Mixing
Two questions about Schubert calculus and Schur functions.
$begingroup$
I am reading the file. I have a question on pae 28. How to prove that $[X_{{2,4}}] = S_{(1)} = x_1 + x_2 + cdots$ and $S_{(1)}^4 = 2 S_{(2,2)} + S_{(3,1)} + S_{(2,1,1)}$? I tried to verify $S_{(1)}^4 = 2 S_{(2,2)} + S_{(3,1)} + S_{(2,1,1)}$ directly. But on the left hand side we have a term $x_1^4$ and it seems that on the right hand side we don't have $x_1^4$. Thank you very much.
geometry algebraic-geometry representation-theory schubert-calculus
edited Jan 2 at 1:03
darij grinberg
10.4k33062
asked Aug 12 '15 at 7:09
LJRLJR
6,57341749
$endgroup$
add a comment |
$begingroup$
I am reading the file. I have a question on pae 28. How to prove that $[X_{{2,4}}] = S_{(1)} = x_1 + x_2 + cdots$ and $S_{(1)}^4 = 2 S_{(2,2)} + S_{(3,1)} + S_{(2,1,1)}$? I tried to verify $S_{(1)}^4 = 2 S_{(2,2)} + S_{(3,1)} + S_{(2,1,1)}$ directly. But on the left hand side we have a term $x_1^4$ and it seems that on the right hand side we don't have $x_1^4$. Thank you very much.
geometry algebraic-geometry representation-theory schubert-calculus
edited Jan 2 at 1:03
darij grinberg
10.4k33062
asked Aug 12 '15 at 7:09
LJRLJR
6,57341749
$endgroup$
add a comment |
$begingroup$
I am reading the file. I have a question on pae 28. How to prove that $[X_{{2,4}}] = S_{(1)} = x_1 + x_2 + cdots$ and $S_{(1)}^4 = 2 S_{(2,2)} + S_{(3,1)} + S_{(2,1,1)}$? I tried to verify $S_{(1)}^4 = 2 S_{(2,2)} + S_{(3,1)} + S_{(2,1,1)}$ directly. But on the left hand side we have a term $x_1^4$ and it seems that on the right hand side we don't have $x_1^4$. Thank you very much.
geometry algebraic-geometry representation-theory schubert-calculus
edited Jan 2 at 1:03
darij grinberg
10.4k33062
asked Aug 12 '15 at 7:09
LJRLJR
6,57341749
$endgroup$
I am reading the file. I have a question on pae 28. How to prove that $[X_{{2,4}}] = S_{(1)} = x_1 + x_2 + cdots$ and $S_{(1)}^4 = 2 S_{(2,2)} + S_{(3,1)} + S_{(2,1,1)}$? I tried to verify $S_{(1)}^4 = 2 S_{(2,2)} + S_{(3,1)} + S_{(2,1,1)}$ directly. But on the left hand side we have a term $x_1^4$ and it seems that on the right hand side we don't have $x_1^4$. Thank you very much.
geometry algebraic-geometry representation-theory schubert-calculus
geometry algebraic-geometry representation-theory schubert-calculus
edited Jan 2 at 1:03
darij grinberg
10.4k33062
asked Aug 12 '15 at 7:09
LJRLJR
6,57341749
edited Jan 2 at 1:03
darij grinberg
10.4k33062
asked Aug 12 '15 at 7:09
LJRLJR
6,57341749
edited Jan 2 at 1:03
darij grinberg
10.4k33062
edited Jan 2 at 1:03
darij grinberg
10.4k33062
edited Jan 2 at 1:03
darij grinberg
10.4k33062
10.4k33062
asked Aug 12 '15 at 7:09
LJRLJR
6,57341749
asked Aug 12 '15 at 7:09
LJRLJR
6,57341749
asked Aug 12 '15 at 7:09
LJRLJR
6,57341749
6,57341749
add a comment |
add a comment |
1 Answer
1
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$begingroup$
That formula is not correct. The correct decomposition is $$S_{(1)}^4 = S_{(4)} + 3 S_{(3,1)} + 2 S_{(2,2)} + 3 S_{(2,1,1)} + S_{(1,1,1,1)}.$$ This can be calculated by the formula $$S_lambda cdot S_{(1)} = sum_{mu} S_{mu}$$ where $mu$ ranges over all partitions obtained from $lambda$ by adding a single box. (Pieri's formula)
answered Aug 12 '15 at 7:52
TedTed
21.5k13260
$endgroup$
add a comment |
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1 Answer
1
active
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1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
That formula is not correct. The correct decomposition is $$S_{(1)}^4 = S_{(4)} + 3 S_{(3,1)} + 2 S_{(2,2)} + 3 S_{(2,1,1)} + S_{(1,1,1,1)}.$$ This can be calculated by the formula $$S_lambda cdot S_{(1)} = sum_{mu} S_{mu}$$ where $mu$ ranges over all partitions obtained from $lambda$ by adding a single box. (Pieri's formula)
answered Aug 12 '15 at 7:52
TedTed
21.5k13260
$endgroup$
add a comment |
$begingroup$
That formula is not correct. The correct decomposition is $$S_{(1)}^4 = S_{(4)} + 3 S_{(3,1)} + 2 S_{(2,2)} + 3 S_{(2,1,1)} + S_{(1,1,1,1)}.$$ This can be calculated by the formula $$S_lambda cdot S_{(1)} = sum_{mu} S_{mu}$$ where $mu$ ranges over all partitions obtained from $lambda$ by adding a single box. (Pieri's formula)
answered Aug 12 '15 at 7:52
TedTed
21.5k13260
$endgroup$
add a comment |
$begingroup$
That formula is not correct. The correct decomposition is $$S_{(1)}^4 = S_{(4)} + 3 S_{(3,1)} + 2 S_{(2,2)} + 3 S_{(2,1,1)} + S_{(1,1,1,1)}.$$ This can be calculated by the formula $$S_lambda cdot S_{(1)} = sum_{mu} S_{mu}$$ where $mu$ ranges over all partitions obtained from $lambda$ by adding a single box. (Pieri's formula)
answered Aug 12 '15 at 7:52
TedTed
21.5k13260
$endgroup$
That formula is not correct. The correct decomposition is $$S_{(1)}^4 = S_{(4)} + 3 S_{(3,1)} + 2 S_{(2,2)} + 3 S_{(2,1,1)} + S_{(1,1,1,1)}.$$ This can be calculated by the formula $$S_lambda cdot S_{(1)} = sum_{mu} S_{mu}$$ where $mu$ ranges over all partitions obtained from $lambda$ by adding a single box. (Pieri's formula)
answered Aug 12 '15 at 7:52
TedTed
21.5k13260
answered Aug 12 '15 at 7:52
TedTed
21.5k13260
answered Aug 12 '15 at 7:52
TedTed
21.5k13260
answered Aug 12 '15 at 7:52
TedTed
21.5k13260
21.5k13260
add a comment |
add a comment |
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