Schemes 3: exactness and sheaves

This lecture is part of an online algebraic geometry course on schemes, based on chapter II of "Algebraic geometry" by Hartshorne.
In it we discuss exactness of morphisms of sheaves over a topological space.

[raymondchou]

Never in my life did I ever think I would be learning Algebraic Geometry from Richard Borcherds himself. Thank you!

[antoinebrgt]

Thanks a lot for these amazing lectures, the format is really good (the writing is clear, these 30 min videos are nice to stay focused) and I like that you illustrate the concepts with concrete examples.

[michaelball]

Thanks for the videos Richard. Great public service!

[hausdorffm]

About 4:56, I show why the entire space is not Hausdorff. I use the notation in 4:56 at which f and g are defined. First of all, I point out that for any neighbourhood U of the origin of R^1, U contains a negative real number x and there is a neighborhood V of x so that V is contained in the intersection of U and the set of all negative numbers. By the definition of f and g, we can say that the restriction of f and g to V are identically zero on V. This implies that the restrictions f|V and g|V coincides as the element of stalk on x, and hence, they represent a same germ. So, therefore for any two open sets U(f) and U(g) containing f and g in etale space, we can find at least one common element contained in both U(f) and U(g). Therefore we cannot distinct two elements of etale space by open sets.

I am not sure why the g has infinite bumps in the definition. In the above argument, I did not use the bumps. Of course, to show that f and g are distinct germs, the bumps are used. But, to me, the infinite bumps are redundant or I guess we can more simple function as g without bumps. I misunderstand something?? Probably, this is as same as the comments of Olivier Bégassat. I completely agree with him. Hum, awww, Using the infinite bumps, can we say that g is not analytic???

I did not prove why the sheaf of the analytic functions is Hausdorff. (4:56)

5:50 13:00 If pre-sheaf is not related explicitly any etale space, then it denoted simply by F. In addition, if sheaf arises as an sections of certain etale space, then it is denoted by F^+.


19:27 Using skyscraper sheaves, two examples are made. In the movie at 19:27, two sheaves are defined, i.e., F is defined as a skyscraper sheaf on reals R which assigns the ring R to each open set U of R, if U contains 0. G denotes the sheaf over reals R, which, maps each connected open set of U to the smooth functions on U. Let x be a fixed real number and at 19:27, two different morphism between sheaves F, G are defined which is different according to whether x is zero or not.
First case, define a map of sheaves 0 -> G -> G -> F ->0 in which the first map from G to itself defined by the multiplication of the fixed real number x. If the fixed multiplier x is not zero, then this map looks 0 -> B -> B -> 0 -> 0 for stalks on any nonzero real number. And for the stalk on zero, the morphism is something to be 0 -> B -> B -> R -> I get stack... what is the definition of the third map, namely, what is the map G->F in the 0 -> G -> G -> F ->0. Probably it is the evaluation map of the point 0? isn't it? If so, then the two morphisms of stalks 0 -> B -> B -> 0 -> 0 and 0 -> B -> B -> R -> 0 are exact and hence the sequence sheaves 0 -> G -> G -> F ->0 is exact. On the other hand, if the multiplier x is not zero, then the sequence is, in view of stalk, 0 -> B -> 0 -> 0 -> 0 and 0 -> B -> 0-> R -> 0 and the later is not exact. So this example gives two sequence, one of which is exact and the other is not according to whether the multiplier is zero or not.



23:40 Exactness of sheaves dose not imply for sections of open sets. Let X be open subset in the set of all complex numbers. Consider the sequence of sheaves on X as follows.

0 -> constant sheaf of 2*pi*i -> Holomorphic functions -> non zero holomorphic functions -> 0,

where the first map is the injection and the third map is exponential map. In this sequence, the induced morphisms on stalks is exact. However, the induced sequence on some open set dose not exact.
If X is the punctured disc, then the identity map from X to complex numbers are contained in the third sheaf. However, its logarithm, i.e., log(x) is not a function on the punctured disc (because it is a multiple valued function on punctured disc or in other words, it is a function on the universal covering of the punctured disc) and thus, the identity map cannot represent as the image of exponential map of any holomorphic function. So, the condition that the sequence of stalks 0 ->F_x ->G_x -> H_x ->0 is exact for any base point x dose not imply the condition that the sequence of groups 0 ->F(U) ->G(U) -> H(U) ->0 is exact for any open subset U in the base space.

[olivierbegassat]

Any reason for choosing a smooth function with infinitely many bumps to the right of zero rather than a smooth bump function à la f(x) = 0 on R- and f(x) = exp(-1/x) on R+* ?

[ipudisciple]

4:40 Wait. So if G is the sheaf of analytic functions on (X = ℝ), then the étalé space A of G is (a) locally homeomorphic to ℝ, and (b) Hausdorff. But that means A is a 1-dimensional manifold. Doesn’t that mean that A = ℝ and the covering is trivial? After much thought, I think I understand. The situation is that A = ℝ[[x]] × X, where ℝ[[x]] has the discrete topology, which is still a 1-dimensional manifold, albeit a weird one. Now if U has c components then G(U) ≅ ℝ[[x]]^c because it has to be locally constant.

[bolzep]

I've not been able to see why in the exponential sequence (from around 21:35), the constant sheaf 2*pi*i*Z gets mapped to zero, since e^(2*pi*i*Z)=1, which is the neutral element of a multiplicative group but zero anywhere and thus not in the kernel of the exponential map, which makes the sequence not exact!?

[hbowman]

For USAns, "stalks of wheat", not corn.