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[This is an expanded re-posting of a question I asked in April 2020, that subsequently got auto-deleted as an "abandoned question." Hopefully launching a second attempt now (over two years after I first asked it, and I believe over a year after it was deleted) is okay, particularly since I've expanded it with variables, equations, etc. to "flesh out" or clarify what I had written earlier.]

I have a BA in Mathematics from the University of Maine at Farmington, but am in no means a professional mathematician and my knowledge of things like tessarines (bicomplex numbers) and split-complex numbers is from Wikipedia. This question is kind of a question of mathematical terminology when dealing with division algebras that are (in a certain sense) "products" of other division algebras. I know what it means when a group is a direct product of two groups of lower order, but when it comes to (hyper)complex numbers I'm not sure of the jargon.

Tessarines are commonly written in the form $w+xi+yj+zk$ (with possible different coefficients than $w$, $x$, $y$ and $z$ or different order of those coefficients, but I'll use these coefficients here), where $ij=ji=k$, $i^2=-1$, and $j^2=+1$. So you have a real number component (possibly $0$), a real coefficient (possibly $0$) of the same $i$ used in the complex numbers, a real coefficient (possibly $0$) of the same $j$ used in the split-complex numbers and a real coefficient (possibly $0$) of $k$ which is the product of $i$ and $j$.

Introducing complex variables $α=w+xi$ and $β=y+zi$ and split-complex variables $γ=w+yj$ and $λ=x+zj$, tessarines can be written in the forms $γ+λi$ or $α+βj$. So, if you substitute split-complex number coefficients for real coefficients in the common form of the complex numbers, or if you substitute complex number coefficients for real coefficients in the common form of the split-complex numbers, you get the tessarines, or at least an algebra isomorphic to the tessarines which is fairly easily converted into the common tessarine form I described above.

So, what does that make the relationship between tessarines, complex numbers and split-complex numbers? The tessarines are the [blank] of the complex numbers and the split-complex numbers. What's the [blank]?

I know tessarines are equivalent to bicomplex numbers, which are sometimes written using the basis {$1$, $h$, $i$, $hi$}, where $i$ can be thought of as the same $i$ used above (it squares to $-1$), $h$ (which also squares to $-1$ and commutes with $i$) is equivalent to $k$ and $hi$ is equivalent to $-j$. These bicomplex numbers have some kind of relationship between the complex numbers and (the complex numbers) themselves. What's that relationship called? The bicomplex numbers are the [blank] of the complex numbers and (the complex numbers) themselves? What's that [blank]?

I think the answer to one of these two questions may be the tensor product, but I'm not sure which one. Wikipedia's article on hypercomplex numbers ( https://en.wikipedia.org/wiki/Hypercomplex_number#Tensor_products ) states that bicomplex numbers (and thus tessarines) are the tensor product of the complex numbers and (the complex numbers) themselves, but I would think of tessarines as more of a product of the complex numbers and the split-complex numbers. But I know algebraic structures often have multiple types of "products."

Thanks to anyone who can help in answering my question here.

