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[This is a slightly tweaked 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.]

Tessarines are a product of sorts of the complex numbers and the split-complex numbers. If you substitute split-complex number coefficients for real coefficients in the form of complex numbers, or substitute complex number coefficients for real coefficients in the form of split-complex numbers, you get the tessarines. Complex numbers can be plotted in $R^2$ (two-dimensional Euclidean space with the normal metric, or the Cartesian plane), while split-complex numbers can be plotted in the pseudo-Euclidean space $R^{1,1}$. $R^{1,1}$ has a metric signature ($+$ $−$) or ($−$ $+$), and I assume that the real axis is generally associated with the $+$ and the "$j$-axis" (assuming the split complex number is written in the form $x+yj$ ) is generally associated with the $−$.

For tessarines (assuming they're of the form $w+xi+yj+zk$ (possible different variables than $w$, $x$, $y$ and $z$ or different order thereof aside) where $ij=ji=k$, $i^2=-1$, and $j^2=+1$ ), I'm fairly certain that they would have to be graphed on the pseudo-Euclidean space $R^{2,2}$, with two $+$s and two $−$s in the metric signature. I'm also fairly certain that the real axis and "$j$-axis" would have to have opposite signs (one $+$ and one $−$), and the same with the "$i$-axis" and "$k$-axis". (It would probably be more correct to label the axes after the variable representing the coefficients, but those variables and the order thereof might be less consistent among mathematicians than the use of $i$, $j$ and $k$ and their relationships to $1$ and $−1$ and to each other.)

I imagine, but am less certain here and my question is basically if my assumption is correct or not, that the real and $i$-axes are generally given have the same sign as each other (probably the $+$), and thus the $j$-and $k$-axes are also given the same sign (probably the $−$). While the $i$ and $k$ are equivalent to each other in mathematical operations involving tessarines, my assumption above would seem consistent with $i$ being the same as the $i$ in complex numbers and $j$ being the same as the $j$ in split-complex numbers, with $k$ being the product of $i$ and $j$.

As axes in non-positive definite coordinate spaces are sometimes thought of as representing either space or time dimensions, you could think of the real axis as representing a real space dimension, the $i$-axis as representing an imaginary space dimension (imaginary in the same sense as in the complex numbers), the $j$ axis as representing a real time dimension and the $k$-axis as an imaginary time dimension. While I have talked above about tessarines being plotted in $R^{2,2}$, it occurs to me that this space of four real dimensions might be more appropriately called $C^{1,1}$, although I don't know if it's kosher to use an indefinite signature with a "base" algebra other than the real numbers to describe a coordinate space.

I could be wrong, however, about there being a standard for how tessarines are plotted on $R^{2,2}$, and maybe plotting tessarines in multi-dimensional space in a way analogous to complex-numbers (the complex plane), quaternions and split-complex numbers just isn't a thing. There are special names for pseudo-Riemannian manifolds with $0$ (Riemannian) or $1$ (Lorentzian) dimensions with the opposite "sign" from all other dimensions, but there doesn't seem to be much attention paid to pseudo-Riemannian manifolds or even pseudo-Euclidean spaces with two or more dimensions of each "sign."

I can't be the only person who's thought of this, however. I'd appreciate whatever help in answering my "question" and confirming or correcting my assumptions that people can give.

  • Interested in or know something about tessarines? Notice how I called them a "product of sorts" of the complex and split-complex numbers? Check out the other question I asked about them today at https://math.stackexchange.com/questions/4489085/relationship-between-tessarines-complex-numbers-and-split-complex-numbers-and . – Kevin M. Lamoreau Jul 09 '22 at 02:45
  • As I understand it, the tessarines' bicomplex norm is that of neither $\Bbb R^{2,,2}$ nor $\Bbb C^{1,,1}$ but $\Bbb C^2$. – J.G. Jul 19 '22 at 07:13
  • I'm not sure of the norm (and perhaps there could be more than one norm for tessarines, one that goes with the bicomplex number formulation and anyother that goes with the formulation of tessarines as the tensor product ("Kronecker product") of the complex numbers and the split-complex numbers) ... anyway I'm not sure the bicomplex norm is what counts when determining how tessarines would be plotted. I mean, one name for $R^{1,1}$ is the "split-complex plane", and the complex plane $C^1$ is a lot like $R^2$, so the space of tessarines would seem to be modeled after $R^{2,2}$ or $C^{1,1}$. – Kevin M. Lamoreau Jul 19 '22 at 22:26
  • It is better to think that the real axis represents time, hyperbolic axes represent Minkowski spatial dimensions (and have the opposite sign in metric), dual axes represent Newtonian (Euclidean) spatial dimensions and imaginary axes... I would like to know :-) – Anixx Aug 06 '22 at 21:39
  • @Anixx, this is interesting. I'll be mulling that over in my spare time in the next day or so. I've always thought of real axis as spacial ones and hyperbolic axes as temporal ones. – Kevin M. Lamoreau Aug 06 '22 at 22:02
  • @KevinM.Lamoreau basically, hyperbolic space is Minkowski space, mathematically. 3D Minkowski space is like space with a real and 3 hyperbolic axes. – Anixx Aug 06 '22 at 22:18

1 Answers1

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In tessarine modulus (which is not a norm as it does not satisfy the norm axioms) real and the two imaginary axes have positive sign before square under root and the hyperbolic axis has negative sign. So, 3 positive and one negative.

In other words,

$|a+bi+cj+dij|=\sqrt{a^2+b^2-c^2+d^2}$.

Hyperbolic axes get negative sign, real and imaginary get positive sign, dual (if any) have zero coefficient (do not contribute to modulus).

Anixx
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  • Thank you for all your replies to my questions/comments today and yesterday, @Anixx, but I'm not ready to accept this answer yet. We've established in the other thread that there are kind of two fundemental ways of looking at this algebra, where the key variable seems to be whether the hyperbolic unit times one of the two imaginary units equals the positive or negative of the other. But if we go the Cockle route, with $j$ being the hyborbolic unit but $i$ and $k$ not being interchangeable ($k$ being $ij$ as you write here), then why can't the modulus be $|a+bi+cj+dij|=\sqrt{a^2+b^2-c^2-d^2}$? – Kevin M. Lamoreau Aug 06 '22 at 21:53
  • @KevinM.Lamoreau the sign before $ij$ (of $d$) is irrelevant here, because the coefficient gets squared under the root. – Anixx Aug 06 '22 at 21:55
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    @KevinM.Lamoreau whether you take $k=ij$ or $k=-ij$, the modulus is the same, because the coefficient of $k$ gets squared. – Anixx Aug 06 '22 at 21:57
  • The question in my first comment may be a silly one, like "why can't 1 + 1 = 3", but I just don't see how tessarines (in the Cockle formation) wouldn't be plottable in $R^{2,2}$ or $C^{1,1}$. I guess I need to take a break from this. Maybe tomorrow I'll have a clearer head and can understand this. Thanks again for trying to explain this to me. – Kevin M. Lamoreau Aug 06 '22 at 21:59