Suppose $T$ is a self-adjoint operator in $B(H)$ with $\sigma(T)$ a spectrum of $T$. $\mu$ is a spectral measure. For the operators having a generally continuous spectrum how to calculate the multiplicity function? Where multiplicity is usually called spectral multiplicity. Up to compact operators we know how to decompose the spectrum and get the multiplicity function on each bit of disjoint chunks. But how to apply these in continuous spectral measure case? Is related to the direct integral decomposition of the von Neumann algebra generated by $\pi(C(\sigma(T))$ or not?

If I am not clear, please help me. Thanks in advance.


A measurable Hilbert bundle is something of the form $\bigcup X_n \times H_n$ where $(X_n)$ is a measurable partition of a $\sigma$-finite measure space $X$ and $H_n$ is a Hilbert space of dimension $n$, for $n = 0, 1, 2, \ldots, \infty$. (I assume we're working with separable Hilbert spaces.) The associated Hilbert space is the $l^2$ direct sum of the spaces $L^2(X_n) \otimes H_n \cong L^2(X_n, H_n)$.

If $T$ is any bounded self-adjoint operator, one form of the spectral theorem says that there is a measurable Hilbert bundle over $X = \sigma(T)$, such that $T$ is unitarily equivalent to multiplication by $x$ acting on the Hilbert space associated to this bundle. The set $X_n$ is the multiplicity $n$ portion of $\sigma(T)$.

Does that help?

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  • $\begingroup$ Sir, but getting the decomposition of the spectrum i.e, getting $X_{n}$ s, without using multiplicity function is not understood by me. Can you please elaborate? $\endgroup$ – user136400 Feb 28 '19 at 9:17
  • $\begingroup$ Too long to explain in detail here, but it is done in detail in either of my books Measure Theory and Functional Analysis or Mathematical Quantization. $\endgroup$ – Nik Weaver Feb 28 '19 at 12:00
  • $\begingroup$ Okk. Thanks I will have a look $\endgroup$ – mathlover Feb 28 '19 at 17:18
  • $\begingroup$ @NikWeaver if there is an example of computing multiplicity function for continuous spectral measure please let me know. I did not see it in any book. $\endgroup$ – user136400 Mar 3 '19 at 15:07
  • $\begingroup$ You keep saying "calculating" and "computing", and I thought it was a language issue but now I wonder. What do you mean by "computing" a multiplicity function? You feed the operator into a computer and it prints out the multiplicity function? $\endgroup$ – Nik Weaver Mar 3 '19 at 15:17

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