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$mathbb P^1(mathbb C)$ inside $Gr(2,4)$
$begingroup$
Let $V$ be a vector space over $mathbb C$ of dimension $4$ and consider the Grassmanian $Gr(2,V)$. Fix an index $I=(1,3)$ and a complete flag of $V=langle e_1, e_2, e_3, e_4 rangle$ given by $ F =langle e_1 rangle subset langle e_1, e_2 rangle subset langle e_1, e_2, e_3 rangle subset V$.
The associated Schubert variety $X_I(F)$ is given by the span of the matrices
$$
begin{bmatrix}
1 & 0 \
0 & * \
0 & 1 \
0 & 0 \
end{bmatrix}
cup
begin{bmatrix}
1 & 0 \
0 & 1 \
0 & 0 \
0 & 0 \
end{bmatrix}
$$
Since it is $mathbb C cup mathbb C^0$ I'm wondering if it is a copy of $mathbb P^1(mathbb C)$ inside the Grassmanian in some way, or if this is not exact.
Thanks!
algebraic-geometry complex-geometry schubert-calculus
edited Jan 5 at 12:38
Matt Samuel
37.8k63665
asked Jul 17 '17 at 0:26
MaffredMaffred
2,680625
$endgroup$
add a comment |
$begingroup$
Let $V$ be a vector space over $mathbb C$ of dimension $4$ and consider the Grassmanian $Gr(2,V)$. Fix an index $I=(1,3)$ and a complete flag of $V=langle e_1, e_2, e_3, e_4 rangle$ given by $ F =langle e_1 rangle subset langle e_1, e_2 rangle subset langle e_1, e_2, e_3 rangle subset V$.
The associated Schubert variety $X_I(F)$ is given by the span of the matrices
$$
begin{bmatrix}
1 & 0 \
0 & * \
0 & 1 \
0 & 0 \
end{bmatrix}
cup
begin{bmatrix}
1 & 0 \
0 & 1 \
0 & 0 \
0 & 0 \
end{bmatrix}
$$
Since it is $mathbb C cup mathbb C^0$ I'm wondering if it is a copy of $mathbb P^1(mathbb C)$ inside the Grassmanian in some way, or if this is not exact.
Thanks!
algebraic-geometry complex-geometry schubert-calculus
edited Jan 5 at 12:38
Matt Samuel
37.8k63665
asked Jul 17 '17 at 0:26
MaffredMaffred
2,680625
$endgroup$
add a comment |
$begingroup$
Let $V$ be a vector space over $mathbb C$ of dimension $4$ and consider the Grassmanian $Gr(2,V)$. Fix an index $I=(1,3)$ and a complete flag of $V=langle e_1, e_2, e_3, e_4 rangle$ given by $ F =langle e_1 rangle subset langle e_1, e_2 rangle subset langle e_1, e_2, e_3 rangle subset V$.
The associated Schubert variety $X_I(F)$ is given by the span of the matrices
$$
begin{bmatrix}
1 & 0 \
0 & * \
0 & 1 \
0 & 0 \
end{bmatrix}
cup
begin{bmatrix}
1 & 0 \
0 & 1 \
0 & 0 \
0 & 0 \
end{bmatrix}
$$
Since it is $mathbb C cup mathbb C^0$ I'm wondering if it is a copy of $mathbb P^1(mathbb C)$ inside the Grassmanian in some way, or if this is not exact.
Thanks!
algebraic-geometry complex-geometry schubert-calculus
edited Jan 5 at 12:38
Matt Samuel
37.8k63665
asked Jul 17 '17 at 0:26
MaffredMaffred
2,680625
$endgroup$
Let $V$ be a vector space over $mathbb C$ of dimension $4$ and consider the Grassmanian $Gr(2,V)$. Fix an index $I=(1,3)$ and a complete flag of $V=langle e_1, e_2, e_3, e_4 rangle$ given by $ F =langle e_1 rangle subset langle e_1, e_2 rangle subset langle e_1, e_2, e_3 rangle subset V$.
The associated Schubert variety $X_I(F)$ is given by the span of the matrices
$$
begin{bmatrix}
1 & 0 \
0 & * \
0 & 1 \
0 & 0 \
end{bmatrix}
cup
begin{bmatrix}
1 & 0 \
0 & 1 \
0 & 0 \
0 & 0 \
end{bmatrix}
$$
Since it is $mathbb C cup mathbb C^0$ I'm wondering if it is a copy of $mathbb P^1(mathbb C)$ inside the Grassmanian in some way, or if this is not exact.
