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Ryan Mickler
Ryan Mickler
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AnotherA homological proof would use the intersection form of the torus. if you consider these paths as based loops on the torus, you see that they are represented as (1,1), and (1,-1), in terms of the standard homology generators. knowing that the intersection form is (0 1; -1 0), we find that the intersection index

Q(v,w) = (1,1)(0 1; -1 0)(1,-1)^t = 2

they already intersect once at the origin, so they must intersect somewhere else in the square. However, you must already have had to compute the intersection form.

Another proof would use the intersection form of the torus. if you consider these paths as based loops on the torus, you see that they are represented as (1,1), and (1,-1), in terms of the standard homology generators. knowing that the intersection form is (0 1; -1 0), we find that the intersection index

Q(v,w) = (1,1)(0 1; -1 0)(1,-1)^t = 2

they already intersect once at the origin, so they must intersect somewhere else in the square. However, you must already have had to compute the intersection form.

A homological proof would use the intersection form of the torus. if you consider these paths as based loops on the torus, you see that they are represented as (1,1), and (1,-1), in terms of the standard homology generators. knowing that the intersection form is (0 1; -1 0), we find that the intersection index

Q(v,w) = (1,1)(0 1; -1 0)(1,-1)^t = 2

they already intersect once at the origin, so they must intersect somewhere else in the square. However, you must already have had to compute the intersection form.

Source Link
Ryan Mickler
Ryan Mickler
  • 111
  • 3

Another proof would use the intersection form of the torus. if you consider these paths as based loops on the torus, you see that they are represented as (1,1), and (1,-1), in terms of the standard homology generators. knowing that the intersection form is (0 1; -1 0), we find that the intersection index

Q(v,w) = (1,1)(0 1; -1 0)(1,-1)^t = 2

they already intersect once at the origin, so they must intersect somewhere else in the square. However, you must already have had to compute the intersection form.