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    • CommentRowNumber1.
    • CommentAuthorsvennik
    • CommentTimeNov 13th 2020
    • (edited Nov 13th 2020)

    In this article it says that the groupoid cardinality of the 2D hypercomplex number systems is 32\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 11,εkε1 \mapsto 1, \epsilon \mapsto k\epsilon (where kk is an arbitrary non-zero real number), so the value of the groupoid cardinality is actually 12+12+1=1\frac 1 2 + \frac 1 2 + \frac 1 \infty = 1.

    Who is right?

    • CommentRowNumber2.
    • CommentAuthorDavidRoberts
    • CommentTimeNov 14th 2020

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

    • CommentRowNumber3.
    • CommentAuthorUrs
    • CommentTimeNov 14th 2020

    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 ϕ k\phi_k : 111 \mapsto 1 and εkε\epsilon \mapsto k \epsilon for k{0}k \in \mathbb{R} \setminus \{0\} is certainly an automorphism of the ring of dual numbers [ε]/(ε 2)\mathbb{R}[\epsilon]/(\epsilon^2):

    Explicitly, check that

    ϕ k(1+aε)ϕ k(1+bε) =(1+kaε)(1+kbε) =(1+k(a+b)ε) =ϕ k(1+(a+b)ε) =ϕ k((1+aε)(1+bε)) \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)

    [ε]/(ε 2)C ()/(ε 2) \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 [ε]/(ε 2)\mathbb{R}[\epsilon]/(\epsilon^2) (given by the diffeomorphism’s first Taylor coefficient, which is the kk from above).

    • CommentRowNumber4.
    • CommentAuthorDavid_Corfield
    • CommentTimeNov 14th 2020

    These cardinality do behave oddly. Sometimes it makes better ’sense’ to take the cardinality of the open interval as 1-1, so that the nonzero reals have cardinality 2-2. See, e.g., here.

    • CommentRowNumber5.
    • CommentAuthorUrs
    • CommentTimeNov 14th 2020
    • (edited Nov 14th 2020)

    [ removed ]

    • CommentRowNumber6.
    • CommentAuthorDavidRoberts
    • CommentTimeNov 14th 2020

    Ah, ok. I was thinking it was something like the function [ε]/(ε 2)[kε]/(k 2ε 2)\mathbb{R}[\varepsilon]/(\varepsilon^2) \to \mathbb{R}[k\varepsilon]/(k^2\varepsilon^2), but now I see what was intended!

    • CommentRowNumber7.
    • CommentAuthorUrs
    • CommentTimeNov 15th 2020

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

    • CommentRowNumber8.
    • CommentAuthorsvennik
    • CommentTimeNov 15th 2020
    • (edited Nov 15th 2020)

    Regarding the claim that the perplex numbers have precisely two automorphisms, this is true. First, it’s easy to see that ϕ 1(a+be)=a+be\phi_1(a + be) = a + be and ϕ 2(a+be)=abe\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 ϕ\phi be an automorphism over the perplex numbers. Observe that ϕ(e) 2=ϕ(e 2)=ϕ(1)=1\phi(e)^2 = \phi(e^2) = \phi(1) = 1. It follows that ϕ(e){1,1,e,e}\phi(e) \in \{1,-1,e,-e\}. But if ϕ(e){+1,1}\phi(e) \in \{+1, -1\} then ϕ\phi is not injective. So therefore ϕ(e){e,e}\phi(e) \in \{e, -e\}.

    • CommentRowNumber9.
    • CommentAuthorUrs
    • CommentTimeNov 15th 2020

    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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