Graded and Filtered Algebras
$begingroup$
This is right out of Kassel's Quantum Groups book which I am self-studying. It is on page 14.
The general set-up is this. Let $A$ be a filtered algebra with filtration $F_0(A) subset F_1(A) subset cdots subset A$, and suppose that $I$ is a two-sided ideal of $A$. The quotient algebra is then filtered with filtration $F_i(A/I) = F_i(A)/F_i(A) cap I$. We know that for a filtered algebra, $A$, there exists an associated graded algebra, called $mathrm{gr}(A) = oplus S_i$ where $S_i = F_i(A)/F_{i-1}(A)$.
Define $M(2)$ as the polynomial algebra $k[a,b,c,d]$ and define $SL(2) = M(2)/(ad - bc -1)$.
His first claim is that $mathrm{gr}(A/I) = oplus_{i in mathbb{N}} F_i(A)/(F_{i-1}(A) + F_i(A) cap I)$. Following this, he claims that $mathrm{gr}(SL(2)) cong k[a,b,c,d]/(ad-bc)$ (note that the ideal $(ad-bc-1)$ is not generated by homogeneous elements so that $SL(2)$ is not graded).
I am unsure how one gets to these results, and any feedback on this would be greatly appreciated!
abstract-algebra graded-rings quantum-groups filtrations
edited Dec 1 '18 at 10:55
user26857
39.3k124183
asked Mar 21 '12 at 22:12
Elden ElmantoElden Elmanto
684315
$endgroup$
add a comment |
$begingroup$
This is right out of Kassel's Quantum Groups book which I am self-studying. It is on page 14.
The general set-up is this. Let $A$ be a filtered algebra with filtration $F_0(A) subset F_1(A) subset cdots subset A$, and suppose that $I$ is a two-sided ideal of $A$. The quotient algebra is then filtered with filtration $F_i(A/I) = F_i(A)/F_i(A) cap I$. We know that for a filtered algebra, $A$, there exists an associated graded algebra, called $mathrm{gr}(A) = oplus S_i$ where $S_i = F_i(A)/F_{i-1}(A)$.
Define $M(2)$ as the polynomial algebra $k[a,b,c,d]$ and define $SL(2) = M(2)/(ad - bc -1)$.
His first claim is that $mathrm{gr}(A/I) = oplus_{i in mathbb{N}} F_i(A)/(F_{i-1}(A) + F_i(A) cap I)$. Following this, he claims that $mathrm{gr}(SL(2)) cong k[a,b,c,d]/(ad-bc)$ (note that the ideal $(ad-bc-1)$ is not generated by homogeneous elements so that $SL(2)$ is not graded).
I am unsure how one gets to these results, and any feedback on this would be greatly appreciated!
abstract-algebra graded-rings quantum-groups filtrations
edited Dec 1 '18 at 10:55
user26857
39.3k124183
asked Mar 21 '12 at 22:12
Elden ElmantoElden Elmanto
684315
$endgroup$
add a comment |
$begingroup$
This is right out of Kassel's Quantum Groups book which I am self-studying. It is on page 14.
The general set-up is this. Let $A$ be a filtered algebra with filtration $F_0(A) subset F_1(A) subset cdots subset A$, and suppose that $I$ is a two-sided ideal of $A$. The quotient algebra is then filtered with filtration $F_i(A/I) = F_i(A)/F_i(A) cap I$. We know that for a filtered algebra, $A$, there exists an associated graded algebra, called $mathrm{gr}(A) = oplus S_i$ where $S_i = F_i(A)/F_{i-1}(A)$.
Define $M(2)$ as the polynomial algebra $k[a,b,c,d]$ and define $SL(2) = M(2)/(ad - bc -1)$.
His first claim is that $mathrm{gr}(A/I) = oplus_{i in mathbb{N}} F_i(A)/(F_{i-1}(A) + F_i(A) cap I)$. Following this, he claims that $mathrm{gr}(SL(2)) cong k[a,b,c,d]/(ad-bc)$ (note that the ideal $(ad-bc-1)$ is not generated by homogeneous elements so that $SL(2)$ is not graded).
I am unsure how one gets to these results, and any feedback on this would be greatly appreciated!
abstract-algebra graded-rings quantum-groups filtrations
edited Dec 1 '18 at 10:55
user26857
39.3k124183
asked Mar 21 '12 at 22:12
Elden ElmantoElden Elmanto
684315
$endgroup$
This is right out of Kassel's Quantum Groups book which I am self-studying. It is on page 14.
