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Stream: deprecated: mathematics

Topic: quaternions

John Baez (Jun 09 2023 at 15:59):

Hi! Since the algebra of quaternions $\mathbb{H}$ is not a field, the question of "can Hilbert spaces be defined over fields other than $\mathbb{R}$ and $\mathbb{C}$?" is not particularly relevant to the study of quaternionic Hilbert spaces.

John Baez (Jun 09 2023 at 16:12):

It sounds like you're wondering about what it means to say $x \ge 0$ when $x$ is a quaternion. By definition, a quaternion $x$ has $x \ge 0$ iff $x = y\overline{y}$ for some quaternion $y$.

(The same definition works if we replace the word "quaternion" here by "real number" or "complex number".)

B. Wilson (Jun 09 2023 at 16:14):

Oh! Thank you for the quick response! AFAIU, the only reason quaterions aren't a field is just because fields axiomatically assert commutivity of the multiplication. Is that correct? I'll have to look more carefully at the MO proof sketch to see where that comes in.

John Baez (Jun 09 2023 at 16:16):

Yes, fields are commutative; quaternions are not a field but rather a [[division ring]] (indeed a [[normed division algebra]]).

Another way to think about this: there's a unique subalgebra of $\mathbb{H}$ isomorphic to the real numbers, given by the real multiples of $1 \in \mathbb{H}$. Furthermore the isomorphism between this subalgebra and $\mathbb{R}$ is unique. So, we can talk about a quaternion being real in an unambiguous way, and $x \ge 0$ means that your quaternion is real and $\ge 0$ in the usual sense for real numbers.

John Baez (Jun 09 2023 at 16:18):

But anyway, I should have defined $\ge 0$ for $\ast$-algebras when I was discussing $\ast$-algebras in that paper; I'm so used to this concept that I forgot!

John Baez (Jun 09 2023 at 16:19):

$\mathbb{H}^n$ becomes a quaternonic Hilbert space via

$\langle v, w\rangle = \sum_{i=1}^n \overline{v_i} w_i$

John Baez (Jun 09 2023 at 16:20):

and it's good to check that with the definition I just gave,

$\langle v, v \rangle \ge 0$

for all $v \in \mathbb{H}^n$.

John Baez (Jun 09 2023 at 16:22):

The MathOverflow proof uses facts about fields here:

Thus $K$ is an extension of $\mathbb{R}$ of degree at most 2 (specifically, of degree equal to the size of the group of automorphisms generated by $a \mapsto a^\ast$), and so it can only be $\mathbb{R}$ or $\mathbb{C}$.

John Baez (Jun 09 2023 at 16:25):

This step of the proof should be expanded quite a bit for people who aren't experts on fields: it's using a fact about Galois theory that's true for fields but fails for division rings.

B. Wilson (Jun 09 2023 at 16:34):

You're immediately bestowing nice things upon me! What did I do to deserve this?
That definition for $x \ge 0$ is nice. I was somewhat vaguely worrying about automorphisms of $\mathbb{H}$ messing up $\mathbb{R}$ embeddings kind of like basis changes or whatever, but I think I see how algebra homomorphisms in this case force $\mathbb{R}$ to be fixed, though I need to think about this more carefully.

B. Wilson (Jun 09 2023 at 16:36):

John Baez said:

This step of the proof should be expanded quite a bit for people who aren't experts on fields: it's using a fact about Galois theory that's true for fields but fails for division rings.

Ah! That was exactly the sentence I stared at for a few minutes and then gave up on. I appreciate you fielding these low-brow questions of mine.

John Baez (Jun 09 2023 at 16:41):

No problem. The answer to "What did I do to deserve this?" is "you asked a question about quaternionic Hilbert spaces, and I'm lying in bed waking up, drinking coffee, looking for interesting things to talk about". :upside_down:

John Baez (Jun 09 2023 at 16:43):

B. Wilson said:

You're immediately bestowing nice things upon me! What did I do to deserve this?
That definition for $x \ge 0$ is nice. I was somewhat vaguely worrying about automorphisms of $\mathbb{H}$ messing up $\mathbb{R}$ embeddings kind of like basis changes or whatever, but I think I see how algebra homomorphisms in this case force $\mathbb{R}$ to be fixed, though I need to think about this more carefully.

