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In Quaternion mathematics, normally $$i ^ {2} = j ^ {2} = k ^ {2} = i*j*k = {-1}$$

But I am looking at a different type of Formula where $$i ^ {j ^ k} = {-1} $$

(edit again, since somehow this remains elusive to explain...)

With functions, $f$ and all permutations of the subset variables $N$:

Where $i$ is used in cases of non-real $f(x)=N$

Where $i ^ j$ is used in cases of non-real $f(x,y)=N$

Where $i ^ {j ^ k}$ is used in cases of non-real $f(x,y,z) = N$

This of course must obey other permutations of $x$, $y$, and $z$ such as:

Where $k ^ i$ is used in cases of non-real $f(z,x) = N$; etc...


I have given a bit more thought about this, and since $x$, $y$ and $z$ are possibly inter-changable depending on the orientation in real space, consider the possibility:

$i ^ {j ^ k} = j ^ {k ^ i} = k ^ {i ^ j} = -1$

Would be an "exponential-type ring".


What are some important fundamental differences (or formulae) I should be looking at in how i, j and k compare? (more importantly is the second formula even possible?)


Consider in the second example a hypothetical anti-DeSitter space, where the real spatial directions X, Y, and Z are real components, and i, j and k are imaginary spatial directions that associate to each one.

$i^2 = -1$; for $x$

$j^2 = -1$; for $y$

$k^2 = -1$; for $z$

Because, I am looking at a different approach to Riemann Tensors, my approach is to use exponentiation, rather than multiply. I am hoping someone that has experience with this type of thing can point out glaring flaws in this type of approach, and apologies in advance if I have errors in the reasoning here, its been some time since I've worked out anything this complex.

I have further details here here and here regarding my problem.


What values are possible or valid for $a$, $b$, and $c$:

$(ai)^{(bj)^{(ck)}} = -1$

if $i^2 = j^2 = k^2 = -1$

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What is the motivation for this formula? At first glance, it seems rather arbitrary... –  Rahul Sep 30 '10 at 0:44
@rahul its a programming project idea I have that needs the math from the second formula. –  Talvi Watia Sep 30 '10 at 0:54
@Talvi Watia: It is not possible to make any sense of your 2nd formula until you specify the meaning of the terms - most esp $i^{j^k}$, since exponentiation is not a ring operation. –  Bill Dubuque Sep 30 '10 at 1:36
@Talvi: in a ring, you have addition and you have multiplication. In general, if $R$ is a ring, and $a,b\in R$, "$a^b$" may have no meaning. (For instance, in the real numbers we do not have a ready meaning for $a^b$ when $a$ is negative and $b$ is not an integer). The only "exponentiation" that you get "for free" in a ring is exponentiation with nonnegative integer exponents, because they are really just abbreviations for repeated multiplications. So... what do you mean by $j^k$, and what do you mean by $i^{j^k}$ in this ring? –  Arturo Magidin Sep 30 '10 at 3:11
@Talvi: I have no idea what "in cases of nonreal $f(x)=N$" means. What is $f$? What is $N$? I don't think your new edits clarify or explain the meaning of $i^{j^k}$; at least, not for me. –  Arturo Magidin Sep 30 '10 at 6:46

1 Answer 1

up vote 2 down vote accepted

This may be a complete stupid thing, for which I'll lose some hundred points of reputation, but here it is.

I don't understand anything at all about your functionss $f$ "and all permutations of the subset variables N...", nor the rest of your question. But I think that expressions like

$$ i^{j^k} \ , \qquad j^{k^i} \ , \qquad \text{and}\qquad k^{i^j} $$

could make sense. However, they don't seem to be equal to $-1$.

Let's remember how we define $x^y$ for real numbers $x, y$, $x>0$. We put

$$ x^y = e^{y\log x} \ , $$

don't we? After choosing a determination for the complex logarithm, we define also

$$ i^z = e^{z \log i }\ . $$

So we need to compute $\log i$. For this we can rely on Euler's formula, which, for a real number $\theta$, says:

$$ e^{i\theta} = \cos\theta + i \sin\theta \ . $$

If we ask a complex number $z = x + iy$ to be a logarithm for $i$, we must have

$$ i = e^z = e^{x + iy} = e^x e^{iy} = e^x (\cos y + i \sin y) \ . $$

This means that

$$ 0 = e^x \cos y \qquad \text{and} \qquad 1 = e^x\sin y \ . $$

So we may take $y = \frac{\pi}{2}$ and then $x= 0$. Hence a logarithm for $i$ is $\log i = i\frac{\pi}{2}$ and

$$ i^z = e^{zi\frac{\pi}{2}} \ . $$

So far so good: what about quaternions? Well, we don't need to define exponentials and logarithms for them, but just notice that Euler's formula only needs two things to be true: (a) the Taylor series for $\exp z$, $\cos z$ and $\sin z$ and (b) "something", called $i$, that can be added with real numbers and itself, multyplied by real numbers and such that $i^2 = -1$. Right? Okay. So, for the same reasons as for $i$, we may take

$$ \log i = i\frac{\pi}{2} \ , \qquad \log j = j\frac{\pi}{2} \ , \qquad \text{and} \qquad \log k = k\frac{\pi}{2} \ . $$

Now, we can compute $j^k$:

$$ j^k = e^{k\log j} = e^{k j\frac{\pi}{2}} = e^{-i \frac{\pi}{2}} = \cos \left( \frac{-\pi}{2} \right) +i \sin \left( \frac{-\pi}{2} \right) = -i \ . $$


$$ i^{j^k} = i^{-i} = e^{-i \log i} = e^{-i i \frac{\pi}{2}} = e^{\frac{\pi}{2}} \ . $$

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Why is $j^k=e^{k\log j}$ rather than $e^{(\log j)k}? –  Robin Chapman Sep 30 '10 at 17:59
@Robin. Good point. Since quaternions' product isn't commutative order matters. Hmmmm. Let's try the dishonest way: could we simply say "by definition"? –  a.r. Sep 30 '10 at 18:10
+1 great answer. but I think it breaks at $e^{(log j)k}$ -> $e^{-i(\pi \over 2)}$ since i, j, k are not necessarily commutative. In the simple case, they are, but the complex They may not be a ring. Ironically, $e^{\pi \over 2}$ is the correct answer! (the threshold limit of the Planck constant to the Yang-Mills gap.) –  Talvi Watia Sep 30 '10 at 22:45
To explain further, the $i^{j^k}=-1$ must balance with the sum of the real value $e^{\pi \over 2}$ for quark tensors to exist correctly through what is seen through observation. perhaps the $=-1$ is misleading as it is a charge, not a distance. –  Talvi Watia Sep 30 '10 at 22:53
@Robin Chapman: Is $e^{(\log j) k}$ the standard definition of quaternion exponentiation $j^k$? If so, could you explain why it is preferred over $e^{k \log j}$? –  Rahul Oct 1 '10 at 4:50

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