# K-groups and the Chern Character

Let $E$ be a complex vector bundle over a compact smooth manifold $M$. Let $\nabla^E$ be a (C-linear) connection on $E$ and let $R^E$ denote its curvature.

The Chern character form associated to $\nabla^E$ is defined by
$ch\left(E,\nabla^E\right)=tr\left[ \exp\left(\frac{\sqrt{-1}}{2\pi}R^E\right)\right]\in\Omega^{even}(M).$
Obviously, $ch\left(E,\nabla^E\right)$ is a closed form. We denote the association cohomology class by $ch(E)$ which is called the Chern character of $E$.

1. $ch(E\oplus F)= ch(E)+ch(F)$; \qquad ($\oplus$ is Whithney direct sum)
2. $ch(E\otimes F)=ch(E)\times ch(F)$;
3. If $E\cong F$, then $ch(E)=ch(F)$.

The crux of the proof is that the calculate does not depend on the representative element.

Denote by Vector($M$) the set of all complex vector bundles over $M$, then under the Whitney direct sum operation, Vect($M$) becomes a semi-abelian group. And $(Vect(M)$, $\otimes$, $\oplus)$ is a semi-ring. Now we introduce an equivalence relation ‘$\sim$’ in Vect($M$) such that
$E \sim F\Leftrightarrow E\cong F.$
So the following map
$ch : Vect(M)\rightarrow H^{even}_dR(M,C)$
is a homomorphism between semi-groups. there is a nature method for extending a semi-abelian group to a abelian group. The fundamental ideal: let $N$ be the natural number($0 \not\in N$). Now we extend semi-abelian group $(N,+)$ to a abelian group.

i) Take
$N\times N=\set{(m,n)|m,n\in N}.$
In $N\times N$, there have a natural sum ‘$+$’ as follows:
$\bigg( \forall m_1,m_2,n_1,n_2 \in N \bigg) \quad (m_1,n_1)+(m_2,n_2)=(m_1+m_2, n_1+n_2).$
One introduces an equivalence relation ‘$\sim$’ in $N\times N$ as follows:
$(m_1,n_1)\sim (m_2,n_2) \Leftrightarrow m_1+n_2=m_2+n_1.$
Please note there is not subtract. Let $Z=N\times N /\sim$, it is easily proved that $Z$ is a abelian group. The zero element is $[(m,m)],m\in N$. And $[(m,n)]^{-1}=[(n,m)]$.

ii)Make a map
$\phi : N \rightarrow Z, \qquad \phi(m)=[(m+1,1)].$
Obviously, the map $\phi$ is isomorphism from $N$ to a semi-subgroup $\set{[(m+1,1)]}$. So $Z$ is a dilation of $N$. then we can denote
$m=[(m+1,1)],\quad 0=[(m,m)],\quad -m=[(1,m+1)],\quad\forall m \in N.$
One can prove this dilation is the smallest.

Now we can extend semi-abelian group $Vect(M)/\sim$ to a group $K(M)$ which is called the K-group of $M$. Naturally, the map $ch$ is extended a group homomorphism,
$ch: K(M) \rightarrow H^{even}_dR(M,\C).$
Atiyah and Hirzebruch prove this homomorphism is an isomorphism if one ignores the torsion elements in $K(M)$. This theory belongs the K-theory.

$\langle \hat{A}(TM)ch(E),[M] \rangle =\int_M \hat{A}(TM,\nabla^{TM})ch(E,\nabla^E)\in \C.$
If $M$ is an even dimensional oriented spin closed manifold(see[milnor]), the characteristic number
$\langle \hat{A}(TM)ch(E),[M] \rangle$
is a integer.(Atiyah, Hirzebruch, Borel theorem)