An automorphism of any algebra needs to preserve $1$, and an automorphism of any algebra over $\mathbb{R}$ needs to preserve multiplication by real numbers, so it preserves the set of real multiples of $1$. So any algebra over the reals has a subalgebra $\{\alpha 1: \alpha \in \mathbb{R}\}$ that's invariant under all automorphisms.

John Baez (Jun 09 2023 at 16:44):

Indeed, there's a god-given copy of the reals sitting inside any real algebra... and the same story works if we replace the reals by any other field.

B. Wilson (Jun 09 2023 at 16:45):

John Baez said:

No problem. The answer to "What did I do to deserve this?" is "you asked a question about quaternionic Hilbert spaces, and I'm lying in bed waking up, drinking coffee, looking for interesting things to talk about". :upside_down:

Haha. I'll store this away. If I urgently need your attention in the future, then maybe I'll just start talking about $E_8$ and the octonions :upside_down:

B. Wilson (Jun 09 2023 at 16:52):

The $\mathbb{R}$-algebra case seems pretty straightforward, as you point out, but if we think of $\mathbb{H}$ as a complex algebra over $\mathbb{C}$, then we just have to notice that it's forgetfully also an $\mathbb{R}$-algebra as well and voilá, I guess?

John Baez (Jun 09 2023 at 16:53):

$\mathbb{H}$ is not an algebra over $\mathbb{C}$.

John Baez (Jun 09 2023 at 16:54):

I'll leave this as a puzzle: look up the concept of "algebra over a field" and figure out why.

B. Wilson (Jun 09 2023 at 17:21):

Yikes. Because the product is not $\mathbb{C}$-linear by any stretch of the imagination. I was thinking a silly scheme like $(a, b) \mapsto a + bj$, and a complex bilinear multiplication would mean that $ik = (i, 0)(0, i) = -(1, 0)(0, 1) = -(1+j)$ which is an abomination. It seems like this holds in general, but I'll think about it more closely as well.

Thanks for the swift correction. It's 2am and time for me to go to bed, but I'll continue reading through your paper this weekend!

B. Wilson (Jun 09 2023 at 17:22):

Thank you again for the warm welcome :man_bowing:

John Baez (Jun 09 2023 at 18:11):

Sure! I'm not sure I get your calculation, but yes, the product is not $\mathbb{C}$-bilinear. An algebra A over a field F needs to have

(fa)b = a(fb)

for all a,b $\in$ A, f $\in$ F. Taking b = 1 we see that elements of the field give elements of the algebra that commute with everything in the algebra. With more work you can show there has to be a god-given copy of the field sitting inside the algebra, consisting of the elements f1 (where 1 $\in A$). This copy of F in A is a subalgebra of A.

John Baez (Jun 09 2023 at 18:12):

There are infinitely many copies of the complex numbers sitting inside the quaternions as subalgebras, but none of them commutes with everything!

B. Wilson (Jun 10 2023 at 00:11):

Aww, you just came out and said it! Falling asleep last night, I realized exactly what you cleanly state about $\mathbb{K}$-linearity implying that $\mathbb{K}$ is in the center of the multiplicitive group. I'll have to be faster next time :P

Cool. And it appears that imaginary quaternions square to a real: $x = (ai + bj + ck)^2 = -(a^2 + b^2 + c^2) + ab(ij + ji) + ac(ik + ki) + bc(jk + kj) = -(a^2 + b^2 + c^2) = -x\bar{x}$. So for any unit-length vector, $x$, we have $(a_0 + b_{0}x)(a_1 + b_{1}x) = a_{0}a_{1} - b_{0}b_{1} + (a_{0}b_{1} + b_{0}a_{1})x$, hence the product lives in the span of $\{1, x\}$. Which, if I'm not missing anything, seems to say that any $\mathbb{C}$ copy in $\mathbb{H}$ fixes $\mathbb{R}$ and just picks a unit vector, the punchline being that the space of $\mathbb{C}$-embeddings looks like $S^2$??

John Baez (Jun 10 2023 at 00:40):

Sorry for jumping the gun. Yes, to get an algebra embedding of $\mathbb{C}$ in $\mathbb{H}$ you just need to pick a square root of -1 in $\mathbb{H}$, which is the same as picking an imginary quaternion of norm 1, so there's an $S^2$ of choices.

