Applying the divided difference operator
This question is about divided difference operators.
How do I perform $partial _2$ or $partial_3$ on $x_1^2x_2$?
$partial_i$ is defined as $frac{p-r_i.p}{x_i-x_{i+1}}$, where $r_i$ is the reflection operator; i.e. $r_i.p(x_1,x_2,dots,x_i,x_{i+1},dots,x_n)=p(x_1,x_2,dots,x_{i+1},x_i,dots,x_n)$.
Do I just assume that $x_1^2x_2=x_1^2x_2x_3^0x_4^0$?
Edit: I have another question. The calculation of Schubert polynomials uss the following formula: $S_{pi}=partial_{pi^{-1}w_0^n}S_{w_0^n}$. What does $partial_{pi^{-1}w_0^n}$ mean? Do I just concatenate $pi^{-1}$ and $w_0^n$ as words? Like if $w_0^3=321$ and $pi=(12)$, we get $partial_{pi^{-1}w_0^n}=partial_{1321}$?
combinatorics representation-theory schubert-calculus
edited 2 days ago
Matt Samuel
37.5k63665
asked May 4 '16 at 12:31
fierydemonfierydemon
4,40512155
add a comment |
This question is about divided difference operators.
How do I perform $partial _2$ or $partial_3$ on $x_1^2x_2$?
$partial_i$ is defined as $frac{p-r_i.p}{x_i-x_{i+1}}$, where $r_i$ is the reflection operator; i.e. $r_i.p(x_1,x_2,dots,x_i,x_{i+1},dots,x_n)=p(x_1,x_2,dots,x_{i+1},x_i,dots,x_n)$.
Do I just assume that $x_1^2x_2=x_1^2x_2x_3^0x_4^0$?
Edit: I have another question. The calculation of Schubert polynomials uss the following formula: $S_{pi}=partial_{pi^{-1}w_0^n}S_{w_0^n}$. What does $partial_{pi^{-1}w_0^n}$ mean? Do I just concatenate $pi^{-1}$ and $w_0^n$ as words? Like if $w_0^3=321$ and $pi=(12)$, we get $partial_{pi^{-1}w_0^n}=partial_{1321}$?
combinatorics representation-theory schubert-calculus
edited 2 days ago
Matt Samuel
37.5k63665
asked May 4 '16 at 12:31
fierydemonfierydemon
4,40512155
See my answer to this mathoverflow question: mathoverflow.net/questions/216242/free-kx-1-dots-x-ns-n-module/…
– David Hill
May 4 '16 at 14:30
add a comment |
This question is about divided difference operators.
How do I perform $partial _2$ or $partial_3$ on $x_1^2x_2$?
$partial_i$ is defined as $frac{p-r_i.p}{x_i-x_{i+1}}$, where $r_i$ is the reflection operator; i.e. $r_i.p(x_1,x_2,dots,x_i,x_{i+1},dots,x_n)=p(x_1,x_2,dots,x_{i+1},x_i,dots,x_n)$.
Do I just assume that $x_1^2x_2=x_1^2x_2x_3^0x_4^0$?
Edit: I have another question. The calculation of Schubert polynomials uss the following formula: $S_{pi}=partial_{pi^{-1}w_0^n}S_{w_0^n}$. What does $partial_{pi^{-1}w_0^n}$ mean? Do I just concatenate $pi^{-1}$ and $w_0^n$ as words? Like if $w_0^3=321$ and $pi=(12)$, we get $partial_{pi^{-1}w_0^n}=partial_{1321}$?
combinatorics representation-theory schubert-calculus
edited 2 days ago
Matt Samuel
37.5k63665
asked May 4 '16 at 12:31
fierydemonfierydemon
4,40512155
This question is about divided difference operators.
How do I perform $partial _2$ or $partial_3$ on $x_1^2x_2$?
$partial_i$ is defined as $frac{p-r_i.p}{x_i-x_{i+1}}$, where $r_i$ is the reflection operator; i.e. $r_i.p(x_1,x_2,dots,x_i,x_{i+1},dots,x_n)=p(x_1,x_2,dots,x_{i+1},x_i,dots,x_n)$.
Do I just assume that $x_1^2x_2=x_1^2x_2x_3^0x_4^0$?
