Fruits of procrastination

Presheaves that are not sheaves

This is a post about pre-sheaves that are not sheaves. The two properties that a sheaf satisfies that a presheaf does not, are the “Gluability” axiom and the identity axiom.

Gluability- Over the open set U, consider the set F(U), which is the set of all bounded continuous functions. Sections which have the same restriction to intersections of open sets sometimes cannot be glued together to give global sections; i.e. they cannot be glued together to give a section defined on the whole topological space that is bounded. For instance, consider the function f(x)=x. Consider the cover \{(k,k+2)\}_{k\in\Bbb{N}} of \Bbb{R}. The restriction of f(x)=x on each such bounded interval is bounded. Hence, f(x)|_{(k,k+2)}\in F((k,k+2)). However, on gluing together over the whole real line, we get an unbounded function.

Another example seems slightly trickier to me, and requires knowledge of complex analysis. Vakil’s book says that “holomorphic functions that admit a square root” form a presheaf and not a sheaf. Note that we just need to consider functions that admit a square root, and not the square roots themselves. Take the function f(z)=z. It admits a square root on both U=\Bbb{C}\setminus \Bbb{R^+} and V=\Bbb{C}\setminus \Bbb{R^-}. Both U and V are open sets. Moreover, the restrictions of the function f(z)=z on the overlap of the two sets agree at each point. However, when we glue the two parts together, we get f(z)=z defined on the whole complex plane, which is holomorphic, but does not admit a square root.

Projective varieties

Today I will try and study projective varieties and their ideals. Although my understanding of these objects has improved over time, there are still a lot of chinks that need to be filled.

Something that has defied complete understanding is what kinds of polynomials are projective varieties the zeroes of? Do these polynomials have to be homogeneous? The point is that each homogeneous component has to satisfy the zeroes independently. Hence, a polynomial corresponding to a projective variety need not be homogeneous. In fact, a sum of homogeneous polynomials, possibly of different degrees, satisfies a smaller algebraic set than each of the individual components do. This follows from what is written above.

What does the ideal corresponding to an affine projective variety look like? Why does it have to be generated by homogeneous polynomials? It does not have to be generated by homogeneous polynomials. However, it *can* be. This is because each of the generators, of which there are finite, can be broken down into their homogeneous components. And it is easy to see that the ideal generated by the homogeneous components is the same as that generated by the generators mentioned above.

What does the coordinate ring of a projective variety look like? It is constructed in exactly the same way that the coordinate ring of an affine variety is constructed. The only difference is that the ideal, which we quotient the polynomial ring k[x_1,x_2,\dots,x_n] by, is homogeneous in this case.

But what about the non-homogeneous polynomials in this coordinate ring? Are they even well defined over a projective variety? This is a question I will get back to and resolve as soon as I can.