# Quaternions and 4-vectors

I recently realised that quaternions could be used to write intervals or norms of vectors in special relativity:

$(t,ix,jy,kz)^2 = t^2 + (ix)^2 + (jy)^2 + (kz)^2 = t^2 - x^2 - y^2 - z^2$

Is it useful? Is it used? Does it bring anything? Or is it just funny?

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+1 for reminding me of having had that question a while ago too –  Tobias Kienzler Mar 21 '11 at 9:09
I would say, it's just funny. –  Dimensio1n0 Jul 19 '13 at 6:23

The object you're talking about is called, in mathematics, a Clifford algebra. The case when the algebra is over the complex field in general has a significantly different structure from the case when the algebra is over the real field, which is important in Physics. In Physics, in the specific case of 4 dimensions, using the Minkowski metric as you have in your Question, and over the complex field, the algebra is called the Dirac algebra. Once you have the name Clifford algebra, you can look them up in Google, where the first entry is, unsurprisingly, Wikipedia, http://en.wikipedia.org/wiki/Clifford_algebra, which gives you a reasonable flavor of the abstract construction methods that mathematicians prefer. The John Baez page that is linked to from the Wikipedia page is well worth reading (if you spent a year learning everything that John Baez has posted over the years, almost always with unusual clarity and engagingly, you would know most of the mathematics that might be useful for Physics).

It's not so much that the Clifford algebras are funny. Their quadratic construction is interrelated, often closely, with many other constructions in mathematics.

There are people who are enthusiastic about Clifford algebras, sometimes very or too much so, and a lot of ink has been spilled (Joel Rice's and Luboš Motl's Answers are rather inadequate to the literature, except that I think they chose to interpret your Question narrowly where I've addressed what your construction has led to in Mathematics more widely), but there are many other fish in the sea to admire.

EDIT: Particularly in light of Marek's comments below, it should be said that I interpreted Isaac's Question generously. There is a somewhat glaring mistake in the OP that is pointed out by Luboš (which I hope you see, Isaac). Nonetheless there is a type of construction that is closely related to what I chose to take to be the idea of the OP, Clifford algebras.

Isaac, this is how I think your derivation ought to go, if we just use quaternions, taking $q=t+ix+jy+kz$, $$q^2=(t+ix+jy+kz)(t+ix+jy+kz)=t^2-x^2-y^2-z^2+2t(ix+jy+kz).$$ The $xy,yz,zx$ terms cancel nicely, but the $tx,ty,tz$ terms don't, unless we do as Luboš did and introduce the conjugate $\overline{q}=t-ix-jy-kz$. This, however, doesn't do what I take you to be trying to do. So, instead, we introduce a fourth object, $\gamma^0$, for which $(\gamma^0)^2=+1$, and which anti-commutes with $i$,$j$, and $k$. Then the square of $\gamma^0t+ix+jy+kz$ is $t^2-x^2-y^2-z^2$. The algebra this generates, however, is more than just the quaternions, it's the Clifford algebra $C(1,3)$.

EDIT(2): Hi, Isaac. I've thought about this way too much overnight. I think now that I was mistaken, you didn't make a mistake. I think you intended your expression $(a,b,c,d)^2$ to mean the positive-definite inner product $a^2+b^2+c^2+d^2$. With this reading, however, we see three distinct structures, the positive-definite inner product, the quaternions, and the Minkowski space inner product that emerges from using the first two together. Part of what made me want to introduce a different construction is that in yours the use of the quaternions is redundant, because you'd get the same result that you found remarkable if you just used $(a,ib,ic,id)^2$ (as Luboš also mentioned). Even the positive-definite inner product is redundant, insofar as what we're really interested in is just the Minkowski space inner product. Also, of course, I know something that looks similar and that has been mathematically productive for over a century, and that can be constructed using just the idea of a non-commutative algebra and the Minkowski space inner product.

To continue the above, we can write $\gamma^1=i$, $\gamma^2=j$, $\gamma^3=k$ for the quaternionic basis elements, together with the basis element $\gamma^0$, then we can define the algebra by the products of basis elements of the algebra, $\gamma^\mu\gamma^\nu+\gamma^\nu\gamma^\mu=2g^{\mu\nu}$. Alternatively, for any vector $u=(t,x,y,z)$ we can write $\gamma(u)=\gamma^0u_0+\gamma^1u_1+\gamma^2u_2+\gamma^3u_3$, then we can define the algebra by the product for arbitrary 4-vectors, $\gamma(u)\gamma(v)+\gamma(v)\gamma(u)=2(u,v)$, where $(u,v)$ is the Minkowski space inner product. Hence, we have $[\gamma(u)]^2=(u,u)$. Now everything is getting, to my eye, and hopefully to yours, rather neat and tidy, and nicely in line with the conventional formalism.

