Let $X$ be a compact topological space, $H$ be a complex Hilbert space and endow $F(H)$, the space of bounded Fredholm operators in $H$, with the uniform norm topology (inherited from $B(H)$).

Let $T: X\to F(H)$, $x\mapsto T_x$, be a continuous map. There exists a closed subspace $V\subseteq H$ of finite codimension, i.e. $\dim H/V<\infty$, such that $V\cap \ker T_x = \{0\}$ for all $x\in X$.

I have proved that $H/T(V) = \bigsqcup\limits_{x\in X} H/T_x(V)$ is a vector bundle over $X$ (of finite rank). In particular, $\dim H/T_x(V)$ is independent of $x$ (here we can assume connectedness of $x$).

For $x\in X$, let $P_x: H\to H$ be the orthogonal projection onto $T_x(V)$. In order to induce a specific map of bundles (see here for details), I need to check the continuity of the map $X\times H\to H$ given by $(x,u)\mapsto P_x(u)$.

Question: Is $(x,u)\mapsto P_x(u)$ continuous?

Looking at the inequality $$ \|P_y(v)-P_x(u)\| \leq \|P_y(v-u)\| + \|(P_y-P_x)(u)\|$$ we conclude it suffices to prove that $x\mapsto P_x$ is continuous when one gives $B(H)$ the strong operator topology, but I could not prove it.

Any help is appreciated. Thanks in advance!


Let $S_x=T_x|_V$. Then $S_x$ coincides with the composition of $T_x$ with the inclusion of $V$ into $H$, both Fredholm operators, so $S_x$ is also Fredholm.

Like every Fredholm operator, $S_x$ has closed range and it is clearly one-to-one. From this and the open mapping Theorem it easily follows that $S_x^*S_x$ is invertible.

Therefore $R_x:=S_x(S_x^*S_x)^{-1/2}$ is a well defined isometry having the same range as $T_x$ and we deduce that $P_x=R_xR_x^*$, from where the norm continuity of $P_x$ follows.

  • $\begingroup$ Thank you Ruy! I was trying to prove that $R$ is an isometry without success. In order to an operator $A:V_1\to V_2$ to be an isometry, it is necessary that $A^*A=id_{V_1}$. We have $R^*R=(S^* S)^{-1}$, which need not to be $id_{V}$. Am I missing something? $\endgroup$ – Rodrigo Dias Aug 14 '20 at 18:52
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    $\begingroup$ @RodrigoDias, I am sorry. The exponent should be $-1/2$. I have just edited my answer to fix this. It should be like in the polar decomposition. $\endgroup$ – Ruy Aug 14 '20 at 19:57
  • $\begingroup$ Right, I got it! $\endgroup$ – Rodrigo Dias Aug 14 '20 at 20:12

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