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I am interested in the concept of direct integrals. The definition I am concerned with can be found in Hall, "Quantum Theory for Mathematicians" (p. 146). Short version: We have a family of separable Hilbert spaces $(H(\omega))_{\omega \in \Omega}$ and a $\sigma$-finite measure space $(\Omega, \mu)$. The author uses sections $s: \Omega \to \cup_{\omega\in\Omega} H(\omega) $ with $s(\omega) \in H(\omega)$ and constructs (if the dimension map is measureable) a measureable set of (pointwise) orthonormal sections, in the sense that the scalar products with these are measureable. Then, the set of square integrable, measureable sections (modulo null sets) define the direct integral.

Later on, the same author uses the fact that direct integrals and direct sums interchange: $$\int_{\Omega}^{\oplus} \bigoplus_{i=1}^\infty H^i_\omega \; d \mu (\omega) = \bigoplus_{i=1}^\infty \int_{\Omega}^{\oplus} H^i_\omega \; d \mu (\omega).$$

Is there a way to see that easily? Or did I get it wrong and the statement is generally not true?

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  • $\begingroup$ Equality of course must be interpreted as "there is an obvious isomorphism," and I think the only map that comes to mind is the one that sends $s(\omega)=(s_i)_{i\in\mathbb N}(\omega)$ to $(s_i(\omega))_i$ (in other words, it doesn't do anything other than slightly reinterpret). Have you tried if this works? $\endgroup$
    – Christian Remling
    Commented Aug 9, 2017 at 17:33
  • $\begingroup$ Also, the statement has to be true. If it isn't, then this just means that one has to revise definitions. $\endgroup$
    – Christian Remling
    Commented Aug 9, 2017 at 17:33
  • $\begingroup$ Yes, i tried that. This isomorphism induces - obviously - a measurable orthonormal section. But you need to show, that the direct integral is independent of the choice of such a section, so that this isomorphism induces the same direct integral, don't you? $\endgroup$
    – MrUser
    Commented Aug 9, 2017 at 18:14
  • $\begingroup$ I'm not sure actually I completely understand the definition you summarize, but for me, the elements of a direct integral are measurable (in a suitable sense) functions $s(\omega)$ whose norm, computed in the obvious way, is finite. Then I don't think there will be any problems, though I might be missing something. $\endgroup$
    – Christian Remling
    Commented Aug 9, 2017 at 18:25
  • $\begingroup$ A section $s$ is measurable with respect to the "measurable structure" (i.e. to the set of measurable orthonormal sections), iff $<s(\omega),e_i(\omega)>$ is measurable for all $i\in\mathbb{N}$. I guess, that was missing. $\endgroup$
    – MrUser
    Commented Aug 9, 2017 at 18:31

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