Mixing
Poincare duality pairing matrix Grassmannian
$begingroup$
I'm trying to prove that for $i+j = 2k(n-k)$ and $i = 2s$, $j = 2l$, ($i neq j$, this is easy) the following Poincaré duality pairing holds:
$$
H^i(mathbb{G}(k, n)) times H^{j}(mathbb{G}(k, n)) longrightarrow mathbb{Z}
$$
where $mathbb{G}$ is the Grassmannian. I can create a basis for the two spaces $H^i(mathbb{G}(k, n))$ and $H^{j}(mathbb{G}(k, n))$ by the Schubert classes $[X_{lambda}]$ with $|lambda| = s$ and $[X_{gamma}]$ with $|gamma| = l$ and with them construct the pairing $s times l$ matrix. I know that this matrix must have determinant $pm 1$ and I was told I can explicitly compute the entries, but I did not manage to do it.
Thank you very much for any help.
algebraic-topology grassmannian schubert-calculus
edited Jan 5 at 15:51
Matt Samuel
37.8k63665
asked Oct 29 '18 at 16:37
Francesco CarzanigaFrancesco Carzaniga
302110
$endgroup$
add a comment |
$begingroup$
I'm trying to prove that for $i+j = 2k(n-k)$ and $i = 2s$, $j = 2l$, ($i neq j$, this is easy) the following Poincaré duality pairing holds:
$$
H^i(mathbb{G}(k, n)) times H^{j}(mathbb{G}(k, n)) longrightarrow mathbb{Z}
$$
where $mathbb{G}$ is the Grassmannian. I can create a basis for the two spaces $H^i(mathbb{G}(k, n))$ and $H^{j}(mathbb{G}(k, n))$ by the Schubert classes $[X_{lambda}]$ with $|lambda| = s$ and $[X_{gamma}]$ with $|gamma| = l$ and with them construct the pairing $s times l$ matrix. I know that this matrix must have determinant $pm 1$ and I was told I can explicitly compute the entries, but I did not manage to do it.
Thank you very much for any help.
algebraic-topology grassmannian schubert-calculus
edited Jan 5 at 15:51
Matt Samuel
37.8k63665
asked Oct 29 '18 at 16:37
Francesco CarzanigaFrancesco Carzaniga
302110
$endgroup$
add a comment |
$begingroup$
I'm trying to prove that for $i+j = 2k(n-k)$ and $i = 2s$, $j = 2l$, ($i neq j$, this is easy) the following Poincaré duality pairing holds:
$$
H^i(mathbb{G}(k, n)) times H^{j}(mathbb{G}(k, n)) longrightarrow mathbb{Z}
$$
where $mathbb{G}$ is the Grassmannian. I can create a basis for the two spaces $H^i(mathbb{G}(k, n))$ and $H^{j}(mathbb{G}(k, n))$ by the Schubert classes $[X_{lambda}]$ with $|lambda| = s$ and $[X_{gamma}]$ with $|gamma| = l$ and with them construct the pairing $s times l$ matrix. I know that this matrix must have determinant $pm 1$ and I was told I can explicitly compute the entries, but I did not manage to do it.
Thank you very much for any help.
algebraic-topology grassmannian schubert-calculus
edited Jan 5 at 15:51
Matt Samuel
37.8k63665
asked Oct 29 '18 at 16:37
Francesco CarzanigaFrancesco Carzaniga
302110
$endgroup$
I'm trying to prove that for $i+j = 2k(n-k)$ and $i = 2s$, $j = 2l$, ($i neq j$, this is easy) the following Poincaré duality pairing holds:
$$
H^i(mathbb{G}(k, n)) times H^{j}(mathbb{G}(k, n)) longrightarrow mathbb{Z}
$$
where $mathbb{G}$ is the Grassmannian. I can create a basis for the two spaces $H^i(mathbb{G}(k, n))$ and $H^{j}(mathbb{G}(k, n))$ by the Schubert classes $[X_{lambda}]$ with $|lambda| = s$ and $[X_{gamma}]$ with $|gamma| = l$ and with them construct the pairing $s times l$ matrix. I know that this matrix must have determinant $pm 1$ and I was told I can explicitly compute the entries, but I did not manage to do it.
Thank you very much for any help.
algebraic-topology grassmannian schubert-calculus
algebraic-topology grassmannian schubert-calculus
edited Jan 5 at 15:51
Matt Samuel
37.8k63665
asked Oct 29 '18 at 16:37
Francesco CarzanigaFrancesco Carzaniga
302110
edited Jan 5 at 15:51
Matt Samuel
37.8k63665
asked Oct 29 '18 at 16:37
Francesco CarzanigaFrancesco Carzaniga
302110
edited Jan 5 at 15:51
Matt Samuel
37.8k63665
edited Jan 5 at 15:51
Matt Samuel
37.8k63665
edited Jan 5 at 15:51
Matt Samuel
37.8k63665
37.8k63665
asked Oct 29 '18 at 16:37
Francesco CarzanigaFrancesco Carzaniga
302110
asked Oct 29 '18 at 16:37
Francesco CarzanigaFrancesco Carzaniga
302110
asked Oct 29 '18 at 16:37
Francesco CarzanigaFrancesco Carzaniga
302110
302110
add a comment |
add a comment |
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