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Are there algebras over real numbers (with exponentiation), such that there is such $z$ that does not include components in $\mathbb{C}$ (or in a subset isomorphic to $\mathbb{C}$), for which $(-1)^z\in \mathbb{R}$ and irrational? What about such $z$ that $z^z\in\mathbb{R}$ and irrational?

I mean, we can rise $(-1)$ to an imaginary power and have an irrational real number. As I know, neither in split-complex numbers, nor tessarines, nor in dual numbers, nor in quaternions you have elements with this property except those that have components in subrings, isomorphic to $\mathbb{C}$.

Are there such examples in other extensions of $\mathbb{R}$, matrices, vectors, tensors, power series, whatever?

Maybe my question is not perfectly formulated and one would be able to come up with some trivial examples, but I hope you understood the idea.

Anixx
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    I'm not sure what you mean by algebra in this context. If e^x is involved one is talking about fundamentally analytic statements. – JoshuaZ Jun 13 '21 at 12:58
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    Very marginally relevant might be The exponential function in linear algebras by Paul Dienes [Quarterly Journal of Mathematics (Oxford) 1 (1930), 300−309], which I discussed in a couple of sci.math posts on 30 January 2006 (1st post and 2nd post). – Dave L Renfro Jun 13 '21 at 15:09
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    @M.G. thanks, fixed. – Anixx Jun 13 '21 at 15:30
  • @DaveLRenfro thanks for the interesting links, but I think your sci.math discussion is more connected with this my question: https://math.stackexchange.com/questions/4075766/why-is-ln-0-ne-ln-infty which also discusses the divergent integral $-\int_0^1 \frac1x dx$, which can be considered in a sense "logarithm of zero", although ascribing any value to logarithm of zero brings algebraic diffdiculties). – Anixx Jun 13 '21 at 15:44
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    @M.G. $(-1)^\varepsilon=1+i \pi \varepsilon$ This is if we extend reals with both complex numbers and duals. In only reals and duals, no solution. This is because of the general rule, $f(a+b \varepsilon)=f(a)+b f'(a) \varepsilon$ – Anixx Jun 13 '21 at 15:46
  • @Anixx: oops, sorry, you are correct. – M.G. Jun 13 '21 at 16:04

1 Answers1

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I will address the question in the case $f(z) = (-1)^z = \exp(z \log(-1))$ and $A$ is a finite-dimensional commutative (associative unital) $\mathbb{R}$-algebra, where $\log(-1)$ is a suitable choice of the logarithm in $A$.

Wlog. $A$ is a local $\mathbb{R}$-algebra $(A,\mathfrak{m})$. There are two possibilities. First possibility is $A / \mathfrak{m} \cong \mathbb{C}$, in which case $A$ has a $\mathbb{C}$-structure extending the $\mathbb{R}$-structure, and the second possibility is $A / \mathfrak{m} \cong \mathbb{R}$. Either way, $A \cong \mathbb{K} \oplus \mathfrak{m}$ as vector spaces (not as algebras though!), where $\mathbb{K} \in \{\mathbb{R},\mathbb{C}\}$.

1. Claim: $\log(-1)$ exists iff $A$ has a complex structure.

Proof: Write $w = r + X \in A$ for $r\in \mathbb{K}$ and $X \in \mathfrak{m}$. We need $\exp(w) = -1$, i.e. $\exp(X) = -\exp(-r) \in \mathbb{K}^\times$. However $$ \exp(X) = 1 + \sum_{k=1}^n \frac{X^k}{k!} \in 1 + \mathfrak{m} \subset A^\times, $$ where $n \in \mathbb{N}$ is the smallest integer such that $X^n = 0$, as all elements of $\mathfrak{m}$ are nilpotent. Hence, in particular $$ -\exp(-r) = 1, $$ which is possible iff $r \in \mathbb{C}$. Moreover, we have $X = 0$ for the following reason: $$ 0 = \sum_{k=1}^n \frac{X^k}{k!} = X \left( 1 + \sum_{k=2}^n \frac{X^{k-1}}{k!} \right), $$ but the expression in the brackets is a unit because $\sum_{k=2}^n \frac{X^{k-1}}{k!} \in \mathfrak{m}$. This argument is important for our second claim.

Thus the function $(-1)^z$ makes sense only if $A$ has a complex structure.

2. Claim: there is no $z \in A$ without complex components such that $(-1)^z \in \mathbb{R}$

Proof: Fix a value/branch $\lambda:= \log(-1) \in \mathbb{C}^\times$. Since $z$ has no complex components, we have $z = X \in \mathfrak{m}$, i.e. $z$ is nilpotent, and $(-1)^z = \exp(\lambda X) =: \exp(Y)$ for $Y:= \lambda X \in \mathfrak{m}$. Same argument as above shows that $\exp(Y) \in \mathbb{R}$ if and only if $Y = 0$.