The general set-up is this. Let $A$ be a filtered algebra with filtration $F_0(A) subset F_1(A) subset cdots subset A$, and suppose that $I$ is a two-sided ideal of $A$. The quotient algebra is then filtered with filtration $F_i(A/I) = F_i(A)/F_i(A) cap I$. We know that for a filtered algebra, $A$, there exists an associated graded algebra, called $mathrm{gr}(A) = oplus S_i$ where $S_i = F_i(A)/F_{i-1}(A)$.
Define $M(2)$ as the polynomial algebra $k[a,b,c,d]$ and define $SL(2) = M(2)/(ad - bc -1)$.
His first claim is that $mathrm{gr}(A/I) = oplus_{i in mathbb{N}} F_i(A)/(F_{i-1}(A) + F_i(A) cap I)$. Following this, he claims that $mathrm{gr}(SL(2)) cong k[a,b,c,d]/(ad-bc)$ (note that the ideal $(ad-bc-1)$ is not generated by homogeneous elements so that $SL(2)$ is not graded).
I am unsure how one gets to these results, and any feedback on this would be greatly appreciated!
abstract-algebra graded-rings quantum-groups filtrations
abstract-algebra graded-rings quantum-groups filtrations
edited Dec 1 '18 at 10:55
user26857
39.3k124183
asked Mar 21 '12 at 22:12
Elden ElmantoElden Elmanto
684315
edited Dec 1 '18 at 10:55
user26857
39.3k124183
asked Mar 21 '12 at 22:12
Elden ElmantoElden Elmanto
684315
edited Dec 1 '18 at 10:55
user26857
39.3k124183
edited Dec 1 '18 at 10:55
user26857
39.3k124183
edited Dec 1 '18 at 10:55
user26857
39.3k124183
39.3k124183
asked Mar 21 '12 at 22:12
Elden ElmantoElden Elmanto
684315
asked Mar 21 '12 at 22:12
Elden ElmantoElden Elmanto
684315
asked Mar 21 '12 at 22:12
Elden ElmantoElden Elmanto
684315
684315
add a comment |
add a comment |
1 Answer
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$begingroup$
We have
$$begin{align*}
text{gr}(A/I) &=& bigoplus F_i(A/I)/F_{i-1}(A/I) \
&cong& bigoplus (F_i(A)/F_i(A) cap I)/(F_{i-1}(A)/F_{i-1}(A) cap I) \
&cong& bigoplus F_i(A)/(F_{i-1}(A) + F_i(A) cap I) end{align*}$$
which follows from the fact that $F_{i-1}(A) cap I subseteq F_i(A) cap I$ and I suppose the third isomorphism theorem.
With $A = k[a, b, c, d]$ and $I = (ad - bc - 1)$ we have that $F_i(A)$ consists of polynomials of total degree at most $i$ and that $F_{i-1}(A) + F_i(A) cap I$ consists of polynomials of total degree at most $i-1$ together with polynomials of the form $f (ad - bc - 1)$ where $f$ has total degree at most $i-2$. Since $f in F_{i-1}(A)$, instead of adding in terms of the form $f(ad - bc - 1)$ we can add in terms of the form $f(ad - bc)$. Do you see how the result follows from here?
answered Mar 21 '12 at 22:38
Qiaochu YuanQiaochu Yuan
279k32586931
$endgroup$
$begingroup$
Yup. Thank you very much!
$endgroup$
– Elden Elmanto
Mar 21 '12 at 23:08
add a comment |
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$begingroup$
We have
$$begin{align*}
text{gr}(A/I) &=& bigoplus F_i(A/I)/F_{i-1}(A/I) \
&cong& bigoplus (F_i(A)/F_i(A) cap I)/(F_{i-1}(A)/F_{i-1}(A) cap I) \
&cong& bigoplus F_i(A)/(F_{i-1}(A) + F_i(A) cap I) end{align*}$$
which follows from the fact that $F_{i-1}(A) cap I subseteq F_i(A) cap I$ and I suppose the third isomorphism theorem.
With $A = k[a, b, c, d]$ and $I = (ad - bc - 1)$ we have that $F_i(A)$ consists of polynomials of total degree at most $i$ and that $F_{i-1}(A) + F_i(A) cap I$ consists of polynomials of total degree at most $i-1$ together with polynomials of the form $f (ad - bc - 1)$ where $f$ has total degree at most $i-2$. Since $f in F_{i-1}(A)$, instead of adding in terms of the form $f(ad - bc - 1)$ we can add in terms of the form $f(ad - bc)$. Do you see how the result follows from here?
answered Mar 21 '12 at 22:38
Qiaochu YuanQiaochu Yuan
279k32586931
$endgroup$
$begingroup$
Yup. Thank you very much!