Edit: I have another question. The calculation of Schubert polynomials uss the following formula: $S_{pi}=partial_{pi^{-1}w_0^n}S_{w_0^n}$. What does $partial_{pi^{-1}w_0^n}$ mean? Do I just concatenate $pi^{-1}$ and $w_0^n$ as words? Like if $w_0^3=321$ and $pi=(12)$, we get $partial_{pi^{-1}w_0^n}=partial_{1321}$?
combinatorics representation-theory schubert-calculus
combinatorics representation-theory schubert-calculus
edited 2 days ago
Matt Samuel
37.5k63665
asked May 4 '16 at 12:31
fierydemonfierydemon
4,40512155
edited 2 days ago
Matt Samuel
37.5k63665
asked May 4 '16 at 12:31
fierydemonfierydemon
4,40512155
edited 2 days ago
Matt Samuel
37.5k63665
edited 2 days ago
Matt Samuel
37.5k63665
edited 2 days ago
Matt Samuel
37.5k63665
37.5k63665
asked May 4 '16 at 12:31
fierydemonfierydemon
4,40512155
asked May 4 '16 at 12:31
fierydemonfierydemon
4,40512155
asked May 4 '16 at 12:31
fierydemonfierydemon
4,40512155
4,40512155
See my answer to this mathoverflow question: mathoverflow.net/questions/216242/free-kx-1-dots-x-ns-n-module/…
– David Hill
May 4 '16 at 14:30
add a comment |
See my answer to this mathoverflow question: mathoverflow.net/questions/216242/free-kx-1-dots-x-ns-n-module/…
– David Hill
May 4 '16 at 14:30
See my answer to this mathoverflow question: mathoverflow.net/questions/216242/free-kx-1-dots-x-ns-n-module/…
– David Hill
May 4 '16 at 14:30
See my answer to this mathoverflow question: mathoverflow.net/questions/216242/free-kx-1-dots-x-ns-n-module/…
– David Hill
May 4 '16 at 14:30
add a comment |
1 Answer
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For the first question: Yes, $x_1^2x_2 = x_1^2x_2^1x_3^0x_4^0$, hence
$$partial_2(x_1^2x_2) = frac{x_1^2x_2 - x_1^2x_3}{x_2-x_3} = x_1^2.$$
For the second question: No, it is not just concatenation. To compute $partial_w$ you need to find a reduced word $I = [i_1, i_2, ldots, i_m]$ for $w$: write $w = r_{i_1}r_{i_2} dotsb r_{i_m}$ with $m$ minimal. Then $partial_w = partial_{i_1}partial_{i_2} dotsb partial_{i_m}.$
In your example, with permutations written in second-line notation, $w_0^3 = 321$ and $pi = 213$, hence $pi^{-1}w_0^3 = 312 = r_2r_1$. Therefore $partial_{pi^{-1}w_0^3} = partial_2partial_1$.
answered May 4 '16 at 14:02
Catalin ZaraCatalin Zara
3,717414
To Ayush: If my answer is not clear enough for you to accept it, please let me know what else should I add. Thank you.
– Catalin Zara
Nov 30 '16 at 21:06
add a comment |
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1 Answer
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
For the first question: Yes, $x_1^2x_2 = x_1^2x_2^1x_3^0x_4^0$, hence
$$partial_2(x_1^2x_2) = frac{x_1^2x_2 - x_1^2x_3}{x_2-x_3} = x_1^2.$$
For the second question: No, it is not just concatenation. To compute $partial_w$ you need to find a reduced word $I = [i_1, i_2, ldots, i_m]$ for $w$: write $w = r_{i_1}r_{i_2} dotsb r_{i_m}$ with $m$ minimal. Then $partial_w = partial_{i_1}partial_{i_2} dotsb partial_{i_m}.$
In your example, with permutations written in second-line notation, $w_0^3 = 321$ and $pi = 213$, hence $pi^{-1}w_0^3 = 312 = r_2r_1$. Therefore $partial_{pi^{-1}w_0^3} = partial_2partial_1$.
answered May 4 '16 at 14:02
Catalin ZaraCatalin Zara
3,717414
To Ayush: If my answer is not clear enough for you to accept it, please let me know what else should I add. Thank you.
