## December 3, 2022

### Neutrino Dark Matter

#### Posted by John Baez

I talked to Neil Turok at a café today. He used to be the head of the Perimeter Institute, but now he’s at the University of Edinburgh.

He coauthored a paper arguing that dark matter is very heavy right-handed neutrinos:

It’s very natural to add right-handed neutrinos to the Standard Model, and if they’re heavy they can make the observed left-handed neutrinos light via the ‘see-saw mechanism’. The problem is to keep them from decaying too fast!

For a heavy neutrino to serve as dark matter, it needs to be quite stable. Apparently this is tough if it interacts with the Higgs—how true is that, exactly? But neutrino that’s its own antiparticle can have a mass without interacting with the Higgs: a so-called ‘Majorana mass’.

In Turok’s theory all the neutrinos have Majorana masses, described by a mass matrix. To make the heaviest right-handed neutrino stable, a bunch of matrix entries must vanish—and this makes the lightest left-handed neutrino massless!

So, out of this theorizing Turok gets a testable prediction: one of the observed neutrinos is massless!

This is interesting, because we only know that two of the three observed neutrinos have a nonzero mass. Neutrino oscillation experiments don’t let us measure neutrino masses: they only tell us differences between squares of neutrino masses!

In 1998, research results at the Super-Kamiokande neutrino detector determined that neutrinos can oscillate from one flavor to another, which requires that they must have a nonzero mass. While this shows that neutrinos have mass, the absolute neutrino mass scale is still not known. This is because neutrino oscillations are sensitive only to the difference in the squares of the masses. As of 2020, the best-fit value of the difference of the squares of the masses of mass eigenstates 1 and 2 is 0.00007 $\mathrm{eV}^2$ , while for eigenstates 2 and 3 it is 0.00251 $\mathrm{eV}^2$. Since this is the difference of two squared masses, at least one of them must have a value which is at least the square root of this value. Thus, there exists at least one neutrino mass eigenstate with a mass of at least 0.05 eV

We need other experiments to measure, or bound, the actual masses of neutrinos! Luckily, some good experiments are in the works now.

For example, Euclid is a space telescope that will study the large-scale structure of the Universe in exquisite detail.

It should be able to measure the sum of all 3 light neutrino masses to within 0.03 eV:

Right now we just know this sum is $\lt$ 0.1 eV. With Euclid, and other forthcoming experiments, we may be able to constrain it much better. Combining this with our knowledge of the differences of squared masses, we should either be able to prove the lightest neutrino isn’t massless, disproving the theory I’m talking about… or get evidence that it’s almost massless, supporting it.

Turok explained a lot more to me, but in case you’re not used to this stuff, here’s a summary of what I just said: if neutrinos have Majorana masses and one is massless, that’s indirect evidence that dark matter is a heavy right-handed neutrino!

Finally, let me state a few of my own prejudices.

I want leptons to be very similar to quarks, since they’re already quite similar in many ways, and the mathematics of theories like the SO(10) GUT make these similarities seem very natural.

This makes me want right-handed neutrinos. So I like the idea of dark matter being heavy right-handed neutrinos - especially since the see-saw mechanism says this is compatible with the left-handed ones being very light.

But I don’t like Majorana masses for neutrinos, since the quarks don’t have those.

So, I want to know if, in the theories I prefer, where neutrinos gets masses in the same way that quarks do, one very heavy and stable (or almost stable) right-handed neutrino is compatible with—or implies—an extremely light or massless left-handed neutrino.

This must have been studied; I just need to read some more papers. Actually I once read some papers by Mohapatra about this, but I need to reread them with a closer eye to the stability of the most massive neutrino.

Finally, I should add that it was very invigorating to talk to someone who is really excited about fundamental physics, with some concrete testable ideas about it. Turok thinks a lot of the big remaining problems in physics can be solved using the Standard Model with very slight tweaks, no new particles. I don’t know if I believe that, but it was great talking to him. I haven’t thought seriously about particle physics for years, because it seems so depressing.

Posted at December 3, 2022 10:27 AM UTC

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### Re: Neutrino Dark Matter

While we’re waiting for someone to answer your question, I – a mathematician with limited knowledge of physics – would like to ask something more basic.

