Skip to main content
4 of 4
deleted 1 character in body
Jon Bannon
  • 7.1k
  • 6
  • 69
  • 112

Why is spectral theory developed for $\mathbb C$

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?

Yaddle
  • 381
  • 1
  • 10