John Baez (Jun 10 2023 at 00:41):

So, each time someone asks you "where'd he go?" and you point and say "thataway!", you're actually specifying an embedding of the complex numbers in the quaternions.

Mike Shulman (Jun 10 2023 at 02:01):

...at least, as long as you're standing in a 3-space with chosen coordinate axes. (-:O

B. Wilson (Jun 10 2023 at 09:21):

Fiddling around with the Cayley-Dickson construction, and unless I messed something up, it looks like the same basic argument holds for any of the algebras in the heirarchy! Thus we have that $\mathbb{C}$ embeds $S^{n-2}$ ways into the $n$-dimensional Cayley-Dickson algebra‽

B. Wilson (Jun 10 2023 at 09:24):

$S^0$ is two points, so $\mathbb{C}$ embeds into itself via the identity and conjugation? Even more degenerately, $S^{-1}$ is the empty set, corresponding with $\mathbb{C}$ not embedding into $\mathbb{R}$ :melting_face:

B. Wilson (Jun 10 2023 at 09:27):

If so, then giving vague directions to your 7-dimensional friends is actually specifying a complex embedding into $\mathbb{O}$!

John Baez (Jun 10 2023 at 14:34):

I haven't thought much about the higher levels of the Cayley-Dickson construction, but I know that the square roots of -1 in the octonions are exactly the imaginary octonions of norm 1, so there's a 6-sphere of algebra embeddings of $\mathbb{C}$ in $\mathbb{O}$.

John Baez (Jun 10 2023 at 15:57):

Mike Shulman said:

...at least, as long as you're standing in a 3-space with chosen coordinate axes. (-:O

Yeah, I thought about that. But there's a functor from 3d real inner product spaces to quaternion algebras. I think the physically useful quaternions are the ones where the 3d inner product space is any tangent space of the Euclidean space we (approximately) inhabit, not $\mathbb{R}^3$ with its standard inner product. We can slap Cartesian coordinates on Euclidean space if we want, and that gives an isomorphism between any of its tangent spaces and $\mathbb{R}^3$, and an isomorphism between its quaternion algebra and $\mathbb{R}^4$, but we can also loftily refrain from doing this, and that's what I was doing, as a pre-emptive maneuver to thwart any jokes like this.

John Baez (Jun 10 2023 at 16:06):

("Any" tangent space because they're all canonically isomorphic, using translations. Note: no "origin" in Euclidean space was required in my pedantic remark: it's an affine space, not a vector space.)

Mike Shulman (Jun 10 2023 at 16:13):

So when you said "an embedding of the complex numbers in the quaternions", by "the complex numbers" you meant a field with a specified square root of -1 called $i$, but by "the quaternions" you didn't mean a division ring with specified square roots of -1 called $i,j,k$?

John Baez (Jun 10 2023 at 20:57):

Right. It may seem weird but while mathematical physicists are somewhat loath to equip physical space with coordinates before they really need to, they rarely think twice about picking a standard square root of -1 in the complex numbers. If we didn't do that we'd have to admit a bunch of things like: the choice of which particles count as positively charged and which count as negative is arbitrary, since they correspond to two representations of U(1) that differ by complex conjugation! Of course this choice is somewhat arbitrary, but it's an easy choice to maintain once made, because we live in a world with lots of electrons and few positrons, etc. (Arguably Benjamin Franklin made the wrong choice by saying electrons are negatively charged: thanks to this convention, the flow of current in wires goes the opposite direction from how the electrons are moving!)

For those who don't see what's going on: say you have any algebras C isomorphic to the complex numbers and H isomorphic to the quaternions. Then the space of homomorphisms from C to H is a 2-sphere, while the space of subalgebras of H isomorphic to C is a projective plane: the sphere with opposite points identified, since thanks to complex conjugation two different homomorphisms from C to H have the same image.

John Baez (Jun 10 2023 at 21:00):

So, if you want to pick out a homomorphism from the (standard mathematical) complex numbers to the (spatially defined) quaternions, you point in some direction. The direction picks out a specific square root of -1. But if you only want to specify a subalgebra of the quaternions isomorphic to the complex numbers, you hold out a symmetrical rod with your hand in the middle of the rod.

