Sheaf Theory

Note that a locally constant sheaf is nothing more than a covering space. The definitions of constant and locally constant sheaf are sometimes given in a confusing and misleading way.  Correct category theoretic notions should be given up to isomorphism. Given a space $U$ and an object $A$ there is a presheaf ${{\mathcal{F}}^{\prime }}_U^A$ which yields $A$ on every open set and whose restriction map is $i{d}_A$.  This generates a sheaf ${\mathcal{F}}_U^A$.  We say that a sheaf $F$ on a space $X$ is a locally constant sheaf if $X$ has a cover such that on each component $U$ of the cover it is isomorphic to a sheaf of the form ${\mathcal{F}}_U^A$. Monodromy is an abstract way at looking at the liftings of a closed path to a covering space.

Exercise: A locally constant sheaf on $[0,1]$ is a constant sheaf, that is, is isomorphic to a sheaf of the form ${\mathcal{F}}_{[0,1]}^A$. More generally: a locally constant sheaf on a simply connected space is a constant sheaf.

For $F$ an abelian sheaf,  we define $supF=\{x\in X:F_x\ne 0\}$. Show that $supF$ need not be closed. Consider the sheaf $F$ and fix a point $p$ in a Hausdorff space $X$. Then define $F(U)=0$ for $p\in U$ and $F(U)={\mathbb{Z}}_2$ otherwise for some abelian group $A\ne 0$.  Then clearly $supF$ is not closed.   Do not confuse the support of an abelian sheaf with the support of a section $s$ over $U$, the points of $U$ in which $s_x\ne 0$. $sups$ is indeed closed and this follows essentially from the algebraic constraint on restriction map homomorphisms where the image of $0$ must be $0$. Thus the stalk of a zero section must be zero.

Exercise: Show that sections of locally constant abelian sheaf $F$ have closed support (hence unique extensions).  Hint: suppose that $s_x\ne 0$. Then there is an open set $V(x)$ and a section $s_V$ which represents $s_x$. But since $F$ is locally constant we can assume, by taking intersections, that $F_{V(x)}$ is isomorphic to a constant sheaf for a given group $A$. But then clearly we must have $s_y=s_x$ for all $y\in V(x)$ and hence the complement of $sups$ is open. 

Thus (locally) constant abelian sheaves are abstract versions of many important properties of holomorphic functions.

A locally constant sheaf on a simply connected topological space is a constant sheaf. This follows from homotopy invariance of unique extensions. This is perhaps the most fundamental result about locally constant sheaves; more generally this would be the result concerning the representation of the fundamental group based at $x$ by automorphisms of the stalk at $x$.

Field Theory

Let a field $K$ have characteristic $p>0$. Then given $a\in K$ we have that $x-a$ is either irreducible or of the form $(x-b{)}^p$ with $b\in K$. Hence $K$ is perfect iff $K=K^p$ where $K^p=\{x\in K:\mathrm{\exists }y\in K.x=y^p\}$.

Let $K<L$ be an algebraic extension. Suppose we have a $K$-monomorphism $\psi :L\to {\overline{K}}^a$. Then we can extend this to a ${\psi }^{\prime }:{\overline{L}}^a\to {\overline{K}}^a$. This follows from an obvious construction using Zorn's lemma and a basic property of extensions to algebraic extensions.

Finite Group theory

Let $G$ be a group with centralizer $\mathcal{C}(G)$ such that $G/\mathcal{C}(G)$ is cyclic. Then $G$ is Abelian.

Can any finite group be realized as the isometries of a convex polyhedron in ${\mathbb{R}}^n$ ?