Thanks!
algebraic-geometry complex-geometry schubert-calculus
algebraic-geometry complex-geometry schubert-calculus
edited Jan 5 at 12:38
Matt Samuel
37.8k63665
asked Jul 17 '17 at 0:26
MaffredMaffred
2,680625
edited Jan 5 at 12:38
Matt Samuel
37.8k63665
asked Jul 17 '17 at 0:26
MaffredMaffred
2,680625
edited Jan 5 at 12:38
Matt Samuel
37.8k63665
edited Jan 5 at 12:38
Matt Samuel
37.8k63665
edited Jan 5 at 12:38
Matt Samuel
37.8k63665
37.8k63665
asked Jul 17 '17 at 0:26
MaffredMaffred
2,680625
asked Jul 17 '17 at 0:26
MaffredMaffred
2,680625
asked Jul 17 '17 at 0:26
MaffredMaffred
2,680625
2,680625
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Yes, this Schubert cycle is the set of $2$-planes containing a fixed line and contained in a fixed $3$-space. So it is, in particular, the set of $2$-planes in $Bbb C^3$ containing a fixed line. This is, alternatively, the set of $Bbb P^1$'s in $Bbb P^2$ containing a fixed point, and that is easily seen to be a $Bbb P^1$. (For example, choose a $Bbb P^1subsetBbb P^2$ not containing the fixed point $p$. Each line through $p$ meets that $Bbb P^1$ in a unique point, so the pencil of lines through $p$ is isomorphic to that $Bbb P^1$.)
answered Jul 17 '17 at 5:26
Ted ShifrinTed Shifrin
63.1k44489
$endgroup$
add a comment |
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1 Answer
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1 Answer
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votes
$begingroup$
Yes, this Schubert cycle is the set of $2$-planes containing a fixed line and contained in a fixed $3$-space. So it is, in particular, the set of $2$-planes in $Bbb C^3$ containing a fixed line. This is, alternatively, the set of $Bbb P^1$'s in $Bbb P^2$ containing a fixed point, and that is easily seen to be a $Bbb P^1$. (For example, choose a $Bbb P^1subsetBbb P^2$ not containing the fixed point $p$. Each line through $p$ meets that $Bbb P^1$ in a unique point, so the pencil of lines through $p$ is isomorphic to that $Bbb P^1$.)
answered Jul 17 '17 at 5:26
Ted ShifrinTed Shifrin
63.1k44489
$endgroup$
add a comment |
$begingroup$
Yes, this Schubert cycle is the set of $2$-planes containing a fixed line and contained in a fixed $3$-space. So it is, in particular, the set of $2$-planes in $Bbb C^3$ containing a fixed line. This is, alternatively, the set of $Bbb P^1$'s in $Bbb P^2$ containing a fixed point, and that is easily seen to be a $Bbb P^1$. (For example, choose a $Bbb P^1subsetBbb P^2$ not containing the fixed point $p$. Each line through $p$ meets that $Bbb P^1$ in a unique point, so the pencil of lines through $p$ is isomorphic to that $Bbb P^1$.)
answered Jul 17 '17 at 5:26
Ted ShifrinTed Shifrin
63.1k44489
$endgroup$
add a comment |
$begingroup$
Yes, this Schubert cycle is the set of $2$-planes containing a fixed line and contained in a fixed $3$-space. So it is, in particular, the set of $2$-planes in $Bbb C^3$ containing a fixed line. This is, alternatively, the set of $Bbb P^1$'s in $Bbb P^2$ containing a fixed point, and that is easily seen to be a $Bbb P^1$. (For example, choose a $Bbb P^1subsetBbb P^2$ not containing the fixed point $p$. Each line through $p$ meets that $Bbb P^1$ in a unique point, so the pencil of lines through $p$ is isomorphic to that $Bbb P^1$.)
answered Jul 17 '17 at 5:26
Ted ShifrinTed Shifrin
63.1k44489
$endgroup$
Yes, this Schubert cycle is the set of $2$-planes containing a fixed line and contained in a fixed $3$-space. So it is, in particular, the set of $2$-planes in $Bbb C^3$ containing a fixed line. This is, alternatively, the set of $Bbb P^1$'s in $Bbb P^2$ containing a fixed point, and that is easily seen to be a $Bbb P^1$. (For example, choose a $Bbb P^1subsetBbb P^2$ not containing the fixed point $p$. Each line through $p$ meets that $Bbb P^1$ in a unique point, so the pencil of lines through $p$ is isomorphic to that $Bbb P^1$.)
answered Jul 17 '17 at 5:26
Ted ShifrinTed Shifrin
63.1k44489
answered Jul 17 '17 at 5:26
Ted ShifrinTed Shifrin
63.1k44489
answered Jul 17 '17 at 5:26
Ted ShifrinTed Shifrin
63.1k44489
answered Jul 17 '17 at 5:26
Ted ShifrinTed Shifrin
63.1k44489
63.1k44489
add a comment |
add a comment |
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