Mike Shulman (Jun 11 2023 at 01:14):

John Baez said:

It may seem weird but while mathematical physicists are somewhat loath to equip physical space with coordinates before they really need to, they rarely think twice about picking a standard square root of -1 in the complex numbers.

That doesn't seem weird to me at all. What seems weird to me is using the phrase "the quaternions" to refer to something related to physical space, which doesn't come equipped with coordinates. I think of "the quaternions" as the result of applying the Cayley-Dickson construction to "the complex numbers", which therefore does have a specified basis just as the complex numbers have a specified imaginary unit.

John Baez (Jun 12 2023 at 18:05):

I guess it is weird. Here's my thinking: Hamilton pursued the quaternions to provide a treatment of geometry that treats a vector in 3-space as an imaginary quaternion $\mathbf{x}$ rather than a triple $(x_1, x_2, x_3)$, and modern folks continuing this line of work like to use a manifestly coordinate-independent approach to quaternions, where you start with a 3d inner product space and work from there. This approach is sometimes called "geometric algebra". They tend to define the quaternions using Clifford algebras rather than the Cayley-Dickson construction, so there's no distinguished copy of $\mathbb{C}$ in $\mathbb{H}$.

Admittedly, it would be retrojecting to pretend Hamilton took a coordinate-free approach: the modern abstract definition of "vector space" was only laid out in 1918, in Weyl's physics book Space, Time, Matter. But it is interesting that Hamilton invented quaternions before people wrote geometry and physics using vectors as unified entities like $\mathbf{x}$ or $\vec{\mathbf{x}}$ rather than lists of numbers. Later people like Heaviside decided it was pointless to combine a vector and scalar into a single thing, the quaternion. This launched a war between the vector-lovers and the quaternionists, and we all know who won.

In the paper where Maxwell introduced the term "curl", he wrote:

The invention of the calculus of Quaternions is a step towards the knowledge of quantities related to space which can only be compared, for its importance, with the invention of triple coordinates by Descartes.

He defined the curl and divergence as the vector and scalar part of the already known quaternion derivative of a vector-vaued function!

Mike Shulman (Jun 12 2023 at 18:15):

To be extra clear, my problem is not with applying the word "quaternions" to such things, but the word "the". In general, we prefer to reserve the [[generalized the]] for objects inhabiting a contractible space, but the space of such "coordinate-independent quaternion algebras", while connected, is not contractible. I would be perfectly happy to call it "a quaternion algebra".

John Baez (Jun 12 2023 at 18:52):

That's better, I agree. This is a really good example of a homotopy type with nonvanishing $\pi_1$ and $\pi_3$ showing up in Nature.

B. Wilson (Jun 13 2023 at 14:05):

Trying to follow along here, but would you mind hand-holding a bit?

I'm thinking that the space of $\mathbb{H}$ s here can be brutishly built as the collection of pairs of CD-constructed $\mathbb{H}$ along with some base-changing automorphism. Since algebra homomorphisms give us $-1 = g(-1) = g(i^2) = g(i)^2$ for any imaginary unit $i$, we're just looking at the $\mathbb{R}$-algebra ${\rm Aut}(\mathbb{H})$. Then something something compact-open topology something something outer morphisms give us a nontrivial $\pi_1$? My grokkiing of homology here is super sketch, and I suspect you guys have cleaner, high-flying ideas in mind.

John Baez (Jun 13 2023 at 15:30):

There's a lot to say here but $\mathrm{Aut}(\mathbb{H})$ is not an algebra, it's a Lie group - a group that's a manifold. So it has a topology, and it's isomorphic to SO(3), the group of rotations in 3d space. This group has $\pi_1 \cong \mathbb{Z}/2$ and $\pi_3 \cong \mathbb{Z}$, and in fact infinitely many other nontrivial homotopy groups.

John Baez (Jun 13 2023 at 15:40):

The Cayley-Dickson construction is surprisingly unhelpful in understanding most things about the quaternions. To see how SO(3) gets into the game it's better to think about how the imaginary quaternions are a 3d real vector space with an inner product, and any rotation of this space gives an automorphism of the quaternions (and vice versa).

