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Spectral theory is a fundamental part of operator theory and the spectrum of many operators is investigated throughout the existing literature. And that is for a good reason: If $A$ is some closed operator on a Banach space $E$ that generates a strongly continuous semigroup $(T(t))_{t \geq 0}$, then the spectral theory of $A$ has a huge influence on the asymptotic behaviour of the semigroup it generates. This is just one example how spectral theory is usefull to answer other questions.

What is also well-known, is that if one wants to study the spectrum of operators one usually supposes that the scalars of the Banach space are complex because this allows for a much more richer spectral theory:

Even if one considers just matrices $A \in \mathbb R^{n \times n}$ and their generated (norm-continuous) semigroups $(\mathrm e^{t A})_{t \geq 0}$, then one needs complex numbers to understand the asymptotic behaviour of their generated semigroups. In particular, $(\mathrm e^{t A})_{t \geq 0}$ converges if and only if each eigenvalue of $A$ has real part less or equal to zero and if $\sigma_p(A) \cap \mathrm i \mathbb R \subseteq \{0\}$ and if $0$ is either a first order pole of the resolvent of $A$ or not an eigenvalue at all. So it is natural to study spectral theory in the context of the complex numbers. For that reason one usally considers the complexification of Banach spaces, even if one is interested in Banach spaces with real scalar field, in order to use tools from spectral theory.

Now here is my question: Why does one stop at $\mathbb C$? $\mathbb C$ is naturally embedded into $\mathbb H$ (the quaternions) and also into $\mathbb O$ (the octonians). Are there situations where the spectral theory of an operator with respect to $\mathbb C$ differs from its spectral theory with respect to $\mathbb H$ and $\mathbb O$? Is there even a meaningful way to study spectral theory with respect to $\mathbb H$ and $\mathbb O$? And if not, why is this the case?

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    $\begingroup$ "Basic" linear algebra in the sense of "theory of vector spaces" works fine over $\mathbb{H}$ and other skew-fields as long as one is willing to keep track of the difference between left- and right-vector spaces. A more interesting question is whether functional analysis still works over $\mathbb{H}$. For example: Does an $\mathbb{H}$-hilbert space have any interesting properties that $\mathbb{C}$-hilbert spaces don't have? $\endgroup$
    – Johannes Hahn
    Commented Nov 11, 2021 at 20:11
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    $\begingroup$ I don't understand why this was downvoted (even twice). The question seems to make sense, is well-motivated, and apparently OP has put some thought into it before asking. $\endgroup$
    – Jochen Glueck
    Commented Nov 11, 2021 at 20:16
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    $\begingroup$ The theory of eigenvalues and eigenvectors is not so nice for quaternionic vector spaces. There is no Jordan normal form or spectral theorem. There are lots of papers about these issues, but no simple statement of how eigentheory should work, even for generic square matrices. $\endgroup$
    – Ben McKay
    Commented Nov 11, 2021 at 20:21
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    $\begingroup$ Commutativity is an important feature of $C$ which makes the theory meaningful and useful. Theory of analytic functions is fundamental for spectral theory. One can define analytic functions over quaternions, but nothing comparable to the theory over $C$ arises from this. So the answer is that over $C$ one obtains a rich mathematical theory with important applications. $\endgroup$
    – Alexandre Eremenko
    Commented Nov 11, 2021 at 20:27
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    $\begingroup$ Unique factorisation of polynomials isn't true over $\mathbb H$ (which are called the quaternions not the "hamiltonians") nor over $\mathbb O$ (which are called the octonions not "octonians"). In particular, $z^2 + 1$ can be factorised infinitely many ways over $\mathbb H$, while only one way over $\mathbb C$. Also observe that since multiplication is not commutative in $\mathbb H$, a linear polynomial can be of the form $f(z) = azb + czd + ezf + \dotsc$, while it can only be of the form $f(z) = az + b$ over $\mathbb C$. Observe that the polynomial $z - izi - j$ has no roots in $\mathbb H$ $\endgroup$
    – wlad
    Commented Nov 12, 2021 at 6:45

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