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There are well-described methods of generalizing arbitrary functions to matrices in a natural way.

Basically, if $A=PD_AP^{-1}$ where $D_A$ is a diagonal matrix, then $f(A)=Pf(D_A)P^{-1}$, where the function $f$ is applied to the diagonal matrix element-wise.

This is automatized in some CAS systems, such as Mathematica, so one can apply arbitrary functions to matrices.

For instance, it can be applied to the Sign function with MatrixFunction[Sign, A]. It should be noted that this function is defined on the complex plane as $z/|z|$ and $\operatorname{sign} 0=0$, which is different from some other definitions (Higham - What Is the Matrix Sign Function? also talks about generalizing the sign function to matrices, but it seems they use a different definition).

For instance, this method gives $$\operatorname{sign}\left( \begin{array}{cc} 1 & -8 \\ 1 & 7 \\ \end{array} \right)=1$$ (e.g., produces an identity matrix) but $$\operatorname{sign}\left( \begin{array}{cc} 1 & -8 \\ -1 & 7 \\ \end{array} \right)=\left( \begin{array}{cc} -\frac{3}{\sqrt{17}} & -\frac{8}{\sqrt{17}} \\ -\frac{1}{\sqrt{17}} & \frac{3}{\sqrt{17}} \\ \end{array} \right).$$

The function can be even applied to some zero divisors.

It also can be applied to hypercomplex numbers represented in matrix form. Now, I noticed that while $\operatorname{sign} z$ can take infinitely-many values on the complex numbers, it can take only 9 values on split-complex numbers: $0$, $1$, $-1$, $j$, $-j$, $1/2+j/2$, $1/2-j/2$, $-1/2+j/2$, $-1/2-j/2$. When applied to dual numbers, it seems to give 5 different values.

The usual rule $\operatorname{sign} (AB)=\operatorname{sign} A\cdot \operatorname{sign} B$ still holds though.

That said, I wonder, whether it has any fundamental importance (telling us about the properties of the ring), whether a ring has finite (as split-complex numbers) or infinite (as complex numbers) set of possible values of the sign function? Can this numerocity be predicted?

What about the $p$-adic rings, can the sign function be generalized there as well?

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  • $\begingroup$ By "split-complex" and "dual" numbers do you mean $\mathbb{R}[j]$, $j^2=1$, and $\mathbb{R}[\varepsilon]$, $\varepsilon^2 = 0$, respectively, or $\mathbb{C}[j]$, $j^2=1$, and $\mathbb{C}[\varepsilon]$, $\varepsilon^2 = 0$? I guess you mean the first two, because you get only 9 values, namely $0$, $(0,\pm 1)$, $(\pm 1,0)$ and $(\pm 1, \pm 1)$. $\endgroup$
    – M.G.
    Commented Nov 7, 2021 at 12:23
  • $\begingroup$ My point is that if the algebras have complex structure, then $\operatorname{sign}$ will always have infinite range for trivial reasons, so it only makes sense to look at real algebras without compatible complex structure. $\endgroup$
    – M.G.
    Commented Nov 7, 2021 at 12:31
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    $\begingroup$ @M.G. well, with dual numbers, yes, I used some handwaving and cheating. For non-zero-divisor numbers the sign is simply the sign of their real part, the sign of 0 is 0, and for $|\varepsilon|$ the matrix method gives $\varepsilon \text{sign}'(0)$. As I currently develop the theory of divergent integrals, $\text{sign}'(0)$ is a very well defined value there, equal to $2\delta(0)=\frac2\pi\int_0^\infty dx$. So, the sign is not a dual number, not a divergent integral, but some weird product of them. Still, I think, such ring is conceivable. $\endgroup$
    – Anixx
    Commented Nov 7, 2021 at 13:09
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    $\begingroup$ yes, the existence or absence of complex structure leads to qualitatively different outcomes. Still it seems that the infinitude of the range reduces to the topology of the group of units of the $\mathbb{R}$-algebra. The very least, one wants $\operatorname{sign}$ to be a multiplicative function with $\operatorname{sign}(\pm1) = \pm1$ and continuous outside of $0$. The first two conditions imply that $\operatorname{sign}$ maps units to units. (cont.) $\endgroup$
    – M.G.
    Commented Nov 7, 2021 at 14:00
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    $\begingroup$ Continuity now implies that connectedness of the group of units puts a restriction on the range of $\operatorname{sign}$: if the group of units is connected, $\operatorname{sign}$ must have uncountably infinite range. Notice that this argument does not care about complex or real. $\endgroup$
    – M.G.
    Commented Nov 7, 2021 at 14:02

1 Answer 1

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Let $\mathcal{A}$ be a finite-dimensional commutative associative unital $\mathbb{R}$-algebra. (This covers split-complex, hyberbolic and a lot of other number "systems".) We want to extend the function $\operatorname{sign}(x)$ to $\mathcal{A}$ in a meaningful way (to be determined) and understand the number of values in its range.