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"The object you're talking about..." -> really? I don't see where he's talking about Clifford algebras. Except, of course, if you mean $C(0,2) \cong \mathbb H$ but how is this relevant to the question? Also, bringing in Dirac algebra into the game just because Minkowski space was mentioned, seems quite off topic. OP's just found one accidental similarity that isn't using any of the structure of those theories (as Luboš correctly says). If he was instead interested in duality between four-vectors and hermitian $2 \times 2$ matrices (and $Spin(1,3) \cong SL(2, \mathbb C)$) then we could talk... –  Marek Mar 20 '11 at 20:13
Hi Marek. You're right of course that Isaac is just using (partly ab using) the quaternions, which as an algebra is isomorphic to $C(0,2)$. But I thought it worthwhile to look at the mathematical structures that are related to the more intrinsic aspects of the Question. Isaac is using the quaternions to construct an object that squares to the inner product in Minkowski space. "Does it bring anything?" yes, this type of construction leads to $C(1,3)$. The formalism used by Isaac is perhaps a little awkward, but I wanted to show some of the context that I think he can place himself in. –  Peter Morgan Mar 20 '11 at 21:16
@Peter: oh, so you interpreted his four-vector as a vector of Clifford algebra and its norm as relating to its defining metric? All right then. I am pretty sure that this is not what Isaac had in mind but it's true that this leads to $C(1,3)$ and more generally to Clifford algebras. –  Marek Mar 20 '11 at 22:40
@Marek Definitely intended and hoping to broaden horizons rather than addressing exactly what Isaac's Question asked. "Does it bring anything?" invites wider horizons. I wish I could phrase Questions so that people would Answer the Question I should have asked, not what I in fact asked. I hope my Answer is a little useful to him. –  Peter Morgan Mar 20 '11 at 23:41
A late down vote, 6 hours ago? There's surely a lot that I could have done differently, but you're holding me to a high standard to say that "This answer is not useful". Just curious. –  Peter Morgan Mar 23 '11 at 16:35

It's just funny. Note that your equation doesn't actually use any single general quaternion. You only use the $i,j,k$ imaginary units in an ad hoc way to get three minus signs whenever you need them.

If you were using an actual quaternion $$q = t + xi + yj + zk,$$ then the only semi-natural real bilinear invariant you may construct out of it is $$q\bar q = (t + xi + yj + zk ) ( t - xi - yj - zk) = t^2+ x^2 +y^2 +z^2$$ so the 4 real components in a quaternion still have the Euclidean, rather than Minkowskian, signature. But even for a 4-dimensional Euclidean space, the quaternions are actually just a game because we haven't really used the main nontrivial structure of the quaternions, their multiplication, in any nontrivial way. Quaternions are not genuine quaternions if you never use the relations $ij=-ji = k$ and its cyclic permutations - and we haven't used them above. We only used the fact that $i,j$ etc. anticommute with each other, but we didn't really care what their product is.

Because we haven't really used those relations, we haven't used full quaternions - except as a meaningless bookkeeping device. In the same way, one may organize 8 real numbers under the umbrella of a single "octonion" except that if the complicated and cool octonion multiplication table - with the $G_2$ automorphism group - is never employed, it's clear that the "octonion" interpretation was just a game to give a name to a collection of 8 numbers. But not every collection of 4 or 8 numbers deserves to be called "quaternion" and "octonion", even though, of course, one may get the individual components out of the "quaternion" and "octonion", too.

In the very same way, a general pair of two real numbers is simply not a complex number. By its very essence, a complex number must act as one number - so there must be a notion of holomorphy required somewhere or everywhere in the formalism - rather than two numbers. The references linked in the other answers don't understand the purpose and relevance of all those mathematical structures, so they lead to incorrect answers to the fundamental question whether the trick is real or just a fun. The right answer is that it is just a fun, and your fun even used a wrong signature that differs from a somewhat more natural fun.

