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## Month: December, 2014

### Algebraic Geometry 4: A short note on Projective Varieties

What is a variety? It is the set of common zeroes for a set of polynomials. For example, for the set of polynomials $\{x+y,x-y\}\in\Bbb{R}[x,y]$, the variety is $(0,0)$.

Now what is a projective variety? Simply put, it is the common set of zeroes of polynomials in which a one-dimensional subspace is effectively considered one point. Hence, for the variety to be well-defined, if one point of a one-dimensional subspace satisies the variety, every point of the one-dimensional subspace has to satisfy that variety. Confused?

Take the polynomial $x+y+z\in\Bbb{C}[x,y,z]$. The point $(1,-1,0)$ satisfies this polynomial. Now note that the points $\lambda(1,-1,0)$ also satisfy this polynomial for every $\lambda\in\Bbb{C}$. Hence this is a projective variety. Now take $x+y+z-1\in\Bbb{C}[x,y,z]$. Here $\lambda(1,0,0)$ satisfies the polynomial for only $\lambda=1$. Hence, this is not a projective variety.

But why? Why would you want to consider a whole line as one point? When you watch the world from your little nest, every line running along your ine of sight becomes a point. Hence, athough it may be a line in “reality” (whatever this means), for you it is a point. This is the origin of projective geometry, although things have gotten sightly complicated since then.

### Algebraic Geometry 3: Some more definitions

Index categories: These are categories in which the objects are essentially elements of a partially ordered set, and there exists at most one morphism between two objects. One example would be $\Bbb{N}$, where $\exists f\in$ Hom$(x,y)$ iff $x\leq y$.

Let $\mathcal{I}$ be an index category. A functor $F:\mathcal{I}\to\mathcal{K}$ is said to be indexed by $\mathcal{I}$. What does such a functor look like? If this functor is fully faithful, then $\mathcal{K}$ will also be an index category. However, even if it is neither faithful nor full, there is some structure that is imposed. For example, if $\mathcal{I}$ is $\Bbb{Z}$, then we can index a subset of objects in $\mathcal{K}$, and ensure that morphisms exist between all such indexed objects (mapping objects of lower indices to those with higher indices).

Limit: Let $\mathcal{K}$ be indexed by $\Bbb{Z}$. Then the limit is an object $\varprojlim\limits_{\mathcal{\Bbb{Z}}} A_i$ such that there exists a morphism between $\varprojlim\limits_{\mathcal{\Bbb{Z}}} A_i$ and every object (including itself), and if $A_m$ and $A_n$ are two such objects with $F(f)\in$ Hom$(A_m,A_n)$ (remember that $F$ is the functor between $\Bbb{Z}$ and $\mathcal{K}$), then the three morphisms under consideration commute.

So what exactly is happening here? It is easy to see that $\varprojlim\limits_{\Bbb{Z}} A$ is the initial object in $F(\Bbb{Z})\subset \mathcal{K}$. For example, let $F:\Bbb{N}\to$Set be a functor which maps $n\to \{1,2,3,\dots,n\}$, and the morhism $m\to n$ is mapped to the morphism $x\to x, \forall x\in\{1,2,3,\dots,m\}$. Then $\emptyset$ or $F(0)$ is the unique limit.

Similarly, the colimit is the final object of $F(P)\subset \mathcal{K}$, where $P$ is any partially ordered set.

Filtered set: A nonempty partially ordered set $S$ is said to be filtered is for any $x,y\in S$, there exists a $z\in S$ such that $x\geq z$ and $y\geq z$. For example, $\Bbb{Z}$ is fitered, and so is any subset of it. However, if $S=\{apple,orange,banana\}$, and $apple is the only relation we know, then this is not a filtered poset (partially ordered set).

Adjoints: Two covariant functors $F:A\to B$ and $G:B\to A$, where $A,B$ are categories, are considered to be adjoint if there is a natural bijection $\tau_{AB}:$ Mor$_A(F(A),B)\to$ Mor$(A,G(B))$, and for all $f:A\to A'$ in $A$ we require

Philosophy behind adjoint functors: These ensure that at least the set of morhisms between objects (both in the domain and range) are “isomorphic” to each other, whilst not being concerned about the objects themselves.

