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1. • CommentRowNumber1.
   • CommentAuthorsvennik
   • CommentTimeNov 13th 2020
   • (edited Nov 13th 2020)
   • PermaLink
   Author: svennik
   Format: MarkdownItex[In this article](https://ncatlab.org/nlab/show/hypercomplex+number) 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?

   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?

2. • CommentRowNumber2.
   • CommentAuthorDavidRoberts
   • CommentTimeNov 14th 2020
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   Author: DavidRoberts
   Format: MarkdownItexAre you sure that's an automorphism, or just an isomorphism to a different algebra?

   Are you sure that’s an automorphism, or just an isomorphism to a different algebra?

3. • CommentRowNumber3.
   • CommentAuthorUrs
   • CommentTimeNov 14th 2020
   • PermaLink
   Author: Urs
   Format: MarkdownItexI 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).

   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).

4. • CommentRowNumber4.
   • CommentAuthorDavid_Corfield
   • CommentTimeNov 14th 2020
   • PermaLink
   Author: David_Corfield
   Format: MarkdownItexThese 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](https://math.ucr.edu/home/baez/counting/counting.pdf).

   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.

5. • CommentRowNumber5.
   • CommentAuthorUrs
   • CommentTimeNov 14th 2020
   • (edited Nov 14th 2020)
   • PermaLink
   Author: Urs
   Format: MarkdownItex[ removed ]

   [ removed ]

6. • CommentRowNumber6.
   • CommentAuthorDavidRoberts
   • CommentTimeNov 15th 2020
   • PermaLink
   Author: DavidRoberts
   Format: MarkdownItexAh, 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!

   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!

7. • CommentRowNumber7.
   • CommentAuthorUrs
   • CommentTimeNov 15th 2020
   • PermaLink
   Author: Urs
   Format: MarkdownItexI have edited the entry, in an attempt to fix the issue. See the edit log in the entry's thread [here](https://nforum.ncatlab.org/discussion/11882/hypercomplex-number/?Focus=87733#Comment_87733).

   I have edited the entry, in an attempt to fix the issue. See the edit log in the entry’s thread here.

8. • CommentRowNumber8.
   • CommentAuthorsvennik
   • CommentTimeNov 15th 2020
   • (edited Nov 15th 2020)
   • PermaLink
   Author: svennik
   Format: MarkdownItexRegarding 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 $\phi(e) \in \{e, -e\}$.

   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 $\phi(e) \in \{e, -e\}$.

9. • CommentRowNumber9.
   • CommentAuthorUrs
   • CommentTimeNov 15th 2020
   • PermaLink
   Author: Urs
   Format: MarkdownItexThanks, that's easy enough. I have edited that into the entry (logs [here](https://nforum.ncatlab.org/discussion/11882/hypercomplex-number/?Focus=87747#Comment_87747)). 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.

   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.