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I give an answer here using only elementary algebraic geometry.

We know that the Grassmannian Gr(2,5) is defined by 5 linearly independent quadrics in P9 and V(5) can be viewed as defined by 5 linearly independent quadrics in P6.

Given a point p=[1,0,,0]X for XP6 defined by Qi,1i5, where Qi are homogeneous of degree 2. Write Qi=qi,2(x1,,x6)+x0qi,1(x1,,x6) with qi,k homogeneous of degree k.

I claim that there there is a line contained in X passing through p if and only if qi,k=0,1i5,k=1,2() have a common zero in P5=P(k[x1,x6]). To see this, note that given a common zero [a1,a2,,a6]P5, the line joining p to [0,a1,a2,,a6] is contained in X and the line joining [1,0,,0] and [1,a1,,an] is the same as the line joining [1,0,,0] and [0,a1,,an].

Since Qi are linearly independent, we see that the linear part qi,1,1i5 are linearly independent. By the geometric version of Krull’s Principal Ideal Theorem[11.3.2, Vakil’s notes], the dimension of the subvariety in P5 defined by () is 0. This means () has finitely many common zeros, that is, there are at most finitely many lines passing through a given point.

Now back to X=V(5), as suggested by Enrico in the comment. We take 3 hyperplanes ''general enough'' to get explicit defining equations of V(5). Consider the Plucker coordinate wij in P9, take w12=w34, w13=w45, w23=w15, the defining equations for V(5) are

{w234w45w24+w14w15=0w245w14w35+w15w34=0w215w45w25+w34w35=0w15w45w24w35+w25w34=0w34w45w14w25+w15w24=0

By the above argument, choose p to be the point with the only nonzero entry wij for {i,j}{3,4},{4,5},{1,5} we can have some lines. For example, we can find two nonintersecting lines: one joining [x14=1,xij=0] and [x12=1,xij=0], the other joining [x24=1,xij=0] and [x25=1,xij=0]. This has solved the question.


Remark. Suppose we are given the fact that for every point, there is at least one line passing through it, we can show that the subvariety in Gr(2,7) characterising lines on V(5) has dimension 2. Consider the following incidence correspondence, as we have discussed above, for any pV(5) the fibre pr11(p) has dimension 0.

[![enter image description here][1]][1]enter image description here

By Upper Semicontinuity of Fibre Dimension, we have dimΦ=dimV(5)=3. Take a line lV(5), dimpr12([l])=1, by Upper Semicontinuity again, dimΦdimpr2Φ=1, dimpr2Φ=2. We conclude that the subvariety of Grass(2,7) characterising lines on V(5) has dimension 2.

A stronger result than dimpr2Φ=2 is that the Hilbert scheme of lines on V(5) is isomorphic to P2 as in Sasha's answer and the reference is [Theorem 1, 2]. And the number of lines passing through a point is 3 when counted with multiplicity [P.26, 1].

[1] Takagi, Hiromichi; Zucconi, Francesco, Geometries of lines and conics on the quintic del Pezzo 3-fold and its application to varieties of power sums, Mich. Math. J. 61, No. 1, 19-62 (2012). ZBL1262.14048.

[2]Furushima, Mikio; Nakayama, Noboru, The family of lines on the Fano threefold (V_ 5), Nagoya Math. J. 116, 111-122 (1989). ZBL0731.14025. [1]: https://i.sstatic.net/M6cjS.png

I give an answer here using only elementary algebraic geometry.

We know that the Grassmannian Gr(2,5) is defined by 5 linearly independent quadrics in P9 and V(5) can be viewed as defined by 5 linearly independent quadrics in P6.

Given a point p=[1,0,,0]X for XP6 defined by Qi,1i5, where Qi are homogeneous of degree 2. Write Qi=qi,2(x1,,x6)+x0qi,1(x1,,x6) with qi,k homogeneous of degree k.

I claim that there there is a line contained in X passing through p if and only if qi,k=0,1i5,k=1,2() have a common zero in P5=P(k[x1,x6]). To see this, note that given a common zero [a1,a2,,a6]P5, the line joining p to [0,a1,a2,,a6] is contained in X and the line joining [1,0,,0] and [1,a1,,an] is the same as the line joining [1,0,,0] and [0,a1,,an].

