Taylor spectrum of commuting operators

Fom the following paper (M. Ch—o, H. Motoyoshi, B. Na¡cevska Nastovska: On the joint spectra of commuting tuples of operators and a conjugation) we have

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Let $A= \begin{pmatrix}0&1\\1&0\end{pmatrix}$ and $I= \begin{pmatrix}1&0\\0&1\end{pmatrix}$. The Harte spectrum of $(I,A)$ is equal to $$\sigma_H(I,A)=\{(1,1);(1,-1)\}.$$ I want to calculate the taylor spectrum of $(I,A)$ denoted $\sigma_T(I,A)$

According to the following paper of Taylor: J. L. Taylor, A joint spectrum for several commuting operators, J. Functional Anal. 6(1970), 172-191.

we have

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  • $\begingroup$ If you don't mind me asking, what text is the first picture from? $\endgroup$ – M.G. Nov 30 '18 at 9:29
  • $\begingroup$ @M.G. Please see my editing. $\endgroup$ – Student Nov 30 '18 at 9:36

The Taylor spectrum happens to be the same $\sigma_T(I,A) = \{(1,1), (1,-1)\}$. Let $R_X(\lambda) = (X-\lambda)^{-1}$ be the resolvent of $X$. You have the identities \begin{gather*} \begin{bmatrix} I-\lambda & A-\mu \end{bmatrix} \begin{bmatrix} R_I(\lambda) \\ 0 \end{bmatrix} = I, \\ \begin{bmatrix} R_I(\lambda) \\ 0 \end{bmatrix} \begin{bmatrix} I-\lambda & A-\mu \end{bmatrix} + \begin{bmatrix} -(A-\mu) \\ I-\lambda \end{bmatrix} \begin{bmatrix} 0 & R_I(\lambda) \end{bmatrix} = \begin{bmatrix} I & 0 \\ 0 & I \end{bmatrix} , \\ \begin{bmatrix} 0 & R_I(\lambda) \end{bmatrix} \begin{bmatrix} -(A-\mu) \\ I-\lambda \end{bmatrix} = I , \end{gather*} whenever the resolvent $R_I(\lambda)$ exists. These identities (in homological algebra, they are known as a contracting homotopy for this complex) imply that, whenever $\lambda$ is not in the spectrum of $I$ (namely, when $R_I(\lambda)$ exists), Taylor's Koszul complex is exact and hence the corresponding value of $(\lambda,\mu)$ does not belong to $\sigma_T(I,A)$. You can write similar formulas, but using $R_A(\mu)$ instead. Hence, you have reduced the calculation to $\sigma_T(I,A) \subseteq \{ (1,\mathbb{C}) \} \cap \{ (\mathbb{C},1), (\mathbb{C},-1) \} = \{ (1,1), (1,-1) \}$. Now it's just a matter of checking that for these values of $(\lambda,\mu)$ the Koszul complex really does fail to be exact, which is easy to see from the known common eigenvectors of $I$ and $A$.

High brow explanation: A cochain map between two complexes descends to a map in cohomology. For example, the identity cochain map induces the identity map in cohomology. For any complex, a homotopy induces a cochain map from the complex to itself, which happens to descend to the zero map in cohomology. So, if the identity cochain map from a complex to itself is a induced by a homotopy, then it descends to cohomology as both the identity map and the zero map. That is only possible when the cohomology vanishes, meaning that the complex is exact.

Low brow explanation: Consider a complex of linear operators $D_i$ (meaning of course that $D_{i+1} D_i = 0$). A homotopy $h_i$ is a sequence of linear maps as illustrated in $$ \require{AMScd} \begin{CD} \cdots V_{-1} @>{D_0}>{\stackrel{\dashleftarrow}{h_0}}> V_0 @>{D_1}>{\stackrel{\dashleftarrow}{h_1}}> V_{1} \cdots , \end{CD} $$ Defining the operators $N_i = h_{i+1} D_{i+1} + D_i h_i$, it is easy to check that $D_i N_i = N_{i+1} D_{i+1}$, meaning that $N_i$ constitute a cochain map, which is induced by the homotopy $h_i$. If $N_i = I_i$ is equal to the identity map for each $i$, then we call $h_i$ a contracting homotopy.

If you try to solve the equation $D_{i} v = u$, where $D_{i+1} u = 0$. Then a contracting homotopy can be used as follows: $$ u = I_{i} u = (h_{i+1} D_{i+1} + D_i h_i) u = D_i (h_i u) . $$ Hence, a solution $v = h_i u$ always exists. This shows that the complex is exact, whenever a contracting homotopy exists.

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  • $\begingroup$ Thank you for your answer but please I don't understand how you get the three idenity and what you mean by $[ ]$? $\endgroup$ – Student Nov 30 '18 at 10:29
  • $\begingroup$ @Student, these are all block matrices. Multiply everything explicitly to check the identities explicitly. As an exercise, to check your understanding, reproduce the same identities by using $R_A(\mu)$. $\endgroup$ – Igor Khavkine Nov 30 '18 at 10:37
  • $\begingroup$ Ok thank you. I think there is a small mistake. You use $\Lambda$ but you mean $\lambda$ $\endgroup$ – Student Nov 30 '18 at 10:38
  • $\begingroup$ @Student, ah yes, a couple of typos. Thanks for noticing. Now fixed! $\endgroup$ – Igor Khavkine Nov 30 '18 at 11:24
  • $\begingroup$ Dear Professor, While the three identities are obviously true, I do not know enough about the Koszul complex to see the connection. I also don't know about the connection between the eigenvalues of I and A and the Koszul complex. I hope that you explain me a bit or to propose me a good reference. Thanks a lot. $\endgroup$ – Student Dec 4 '18 at 5:46

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