Thus there aren't any non-trivial $z \in A$ such that $(-1)^z \in \mathbb{R}$.

I suspect one could show something similar in the case of infinite-dimensional commutative (associative unital) Banach algebras.

I don't have an answer for non-commutative algebras, finite-dimensional or not.

M.G.
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  • How do you conclude that if $z$ has no components in $\mathbb{C}$ it is a nilpotent? For instance, $j$ of split-complex numbers is not a nilpotent, though it has no components in $\mathbb{C}$. – Anixx Jun 13 '21 at 17:08
  • And, by the way, it is a counter-example, as $(-1)^j=-1$, so I would have to modify the question to explicitely exclude such trivial cases (I am interested in irrational or at least, non-integer results). – Anixx Jun 13 '21 at 17:16
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    The split-complex numbers are not local, they are a direct sum of two local algebras, namely $\mathbb{R} \oplus \mathbb{R}$. Every finite-dimensional commutative algebra is a direct sum of local ones. If $f(z)$ is an analytic function, then it extends to a direct sum via $f(z_1 \oplus z_2) := f(z_1) \oplus f(z_2)$, so it suffices to consider only local ones. If $A$ is local, then additionally $A = \mathbb{K} \oplus \mathbb{m}$ as stated in the answer, and $\mathfrak{m}$ contains the nilpotent elements. – M.G. Jun 13 '21 at 17:18
  • Definitely, it contains nilpotents, but $j$ is not nilpotent, and has no complex components. It actually gives a real number in $(-1)^j$. – Anixx Jun 13 '21 at 17:21
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    The only nilpotent element in the split-complex numbers is $0 = (0,0)$. $j$ is simply $(1,-1)$, so $(-1)^j = (-1)^{(1,-1)} = (-1,-1) = -1$ as you say. Direct sums of $\mathbb{R}$ are trivial setting to answer the question. Choosing a fancy basis of $\mathbb{R}^{\oplus N}$ does not change the nature of the algebra. – M.G. Jun 13 '21 at 17:26
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    No, they are zero divisors as you say, but they are not nilpotent. For what $n \in \mathbb{N}$ is $(j+1)^n = 0$ :-) ? – M.G. Jun 13 '21 at 17:29
  • Ah, oops. Yes... – Anixx Jun 13 '21 at 17:30
  • Any other questions? Anything I have missed or not answered (aside from the other cases not addressed in the answer)? :-) – M.G. Jun 13 '21 at 17:34
  • In you last line did you mean $(-1)^z \in \mathbb{R}$? – Anixx Jun 13 '21 at 17:34
  • Yes, sorry, a typo, now fixed :-) – M.G. Jun 13 '21 at 17:35
  • What do you mean by non-trivial $z$? I mean did you account for the case of $j$? – Anixx Jun 13 '21 at 17:38
  • The claim as stated "Claim:* there is no $z \in A$ without complex components such that $(-1)^z \in \mathbb{R}$" seems not to account for the case of $j$ – Anixx Jun 13 '21 at 17:39
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    @Anixx: once again, as I said, $A$ is assumed local. $j$ is not an element of a local algebra. The split complex numbers are not a local algebra, but a direct sum of two local algebras, namely $\mathbb{R}$. Once you know the solutions for $z$ in local algebras, you can construct all solutions for all finite-dimensional commutative algebras by using the direct sum property - it works component-wise, so that's completely trivial. – M.G. Jun 13 '21 at 17:41
  • Okay, it seems I have to read what is local algebra. So, do you say that from $(-1)^1=-1$ we automatically get $(-1)^j=-1$ so we just discard this result as trivial and searching for something special, yes? – Anixx Jun 13 '21 at 17:45
  • Yes, precisely. – M.G. Jun 13 '21 at 17:46
  • Thanks, so now we have to consider only infinitely-dimensional and non-commutative cases. – Anixx Jun 13 '21 at 17:47
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    @Anixx: non-commutative as well. It seems to be much more interesting / difficult. I strongly suspect that the answer in the infinite-dimensional commutative setting is negative, i.e. no interesting $z$, because many arguments like the above go through to the commutative Banach algebra setting. – M.G. Jun 13 '21 at 17:50
  • Since the majority of Clifford algebras are non-commutative, can one say that your answer only covers split-dual-tessarines and various other multi-dimensional combinations of split, dual and complex numbers? – Anixx Jun 13 '21 at 17:56
  • @Anixx: btw, regarding your comment about local algebras... As far as finite-dimensional commutative (associative unital) algebras are concerned, there are just two important algebraic facts to get you started: they are Artinian rings by virtue of their finite-dimensionality + the structure theorem for Artinian rings: https://en.wikipedia.org/wiki/Artinian_ring A lot of useful stuff follows then from that. – M.G. Jun 13 '21 at 17:58
