. Finally there is a last type of relations, the quantum Serre relation and this is basically the only thing you need to check.



For example, in type $A_2$ you need to check that $$ limlimits_{q to 1} (E_1^2E_2 - (q+q^{-1})E_1E_2E_1 + E_2E_1^2) = e_1^2e_2 - 2e_1e_2e_1 + e_2e_1^2 $$ which is clear.



Here I took the Lusztig presentation indexed by a set of simple roots but of course it also holds with different generators.



Just a remark about representation theory, if $q$ is not a root of unity $U_q$ and $U$ basically have the same representation theory. When $q$ is a root of unity, this is more complicated and look like representation of $U$ but in positive characteristic. This is well explained in Jantzen's book on quantum groups.






share|cite|improve this answer























  • Thank you for your answer, do you know the converse case?
    – Daisy
    Dec 6 at 2:22










  • @Daisy : What do you mean about the converse case ?
    – Nicolas Hemelsoet
    Dec 6 at 11:52










  • ， How to get $U_{q}(g)$ from $U(g)$? I just know the case of $sl_2$, the other cases is similar?
    – Daisy
    Dec 7 at 2:05










  • @Daisy : I edited my answer, I hope this is more clear.
    – Nicolas Hemelsoet
    Dec 7 at 7:57













up vote
2
down vote



accepted







up vote
2
down vote



accepted






Let me assume $mathfrak{g}$ semisimple for simplicity (I think that also works in general).



The answer to your question is yes, we always have $limlimits_{q to 1} U_q = U$.



This is what Qiaochu said in the comments section. The structure of $U_q$ is richer mainly because of the interesting coproduct, leading to the notion of $R$-matrix.



To check it, $U_q$ has generators $E_i, F_i, K_i^{pm1}$ and different scalars $q_i$ so that each $(E_i,F_i,K_i^{pm 1},q_i)$ is isomorphic to $U_q(mathfrak{sl}_2)$. You also want the $K_i$ to commute and same with the $q_i$. Finally there is a last type of relations, the quantum Serre relation and this is basically the only thing you need to check.



For example, in type $A_2$ you need to check that $$ limlimits_{q to 1} (E_1^2E_2 - (q+q^{-1})E_1E_2E_1 + E_2E_1^2) = e_1^2e_2 - 2e_1e_2e_1 + e_2e_1^2 $$ which is clear.



Here I took the Lusztig presentation indexed by a set of simple roots but of course it also holds with different generators.



Just a remark about representation theory, if $q$ is not a root of unity $U_q$ and $U$ basically have the same representation theory. When $q$ is a root of unity, this is more complicated and look like representation of $U$ but in positive characteristic. This is well explained in Jantzen's book on quantum groups.






share|cite|improve this answer














Let me assume $mathfrak{g}$ semisimple for simplicity (I think that also works in general).



The answer to your question is yes, we always have $limlimits_{q to 1} U_q = U$.



This is what Qiaochu said in the comments section. The structure of $U_q$ is richer mainly because of the interesting coproduct, leading to the notion of $R$-matrix.



To check it, $U_q$ has generators $E_i, F_i, K_i^{pm1}$ and different scalars $q_i$ so that each $(E_i,F_i,K_i^{pm 1},q_i)$ is isomorphic to $U_q(mathfrak{sl}_2)$. You also want the $K_i$ to commute and same with the $q_i$. Finally there is a last type of relations, the quantum Serre relation and this is basically the only thing you need to check.



For example, in type $A_2$ you need to check that $$ limlimits_{q to 1} (E_1^2E_2 - (q+q^{-1})E_1E_2E_1 + E_2E_1^2) = e_1^2e_2 - 2e_1e_2e_1 + e_2e_1^2 $$ which is clear.



Here I took the Lusztig presentation indexed by a set of simple roots but of course it also holds with different generators.



Just a remark about representation theory, if $q$ is not a root of unity $U_q$ and $U$ basically have the same representation theory. When $q$ is a root of unity, this is more complicated and look like representation of $U$ but in positive characteristic. This is well explained in Jantzen's book on quantum groups.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 7 at 7:56

























answered Dec 5 at 17:48









Nicolas Hemelsoet

5,7452417




5,7452417












  • Thank you for your answer, do you know the converse case?
    – Daisy
    Dec 6 at 2:22










  • @Daisy : What do you mean about the converse case ?
    – Nicolas Hemelsoet
    Dec 6 at 11:52










  • ， How to get $U_{q}(g)$ from $U(g)$? I just know the case of $sl_2$, the other cases is similar?
    – Daisy
    Dec 7 at 2:05










  • @Daisy : I edited my answer, I hope this is more clear.
    – Nicolas Hemelsoet
    Dec 7 at 7:57


















  • Thank you for your answer, do you know the converse case?
    – Daisy
    Dec 6 at 2:22










  • @Daisy : What do you mean about the converse case ?
    – Nicolas Hemelsoet
    Dec 6 at 11:52










  • ， How to get $U_{q}(g)$ from $U(g)$? I just know the case of $sl_2$, the other cases is similar?
    – Daisy
    Dec 7 at 2:05










  • @Daisy : I edited my answer, I hope this is more clear.
    – Nicolas Hemelsoet
    Dec 7 at 7:57
















Thank you for your answer, do you know the converse case?
– Daisy
Dec 6 at 2:22




Thank you for your answer, do you know the converse case?
– Daisy
Dec 6 at 2:22












@Daisy : What do you mean about the converse case ?
– Nicolas Hemelsoet
Dec 6 at 11:52




@Daisy : What do you mean about the converse case ?
– Nicolas Hemelsoet
Dec 6 at 11:52












， How to get $U_{q}(g)$ from $U(g)$? I just know the case of $sl_2$, the other cases is similar?
– Daisy
Dec 7 at 2:05




， How to get $U_{q}(g)$ from $U(g)$? I just know the case of $sl_2$, the other cases is similar?
– Daisy
Dec 7 at 2:05












@Daisy : I edited my answer, I hope this is more clear.
– Nicolas Hemelsoet
Dec 7 at 7:57




@Daisy : I edited my answer, I hope this is more clear.
– Nicolas Hemelsoet
Dec 7 at 7:57


















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