Since Qi are linearly independent, we see that the linear part qi,1,1i5 are linearly independent. By the geometric version of Krull’s Principal Ideal Theorem[11.3.2, Vakil’s notes], the dimension of the subvariety in P5 defined by () is 0. This means () has finitely many common zeros, that is, there are at most finitely many lines passing through a given point.

Now back to X=V(5), as suggested by Enrico in the comment. We take 3 hyperplanes ''general enough'' to get explicit defining equations of V(5). Consider the Plucker coordinate wij in P9, take w12=w34, w13=w45, w23=w15, the defining equations for V(5) are

{w234w45w24+w14w15=0w245w14w35+w15w34=0w215w45w25+w34w35=0w15w45w24w35+w25w34=0w34w45w14w25+w15w24=0

By the above argument, choose p to be the point with the only nonzero entry wij for {i,j}{3,4},{4,5},{1,5} we can have some lines. For example, we can find two nonintersecting lines: one joining [x14=1,xij=0] and [x12=1,xij=0], the other joining [x24=1,xij=0] and [x25=1,xij=0]. This has solved the question.


Remark. Suppose we are given the fact that for every point, there is at least one line passing through it, we can show that the subvariety in Gr(2,7) characterising lines on V(5) has dimension 2. Consider the following incidence correspondence, as we have discussed above, for any pV(5) the fibre pr11(p) has dimension 0.

[![enter image description here][1]][1]

By Upper Semicontinuity of Fibre Dimension, we have dimΦ=dimV(5)=3. Take a line lV(5), dimpr12([l])=1, by Upper Semicontinuity again, dimΦdimpr2Φ=1, dimpr2Φ=2. We conclude that the subvariety of Grass(2,7) characterising lines on V(5) has dimension 2.

A stronger result than dimpr2Φ=2 is that the Hilbert scheme of lines on V(5) is isomorphic to P2 as in Sasha's answer and the reference is [Theorem 1, 2]. And the number of lines passing through a point is 3 when counted with multiplicity [P.26, 1].

[1] Takagi, Hiromichi; Zucconi, Francesco, Geometries of lines and conics on the quintic del Pezzo 3-fold and its application to varieties of power sums, Mich. Math. J. 61, No. 1, 19-62 (2012). ZBL1262.14048.

[2]Furushima, Mikio; Nakayama, Noboru, The family of lines on the Fano threefold (V_ 5), Nagoya Math. J. 116, 111-122 (1989). ZBL0731.14025. [1]: https://i.sstatic.net/M6cjS.png

I give an answer here using only elementary algebraic geometry.

We know that the Grassmannian Gr(2,5) is defined by 5 linearly independent quadrics in P9 and V(5) can be viewed as defined by 5 linearly independent quadrics in P6.

Given a point p=[1,0,,0]X for XP6 defined by Qi,1i5, where Qi are homogeneous of degree 2. Write Qi=qi,2(x1,,x6)+x0qi,1(x1,,x6) with qi,k homogeneous of degree k.

I claim that there there is a line contained in X passing through p if and only if qi,k=0,1i5,k=1,2() have a common zero in P5=P(k[x1,x6]). To see this, note that given a common zero [a1,a2,,a6]P5, the line joining p to [0,a1,a2,,a6] is contained in X and the line joining [1,0,,0] and [1,a1,,an] is the same as the line joining [1,0,,0] and [0,a1,,an].

Since Qi are linearly independent, we see that the linear part qi,1,1i5 are linearly independent. By the geometric version of Krull’s Principal Ideal Theorem[11.3.2, Vakil’s notes], the dimension of the subvariety in P5 defined by () is 0. This means () has finitely many common zeros, that is, there are at most finitely many lines passing through a given point.

Now back to X=V(5), as suggested by Enrico in the comment. We take 3 hyperplanes ''general enough'' to get explicit defining equations of V(5). Consider the Plucker coordinate wij in P9, take w12=w34, w13=w45, w23=w15, the defining equations for V(5) are

{w234w45w24+w14w15=0w245w14w35+w15w34=0w215w45w25+w34w35=0w15w45w24w35+w25w34=0w34w45w14w25+w15w24=0

By the above argument, choose p to be the point with the only nonzero entry wij for {i,j}{3,4},{4,5},{1,5} we can have some lines. For example, we can find two nonintersecting lines: one joining [x14=1,xij=0] and [x12=1,xij=0], the other joining [x24=1,xij=0] and [x25=1,xij=0]. This has solved the question.