If $\mathcal{A}$ carries a compatible $\mathbb{C}$-structure, then any extension of $\operatorname{sign}(z) = z/|z|$ takes infinitely many values ($\mathbb{T}$ is in the range) for trivial reasons since $\mathbb{C} \hookrightarrow \mathcal{A}$.

Furthermore, by virtue of finite dimensionality $\mathcal{A}$ is Artinian and hence decomposes into a direct sum of local Artinian algebras $\mathcal{A} = \bigoplus_{i=1}^n (\mathcal{A}_i, \mathfrak{m_i})$. Thus it suffices to understand the extension of $\operatorname{sign}(x)$ to such a local Artinian algebra $(\mathcal{A},\mathfrak{m})$. Moreover, by the previous considerations we can assume that $(\mathcal{A},\mathfrak{m})$ does not carry a compatible complex structure, hence $\mathcal{A} / \mathfrak{m} \cong \mathbb{R}$ as algebras and $\mathcal{A} = \mathbb{R} \oplus \mathfrak{m}$ as $\mathbb{R}$-vector spaces.

Write $Z = \lambda \oplus X \in \mathcal{A}$ with $\lambda \in \mathbb{R}$ and $X \in \mathfrak{m}$. Thus, in particular, $r$ is the eigenvalue (multiple) of $Z$ and $X$ is nilpotent. A possible extension of $\operatorname{sign}(x)$ is an element of the $\mathcal{A}$-module $\mathscr{D}'(\mathbb{R})[X]$ via the "Taylor series": $$ \operatorname{sign}(Z) = \operatorname{sign}(\lambda \oplus X) := \operatorname{sign}(\lambda) + 2 \sum_{k=1}^{N-1} \frac{\delta^{(k-1)}(\lambda)}{k!} X^k $$ where $N \in \mathbb{N}$ is the smallest integer such that $\forall X \in \mathfrak{m}: X^N = 0$. Note that $\operatorname{sign}(x) \in L^1_{\mathrm{loc}}$ and $0\cdot\infty = 0$. Alternatively, one can look at it as a mapping $$ \mathfrak{m} \to \mathscr{D}'(\mathbb{R}) \otimes_\mathbb{R} \mathcal{A} $$

One can have an honest function when $\lambda \neq 0$, namely $\operatorname{sign}(Z) := \operatorname{sign}(\lambda)$, and so $\operatorname{sign}(Z)$ takes exactly two values on $\mathcal{A}^\times$. If $\lambda = 0$, I am currently unsure what would mean to "count the values of its range". In any case, there are infinitely many nilpotents, unless $\mathfrak{m} = 0$. Lastly, if $\mathcal{A} = \bigoplus_{k=1}^n \mathcal{A}_k$, we have $\mathcal{A}^\times = \prod_{k=1}^n \mathcal{A}_k^\times$ and so the possible-values count on $\mathcal{A}^\times$ is $2^n$.

Summary:

  1. If $\mathcal{A}$ has a complex structure, then $\operatorname{sign}(z)$ would have infinitely many distinct values in its range. In particular, it contains the circle $\mathbb{T}$.

  2. If $\mathcal{A}$ has no complex structure and is not a product of other algebras, then $\operatorname{sign}(x)$ has a boring extension as a function to $\operatorname{sign}(\lambda + X) := \operatorname{sign}(\lambda)$, where $\lambda$ is the eigenvalue part and $X$ is the nilpotent part. In this case, its range contains 3 distinct values inherited from $\mathbb{R}$. And if $\mathcal{A}$ is a product of $n$ such algebras without complex structure, then $\operatorname{sign}$ has $3^n$ distinct values in its range.

  3. If $\mathcal{A}$ has no complex structure and is not a product of other algebras, then the distribution $\operatorname{sign}(x)$ admits an extension as a distribution depending on the nilpotent parameter $X$ and whose integrals are $\mathcal{A}$-valued. In this case, we get an uncountable family of different distributions, parametrized by $X$.

Remaining Parts:

  1. Find an extension to general zero divisors and general idempotents.

  2. Make sense of "continuity" of the distributional extension $\operatorname{sign}(Z)$ over $\mathcal{A}$ (see discussion in the comments below the original question).

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  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ Commented Nov 8, 2021 at 1:34
  • $\begingroup$ By the way, due to isomorphism $a + bj \leftrightarrow (a + b, a - b)$, the split-complex numbers represent intervals of real numbers with element-wise operations. And this intuitively explains why they have 9 signs: if the both ends of an interval are positive, the sign is 1, if the both ends are negative, the sign is -1, and overall there are 9 options (including the both ends being zero, and intervals with one end at zero, which are zero divisors). So, I think, we CAN speak about POSITIVE and NEGATIVE split-complex numbers in this light. $\endgroup$
    – Anixx
    Commented Dec 12, 2021 at 8:07

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