Your videos are such a treat for us poor maths students trying to warp our heads around these concepts, thanks a lot for sharing your knowledge and intuition in a very understandable way!
I know the word “toric variety” but some book uses many notations and I could not understand at all. I watch this lecture, then I understand why cone is used for constructions of varieties. This lecture is amazing. Of course, the other lectures are also great. I really happy to understand toric variety. Let Z be the set of integers. A point (m,n) of Z^2 can be regarded as a monomial x^m*y^n of polynomial ring k[x, 1/x. y, 1/ y]. For example, (-1,2) means (1/x)^2*y^2. The set of all Lattice points in a cone of Z^2 means the set of monomials. We can use these monomials of cone as a generator of a ring. Sometimes, this ring is a coordinate ring of variety. So, for a cone C, we get a variety V( C). Unfortunately, the correspondence of cone C and a variety V(C ) is kind of annoying, that is, for an inclusion of two cones C(1) < C(2), we get the “inverse” inclusion V(C(1)) > V(C(2)). So, we do not use this correspondence, instead, we use the correspondence from a cone C to the variety V(C*), where C* denotes the dual cone of cone C. Then, the correspondence preserve the direction of inclusions, that is, C(1) < C(2) implies V(C(1)*) < V(C(2)*). Here, we note that inclusion of cones C(1) < C(2) implies inclusion of dual cones C(1)* > C(2)*. 12:16 Construction of projective line as a toric varitety. Consider three cones C(1), C(2), C(3) in Z^1 whose rational points is as follows. C(1) = {0,x,x^2,x^3,…}, C(2) = { 0,1/x, 1/x^2, 1/x^3,…}, C(0) = {0} = intersection of C(1) and C(2), From which we get dual cones C(1)* = C(1), C(2)*= C(2), C(0)* = {0,x,x^2,x^3,…, ,1/x, 1/x^2, 1/x^3,…}. Using these sets of monomials as generator, we get coordinate rings K[x. 1/x], K[x], K[1/x]. Thus, for above rings, we get varieties V(C(0)*) = A - {0}, V(C(1)*) = A, V(C(2)*) = A, where A denotes one dimensional affine line. Because there are inclusions C(0) -> C(1), C(0) -> C(2), we get inclusions of varieties V(C(0)*) -> V(C(1)*), V(C(0)*) -> V(C(2)*), which are regarded as gluing maps. Thus we get a projective line as a toric manifold. Product of Two Copies of Projective Line as Toric Variety. At 15:44, because the all cone is self-dual, it is easy to see the corresponding rings. The green cone corresponds to an affine plane with coordinate ring k[x, y] and the red cone corresponds an affine plane with coordinate ring k[x, 1/y]. The dual cone of the cone defined by the intersection of green and red cone is the cone of union of green and red, which has coordinate ring k[x, y, 1/y]. So, in this intersection, the gluing map sends (x, y) to (x, 1/y). The first and the fourth quadrants corresponds two affine plane with coordinate rings k[x,y] and k[1/x, 1/y]. The dual cone of their intersection corresponds to products of two punctured affine planes with coordinate ring k[x, 1/x, y, 1,y]. So, in this intersection, the gluing mapping maps (x,y) to (1/x, 1/y). I do not understand how to get gluing map of affine varieties from inclusions of their rings. 17:24 ~ 18::00 Audio is …. 18:56 Two dimensional projective space as toric variety Two dimensional projective space has 3 neighborhoods, so, there should be 3 cones representing three affine planes. But, in the last figure of the slide at 18:56, there are 5 cones, why? What does 5 cone mean? Infinitely many cones is not projective variety but abstract variety…?? Does abstract variety mean scheme? Probably non-compact scheme? 22:59 I still do not understand why a variety constructed from cones are called a “toric” variety.
Right now, I am working with Toric-Variety that are objects of a Borisov dual. Here, for example, the Group $T$ of isometries of an $Img Z^{n}$ must be replaced by a Borisov Duality of $T/ G:= \mathbb{C}^{3}$ (with $T$ a parameter) The question is, does the dual-space Borisov "replace" the non-empty dual-Poincare of fixed-cones in the $Img_{*} .Z^{n}$ ?
If we view the lattice L=Z^2 as sitting in the vector space V=R^2 (=~ L \otimes R) say, then its dual L* is the lattice in V* given by L* = {v in V* | is in Z for all x in L} =~ Z^2, where is the usual pairing between V and V*. Then if C is a cone in L, its dual cone is C*={v in V* | >= 0 for all x in C}. So for example if you take the cone C spanned by (1,0) and (1,1) (like in the 8:00 mark in the vide), then (x,y) will belong to its dual C* iff x>=0 and x+y>=0.
Said a bit more succinctly also: if you think of Z^2 living inside R^2, the boundary of the dual cone is given by the points orthogonal to the original cone (considered as vectors)
I figure it out. The point is that computing dual cone is different from computing dual lattice, so comment below is useless. 8:29, I follow the computation in kconrad.math.uconn.edu/blurbs/gradnumthy/different.pdf. It seems dual of (1, 0), (1,1) should just be Z^2. But your picture seems to make sense too. So I do not quite understand.
Your videos are such a treat for us poor maths students trying to warp our heads around these concepts, thanks a lot for sharing your knowledge and intuition in a very understandable way!
I know the word “toric variety” but some book uses many notations and I could not understand at all. I watch this lecture, then I understand why cone is used for constructions of varieties. This lecture is amazing. Of course, the other lectures are also great. I really happy to understand toric variety.