I think I understand the field content of the theory you want to consider: we have the fields from the 1970s version of the Standard Model (SM), plus a three-generation right-handed Weyl spinor field $\nu^R$. I.e., the field $\nu^R$ is a section in a complex vector bundle over spacetime whose typical fiber is $\mathbb{C}^2\otimes\mathbb{C}^3$: spinor part $\mathbb{C}^2$ tensor generation part $\mathbb{C}^3$. Let $G$ be the SM gauge group. The physically relevant group $\SL(2,\mathbb{C})\times G$ acts on $\mathbb{C}^2\otimes\mathbb{C}^3$ by $(A,g)(v\otimes w) = A(v)\otimes w$. I agree that this field content has a much prettier description (that I don’t have to spell out here) than the 1970s version in which $\nu^R$ is omitted.

According to the usual philosophy, the SM is an effective field theory. At length scales much larger than the fundamental scale, we expect to see the effects of all renormalizable Lagrangian terms that we can build from the given field content (because we don’t believe in coincidences that make the value of some physical constant experimentally indistinguishable from $0$ without good reason). In your preferred theory, we can build all terms we had in the 1970s version of the SM, plus a few obvious lepton analogues of quark terms we know from the 1970s version. They include Yukawa-type terms that give rise to neutrino masses, analogously to how the quark masses arise in the SM. So far, so good.

But we can also build an additional (super-)renormalizable term, involving only $\nu^R$, that has no quark analogue. This works as follows. We fix an antisymmetric complex-bilinear form $M\colon \mathbb{C}^3\times\mathbb{C}^3 \to \mathbb{C}$ on the generation tensor factor. (We can think of $M$ as a bunch of new physical constants.) The antisymmetric complex-bilinear form $B\colon \mathbb{C}^2\times\mathbb{C}^2 \to \mathbb{C}$ given by

(1)$B(v,w) := \det(v\wedge w)$

(where the determinant $\det$ is interpreted as a map $\bigwedge^2\mathbb{C}^2 \to \mathbb{C}$) is $\SL(2,\mathbb{C})$-invariant. Hence $B\otimes M$ is a symmetric complex-bilinear form on $\mathbb{C}^2\otimes\mathbb{C}^3$, invariant under our action of $\SL(2,\mathbb{C})\times G$. We can therefore build the Lagrangian

(2)$\mathscr{L}(\nu^R) = \text{Re}\big((B\otimes M)(\nu^R,\nu^R)\big) ,$

where $\text{Re}$ denotes the real part.

So, in your beautified Standard Model, we have automatically neutrino terms without quark analogue. (There is no quark analogue because the nontrivial action of $G$ on the quark parts spoils the invariance.)

What is the relation (if any) between these terms and Majorana mass terms? Majorana mass terms as usually discussed by physicists refer to a different field content than we have in your theory, don’t they?

Which observable consequences do the additional terms have? Which experimental constraints are there on $M$?

Posted by: Newtino on December 4, 2022 9:48 AM | Permalink

### Re: Neutrino Dark Matter

Hi! Thanks very much for digging into this. I could use some more conversations like this, since when people who self-identify as physicists start talking about these things, I need to slow them down by making them fill in a lot of details—just as you are doing here.

From what I can tell, your quadratic Lagrangian $\nu^R$ is what physicists, e.g. Turok in my conversation with him, call a “Majorana mass term for a right-handed neutrino”. I’ll admit I haven’t checked this carefully: I find it irksome to translate between different people’s notations, so if there were some other invariant quadratic expression you can construct from $\nu^R$ (and not its derivatives) then maybe that could be the Majorana mass for right-handed neutrinos.

But let me do a tiny bit of actual work. In the paper by Boyle, Finn and Turok they write down a Lagrangian including a Majorana mass term for right-handed neutrinos in equation (106), and that term is

$- \frac{1}{2}(\overline{\nu}_R^c M^\dagger \nu_R + \text{h.c.})$

They say $M$ is a $3 \times 3$ complex matrix and $\overline{\nu}_R^c = - i \gamma^2 \nu_R^*$ is the charge conjugate of $\nu_R$. Here of course $\gamma^2$ is a gamma matrix.

