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This questions concerns justification of a statement made by Folland in A Course in Abstract Harmonic Analysis, 2ed.

Let $\mathcal{H}$ be some finite-dimensional Hilbert space, and let $\mathcal{A}$ be a commutative $\mathbf{C}^\ast$ subalgebra of $\mathcal{L}(\mathcal{H})$ containing $I$, and let $\Sigma=\sigma(\mathcal{A})$ be its spectrum. We write the Gelfand transform by $A\rightarrow C(\Sigma):T\mapsto\widehat{T}$, and the inverse Gelfand transform by $B(\Sigma)\rightarrow \mathcal{L}(\mathcal{H}):f\mapsto T_f$, where $B(\Sigma)$ are bounded Borel functions on the spectrum of $\mathcal{A}$. Similarly, we may define projection valued measures and as in this earlier question.

Suppose our $\mathcal{A}$ is the algebra generated by $\{T,T^\ast,I\}$ for some operator $T\in\mathcal{L}(\mathcal{H})$. I wish to show that we may decompose $\mathcal{H}$ into eigenspaces associated to the spectrum of $T$. There is a theorem in the text that notes $\sigma(A)\approx\sigma(T)$ by homeomorphism, from which it follows that $\sigma(A)$ is finite. If we write $\sigma({\mathcal{A}})=\{\sigma_1,\sigma_2,...,\sigma_n\}$, we see that

$$\sum_iP(\{\sigma_i\})=P(\cup_i\{\sigma_i\})=P(\sigma(\mathcal{A}))=I$$

which seems to imply that we may decompose $I$ into mutually orthogonal projections $P(\{\sigma_i\})$. But I am struggling to see the connection between $P(\{\sigma_i\})$ and the projection onto the eigenspace associated to $\sigma_i$. Something tells me they are one in the same... but I am having trouble making the connection and grasping what exactly $P(\{\sigma_i\})$ does to $\mathcal{H}$.

Any insight would be helpful, this stuff is quite new to me. If anything is unclear I am happy to clarify; I found this question a bit difficult to articulate.

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You use but you don't mention explicitly that $T$ is normal. As such, the Spectral Theorem applies and you have $$\tag1T=\sum_{j=1}^m \sigma_j P_j$$ where $n$ is the dimension of $H$ (and $T$ can be canonically seen as an $n\times n$ matrix), and $m\leq n$. In $(1)$ I'm not allowing repetitions, at the cost that some of the $P_j$ may have rank greater than one.

The spectral projection corresponding to $\sigma_j$ is $P_j$; we may obtain it by functional calculus from $T$ by $P_j=1_{\{\sigma_j\}}(T)$. But, since the spectrum is discrete, we may simply take $f$ to be a polynomial with $f(\sigma_j)=1$ and $f(\sigma_k)=0$ for all $k\neq j$. Then $P_j=f(T)=P(\{\sigma_j\})$ (note that $f$ and $1_{\{\sigma_j\}}$ agree on the spectrum of $T$, that's why $f(T)=1_{\{\sigma_j\}}(T)$).

In summary: yes, they are the same.

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