K-theory

First we will define K-theory of a ring $A$ in gradation 0, that is $\rK_0(A)$. We say that a finitely generated module over $A$ is free if it is isomorphic to the product $A^n$ for some $n$. A finitely generated $A$-module $P$ is projective if it is a direct summand in a free $A$-module, that is there exists an $A$-module $Q$ such that $P\oplus Q\isom A^n$ for some $n$. Such projective module $P$ corresponds to idempotent in the matrix algebra $M_n(A)$. The set of isomorphism classes of finitely generated projective modules over $A$ is a monoid with respect to direct sum of classes defined by

$$[P]+[Q]=:[P\oplus Q].$$

There is an universal abelian group for this monoid (called the Grothendieck group), and we take it as the definition of the K-theory of $A$, denoted by $\rK_0(A)$.

Let $A$ be a commutative algebra over $k$. Suppose we want to construct a map

$$\ch\:\rK_0(A)\to \drH^2(A).$$

First consider an example of a map from a tori $S^1\times S^1$ to a sphere $S^2$ given by contracting the boundary of a square with opposite edges identified.

Figure: $f\:S^1\times S^1\to S^2$

./Chapter3/S1xS1toS2.eps

This map has degree 1 and induces an isomorphism

$$\rH^2(S^2)\xrightarrow{\deg(f)} \rH^2(S^1\times S^1).$$

If we want to find an algebraic map of corresponding coordinate rings

$$S^2_a:=\bC[X,Y,Z]/(X^2+Y^2+Z^2-1)\to \bC[U, U^{-1}, V, V^{-1}]=:S^1\times S^1_a$$

then we will not succeed, because any algebraic map $S^1\times S^1\to S^2$ is homotopic to the constant map. The situation is very different now than it was in case of maps $S^3\to S^2$. Indeed, assume we have the map

$$f^*\: S^2_a\to S^1\times S^1_a.$$

Then it induces a map on K-theory

$$\rK_0(S^2_a)\to \rK_0(S^1\times S^1_a),$$

and we would have a commutative diagram

$$\xymatrix{ \bZ \ar[d] \ar@{=}[r] & \Tilde{\rK}_0(S_a^2) \ar[... ...\ar[r]^-{\deg(f)} & \drH^2(S^1\times S^1_a) & \bC \ar@{=}[l] }$$

which gives a contradiction, because a generator of $\bZ=\Tilde{\rK}_0(S_a^2)$ goes to generator of $\bC=\drH^2(S^2_a)$.

Define a projector $p$ and idempotent $e$ in $M_2(S_a^2)$ by the formulas

$$p:= \twobytwo{X}{Y+iZ}{Y-iZ}{-X},\quad p^2=1,\quad e:=\frac{p+1}{2},\quad e^2=e.$$