Is this the same as what you’re talking about, modulo change in notations, conventions and general aesthetics? I’m guessing it is. Physicists like to use the Pauli matrix $i\sigma_2$ to stand for a symplectic structure, and your $B$ is a symplectic structure, and the gamma matrix $\gamma^2$ has two blocks that look like $\sigma_2$, so maybe it all works out.

Posted by: John Baez on December 4, 2022 12:22 PM | Permalink

### Re: Neutrino Dark Matter

Unfortunately I don’t have the time right now to dig into this, but there is one thing that lets me doubt it’s all just conventions and notation. When physicists discuss Majorana masses, they usually point out that that stuff works also with just one particle generation. The one-generation analogue of the term I wrote down vanishes, though, because the bilinear form $M$ must be antisymmetric (or rather, its symmetric part does not contribute to $\mathscr{L}(\nu^R)$).

I think we have an additional real structure in the Majorana context; hence, in my language, the “field content” – by which I mean everything about the theory that does not involve the Lagrangian – is different. That allows us to build terms that we cannot build without the additional structure. But the term I wrote down can be built anyway.

Posted by: Newtino on December 4, 2022 9:36 PM | Permalink

### Re: Neutrino Dark Matter

We fix an antisymmetric complex-bilinear form $M:\mathbb{C}^3\times\mathbb{C}^3\to \mathbb{C}$ on the generation tensor factor. (We can think of $M$ as a bunch of new physical constants.) The antisymmetric complex-bilinear form $B:\mathbb{C}^2\times \mathbb{C}^2\to\mathbb{C}$ given by

(1)$B(v,w)\coloneqq \det(v\wedge w)$

(where the determinant det is interpreted as a map $\bigwedge^2\mathbb{C}^2\to \mathbb{C}$) is $SL(2,\mathbb{C})$-invariant. Hence $B\otimes M$ is a symmetric complex-bilinear form on $\mathbb{C}^2\otimes\mathbb{C}^3$

Close, but no cigar.

Neutrinos are fermions, so they are represented by Grassmann-valued fields. Hence you want to fix $M:\mathbb{C}^3\times\mathbb{C}^3\to \mathbb{C}$ a symmetric bilinear form.

With that modification, your equation (2) is correct.

What is the relation (if any) between these terms and Majorana mass terms? Majorana mass terms as usually discussed by physicists refer to a different field content than we have in your theory, don’t they?

I would banish that term from the lexicon, as it leads to nothing but confusion. A “Dirac” mass term is one that respects a $U(1)$ symmetry (broadly interpreted as “fermion number”). A “Majorana” mass term is any mass term that doesn’t (i.e., it’s the “generic” situation).

What’s relevant here is something completely different: whether the mass term in question is compatible with unbroken $SU(2)\times U(1)$ Standard Model gauge symmetry. The usual mass terms for the quarks and leptons are not (they are written as Yukawa couplings to the Higgs field, which only become mass terms when the Higgs field gets a VEV, breaking $SU(2)\times U(1)\to U(1)_{\text{EM}}$).

Here, the mass terms for $\nu^R$ are compatible with unbroken $SU(2)\times U(1)$, which is why the eigenvalues of $M$ can be as large as we want.

Posted by: Jacques Distler on December 6, 2022 9:05 AM | Permalink | PGP Sig

### Re: Neutrino Dark Matter

I wrote:

Neutrinos are fermions, so they are represented by Grassmann-valued fields. Hence you want to fix $M:\mathbb{C}^3\otimes \mathbb{C}^3\to\mathbb{C}$ a symmetric bilinear form.

Note that this should be obvious if you don’t get all fancy and try to write down a mass matrix for three fermions, but rather try first to write down a mass for a single fermion.

Then you have “bilinear form” $M:\mathbb{C}\otimes \mathbb{C}\to\mathbb{C}$ which would necessarily vanish if it were anti-symmetric. It’s obvious that if $M$ is symmetric (as it must be) in the $1\times1$ case, it must also be symmetric in the $n\times n$ case.

In fact the first thing Turok and company do is work in a basis where $M$ is diagonal. They then impose a $\mathbb{Z}_2$ symmetry under which one of the right-handed neutrinos (say $\nu_1^R$) is odd. That suffices to ensure that the effective mass matrix for the SM neutrinos has a zero eigenvalue.

