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I will call it "hypermodulus". In simple words, hypermodulus is the exponent of the scalar part of the finite part of the logarithm of the object: $H(A)=\exp (\operatorname{scal} \operatorname{f.p.} \ln A)$.

The "finite part" is meant to mean regularization: Hadamard, Dirichlet, Ramanujan, etc.

The "scalar part" is meant to mean the coefficient of the basis vector $1$ in rings over reals (real part), the arithmetic mean of the diagonal elements in square real matrices (trace divided by the matrix rank) or arithmetic mean of components in $\mathbb R^n$.

Now, it seems hypermodulus can be defined for various objects. Let's see.

  1. For all complex numbers except zero the hypermodulus is equal to the modulus (absolute value).

  2. Zero. The Taylor series for logarithmic function (expanded at $1$), at zero becomes the harmonic series with opposite sign:$-\sum_{k=1}^\infty \frac1k$, this has the regularized value of $-\gamma$ (there are other ways to see this as well). Thus, $H(0)=e^{-\gamma}$. We will use this result further.

  3. Zero divisors in dual numbers. $\ln (a+b\varepsilon)=\ln a+\varepsilon b/a$. So, all zero divisors have hypermodulus $H(b \varepsilon)=e^{-\gamma}$.

  4. Zero divisors in split-complex numbers. $\ln (a+bj)=\frac{1}{2} j (\log (a+b)-\log (a-b))+\frac{1}{2} (\log (a-b)+\log (a+b))$, so $H(\pm a\pm a j)=\sqrt{2a}e^{-\gamma/2}$. Particularly, the idempotents $1/2+j/2$ and $1/2-j/2$ have hypermodulus $e^{-\gamma/2}$.

  5. Other split-complex numbers. For non-zero-divisors, $H(a+bj)=\sqrt{|a^2-b^2|}$. This is always a positive real number.

  6. Infinity (extended real line). Since $\gamma=\lim_{n\to\infty}\left(\sum_{k=1}^n \frac1{k}-\int_1^n \frac 1t dt\right)$, the integral $\ln\infty=\int_1^\infty \frac 1t dt$ has regularized value of $0$. Thus, $H(\infty)=1$.

  7. Other improper elements. In $\mathbb{\overline{R}}^2$ the improper elements $(0,\pm\infty)$ and $(\pm\infty,0)$ have the hypermodulus $H=e^{-\gamma/2}$. In split-complex numbers the elements correspond to improper elements $\pm\infty\pm j\infty$.

  8. Umbral calculus (Bernoulli umbra). In umbral calculus we will use the index lowering (evaluation) operator as "finite part". Thus, $H(B+a)=e^{\psi(a)}$, where $B$ is Bernoulli umbra and $\psi$ is digamma function. Particularly, $H(B)=H(B+1)=e^{-\gamma}$. (the later result can be seen also from here), $H(e^{B})=\frac1{\sqrt{e}}$, $H(e^{B+1})=\sqrt{e}$.

  9. Euler's umbra. $H(E+a)=\exp (-\Phi (-i,1,a+1)-\Phi (i,1,a+1))$, where $\Phi$ is Lerch transcendent. Particularly, $H(E)=e^{-\pi/2}$, $H(E+1)=1/2$.

  10. Divergent integrals. Multiplication of divergent integrals can be defined in different ways. Trivial multiplication of divergent integrals gives hypermodulus equivalent to the hypermodulus of their regularized value. On the other hand, umbral multiplication gives a system that is isomorphic to umbral calculus, with Bernoulli umbra $B=\int_{-1/2}^\infty dx$ and very interesting expressions: $H(\int_0^\infty dx)=\frac{e^{-\gamma}}4$, $H(\int_0^\infty xdx)=e^{\psi\left(\frac12+\frac{i}{2\sqrt{3}}\right)+\psi\left(\frac12-\frac{i}{2\sqrt{3}}\right)}$, $H(\sum_{k=0}^\infty k)=e^{\psi\left(\frac12+\frac{1}{2\sqrt{3}}\right)+\psi\left(\frac12-\frac{1}{2\sqrt{3}}\right)}$ and others.

  11. Formal power series. Formal power series have subsets, isomorphic to umbras with any moments, including Bernoulli umbra and Euler's umbra and many more.

