K-Theory and Fredholm Operators

K-THEORY AND FREDHOLM OPERATORS

ISHAN LEVY

1. Introduction

Given a compact space \(X\), we can consider the commutative monoid of complex vector bundles on \(X\) under the Whitney sum operation. We can force it to become a group in the universal way, giving a group called \(K(X)\). This is a functor, so we can ask whether it is representable. If it is, the representing object should be a group up to homotopy. But in this case, it is represented by the space of Fredholm operators \(\sF \) on Hilbert space \(H\). In particular the composition operation is a commutative group operation on \(\sF \) up to homotopy.

\(K(X)\) for a point is just \(\ZZ \), so in particular the components of the space of Fredholm operators should correspond to the integers, and this function from Fredholm operators to \(\ZZ \) is exactly the index of an operator. Hence in general we will want to construct a natural isomorphism \(\ind : [-,\sF ] \to K(-)\).

To do this, ﬁrst suppose that we have a map \(f:X \to \sF \), and we ﬁx a point \(x \in X\). We should note that the kernel of \(f(x) = f_x\) varies continuously in \(X\), in the sense that if \(V\) is a subspace on which \(f_x\) is injective, then nearby points will also be injective on \(V\). In particular, as \(X\) is compact, we can ﬁnd a \(V\) of ﬁnite codimension in \(H\) such that \(f_x\) is injective on \(V\) for all \(x \in X\). Then \(f_x(V)\) is a subbundle of \(X \times H\), and we can consider \(H/f_X(V)\), the bundle where the ﬁbre of \(x\) is \(H/f_x(V)\). If \(V\) is codimension \(k\), we deﬁne \(\ind _V(f) = k-[H/f_X(V)]\). To see this is well deﬁned, we only need to observe that if we choose a codimension \(n\) subspace \(U \subset V\) then \(H/f_X(U) \cong f_X(U)^\perp \cong f_X(V)^\perp \oplus f_X(U ^\perp \cap V) \cong H/f_X(V) \oplus \CC ^n \times X\). We thus have \(k-[H/f_X(V)] = k+n - [H/f_X(U)]\), so for any \(V,W\), \(\ind _V(f) = \ind _{V\cap W}(f) = \ind _{W}(f)\). Moreover index is clearly natural.

Homotopy invariance comes from the commutative diagram:

[[XX, ℱ× ]I,ℱ ] KK ((XX )× I)

[X, ℱ ] K (X )

ii((((indnd1111ndeeexx××××x((0101(ff))))f))∗∗∗∗)
XXXX 0t1

The vertical maps are isomorphisms and their composites are the identity on \([X,\sF ]\) and \(K(X)\), so \(\ind (f_0) = \ind (f_1)\). To see \(\ind \) is a homomorphism, let \(f,g: X \to \sF \). Then choose \(U,V\) to be ﬁnite codimension subspaces on which \(f,g\) are respectively injective and such that \(g_X(V) \subset U\), so that \(fg\) is injective on \(V\). Then there is an exact sequence \(0 \to U/g_X(V) \to H/fg_X(V) \to H/f_X(U) \to 0\), so \(\ind (fg) = \codim V - [H/fg_X(V)] = \codim V - [U/g_X(V)] - [H/f_X(U)]\) \(= \codim U + \codim V -[H/g_X(U)] - [H/f_X(U)] = \ind (f) + \ind (g)\).

For surjectivity of the index, since it is a homomorphism we need every vector bundle \(V\) to be in the image, as well as \(n \in \ZZ \). The latter can be done by sending \(X\) to a single operator of index \(n\). To see \(V\) is in the image, ﬁnd a vector bundle \(W\) such that \(V \oplus W = \CC ^n \times X\) and let \(\pi _V, \pi _W\) be the projection maps. Now send \(x\) to the operator on \(H\otimes \CC ^n \cong H\) sending \(e_i \otimes v\) to \(e_{i+1} \otimes{\pi _W(v)} + e_i \otimes \pi _V(v)\). The index is \(-W\) which is \(V\) up to a trivial bundle.

Finally let’s examine the kernel of the index. If something is in the kernel, there must be a ﬁnite codimension subspace \(U \subset H\) with \(H/f_X(U)\) trivial. Let \(e_1,\dots e_n\) be a basis of \(U^\perp \) and let \(s_1,\dots s_n\) be trivializing sections of \(f_X(U)^\perp \). Consider the homotopy that has \(f_{x,t}(e_i) = f_x(e_i)(1-t) + s_i t\). It homotopes \(f\) to something that is an isomorphism on \(H\). Thus we have an exact sequence \([X ,\GL (H)] \to [X,\sF ] \to K(X) \to 0\). However, by Kuiper’s theorem, \(\GL (H)\) is contractible, so the index is a natural isomorphism.