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Euler's identity is $e^{i \pi} + 1 = 0$, a special case of the Euler's formula $e^{i\theta} = \cos\theta + i\sin\theta$, where $\theta = \pi$ (half-turn of the unit circle).

It is commonly described as "beautiful" because it is simple and includes a bunch of fundamental mathematical constants and operators, including the complex numbers.

Is there a corresponding Euler's identity for quaternions?

Speculation:

The Tau Manifesto proposes a different fundamental circle constant, $\tau = 2 \pi$, and a corresponding alternate Euler identity, based on a complete turn, the special case of $\theta = \tau$:

$e^{i\tau} = 1$

If both identities are extended to quaternions, is one any simpler or more "beautiful" than the other?

endolith
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    I um... barely understand any of these answers. I'm just going to choose the highest-voted and slink away. – endolith May 28 '11 at 04:21

5 Answers5

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One can define $e^x$, $\sin x$ and $\cos x$ in any Banach algebra by means of power series, e.g. $e^x = \sum x^n/n!$. Then, for any square root of $-1$ in the algebra (which we will denote by $i$, although there could be many, as there are in the quaternions), it is a formal consequence of the power series that $e^{ix}=\cos x + i \sin x$. Because these functions are defined via power series, this is all tautological, and there isn't really any deep significance.

There are things two that make Euler's formula interesting over the reals. First, $e^x$, $\cos x$ and $\sin x$ are usually defined by non-power series methods, and so Euler's formula expresses some sort of non-obvious relation. Second, these functions satisfy all sorts of identities which can be better understood in terms of the identity. However, when we are working over a noncommutative Banach algebra (like the quaternions or $n \times n$ matrices), we generally have that $e^{a+b}=e^a e^b$ only when $ab=ba$, and so the most useful algebraic property of $e^x$ is gone, which means that so are the interesting relations you might want.

Edit: I would like to explain the calculation done in Robert's answer in terms of what I have written above.

Let $q=a+bi+cj+dk=a+r\frac{bi+cj+dk}{r}$ where $r^2=b^2+c^2+d^2$, and we have assumed $r>0$. Note that $\frac{bi+cj+dk}{r}$ is a square root of $-1$, which we will temporarily write as $\sqrt{-1}$. Since $a$ commutes with everything, we have

$$e^q=e^{a+r\sqrt{-1}}=e^a e^{r\sqrt{-1}} = e^a( \cos(r) + \sqrt{-1} \sin(r))= e^a(\cos(r) + \frac{\sin(r)}{r}(bi+cj+dk)).$$

So we see that everything follows from the standard formula $e^{ix}=\cos x + i \sin x.$

Community
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Aaron
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  • What would be some concrete examples of Banach Algebra that you speak of? – Your neighbor Todorovich Aug 09 '15 at 08:43
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    @FemaleTank https://en.wikipedia.org/wiki/Banach_algebra#Examples – Aaron Aug 09 '15 at 18:21
  • recently i've been studying geometric algebra, and the book i've been using offhandedly mentions that e^A*e^B≠AB for general multivectors A and B, and didn't provide any context or understanding. lo and behold i find somewhat of an answer in an almost unrelated stackexchange thread – wyboo Mar 13 '23 at 23:18
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    @wyboo You can understand it by using Taylor series. $e^(tA)=1+tA+\frac{t^2A^2}{2}+O(t^3)$, similarly for $B$, so $e^{tA}e^{tB}=1+t(A+B)+t^2(A^2/2 + AB+B^2/2)+O(t^3)$, but $e^{t(A+B)}=1+t(A+B)+t^2(A+B)^2/2+O(t^3)$. The difference between these two power series is $(AB-BA)t^2/2$ plus higher order terms. if $AB-BA\neq 0$, the series cannot be the same. This doesn't rule out coincidences with specific values of A, B, t, but for fixed non-commuting $A,B$, it shows non-equality for sufficiently small $t$. – Aaron Mar 14 '23 at 02:17
  • For those like me who don't understand why "$\frac{bi+cj+dk}{r}$ is a square root of $-1$", note that $i, j$ and $k$ are all square roots of $−1$, such that $ij = k = −ji, jk = i = −kj, ki = j = −ik$. Then, use these identities to show that $\left(\frac{bi+cj+dk}{r}\right)^2 = -1$. – Mahmoud May 19 '23 at 19:44
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    @mhdadk A very clean proof is to first define the conjugate: if $z=a+bi+cj+dk$, then $\overline{z}=a-bi-cj-dk$. Using the identities, show that $z\overline{z}=a^2+b^2+c^2+d^2$, which is a positive real number. Now, if $a=0$, then $\overline{z}=-z$, and the equation simplifies to $-z^2=b^2+c^2+d^2=r^2$, so $z^2=-r^2$. While the computational part of this approach is the same as that of yours, taking about conjugates yields you things that you can use elsewhere and makes for nice analogy with the complex numbers. – Aaron May 20 '23 at 19:47
  • @Aaron thanks for the help! – Mahmoud May 20 '23 at 20:05
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For a quaternion $q = a + b i + c j + d k$ with $a,b,c,d$ real and $\sqrt{b^2 + c^2 + d^2} = r > 0$, we have $e^q = e^a (\cos(r) + \frac{\sin(r)}{r} (b i + c j + d k))$.

Robert Israel
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    More concisely, let $q$ be a fully imaginary quaternion. Then $\exp q = \cos |q| + \hat{q} \sin |q|$, where $\hat{q} = q |q|^{-1}$. – user76284 Nov 10 '18 at 00:24
  • Another viewpoint combining the above would be: $e^q = e^{a+bi+cj+dk} = e^a\left(cos(|q-a|) + \frac{q-a}{|q-a|} sin(|q-a|)\right)$. – milia Nov 25 '22 at 23:58
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For any non-real element $\theta\in\mathbb H$ the subalgebra generated by $1$ and $\theta$ is isomorphic to complex numbers. So, yes, Euler's formula holds (it's spelled out in Robert Israel's answer) but it gives nothing new, essentially.

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Grigory M
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(It is rather a comment and directly follows from previous answers)

We could say briefly using Hamilton terminology:

For any “unit right quaternion” $r$ (i.e., $r=xi +yj+zk$, $x^2+y^2+z^2=1$) it is true an analogue of the Euler's identity $e^{r\pi}+1 = 0$.

Right quaternions now also called “pure imaginary”.

Alex 'qubeat'
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Regarding the second question, the revised form of Euler's identity using $\tau$ conveys strictly less information than the original identity, and still suffers from an unnatural choice of square root of $-1$ (not a big deal, except when people make claims of beauty or simplicity). In the complex world, it's more natural (to me) to define $\mathbb{Z}(1)$ as the kernel of exponentiation, and phrase Euler's identity as the statement that $\mathbb{Z}(1)$ is generated by $2\pi$ (or $\tau$) times a square root of minus one.

(One problem with the general use of this $\tau$ is that it is very cumbersome in the theory of modular forms, where $\tau$ is already reserved as the coordinate on the complex upper half-plane. One often sets $q = e^{2 \pi i \tau}$ as a Fourier series variable, and this looks very bad when you replace $2\pi$ with $\tau$.)

Regarding your first question, in a noncommutative topological ring like the quaternions, exponentiation is not a homomorphism (and the kernel is not a group). However, the real span of $1$ and $i$ (or any other choice of square root of $-1$) is isomorphic to the complex numbers, and the restriction of exponentiation to this subring is a homomorphism from the additive group to the multiplicative group. Euler's identity therefore holds, but for possibly silly reasons.

Scott Carnahan
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