Equivalence of families indexes of Fredholm operators

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 $$k$$?

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.
• 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? Jul 24, 2020 at 14:43
• @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. Jul 26, 2020 at 2:30