Let Z be the set of integers.
A point (m,n) of Z^2 can be regarded as a monomial x^m*y^n of polynomial ring k[x, 1/x. y, 1/ y]. For example, (-1,2) means (1/x)^2*y^2.
The set of all Lattice points in a cone of Z^2 means the set of monomials. We can use these monomials of cone as a generator of a ring. Sometimes, this ring is a coordinate ring of variety. So, for a cone C, we get a variety V( C).
Unfortunately, the correspondence of cone C and a variety V(C ) is kind of annoying, that is, for an inclusion of two cones C(1) < C(2), we get the “inverse” inclusion V(C(1)) > V(C(2)). So, we do not use this correspondence, instead, we use the correspondence from a cone C to the variety V(C*), where C* denotes the dual cone of cone C. Then, the correspondence preserve the direction of inclusions, that is, C(1) < C(2) implies V(C(1)*) < V(C(2)*). Here, we note that inclusion of cones C(1) < C(2) implies inclusion of dual cones C(1)* > C(2)*.
12:16 Construction of projective line as a toric varitety.
Consider three cones C(1), C(2), C(3) in Z^1 whose rational points is as follows.
C(1) = {0,x,x^2,x^3,…},
C(2) = { 0,1/x, 1/x^2, 1/x^3,…},
C(0) = {0} = intersection of C(1) and C(2),
From which we get dual cones
C(1)* = C(1),
C(2)*= C(2),
C(0)* = {0,x,x^2,x^3,…, ,1/x, 1/x^2, 1/x^3,…}.
Using these sets of monomials as generator, we get coordinate rings
K[x. 1/x],
K[x],
K[1/x].
Thus, for above rings, we get varieties
V(C(0)*) = A - {0},
V(C(1)*) = A,
V(C(2)*) = A,
where A denotes one dimensional affine line.
Because there are inclusions C(0) -> C(1), C(0) -> C(2), we get inclusions of varieties
V(C(0)*) -> V(C(1)*), V(C(0)*) -> V(C(2)*), which are regarded as gluing maps. Thus we get a projective line as a toric manifold.
Product of Two Copies of Projective Line as Toric Variety.
At 15:44, because the all cone is self-dual, it is easy to see the corresponding rings.
The green cone corresponds to an affine plane with coordinate ring k[x, y] and the red cone corresponds an affine plane with coordinate ring k[x, 1/y]. The dual cone of the cone defined by the intersection of green and red cone is the cone of union of green and red, which has coordinate ring k[x, y, 1/y]. So, in this intersection, the gluing map sends (x, y) to (x, 1/y).
The first and the fourth quadrants corresponds two affine plane with coordinate rings k[x,y] and k[1/x, 1/y]. The dual cone of their intersection corresponds to products of two punctured affine planes with coordinate ring k[x, 1/x, y, 1,y]. So, in this intersection, the gluing mapping maps (x,y) to (1/x, 1/y). I do not understand how to get gluing map of affine varieties from inclusions of their rings.
17:24 ~ 18::00 Audio is ….
18:56
Two dimensional projective space as toric variety
Two dimensional projective space has 3 neighborhoods, so, there should be 3 cones representing three affine planes.
But, in the last figure of the slide at 18:56, there are 5 cones, why? What does 5 cone mean?
Infinitely many cones is not projective variety but abstract variety…?? Does abstract variety mean scheme? Probably non-compact scheme?
22:59 I still do not understand why a variety constructed from cones are called a “toric” variety.
I'm puzzled by the sketch of the dual cone at 17:00 - shouldn't it be the 0
I'm trying to study singularity theory and I find your videos helpful. Thanks!
Right now, I am working with Toric-Variety that are objects of a Borisov dual. Here, for example, the Group $T$ of isometries of an $Img Z^{n}$ must be replaced by a Borisov Duality of $T/ G:= \mathbb{C}^{3}$ (with $T$ a parameter)
The question is, does the dual-space Borisov "replace" the non-empty dual-Poincare of fixed-cones in the $Img_{*} .Z^{n}$ ?
Professor, can you explain more specifically what's the dual cone of a cone in Z^2?
If we view the lattice L=Z^2 as sitting in the vector space V=R^2 (=~ L \otimes R) say, then its dual L* is the lattice in V* given by L* = {v in V* | is in Z for all x in L} =~ Z^2, where is the usual pairing between V and V*. Then if C is a cone in L, its dual cone is C*={v in V* | >= 0 for all x in C}.
So for example if you take the cone C spanned by (1,0) and (1,1) (like in the 8:00 mark in the vide), then (x,y) will belong to its dual C* iff x>=0 and x+y>=0.
Said a bit more succinctly also: if you think of Z^2 living inside R^2, the boundary of the dual cone is given by the points orthogonal to the original cone (considered as vectors)
in a Toric-manifold, the dual is not cubic. More General There is a dual over Groups of $\mathbb{C}^{3}$ Which are cubes in $
ho{} x_{12}+ x_{12}$
This dude is always having connection issues lmao
I figure it out. The point is that computing dual cone is different from computing dual lattice, so comment below is useless.
8:29, I follow the computation in kconrad.math.uconn.edu/blurbs/gradnumthy/different.pdf. It seems dual of (1, 0), (1,1) should just be Z^2. But your picture seems to make sense too. So I do not quite understand.