Remark. Suppose we are given the fact that for every point, there is at least one line passing through it, we can show that the subvariety in Gr(2,7) characterising lines on V(5) has dimension 2. Consider the following incidence correspondence, as we have discussed above, for any pV(5) the fibre pr11(p) has dimension 0.

enter image description here

By Upper Semicontinuity of Fibre Dimension, we have dimΦ=dimV(5)=3. Take a line lV(5), dimpr12([l])=1, by Upper Semicontinuity again, dimΦdimpr2Φ=1, dimpr2Φ=2. We conclude that the subvariety of Grass(2,7) characterising lines on V(5) has dimension 2.

A stronger result than dimpr2Φ=2 is that the Hilbert scheme of lines on V(5) is isomorphic to P2 as in Sasha's answer and the reference is [Theorem 1, 2]. And the number of lines passing through a point is 3 when counted with multiplicity [P.26, 1].

[1] Takagi, Hiromichi; Zucconi, Francesco, Geometries of lines and conics on the quintic del Pezzo 3-fold and its application to varieties of power sums, Mich. Math. J. 61, No. 1, 19-62 (2012). ZBL1262.14048.

[2]Furushima, Mikio; Nakayama, Noboru, The family of lines on the Fano threefold (V_ 5), Nagoya Math. J. 116, 111-122 (1989). ZBL0731.14025.

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L. Xie
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I give an answer here using only elementary algebraic geometry.

We know that the Grassmannian Gr(2,5) is defined by 5 linearly independent quadrics in P9 and V(5) can be viewed as defined by 5 linearly independent quadrics in P6.

Given a point p=[1,0,,0]X for XP6 defined by Qi,1i5, where Qi are homogeneous of degree 2. Write Qi=qi,2(x1,,x6)+x0qi,1(x1,,x6) with qi,k homogeneous of degree k.

I claim that there there is a line contained in X passing through p if and only if qi,k=0,1i5,k=1,2() have a common zero in P5=P(k[x1,x6]). To see this, note that given a common zero [a1,a2,,a6]P5, the line joining p to [0,a1,a2,,a6] is contained in X and the line joining [1,0,,0] and [1,a1,,an] is the same as the line joining [1,0,,0] and [0,a1,,an].

Since Qi are linearly independent, we see that the linear part qi,1,1i5 are linearly independent. By the geometric version of Krull’s Principal Ideal Theorem[11.3.2, Vakil’s notes], the dimension of the subvariety in P5 defined by () is 0. This means () has finitely many common zeros, that is, there are at most finitely many lines passing through a given point.

Now back to X=V(5), as suggested by Enrico in the comment. We take 3 hyperplanes ''general enough'' to get explicit defining equations of V(5). Consider the Plucker coordinate wij in P9, take w12=w34, w13=w45, w23=w15, the defining equations for V(5) are

{w234w45w24+w14w15=0w245w14w35+w15w34=0w215w45w25+w34w35=0w15w45w24w35+w25w34=0w34w45w14w25+w15w24=0

By the above argument, choose p to be the point with the only nonzero entry wij for {i,j}{3,4},{4,5},{1,5} we can have some lines. For example, we can find two nonintersecting lines: one joining [x14=1,xij=0] and [x12=1,xij=0], the other joining [x24=1,xij=0] and [x25=1,xij=0]. This has solved the question.


Remark. Suppose we are given the fact that for every point, there is at least one line passing through it, we can show that the subvariety in Gr(2,7) characterising lines on V(5) has dimension 22. Consider the following incidence correspondence, as we have discussed above, for any pV(5) the fibre pr11(p) has dimension 0.

[![enter image description here][1]][1]

By Upper Semicontinuity of Fibre Dimension, we have dimΦ=dimV(5)=3. Take a line lV(5), dimpr12([l])=1, by Upper Semicontinuity again, dimΦdimpr2Φ1dimΦdimpr2Φ=1, dimpr2Φ2dimpr2Φ=2. We conclude that the subvariety of Grass(2,7) characterising lines on V(5) has dimension 22.

A stronger result than dimpr2Φ2dimpr2Φ=2 is that the Hilbert scheme of lines on V(5) is isomorphic to P2 as in Sasha's answer and the reference is [Theorem 1, 2]. And the number of lines passing through a point is 3 when counted with multiplicity [P.26, 1].