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+1 This is very a true - and often overlooked - concept. –  Sklivvz Mar 20 '11 at 20:45
It's strange. I wrote almost the same answer! and I got 2 negative votes and I had to delete my answer. @Deepak where's your comment now? –  user1355 Mar 21 '11 at 2:20
@sb1 I was out for the evening so I'm late in posting this comment. @Lubos' your answer also seems to dismiss the value of quaternions. I realize that you are sticking to strictly what the OP said. But your language seems to suggest that you find them to be a just a funny curiosity. You clearly don't believe this to be true since you have a preprint with "quaternion" in the title (arxiv.org/abs/hep-th/9612198)! A good and worthwhile answer would be one that pointed out actual uses of quaternions rather than just pointing out the obvious error in the OP's question. –  user346 Mar 21 '11 at 6:08
@Deepak: the question is about (non-)importance of connection between quaternions and four-vectors. No one (and in particular Luboš) dismisses quaternions per se, as they are obviously very important mathematical structure. You need to read this answer more carefully ;) –  Marek Mar 21 '11 at 11:40
Dear @sb1, I would up-vote your answer - which I see even when it's deleted - but since it was deleted, I can't vote on it. You shouldn't retract your valid answer just because someone finds a reference that disagrees with your answer. ... Let me just confirm Marek's points that all of us, including me, seem to love quaternions. My love is really the reason why I don't want to cheapen them and see them even at places where they don't really play a role. –  Luboš Motl Mar 23 '11 at 9:52

Cornelius Lanczos has a chapter on quaternions and special relativity in his "The Variational Principles of Mechanics". So, is has been used. But it seems more straightforward to consider the multivector algebra of spacetime so t,x,y,z really are on the same footing.

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There is a book: "Quaternions, Clifford Algebras and Relativistic Physics." by Patrik R. Girard. Find this if you want to learn more -- very good reading, not very complex and not very long. I'll just cite the first paragraph of chapter 3.

From the very beginning of special relativity, complex quaternions have been used to formulate that theory [45]. This chapter establishes the expression of the Lorentz group using complex quaternions and gives a few applications. Complex quaternions constitute a natural transition towards the Clifford algebra H ⊗ H.

Well and the reference:

[45] L. Silberstein, The Theory of Relativity, Macmillan, London, 1914.

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Once again, while Clifford algebras are certainly nice gadgets I fail to see any relevance to the question. –  Marek Mar 20 '11 at 20:18

You've stumbled on a fertile area. Though not strictly what you were asking about, I can tell you that perhaps the most interesting relationship between orthogonal groups and quaternions comes from looking at spinors. As you may know, the symmetry group called $Spin$ double covers the group of rotations, and is the more relevant group for physics since spinors transform under this larger group. A useful example is the double cover $SU(2) \rightarrow SO(3),$ that is, $SU(2) = Spin(3)$ in the Euclidean signature.

Topologically, $SU(2)$ is a 3-sphere, which we can think of as the unit quaternions (remember, the norm is Euclidean, as pointed out by others). To understand the map $SU(2) \rightarrow SO(3),$ let $v$ be an imaginary quaternion (which we can think of as a 3-vector), and let $q$ be in $SU(2).$ Then since multiplication of quaternions preserves the norm,

$$\overline{q} v q$$

has the same norm as v, and you will note that it is still imaginary. Thus, the action $v\stackrel{q}{\mapsto} \overline{q}vq$ of $SU(2)$ on $R^3$ (the imaginary quaternions) is by a rotation. Furthermore, $q$ and $-q$ act the same way. So we have described a double cover of $SU(2)$ onto 3-dimensional rotations.

Maybe this is not what you immediately inquired about or discovered, but it's probably worth knowing.

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First note that $q = t + xi + yj +zk$ and $q\prime = t - xi - yj - zk$

So $qq\prime = t^2 + x^2 + y^2 + z^2$

Hence, it should be $t^2 + x^2 + y^2 +z^2$ and hence it is not the right signature of special relativity. Secondly since quaternion does not have an analog of a notion like holomorphic function (since its conjugate is not independent) it is not physically ideal or useful as far as I know. You can generalize it and have excellent usefulness but you have to give up its division algebra property. Clifford algebra is an example.

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quaternions are extremely helpful in relativity and computer graphics. Rotation matrices for instance can be represented as quaternions. So the effect of a rotation on a quaternion can be obtained by multiplying it with another quaternion. Also see [Doug Sweetser's] page for further details, illustrations and examples. You also don't have to care about his thoughts on unification of GR and EM, in order to appreciate the info on quaternions on his page. –  user346 Mar 20 '11 at 14:10
@Deepak: As far as I know pure quaternions are not useful unless you sacrifice its division algebra property. Please give me some reference where it is used in relativity. –  user1355 Mar 20 '11 at 14:15
@Deepak: I shall delete my answer as soon as you will provide any reference. –  user1355 Mar 20 '11 at 14:18
I'm glad you chose to undelete your answer. The moral burden of being responsible for its initial deletion was getting heavy ;) As for the use of quaternions in relativity, check out the wikipedia page on biquaternions. Though I suppose you're right that one must sacrifice the division algebra property in this case. –  user346 Mar 23 '11 at 17:16
@Deepak: Thanks for your comment :) Nothing personal of course :) –  user1355 Mar 23 '11 at 17:21