John Baez (Jun 13 2023 at 15:41):

Here's why: the famous quaternions i, j, k are one orthonormal basis of the imaginary quaternions, and an automorphism of the quaternions can rotate these vectors into any other right-handed orthonormal basis.

John Baez (Jun 13 2023 at 16:11):

I could explain bits of this in greater detail if you want.

David Egolf (Jun 13 2023 at 16:36):

John Baez said:

Here's why: the famous quaternions i, j, k are one orthonormal basis of the imaginary quaternions, and an automorphism of the quaternions can rotate these vectors into any other right-handed orthonormal basis.

That reminds me how we used i, j, and k for basis vectors (for modelling some physical 3D space) in some of my engineering classes!

John Baez (Jun 13 2023 at 17:41):

Yes, I believe Hamilton was the first to use i, j, k this way.

John Baez (Jun 13 2023 at 17:43):

He thought of i, j, and k and their linear combinations as imaginary quaternions - that is, special case of more general expressions like

$a + b i + c j + d k$

but later Heaviside and Gibbs and others went ahead and chopped quaternions into their scalar part $a$ and their vector part $b i + c j + d k$ and decided to never talk about quaternions as a whole.

John Baez (Jun 13 2023 at 17:44):

So if your teachers ever said "never add a scalar and a vector", they were basically saying "don't dare to think about quaternions!"

John Baez (Jun 13 2023 at 17:46):

I really recommend Russell Crowe's A History of Vector Analysis for the full story of the rise and fall of quaternions.

John Baez (Jun 13 2023 at 17:46):

(There's a joke in there.)

David Michael Roberts (Jun 14 2023 at 04:36):

Can't help but think of the actor, and hence the movie Gladiator, and hence Gibbon's famous book on the Roman empire...

John Baez (Jun 14 2023 at 04:58):

And then this line from the bible, about events in a certain prison during the Roman Empire:

And when he had apprehended him, he put him in prison, and delivered him to four quaternions of soldiers to keep him; intending after Easter to bring him forth to the people.

Mike Shulman (Jun 14 2023 at 05:20):

One reason for chopping quaternions into the scalar part and the vector part, of course, is that you can work with vectors in any dimension, but imaginary quaternions are fundamentally 3-dimensional.

Mike Shulman (Jun 14 2023 at 05:25):

Many of the "pieces" of the quaternions generalize to other dimensions, but in different ways that split them up whereas the quaternions "package" them all together. E.g. the quaternion product wraps up together the dot product and cross product of 3D vectors; the dot product generalizes to any dimension, while the cross product generalizes to things like wedge products that make sense in any dimension but don't in general give out another vector (rather a "bivector"). Similarly, the quaternions of norm 1 form a group $SU(2)$ that's a "double cover" of $SO(3)$; and you can talk about $SU(n)$, $SO(n)$, double covers, etc. in any dimension, but as far as I know they don't come from a quaternion-like algebra any more.

John Baez (Jun 14 2023 at 05:36):

A modern reason to split quaternions into two parts is that these two parts are each irreducible representations of SO(3), and direct summing two irreducible representations gives a thing that's not a "fundamental" sort of thing.

Starting from an $n$-dimensional vector space we often build the [[exterior algebra]] $\Lambda V$, which is a graded algebra where each grade is an irreducible representation of $\mathrm{SO}(n)$ . Physicists call this exterior algebra a "Grassmann algebra", because Grassmann came up with this idea - in 1848, after Hamilton's quaternions but I'm not sure how influenced he was.

Starting from an $n$-dimensional inner product space we also build the [[Clifford algebra]] $\mathrm{Cliff}(V)$, which is a filtered algebra whose associated graded algebra is $\Lambda V$. Clifford came up with this idea in 1878, and he must have been influenced by quaternions, since Clifford algebras are probably the best generalization of quaternions to higher dimensions.

John Baez (Jun 14 2023 at 05:38):

However when $V$ is 3-dimensional $\mathrm{Cliff}(V)$ is not the quaternions; Clifford algebras are $\mathbb{Z}/2$ graded and the even part of $\mathrm{Cliff}(V)$ is isomorphic to the quaternions when $V$ is 3-dmensional!

John Baez (Jun 14 2023 at 05:38):

This seems weird at first but ultimately it makes perfect sense.