Kevin M. Lamoreau
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  • Interested in or know something about tessarines? Check out the other question I asked about them today at https://math.stackexchange.com/questions/4489098/plotting-tessarines-in-the-pseudo-euclidean-space-r2-2-conventions-aroun . – Kevin M. Lamoreau Jul 09 '22 at 02:46
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    Some notes: 1. Tessarines are not a division algebra, 2. Usually we take $k=-ij$. This has to do with choice of logarithm branch at branch cut, so that $\ln j=\frac{i\pi}2+\frac{k\pi}2$ had the both terms with plus sign and to be symmetric against change of $k$ and $i$. – Anixx Aug 05 '22 at 09:06
  • @Anixx, so basically Corrodo Segre's bicomplex number "configuration" (probably not the right word there), with $h$ renamed $k$ (as with my configuation, which seems to be the one used on Wikipedia) and $hi$ renamed $j$ (rather than $-j$). I had gotten the sense that Segre's name for the algebra had triumphed, but that Jame's Cockle's configuration had triumphed. But maybe the only thing of Cockle's that prevailed was the choice of letters for the imaginary elements (to be consistent with Hamilton's quaterions). – Kevin M. Lamoreau Aug 06 '22 at 20:53
  • $k$ = $ij$ seems to work better than $k$ = $-ij$ to describe tessarines as the tensor product of the complex and split-complex numbers, but given that I guess they're more often called bicomplex numbers (being the tensor product of complex numbers and themselves), perhaps that's not very important. I'm weak on logarithms but will bow to your obvious knowledge and your reputation there, and I can see how that property would be valuable. I guess it all comes down to which imaginary element is seen as the product of the other two (with no one thinking $i$ as the product). – Kevin M. Lamoreau Aug 06 '22 at 21:06
  • Yes. For myself, I am still not sure what I do prefer more. On the one hand, $j^2=1$ coincides with split-complex numbers and it is good to have a special name for a constant with unique properties. On the other hand, the letters $i$ and $j$ are similar, so it is good idea to name two imaginary units $i^2=-1$ and $j^2=-1$ with the same properties with similar letters (and electricians already use $j$ for an imaginary unit). – Anixx Aug 06 '22 at 21:09
  • Oh, I didn't realize that in what you were saying with your first comment, that $k$ rather than $j$ would square to $+1$ (the other two imaginary units both squaring to $-1$). So the difference from what I wrote in my question is just a swapping of $j$ and $k$. Never seen that before, but my knowledge of hypercomplex numbers is largely from Wikipedia which anyone can edit (although who would want to lie about math?). Thank you for your information. – Kevin M. Lamoreau Aug 06 '22 at 21:13
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    Well, the argument is: if we have two imaginary units $i_1$ and $i_2$ and one hyperbolic unit $j$, then logarithm of $j$ should be absolutely agnostic about swap of $i_1$ and $i_2$. So, we should chose the branch where $\ln j=\frac{i_1\pi}2+\frac{i_2\pi}2$ that is, the terms have the same sign. Otherwise we would have $\ln i_1=i_1\pi$ and $\ln i_2=-i_2\pi$, and the properties thus would be different. – Anixx Aug 06 '22 at 21:14
  • Okay, so $j^2$ still equals $+1$. A lot of this is over my head. Because of my interest in psuedo-Euclidean spaces (as shown in the question I asked the same day that I've linked to here), having the two imaginary units (I guess it's called a hyperbolic unit if it squares to $+1$) be completely interchangeable is NOT necessarily desirable. $k$ under my configuartion could be an "imaginary time" unit, with $j$ being a real time unit. – Kevin M. Lamoreau Aug 06 '22 at 21:21

1 Answers1

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The tessarines are the [blank] of the complex numbers and the split-complex numbers. What's the [blank]?

The tensor product (of algebras), over $\mathbb{R}$.

The bicomplex numbers are the [blank] of the complex numbers and (the complex numbers) themselves? What's that [blank]?

It's still the tensor product again. This may be a little surprising; it shows that the tensor product isn't cancellative.

Qiaochu Yuan
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  • Wow! That's interesting. Your answer prompted me to read up on the cancellation property. The Wikipedia article had a focus on group-like structures (magmas) rather than algebras, but seeing how (separate left- and right-) cancellation properties were defined there was enough for me to get how the two answers you gave show that the tensor product isn't cancellative. – Kevin M. Lamoreau Jul 19 '22 at 21:40
  • Your first answer matches what I thought of as a "natural product", and your second answer jives with the tensor product section of the Wikipedia article on hypercomplex numbers.

    Am I correct in assuming that the tensor product (complex numbers) * (complex numbers) = (bicomplex numbers) is also over R? That's what that section of the Wikipedia article on hypercomplex numbers says.

    Anyway, thank you for answering my question.

     – Kevin M. Lamoreau Jul 19 '22 at 21:49
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    @Kevin: yes, it's also over $\mathbb{R}$. And you're welcome! – Qiaochu Yuan Jul 20 '22 at 00:01
  • Also, tessarines are the algebraic closure of split-complex numbers. (there is no solution among split-complexes for $x^2+1=0$) – Michael Ejercito Jul 13 '24 at 18:44