  12. Update. Infinite products. Some sources claim that the regularized value of infinite products $\prod_{k=0}^\infty k$ is $\sqrt{2\pi}$ and of $\prod_p p$ (where $p$ is prime) is $4 \pi^2$. But in fact, their calculations find not the finite part, but hypermodulus. They convert the products to series using logarithm, regularize the series and then exponentiate the finite part.

So, we can see that the concept is applicable in various areas of mathematics, where the expressions for hypermodulus often involve the $e^{-\gamma}$ constant.

I wonder,

  1. What algebraic role the $e^{-\gamma}$ plays in all these fields?

  2. Can the concept of hypermodulus be used as some indicator of algebraic role of an element of a set, such as showing whether an element a zero divisor, nilpotent, idempotent or whatever else?

  3. Can it help to somehow define or generalize functions on those elements?

  4. What's in common between Bernoulli umbra and zero divisors?

Anixx
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  • The way you write it, the "scalar part" is not quite well-defined, but I think I see what you mean. But for clarity, what would be the scalar part of a general element of the direct product $\mathbb R \times \mathbb C$? Or of a general element of the $5$-dimensional $\mathbb R$-algebra given by the matrices $\pmatrix{a & b+ci\0& d+ei}$? – Torsten Schoeneberg Jun 15 '22 at 17:52
  • @TorstenSchoeneberg $\mathbb R \times \mathbb C$ can be canonically represented with $3\times3$ matrices of the form $M=\left( \begin{array}{ccc} a & b & c \ c & a & b \ b & c & a \ \end{array} \right)$ The scalar part would be $\text{Tr}M/3=a$. For the other example, we should represent it in the form of a real matrix of rank 5, find its trace and divide by 5. – Anixx Jun 15 '22 at 18:09
  • @TorstenSchoeneberg I think it would be $(a+d)/2$. – Anixx Jun 15 '22 at 18:17
  • @TorstenSchoeneberg well, a rank 4 real matrix would suffice to represent it: $\left( \begin{array}{cccc} a & 0 & b & c \ 0 & a & -c & b \ 0 & 0 & d & e \ 0 & 0 & -e & d \ \end{array} \right)$. Indeed it will be $(a+a+d+d)/4=(a+d)/2$. – Anixx Jun 15 '22 at 18:20
  • I am bound to agree in the second example, but I fail to see that canonical representation for the first. What element in $\mathbb R \times \mathbb C$ corresponds to the matrix $M$? Or conversely, what would be $a,b,c$ for the general element $(x, y+iz) \in \mathbb R \times \mathbb C$? – Torsten Schoeneberg Jun 15 '22 at 18:43
  • @TorstenSchoeneberg $\left( \begin{array}{ccc} 0 & 1 & 0 \ 0 & 0 & 1 \ 1 & 0 & 0 \ \end{array} \right) \to (1, e^{2 i \pi/3}), \left( \begin{array}{ccc} 0 & 0 & 1 \ 1 & 0 & 0 \ 0 & 1 & 0 \ \end{array} \right) \to (1, e^{-2 i \pi/3})$ – Anixx Jun 15 '22 at 19:34
  • @TorstenSchoeneberg well, $(x,a+bi)\to\frac{1}{3} \left(x+2 a\right)+j \left(\frac{b}{\sqrt{3}}-\frac{a}{3}+\frac{x}{3}\right)+k \left(-\frac{b}{\sqrt{3}}-\frac{a}{3}+\frac{x}{3}\right)$, so the scalar part of $(x,a+bi)$ is $\frac{1}{3} \left(x+2 a\right)$. – Anixx Jun 15 '22 at 20:11
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    Thanks, that seems consistent. – Torsten Schoeneberg Jun 16 '22 at 01:57
  • @TorstenSchoeneberg thanks for the question. Indeed, the result seems non-trivial. – Anixx Jun 16 '22 at 11:06
  • FYI: https://math.stackexchange.com/q/4475149/96384 – Torsten Schoeneberg Jun 19 '22 at 05:44
  • @TorstenSchoeneberg by the way, another representation for $(a,b+ci)$ is $\left( \begin{array}{ccc} a & 0 & 0 \ 0 & b & c \ 0 & -c & b \ \end{array} \right)$, which gives the same result for scalar part. – Anixx Jun 19 '22 at 12:11

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