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In this article it says that the groupoid cardinality of the 2D hypercomplex number systems is $\frac 3 2$ because there are three 2D hypercomplex number systems (up to isomorphism) each equipped with one non-trivial algebra automorphism. But surely, the dual numbers have infinitely many automorphisms satisfying $1 \mapsto 1, \epsilon \mapsto k\epsilon$ (where $k$ is an arbitrary non-zero real number), so the value of the groupoid cardinality is actually $\frac 1 2 + \frac 1 2 + \frac 1 \infty = 1$.
Who is right?
Are you sure that’s an automorphism, or just an isomorphism to a different algebra?
I haven’t thought about what the entry claims (or even looked at it until a minute ago), but, in response to David:
The operation $\phi_k$ : $1 \mapsto 1$ and $\epsilon \mapsto k \epsilon$ for $k \in \mathbb{R} \setminus \{0\}$ is certainly an automorphism of the ring of dual numbers $\mathbb{R}[\epsilon]/(\epsilon^2)$:
Explicitly, check that
$\begin{aligned} \phi_k(1 + a \epsilon) \cdot \phi_k(1 + b \epsilon) & =\; (1 + k a \epsilon) (1 + k b \epsilon) \\ & =\; (1 + k (a + b) \epsilon) \\ & =\; \phi_k\big( 1 + (a + b) \epsilon \big) \\ & =\; \phi_k\big( (1 + a \epsilon) \cdot (1 + b \epsilon) \big) \end{aligned}$More conceptually, notice that (Hadamard’s lemma)
$\mathbb{R}[\epsilon]/(\epsilon^2) \;\simeq\; C^\infty(\mathbb{R})/(\epsilon^2)$so that every diffeomorphism of $\mathbb{R}$ which fixes the origin induces an automorphism of $\mathbb{R}[\epsilon]/(\epsilon^2)$ (given by the diffeomorphism’s first Taylor coefficient, which is the $k$ from above).
These cardinality do behave oddly. Sometimes it makes better ’sense’ to take the cardinality of the open interval as $-1$, so that the nonzero reals have cardinality $-2$. See, e.g., here.
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Ah, ok. I was thinking it was something like the function $\mathbb{R}[\varepsilon]/(\varepsilon^2) \to \mathbb{R}[k\varepsilon]/(k^2\varepsilon^2)$, but now I see what was intended!
I have edited the entry, in an attempt to fix the issue. See the edit log in the entry’s thread here.
Regarding the claim that the perplex numbers have precisely two automorphisms, this is true. First, it’s easy to see that $\phi_1(a + be) = a + be$ and $\phi_2(a + be) = a - be$ are two distinct automorphisms. It remains to prove that all automorphisms over the perplex numbers are of this form. Let $\phi$ be an automorphism over the perplex numbers. Observe that $\phi(e)^2 = \phi(e^2) = \phi(1) = 1$. It follows that $\phi(e) \in \{1,-1,e,-e\}$. But if $\phi(e) \in \{+1, -1\}$ then $\phi$ is not injective. So therefore $\phi(e) \in \{e, -e\}$.
Thanks, that’s easy enough. I have edited that into the entry (logs here).
So thanks for bringing up this issue and helping to fix it. I’d like to bow out now. Please feel invited to edit the entry further, as need be.
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