  • @Anixx: no, the structure of finite-dimensional commutative (local) algebras is much more complicated than that with increasing dimension. IIRC, already in dimension 7 you get infinitely many non-isomorphic algebras! But even in lower dimensions, there are other more interesting examples beyond bi-/split-/complex numbers or combinations thereof. – M.G. Jun 13 '21 at 18:02
  • I thought, there is nothing interesting in non-even dimensions. I based my views on this chart: https://i.stack.imgur.com/cOmQT.png – Anixx Jun 13 '21 at 18:06
  • @Anixx: btw, have you tried the 2x2 matrix algebra? The explicit analytic functional calculus is doable there as there is an explicit formula for the dependence of the eigenvalues on the matrix entries. – M.G. Jun 13 '21 at 18:06
  • I think the 2x2 matrix algebra are split-quaternions. I did not try it explicitely. – Anixx Jun 13 '21 at 18:08
  • @Anixx: yes, there is nothing interesting in terms of $(-1)^z \in \mathbb{R}$. I only meant the algebras are interesting on their own for variety of other reasons. – M.G. Jun 13 '21 at 18:08
  • @Anixx: yes, the split-quaternions are isomorphic to the 2x2 real matrices. – M.G. Jun 13 '21 at 18:09
  • Well, in split-quaternions when applying the function to the basis vectors, we have complex unity $i$ and split-complex unity $j$, so we get the same result. $(-1)^k$ gives $-1$. Maybe, non-unity vectors may give some non-trivial results, I don't know. – Anixx Jun 13 '21 at 18:15
  • It is also interesting to see if the case $z^z$ can be reduced to $(-1)^z$. – Anixx Jun 13 '21 at 18:18
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    @Anixx: I think $z^z$ is more subtle, basically b/c it's not holomorphic at $0$, which otherwise helps with nilpotent arguments. It would probably require more work as you have to shift the function around. – M.G. Jun 13 '21 at 18:34
  • @Anixx: in the case of 2x2 matrices, one should be able to deduce a statement for $(-1)^z$ by means of https://en.wikipedia.org/wiki/Analytic_function_of_a_matrix#Arbitrary_function_of_a_2%C3%972_matrix It basically boils down to a system of 4 equations and 4 variables. But before that, you have to give meaning to $\log (-1)$ in the matrix algebra as you have more choice there... – M.G. Jun 13 '21 at 18:45
  • Well, in split-quaternions we still have the complex $i$, so giving meaning to $\ln(-1)$ should not be difficult. – Anixx Jun 13 '21 at 18:49
  • @Anixx: yes, the 2x2 matrices contain (many copies of) the complex numbers, e.g. any conjugation of the usual representation. But idk, you may want to pick a more exciting value for it just to spice up things :-) – M.G. Jun 13 '21 at 18:51
  • Why I asked this question, is mostly to see if we can get some other real mathematical constants (such as $e^{-\gamma}$ for instance) by applying simple operations to non-scalar elements. I believe $\gamma$ should have some algebraic role, not only analytic. Moreover, in the (infinite-dimensional) algebra of divergent integrals I already established that $\operatorname{reg} \ln \omega_+=-\gamma$ (where $\omega_+=\sum_0^\infty 1$). An interesting relation, but still requires regularization (taking scalar part). – Anixx Jun 13 '21 at 18:59
  • It would be interesting to see your opinion as an algebraist on this. Here is some of the latest versions: https://exnumbers.miraheze.org/wiki/Main_Page It needs a browser that supports mathml. Here is a table of some constants as well: https://mathoverflow.net/questions/392295/what-intuitive-meaning-determinant-of-a-divergency-divergent-integral-series – Anixx Jun 13 '21 at 19:04
  • @Anixx: yes, I agree, it would be interesting to see if $\gamma$ comes from algebra in that sense. Periods are yet another connection between the algebraic and transcendental world. Having said that, I don't know anything about divergent integrals, but I will have a look. EDIT: good point about mathml as I am using Edge... – M.G. Jun 13 '21 at 19:06
  • @Anixx: hah, that's pretty cool. – M.G. Jun 13 '21 at 19:20
  • Here is my older post here about the topic: https://mathoverflow.net/questions/115743/an-algebra-of-integrals/342651#342651 and a link to the older wiki that does not require mathml: http://exnumbers.wiki-site.com/index.php/Algebra_of_divergent_integrals_and_series – Anixx Jun 13 '21 at 19:24
  • @Anixx: yes, thanks, I have already seen your older MO post about this. That thread definitely generated a lot of interesting input. – M.G. Jun 13 '21 at 19:34
  • That post is still mostly correct. It has a minor typo in the table, lacks list of determinants/moduluses in the table, less integral representations and no Laplace-transform based equivalence class rule for divergent integrals. – Anixx Jun 13 '21 at 19:37
  • For what is worth, in dual numbers the expression $\varepsilon^\varepsilon$ with a reasonable extension of the set, gives an expression, connected to the Euler-Mascheroni constant: https://mathoverflow.net/q/432396/10059 – Anixx Oct 26 '22 at 10:50