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It seems that by the Gelfand–Mazur theorem quaternions are isomorphic to complex numbers. That is clearly wrong. So where is the catch?

I think that I found the problem but it seams so subtle that would like to have confirmation from someone else and I would like to know where the proof of Gelfand–Mazur blows up.

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The quaternions are certainly a real Banach algebra, but are you sure they are a complex Banach algebra? If you identify $i$ with $i$ to get the obvious action, the scalar multiplication doesn't seem to be bilinear . . .

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  • $\begingroup$ I think probably the same but I don't know what means for scalar multiplication to be bilinear. The algebra multiplication has to be $\mathbb{C}$-bilinear which is not in the case of quaterions. $\endgroup$ – tom Feb 24 '15 at 21:35
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    $\begingroup$ Bilinearity here just means that we must have $r\cdot(a\times b)=a\times (r\cdot b)$ for $r$ a scalar (here, complex number) and $a, b$ elements of the algebra; here $\cdot$ denotes scalar multiplication and $\times$ denotes multiplication in the algebra. If we take the obvious guess at scalar multiplication - which is to say, we pretend $i_\mathbb{C}=i_\mathbb{H}$ - then this breaks down: $i\cdot (j\times k)=i\times i=-1$, but $j\times (i\cdot k)=j\times (-j)=1$. $\endgroup$ – Noah Schweber Feb 24 '15 at 21:37
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    $\begingroup$ I'd say that part of the point of Gelfand-Mazur is to show that this bilinearity condition is really not so subtle, when you get down to it - it's extremely restrictive and powerful! $\endgroup$ – Noah Schweber Feb 24 '15 at 21:38
  • $\begingroup$ @NoahSchweber, I wouldn't say it's the bilinearity that is so powerful; it is the submultiplicative norm. There is a real version of the Gelfand–Mazur theorem (see my answer here for a formulation and a literature reference), where the quaternions are explicitly mentioned as one of the three options. However, apparently infinite field extensions of $\mathbb{R}$ or $\mathbb{C}$ (such as the field of rational functions or the field of Laurent series) do not admit a submultiplicative norm! $\endgroup$ – Josse van Dobben de Bruyn Mar 2 '18 at 22:37

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