# A doubt about the parts of the spectrum of tensor products

Let $\mathcal{H}$ be any complex Hilbert space of infinite dimensional. By an operator $T$ I mean a linear bounded transformation from $\mathcal{H}$ into $\mathcal{H}$, i.e, $T:\mathcal{H}\rightarrow\mathcal{H}$ is a linear transformation, such that $\left\|Tx\right\|\leq \beta \left\|x\right\|$, for some $\beta\geq 0$ in $\mathbb{C}$. The sets $\mathcal{N}(T)$ and $\mathcal{R}(T)$ denote the kernel and the range of $T$. Also, $\mathcal{B}[\mathcal{H}]$ is the set the of all operators defined in $\mathcal{H}$.

The spectrum of $T$ is the set $\sigma(T) = (\lambda \in\mathbb{C}: \mathcal{N}(\lambda I - T)\neq 0$ or $\mathcal{R}(\lambda I - T)\neq \mathcal{H} )$, where $I\in\mathcal{B}[\mathcal{H}]$ is the identity operator (i.e., the spectrum of $T$ is the set of all $\lambda$ such that $(\lambda I - T)$ fail to have a bounded inverse on $\mathcal{R}(\lambda I - T)=\mathcal{H}$). A classical partition of the spectrum is:

i. Point Spectrum: $\sigma_{P}(T) = (\lambda \in \mathbb{C}: \mathcal{N}(\lambda I - T) \neq 0)$.

ii. Continuous Spectrum: $\sigma_{C}(T) = (\lambda \in\mathbb{C}: \mathcal{N}(\lambda I - T)= 0 , \mathcal{R}(\lambda I - T)^{-}=\mathcal{H}$ and $\mathcal{R}(\lambda I - T)\neq \mathcal{H} )$, where $\mathcal{R}(.)^{-}$ denotes the closure of $\mathcal{R}(.)$.

iii. Residual Spectrum: $\sigma_{R}(T) = (\lambda \in\mathbb{C}: \mathcal{N}(\lambda I - T)= 0$ and $\mathcal{R}(\lambda I - T)^{-}=\mathcal{H})$.

Let $\otimes$ denote the tensor product. Consider $T=(A\otimes B)\in\mathcal{B}[\mathcal{H}\otimes\mathcal{H}]$, where $A\in\mathcal{B}[\mathcal{H}]$ and $B\in\mathcal{B}[\mathcal{H}]$. It is well known that $\sigma(T) = \sigma(A\otimes B) = \sigma(A)\sigma(B)$. My questions are:

a. $\sigma_{C}(T) = \sigma_{C}(A)\otimes \sigma_{C}(B)$ ?

b. $\sigma_{R}(T) = \sigma_{R}(A)\otimes \sigma_{R}(B)$ ?

Thank you very much for your attention.

• Do you mean the Hilbert space tensor product (which is another Hilbert space, with Hilbert basis $\{e_i\otimes e_j\}$ if $\{e_j\}$ is a basis of $H$), or the Banach space tensor product (a Banach space isometric to the space of nuclear operators on $H$) ? Oct 28, 2012 at 17:35
• I mean the Hilbert space tensor product. Oct 28, 2012 at 18:11
• @portella, your definition of "residual spectrum" seems to have a typo, it should be $... \neq \mathcal{H}$. Jul 4, 2013 at 18:10

Let me first show that there are problems with various parts of the spectrum here, hence your concerns are well-grounded.

The first is, citing Ichinose in the Transactions from 1978 here and here, that

$$\sigma_{ess}(A\otimes B) = \sigma_{ess}(A)\sigma(B)\cup\sigma(A)\sigma_{ess}(B)$$

showing that equalities like your a. and b. may not always be true.

Further, here is an example of Ichinose showing a funny fact: Let $H=l^2$, $Ae_n=-e_{n-1}$ be the negated left-shift $(Ae_1=0)$. Let $\tau_n\to 0$, $\max |\tau_n|<1$ and define $Be_n:=\tau_n e_{n+1}$.

Then the point spectrum of $B$ is empty, the point spectrum of $A$ is the open circle $|\lambda|<1$. There will be however many elements in the kernel of $(A+I)\otimes(B+I)$: in fact, for every $y\in H$, $$\sum_{n=1}^{\infty}e_n\otimes B^{n-1}y$$ belongs to the kernel of this operator.

Hence, the point spectrum of the tensor product cannot be recovered from the point spectra of the original operators. Since the residual spectrum is the (conjugate of) the point spectrum of the adjoint, it follows that your claim b. cannot hold, i.e., in general $$\sigma_R(A\otimes B) \neq \sigma_R(A)\cdot\sigma_R(B),$$ or even $$\sigma_R(A\otimes B) \neq \sigma_R(A)\cdot\sigma(B)\cup\sigma(A)\cdot\sigma_R(B).$$