[1] Takagi, Hiromichi; Zucconi, Francesco, Geometries of lines and conics on the quintic del Pezzo 3-fold and its application to varieties of power sums, Mich. Math. J. 61, No. 1, 19-62 (2012). ZBL1262.14048.

[2]Furushima, Mikio; Nakayama, Noboru, The family of lines on the Fano threefold (V_ 5), Nagoya Math. J. 116, 111-122 (1989). ZBL0731.14025. [1]: https://i.sstatic.net/M6cjS.png

I give an answer here using only elementary algebraic geometry.

We know that the Grassmannian Gr(2,5) is defined by 5 linearly independent quadrics in P9 and V(5) can be viewed as defined by 5 linearly independent quadrics in P6.

Given a point p=[1,0,,0]X for XP6 defined by Qi,1i5, where Qi are homogeneous of degree 2. Write Qi=qi,2(x1,,x6)+x0qi,1(x1,,x6) with qi,k homogeneous of degree k.

I claim that there there is a line contained in X passing through p if and only if qi,k=0,1i5,k=1,2() have a common zero in P5=P(k[x1,x6]). To see this, note that given a common zero [a1,a2,,a6]P5, the line joining p to [0,a1,a2,,a6] is contained in X and the line joining [1,0,,0] and [1,a1,,an] is the same as the line joining [1,0,,0] and [0,a1,,an].

Since Qi are linearly independent, we see that the linear part qi,1,1i5 are linearly independent. By the geometric version of Krull’s Principal Ideal Theorem[11.3.2, Vakil’s notes], the dimension of the subvariety in P5 defined by () is 0. This means () has finitely many common zeros, that is, there are at most finitely many lines passing through a given point.

Now back to X=V(5), as suggested by Enrico in the comment. We take 3 hyperplanes ''general enough'' to get explicit defining equations of V(5). Consider the Plucker coordinate wij in P9, take w12=w34, w13=w45, w23=w15, the defining equations for V(5) are

{w234w45w24+w14w15=0w245w14w35+w15w34=0w215w45w25+w34w35=0w15w45w24w35+w25w34=0w34w45w14w25+w15w24=0

By the above argument, choose p to be the point with the only nonzero entry wij for {i,j}{3,4},{4,5},{1,5} we can have some lines. For example, we can find two nonintersecting lines: one joining [x14=1,xij=0] and [x12=1,xij=0], the other joining [x24=1,xij=0] and [x25=1,xij=0]. This has solved the question.


Remark. Suppose we are given the fact that for every point, there is at least one line passing through it, we can show that the subvariety in Gr(2,7) characterising lines on V(5) has dimension 2. Consider the following incidence correspondence, as we have discussed above, for any pV(5) the fibre pr11(p) has dimension 0.

[![enter image description here][1]][1]

By Upper Semicontinuity of Fibre Dimension, we have dimΦ=dimV(5)=3. Take a line lV(5), dimpr12([l])=1, by Upper Semicontinuity again, dimΦdimpr2Φ1, dimpr2Φ2. We conclude that the subvariety of Grass(2,7) characterising lines on V(5) has dimension 2.

A stronger result than dimpr2Φ2 is that the Hilbert scheme of lines on V(5) is isomorphic to P2 as in Sasha's answer and the reference is [Theorem 1, 2]. And the number of lines passing through a point is 3 when counted with multiplicity [P.26, 1].

[1] Takagi, Hiromichi; Zucconi, Francesco, Geometries of lines and conics on the quintic del Pezzo 3-fold and its application to varieties of power sums, Mich. Math. J. 61, No. 1, 19-62 (2012). ZBL1262.14048.

[2]Furushima, Mikio; Nakayama, Noboru, The family of lines on the Fano threefold (V_ 5), Nagoya Math. J. 116, 111-122 (1989). ZBL0731.14025. [1]: https://i.sstatic.net/M6cjS.png

I give an answer here using only elementary algebraic geometry.

We know that the Grassmannian Gr(2,5) is defined by 5 linearly independent quadrics in P9 and V(5) can be viewed as defined by 5 linearly independent quadrics in P6.

Given a point p=[1,0,,0]X for XP6 defined by Qi,1i5, where Qi are homogeneous of degree 2. Write Qi=qi,2(x1,,x6)+x0qi,1(x1,,x6) with qi,k homogeneous of degree k.