$endgroup$
– Elden Elmanto
Mar 21 '12 at 23:08
add a comment |
$begingroup$
We have
$$begin{align*}
text{gr}(A/I) &=& bigoplus F_i(A/I)/F_{i-1}(A/I) \
&cong& bigoplus (F_i(A)/F_i(A) cap I)/(F_{i-1}(A)/F_{i-1}(A) cap I) \
&cong& bigoplus F_i(A)/(F_{i-1}(A) + F_i(A) cap I) end{align*}$$
which follows from the fact that $F_{i-1}(A) cap I subseteq F_i(A) cap I$ and I suppose the third isomorphism theorem.
With $A = k[a, b, c, d]$ and $I = (ad - bc - 1)$ we have that $F_i(A)$ consists of polynomials of total degree at most $i$ and that $F_{i-1}(A) + F_i(A) cap I$ consists of polynomials of total degree at most $i-1$ together with polynomials of the form $f (ad - bc - 1)$ where $f$ has total degree at most $i-2$. Since $f in F_{i-1}(A)$, instead of adding in terms of the form $f(ad - bc - 1)$ we can add in terms of the form $f(ad - bc)$. Do you see how the result follows from here?
answered Mar 21 '12 at 22:38
Qiaochu YuanQiaochu Yuan
279k32586931
$endgroup$
$begingroup$
Yup. Thank you very much!
$endgroup$
– Elden Elmanto
Mar 21 '12 at 23:08
add a comment |
$begingroup$
We have
$$begin{align*}
text{gr}(A/I) &=& bigoplus F_i(A/I)/F_{i-1}(A/I) \
&cong& bigoplus (F_i(A)/F_i(A) cap I)/(F_{i-1}(A)/F_{i-1}(A) cap I) \
&cong& bigoplus F_i(A)/(F_{i-1}(A) + F_i(A) cap I) end{align*}$$
which follows from the fact that $F_{i-1}(A) cap I subseteq F_i(A) cap I$ and I suppose the third isomorphism theorem.
With $A = k[a, b, c, d]$ and $I = (ad - bc - 1)$ we have that $F_i(A)$ consists of polynomials of total degree at most $i$ and that $F_{i-1}(A) + F_i(A) cap I$ consists of polynomials of total degree at most $i-1$ together with polynomials of the form $f (ad - bc - 1)$ where $f$ has total degree at most $i-2$. Since $f in F_{i-1}(A)$, instead of adding in terms of the form $f(ad - bc - 1)$ we can add in terms of the form $f(ad - bc)$. Do you see how the result follows from here?
answered Mar 21 '12 at 22:38
Qiaochu YuanQiaochu Yuan
279k32586931
$endgroup$
We have
$$begin{align*}
text{gr}(A/I) &=& bigoplus F_i(A/I)/F_{i-1}(A/I) \
&cong& bigoplus (F_i(A)/F_i(A) cap I)/(F_{i-1}(A)/F_{i-1}(A) cap I) \
&cong& bigoplus F_i(A)/(F_{i-1}(A) + F_i(A) cap I) end{align*}$$
which follows from the fact that $F_{i-1}(A) cap I subseteq F_i(A) cap I$ and I suppose the third isomorphism theorem.
With $A = k[a, b, c, d]$ and $I = (ad - bc - 1)$ we have that $F_i(A)$ consists of polynomials of total degree at most $i$ and that $F_{i-1}(A) + F_i(A) cap I$ consists of polynomials of total degree at most $i-1$ together with polynomials of the form $f (ad - bc - 1)$ where $f$ has total degree at most $i-2$. Since $f in F_{i-1}(A)$, instead of adding in terms of the form $f(ad - bc - 1)$ we can add in terms of the form $f(ad - bc)$. Do you see how the result follows from here?
answered Mar 21 '12 at 22:38
Qiaochu YuanQiaochu Yuan
279k32586931
answered Mar 21 '12 at 22:38
Qiaochu YuanQiaochu Yuan
279k32586931
answered Mar 21 '12 at 22:38
Qiaochu YuanQiaochu Yuan
279k32586931
answered Mar 21 '12 at 22:38
Qiaochu YuanQiaochu Yuan
279k32586931
279k32586931
$begingroup$
Yup. Thank you very much!
$endgroup$
– Elden Elmanto
Mar 21 '12 at 23:08
add a comment |
$begingroup$
Yup. Thank you very much!
$endgroup$
– Elden Elmanto
Mar 21 '12 at 23:08
$begingroup$
Yup. Thank you very much!
$endgroup$
– Elden Elmanto
Mar 21 '12 at 23:08
$begingroup$
Yup. Thank you very much!
$endgroup$
– Elden Elmanto
Mar 21 '12 at 23:08
add a comment |
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