Let $F=F(H,H)$ be the space of bounded Fredholm operators in a Hilbert space $H$ with topology inherited from the norm operator topology, and let $X$ be a compact topological space.

For a continuous map $T\colon X\to F$, there exists a closed subspace $W\subseteq H$ with $\dim H/W<\infty$ such that $W\cap\ker T_x=0$ for all $x\in X$ and $H/T(W) =\bigcup_{x\in X} H/T_x(W)$ is a vector bundle over $X$ (See appendix of K-Theory, Anderson & Atiyah). Then one can show that $$\mbox{Ind}_1(T) = [X\times H/W] - [H/T(W)] \in K(X)$$ does not depend on $W$.

On the other hand, there exists a finite dimensional subspace $V\subseteq H$ such that $V+T_x(H) = H$ for all $x\in X$, so we can define $T^V\colon X\to F(H\oplus V, H)$ by $T^V_x(u,v) = T_x u + v$. Then $T^V_x$ is surjective and $\dim\ker T^V_x$ is constant on $x$. Thus $\ker T^V = \bigcup_{x\in X} \ker T_x$ is also a vector bundle over $X$. One can show that $$ \mbox{Ind}_2(T) = [\ker T^V] - [X\times V] \in K(X)$$ does not depend on $V$.

These index maps are called the family index of families of Fredholm operators in $H$, and it made me suspect that they are equal.

Question: Is it true that $$[X\times H/W] - [H/T(W)] = [\ker T^V] - [X\times V] \qquad (1)$$ in $K(X)$ ? Is there any reference that proves the equivalence of these indexes?

Edit: We can shrink $W$ or augment $V$ in order to have $\dim H/W = \dim V$. Say $H/W \cong V \cong \mathbb{C}^N$, so that $X\times H/W \cong X\times V \cong X\times\mathbb{C}^N$, and therefore $$\mbox{Ind}_1(T) = [X\times\mathbb{C}^N] - [H/T(W)]\ ,$$ $$\mbox{Ind}_2(T) = [\ker T^V] - [X\times\mathbb{C}^N]\ .$$

Equation $(1)$ becomes $$[X\times\mathbb{C}^N] - [H/T(W)] = [\ker T^V] - [X\times\mathbb{C}^N]$$ and it holds iff there exists $k\geq0$ such that

$$\ker T^V \oplus H/T(W) \oplus (X\times\mathbb{C}^k) \cong X\times\mathbb{C}^{2N+k}$$ But why does there exists such $k$?


1 Answer 1


Notice that $V^\perp\cap\ker T_x^*=0$ for every $x$, because for every $w\in V^\perp\cap\ker T_x^*$, $u\in H$, and $v\in V$, one has $$\langle w,T_x(u)+v \rangle = \langle w,T_x(u) \rangle = 0.$$ By composing two isomorphisms $\ker P_{V^\perp}T \ni u \mapsto u\oplus T(-u) \in \ker T^V$ and $\ker P_{V^\perp}T\cong H/T^*(V^\perp)$, one sees $\mathrm{Ind}_2(T)=-\mathrm{Ind}_1(T^*)$. Thus the quality of two indices follows from the fact that self-adjoint Fredholm operator $\left[\begin{smallmatrix} & T^*\\ T & \end{smallmatrix}\right]$ has index zero in either definition.

  • $\begingroup$ Thank for your answer, Narutaka! I've already understood the isomorphisms, but I am struggling to prove that $\mbox{Ind}_2(T\oplus T^*)=0$ in $K(X)$. Do you have any hint? $\endgroup$ Jul 24, 2020 at 14:43
  • $\begingroup$ @Rodrigo Dias: $T\oplus T^*$ is homotopic to $T^*T\oplus I$ and then to $(I+T^*T)\oplus I$, which has index zero because it is invertible. $\endgroup$ Jul 26, 2020 at 2:30

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