I claim that there there is a line contained in X passing through p if and only if qi,k=0,1i5,k=1,2() have a common zero in P5=P(k[x1,x6]). To see this, note that given a common zero [a1,a2,,a6]P5, the line joining p to [0,a1,a2,,a6] is contained in X and the line joining [1,0,,0] and [1,a1,,an] is the same as the line joining [1,0,,0] and [0,a1,,an].

Since Qi are linearly independent, we see that the linear part qi,1,1i5 are linearly independent. By the geometric version of Krull’s Principal Ideal Theorem[11.3.2, Vakil’s notes], the dimension of the subvariety in P5 defined by () is 0. This means () has finitely many common zeros, that is, there are at most finitely many lines passing through a given point.

Now back to X=V(5), as suggested by Enrico in the comment. We take 3 hyperplanes ''general enough'' to get explicit defining equations of V(5). Consider the Plucker coordinate wij in P9, take w12=w34, w13=w45, w23=w15, the defining equations for V(5) are

{w234w45w24+w14w15=0w245w14w35+w15w34=0w215w45w25+w34w35=0w15w45w24w35+w25w34=0w34w45w14w25+w15w24=0

By the above argument, choose p to be the point with the only nonzero entry wij for {i,j}{3,4},{4,5},{1,5} we can have some lines. For example, we can find two nonintersecting lines: one joining [x14=1,xij=0] and [x12=1,xij=0], the other joining [x24=1,xij=0] and [x25=1,xij=0]. This has solved the question.


Remark. Suppose we are given the fact that for every point, there is at least one line passing through it, we can show that the subvariety in Gr(2,7) characterising lines on V(5) has dimension 2. Consider the following incidence correspondence, as we have discussed above, for any pV(5) the fibre pr11(p) has dimension 0.

[![enter image description here][1]][1]

By Upper Semicontinuity of Fibre Dimension, we have dimΦ=dimV(5)=3. Take a line lV(5), dimpr12([l])=1, by Upper Semicontinuity again, dimΦdimpr2Φ=1, dimpr2Φ=2. We conclude that the subvariety of Grass(2,7) characterising lines on V(5) has dimension 2.

A stronger result than dimpr2Φ=2 is that the Hilbert scheme of lines on V(5) is isomorphic to P2 as in Sasha's answer and the reference is [Theorem 1, 2]. And the number of lines passing through a point is 3 when counted with multiplicity [P.26, 1].

[1] Takagi, Hiromichi; Zucconi, Francesco, Geometries of lines and conics on the quintic del Pezzo 3-fold and its application to varieties of power sums, Mich. Math. J. 61, No. 1, 19-62 (2012). ZBL1262.14048.

[2]Furushima, Mikio; Nakayama, Noboru, The family of lines on the Fano threefold (V_ 5), Nagoya Math. J. 116, 111-122 (1989). ZBL0731.14025. [1]: https://i.sstatic.net/M6cjS.png

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L. Xie
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I give an answer here using only elementary algebraic geometry.

We know that the Grassmannian Gr(2,5) is defined by 5 linearly independent quadrics in P9 and V(5) can be viewed as defined by 5 linearly independent quadrics in P6.

Given a point p=[1,0,,0]X for XP6 defined by Qi,1i5, where Qi are homogeneous of degree 2. Write Qi=qi,2(x1,,x6)+x0qi,1(x1,,x6) with qi,k homogeneous of degree k.

I claim that there there is a line contained in X passing through p if and only if qi,k=0,1i5,k=1,2() have a common zero in P5=P(k[x1,x6]). To see this, note that given a common zero [a1,a2,,a6]P5, the line joining p to [0,a1,a2,,a6] is contained in X and the line joining [1,0,,0] and [1,a1,,an] is the same as the line joining [1,0,,0] and [0,a1,,an].

Since Qi are linearly independent, we see that the linear part qi,1,1i5 are linearly independent. By the geometric version of Krull’s Principal Ideal Theorem[[11Theorem[11.3.2, Vakil’s notes], the dimension of the subvariety in P5 defined by () is 0. This means () has finitely many common zeros, that is, there are at most finitely many lines passing through a given point.