– Catalin Zara
Nov 30 '16 at 21:06
add a comment |
For the first question: Yes, $x_1^2x_2 = x_1^2x_2^1x_3^0x_4^0$, hence
$$partial_2(x_1^2x_2) = frac{x_1^2x_2 - x_1^2x_3}{x_2-x_3} = x_1^2.$$
For the second question: No, it is not just concatenation. To compute $partial_w$ you need to find a reduced word $I = [i_1, i_2, ldots, i_m]$ for $w$: write $w = r_{i_1}r_{i_2} dotsb r_{i_m}$ with $m$ minimal. Then $partial_w = partial_{i_1}partial_{i_2} dotsb partial_{i_m}.$
In your example, with permutations written in second-line notation, $w_0^3 = 321$ and $pi = 213$, hence $pi^{-1}w_0^3 = 312 = r_2r_1$. Therefore $partial_{pi^{-1}w_0^3} = partial_2partial_1$.
answered May 4 '16 at 14:02
Catalin ZaraCatalin Zara
3,717414
To Ayush: If my answer is not clear enough for you to accept it, please let me know what else should I add. Thank you.
– Catalin Zara
Nov 30 '16 at 21:06
add a comment |
For the first question: Yes, $x_1^2x_2 = x_1^2x_2^1x_3^0x_4^0$, hence
$$partial_2(x_1^2x_2) = frac{x_1^2x_2 - x_1^2x_3}{x_2-x_3} = x_1^2.$$
For the second question: No, it is not just concatenation. To compute $partial_w$ you need to find a reduced word $I = [i_1, i_2, ldots, i_m]$ for $w$: write $w = r_{i_1}r_{i_2} dotsb r_{i_m}$ with $m$ minimal. Then $partial_w = partial_{i_1}partial_{i_2} dotsb partial_{i_m}.$
In your example, with permutations written in second-line notation, $w_0^3 = 321$ and $pi = 213$, hence $pi^{-1}w_0^3 = 312 = r_2r_1$. Therefore $partial_{pi^{-1}w_0^3} = partial_2partial_1$.
answered May 4 '16 at 14:02
Catalin ZaraCatalin Zara
3,717414
For the first question: Yes, $x_1^2x_2 = x_1^2x_2^1x_3^0x_4^0$, hence
$$partial_2(x_1^2x_2) = frac{x_1^2x_2 - x_1^2x_3}{x_2-x_3} = x_1^2.$$
For the second question: No, it is not just concatenation. To compute $partial_w$ you need to find a reduced word $I = [i_1, i_2, ldots, i_m]$ for $w$: write $w = r_{i_1}r_{i_2} dotsb r_{i_m}$ with $m$ minimal. Then $partial_w = partial_{i_1}partial_{i_2} dotsb partial_{i_m}.$
In your example, with permutations written in second-line notation, $w_0^3 = 321$ and $pi = 213$, hence $pi^{-1}w_0^3 = 312 = r_2r_1$. Therefore $partial_{pi^{-1}w_0^3} = partial_2partial_1$.
answered May 4 '16 at 14:02
Catalin ZaraCatalin Zara
3,717414
edited May 4 '16 at 16:04
answered May 4 '16 at 14:02
Catalin ZaraCatalin Zara
3,717414
answered May 4 '16 at 14:02
Catalin ZaraCatalin Zara
3,717414
answered May 4 '16 at 14:02
Catalin ZaraCatalin Zara
3,717414
3,717414
To Ayush: If my answer is not clear enough for you to accept it, please let me know what else should I add. Thank you.
– Catalin Zara
Nov 30 '16 at 21:06
add a comment |
To Ayush: If my answer is not clear enough for you to accept it, please let me know what else should I add. Thank you.
– Catalin Zara
Nov 30 '16 at 21:06
To Ayush: If my answer is not clear enough for you to accept it, please let me know what else should I add. Thank you.
– Catalin Zara
Nov 30 '16 at 21:06
To Ayush: If my answer is not clear enough for you to accept it, please let me know what else should I add. Thank you.
– Catalin Zara
Nov 30 '16 at 21:06
add a comment |
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See my answer to this mathoverflow question: mathoverflow.net/questions/216242/free-kx-1-dots-x-ns-n-module/…
– David Hill
May 4 '16 at 14:30