### Algebraic Geometry 2: Philosophizing Categories

Faithful functor: Let $F:A\to B$ be a functor between two categories. If the map Mor$(A,A')\to$ Mor$(F(A),F(A'))$ is injective, then the functor is called faithful. For example, the functor between the category of sets with only bijective mappings qualified to be morphisms between objects, to the category of sets with all kinds of mappings allowed to be morphisms, carrying objects and morphisms to themselves, is faithful.
Full functor: If Mor$(A,B)\to$ Mor$(F(A),F(B))$ is surjective, then the functor is called full. For example, the functor defined above is not full.

Natural transformation of covariant functors: Let $F,G$ be functors between categories $A$ and $B$. Let $M,M'\in A$ be objects, and let $f:M\to M'$ be morphisms between them. Then $m:F\to G$ is a natural transformation if the following diagram commutes:

Philosophical point: Why are commutative diagrams so omnipresent and important in Mathematics? It seems to be a fairly arbitrary condition to satisfy! Commuting diagrams essentially signify that “similar things” are happening at “different places”, and the “similar things” can easily be inter-converted. Too hand-wavy? Please bear with me.

Say $M,M',F(M),F(M'),G(M),G(M')$ are all $\Bbb{Z}$, and $m(\Bbb{Z})$ is idenity. Also assume that  $F(f)$ maps $k\to k+1$. Then if $G(f)$ also maps $k\to k+1$, then $m$ is a natural transformation. However, if $G(f)$ is of any other description, then $m$ is not a natural transformation.

Forming other examples should convince you of the fact above.

### Algebraic Geometry Series 1: An introduction to Category Theory

This is intended to be a series of articles, that fleshes out the bare skeleton of Algebraic Geometry. I will be closely following Ravi Vakil’s treatment of the topic, supplemented with other treatments.

A category $\mathcal{C}$ consists of a “collection” of objects, and for each pair of objects, a set of morphisms (or arrows) between them.

Note that the objects being considered need not be distinct. In the category of sets, for instance, the identity morphism $\iota:\{1\}\to\{1\}$ is also a legal morphism. The set of morphisms between objects $A$ and $B$ is referred to as Mor$(A,B)$. Morphisms compose as expected: if $f\in$ Mor$(A,B)$ and $g\in$ Mor$(B,C)$, then $g\circ f\in$ Mor$(A,C)$. Composition is associative: $f\circ (g\circ h)=(f\circ g)\circ h$. Also, for each object $A\in$ Obj$(\mathcal{C})$, there exists an identity morphism id$_A:A\to A$. If $f\in$ Mor$(A,B)$, then $f\circ$ id$_A=f$. Also, if $g\in$ Mor$(B,A)$, then id$_A\circ g=g$.

Fibered products: These can be best studied with a diagram. Suppose we have $\alpha: X\to Z$ and $\beta:Y\to Z$, then the fibered product is an object $X\times_Z Y$ such that it is universal for any object $A$ mapping to both $X$ and $Y$. For example, let us consider the category comprising of all ideals of $\Bbb{Z}$, and a morphism $f:A\to B$ exists if $A\subseteq B$. Then, if $X=(4)$ and $Y=(6)$, we can see both lie inside $(2)$. Hence, we can choose $Z=(2)$. Now every object that maps to both $(4)$ and $(6)$ needs to be generated by a multiple of $lcm(4,6)$. Hence, $X\times_Z Y$ will be generated by $lcm(4,6)$, or $12$.

Functors: A covariant functor $F$ is like a homomorphism between categories $A$ and $B$. A contravariant functor $G$ is slightly different. It maps $f\in$ Mor$(A,B)$ to $G(f)\in$ Mor$(G(B),G(A))$. For example, the identity functor is covariant, while the functor mapping topological open sets ($\exists f:A\to B$ implies $A\subseteq B$) to differentiable functions on them is a contravariant functor.

I would like to discuss the contravariant functor mentioned above. Let us consider the category of open sets in $\Bbb{R}$, such that morphisms exist between two objects only when the domain is a subset (not necessarily proper) of the range. For example, there exists a morphism between $(0,1)$ and $(0,2)$. Now consider a differentiable function on $(0,2)$, like $x^2$. Clearly, it is also differentiable on every subset of $(0,2)$, including $(0,1)$. Hence, there is a mapping from the set of differentiable functions on $(0,2)$ to those on $(0,1)$ (functions map to themselves). However, this mapping may not be surjective. For example, $\frac{1}{x-1}$ is differentiable on $(0,1)$ but not on $(0,2)$.