Now back to X=V(5), as suggested by Enrico in the comment. We take 3 hyperplanes ''general enough'' to get explicit defining equations of V(5). Consider the Plucker coordinate wij in P9, take w12=w34, w13=w45, w23=w15, the defining equations for V(5) are

{w234w45w24+w14w15=0w245w14w35+w15w34=0w215w45w25+w34w35=0w15w45w24w35+w25w34=0w34w45w14w25+w15w24=0

By the above argument, choose p to be the point with the only nonzero entry ijwij for {i,j}{3,4},{4,5},{1,5} we can have some lines. For example, we can find two nonintersecting lines: one joining [x14=1,xij=0] and [x12=1,xij=0], the other joining [x24=1,xij=0] and [x25=1,xij=0]. This has solved the question.


Remark. Suppose we are given the fact that for every point, there is at least one line passing through it, we can show that the subvariety in Gr(2,7) characterising lines on V(5) has dimension 2. Consider the following incidence correspondence, as we have discussed above, for any pV(5) the fibre pr11(p) has dimension 0.

[![enter image description here][1]][1]

By Upper Semicontinuity of Fibre Dimension, we have dimΦ=dimV(5)=3. Take a line lV(5), dimpr12([l])=1, by Upper Semicontinuity again, dimΦdimpr2Φ1 thus, dimpr2Φ2. Now weWe conclude that the subvariety of Grass(2,7) characterising lines on V(5) has dimension 2.

A stronger result than dimpr2Φ2 is that the Hilbert scheme of lines on V(5) is isomorphic to P2 as in Sasha's answer and the reference is [Theorem 1, 2]. And the number of lines passing through a point is 3 when counted with multiplicity [P.26, 1].

[1] Takagi, Hiromichi; Zucconi, Francesco, Geometries of lines and conics on the quintic del Pezzo 3-fold and its application to varieties of power sums, Mich. Math. J. 61, No. 1, 19-62 (2012). ZBL1262.14048.

[2]Furushima, Mikio; Nakayama, Noboru, The family of lines on the Fano threefold (V_ 5), Nagoya Math. J. 116, 111-122 (1989). ZBL0731.14025. [1]: https://i.sstatic.net/M6cjS.png

I give an answer here using only elementary algebraic geometry.

We know that the Grassmannian Gr(2,5) is defined by 5 linearly independent quadrics in P9 and V(5) can be viewed as defined by 5 linearly independent quadrics in P6.

Given a point p=[1,0,,0]X for XP6 defined by Qi,1i5, where Qi are homogeneous of degree 2. Write Qi=qi,2(x1,,x6)+x0qi,1(x1,,x6) with qi,k homogeneous of degree k.

I claim that there there is a line contained in X passing through p if and only if qi,k=0,1i5,k=1,2() have a common zero in P5=P(k[x1,x6]). To see this, note that given a common zero [a1,a2,,a6]P5, the line joining p to [0,a1,a2,,a6] is contained in X and the line joining [1,0,,0] and [1,a1,,an] is the same as the line joining [1,0,,0] and [0,a1,,an].

Since Qi are linearly independent, we see that the linear part qi,1,1i5 are linearly independent. By the geometric version of Krull’s Principal Ideal Theorem[[11.3.2, Vakil’s notes], the dimension of the subvariety in P5 defined by () is 0. This means () has finitely many common zeros, that is, there are at most finitely many lines passing through a given point.

Now back to X=V(5), as suggested by Enrico in the comment. We take 3 hyperplanes ''general enough'' to get explicit defining equations of V(5). Consider the Plucker coordinate wij in P9, take w12=w34, w13=w45, w23=w15, the defining equations for V(5) are

{w234w45w24+w14w15=0w245w14w35+w15w34=0w215w45w25+w34w35=0w15w45w24w35+w25w34=0w34w45w14w25+w15w24=0

By the above argument, choose p to be the point with the only nonzero entry ij for {i,j}{3,4},{4,5},{1,5} we can have some lines. For example, we can find two nonintersecting lines: one joining [x14=1,xij=0] and [x12=1,xij=0], the other joining [x24=1,xij=0] and [x25=1,xij=0]. This has solved the question.


Remark. Suppose we are given the fact that for every point, there is at least one line passing through it, we can show that the subvariety in Gr(2,7) characterising lines on V(5) has dimension 2. Consider the following incidence correspondence, as we have discussed above, for any pV(5) the fibre pr11(p) has dimension 0.

[![enter image description here][1]][1]

By Upper Semicontinuity of Fibre Dimension, we have dimΦ=dimV(5)=3. Take a line lV(5), dimpr12([l])=1, by Upper Semicontinuity again, dimΦdimpr2Φ1 thus dimpr2Φ2. Now we conclude that the subvariety of Grass(2,7) characterising lines on V(5) has dimension 2.

A stronger result than dimpr2Φ2 is that the Hilbert scheme of lines on V(5) is isomorphic to P2 as in Sasha's answer and the reference is [Theorem 1, 2]. And the number of lines passing through a point is 3 when counted with multiplicity [P.26, 1].

[1] Takagi, Hiromichi; Zucconi, Francesco, Geometries of lines and conics on the quintic del Pezzo 3-fold and its application to varieties of power sums, Mich. Math. J. 61, No. 1, 19-62 (2012). ZBL1262.14048.

[2]Furushima, Mikio; Nakayama, Noboru, The family of lines on the Fano threefold (V_ 5), Nagoya Math. J. 116, 111-122 (1989). ZBL0731.14025. [1]: https://i.sstatic.net/M6cjS.png

I give an answer here using only elementary algebraic geometry.

We know that the Grassmannian Gr(2,5) is defined by 5 linearly independent quadrics in P9 and V(5) can be viewed as defined by 5 linearly independent quadrics in P6.

Given a point p=[1,0,,0]X for XP6 defined by Qi,1i5, where Qi are homogeneous of degree 2. Write Qi=qi,2(x1,,x6)+x0qi,1(x1,,x6) with qi,k homogeneous of degree k.

I claim that there there is a line contained in X passing through p if and only if qi,k=0,1i5,k=1,2() have a common zero in P5=P(k[x1,x6]). To see this, note that given a common zero [a1,a2,,a6]P5, the line joining p to [0,a1,a2,,a6] is contained in X and the line joining [1,0,,0] and [1,a1,,an] is the same as the line joining [1,0,,0] and [0,a1,,an].

Since Qi are linearly independent, we see that the linear part qi,1,1i5 are linearly independent. By the geometric version of Krull’s Principal Ideal Theorem[11.3.2, Vakil’s notes], the dimension of the subvariety in P5 defined by () is 0. This means () has finitely many common zeros, that is, there are at most finitely many lines passing through a given point.

Now back to X=V(5), as suggested by Enrico in the comment. We take 3 hyperplanes ''general enough'' to get explicit defining equations of V(5). Consider the Plucker coordinate wij in P9, take w12=w34, w13=w45, w23=w15, the defining equations for V(5) are

{w234w45w24+w14w15=0w245w14w35+w15w34=0w215w45w25+w34w35=0w15w45w24w35+w25w34=0w34w45w14w25+w15w24=0

By the above argument, choose p to be the point with the only nonzero entry wij for {i,j}{3,4},{4,5},{1,5} we can have some lines. For example, we can find two nonintersecting lines: one joining [x14=1,xij=0] and [x12=1,xij=0], the other joining [x24=1,xij=0] and [x25=1,xij=0]. This has solved the question.


Remark. Suppose we are given the fact that for every point, there is at least one line passing through it, we can show that the subvariety in Gr(2,7) characterising lines on V(5) has dimension 2. Consider the following incidence correspondence, as we have discussed above, for any pV(5) the fibre pr11(p) has dimension 0.

[![enter image description here][1]][1]

By Upper Semicontinuity of Fibre Dimension, we have dimΦ=dimV(5)=3. Take a line lV(5), dimpr12([l])=1, by Upper Semicontinuity again, dimΦdimpr2Φ1, dimpr2Φ2. We conclude that the subvariety of Grass(2,7) characterising lines on V(5) has dimension 2.

A stronger result than dimpr2Φ2 is that the Hilbert scheme of lines on V(5) is isomorphic to P2 as in Sasha's answer and the reference is [Theorem 1, 2]. And the number of lines passing through a point is 3 when counted with multiplicity [P.26, 1].

[1] Takagi, Hiromichi; Zucconi, Francesco, Geometries of lines and conics on the quintic del Pezzo 3-fold and its application to varieties of power sums, Mich. Math. J. 61, No. 1, 19-62 (2012). ZBL1262.14048.

[2]Furushima, Mikio; Nakayama, Noboru, The family of lines on the Fano threefold (V_ 5), Nagoya Math. J. 116, 111-122 (1989). ZBL0731.14025. [1]: https://i.sstatic.net/M6cjS.png

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