Posted by: Jacques Distler on December 6, 2022 10:33 PM | Permalink | PGP Sig

### Re: Neutrino Dark Matter

You say: “Neutrinos are fermions, so they are represented by Grassmann-valued fields. Hence you want to fix $M\colon \mathbb{C}^3\times\mathbb{C}^3 \to \mathbb{C}$ a symmetric bilinear form.”

For simplicity, let’s consider fields on Minkowski spacetime $S$, and let’s fix trivializations of our then trivializable bundles.

In my formula, the symbol “$\nu^R$” stands for a function $S\to V:= \mathbb{C}^2\otimes\mathbb{C}^3$. The target is a six-dimensional complex vector space $V$. There is no algebra structure on $V$ (field values don’t get “multiplied”, whatever that might mean).

The phrase “Grassmann-valued fields” seems to indicate that in your interpretation of my formula, $\nu^R$ is again a function from $S$ to some target space $X$. What mathematical object precisely is $X$ in your interpretation? (Is it an algebra? In any case, if it’s a complex vector space, which dimension does it have?)

Posted by: Newtino on December 6, 2022 11:40 PM | Permalink

### Re: Neutrino Dark Matter

The phrase “Grassmann-valued fields” seems to indicate that in your interpretation of my formula, $ν^R$ is again a function from $S$ to some target space $X$. What mathematical object precisely is $X$ in your interpretation?

I don’t think I am going to explain the theory of supermanifolds in a blog comment section. See Manin’s book, or Leites’s review (or probably other more recent sources).

Suffice to say that, for $x,y$ non-coincident points in $S$, $\nu^R(x)\nu^R(y)=-\nu^R(y)\nu^R(x)$. This is completely standard to the physicists.

For the mathematicians, the ring of functions on a supermanifold (here, $S$ is a supermanifold of dimension $4|0$) is $\mathbb{Z}_2$ graded-commutative. The “target” $X$ (in your case) is a complex vector space of dimension $0|6$. But to really reproduce the physicists’ fermi fields, you need to work in families $\mathcal{S}\to B$, where $B$ has dimension $0|n$ for arbitrarily large $n$.

As I said, the “3” here is a complete red herring. You need to be able to write down a mass term for a single fermion before you start trying to write down a mass term for three of them.

Posted by: Jacques Distler on December 7, 2022 4:32 AM | Permalink | PGP Sig

### Re: Neutrino Dark Matter

You write: “I don’t think I am going to explain the theory of supermanifolds in a blog comment section.”

You don’t have to. When you bring up any “super” concept in the lowbrow context of the Standard Model plus sterile neutrinos – a context that involves no $\mathbb{Z}_2$-grading or anything else “super” –, you can at best create an unnecessarily complicated equivalent description of a simple matter. At worst – and that is apparently what happens here – you translate the situation wrongly and obtain something that has nothing to do with the matter at hand.

Have you ever read an article about the Standard Model plus sterile neutrinos that mentions supermanifolds? No? There is a simple reason for that. For the same reason, no such article ever mentions $p$-adic varieties or inaccessible cardinals.

If you cannot describe the simple situation at hand without mentioning supermanifolds, the problem lies on your side. Everyone likes cool mathematical concepts, but one should be able to simplify one’s language when appropriate.

You write: “Suffice to say that, for $x,y$ non-coincident points in $S$, $\nu^R(x)\nu^R(y) = −\nu^R(y)\nu^R(x)$. This is completely standard to the physicists.”

Obviously we are talking about very different things. I have no idea what the source of your misunderstanding is, but the anticommutation rule you wrote down lets me suspect that you are talking about a quantum field $\nu^R$, whose “value” at each spacetime point is an operator on a Hilbert space. (In which case your anticommutation rule is required if and only if $x-y\neq0$ is spacelike.) In contrast, I am talking about the argument $\nu^R$ of the Lagrangian I wrote down, which is a map $\nu^R\colon S\to V$. We may call these maps “classical fields” if that helps.

When particle phenomenologists write down a Lagrangian, they write it down for classical fields. Quantization comes later in the story. The classical fields are (in all cases we have to care about here) arbitrary sections in a given bundle $E$ over spacetime. The values $\nu^R(x),\nu^R(y)$ at different spacetime points are completely unrelated. An expression like $\nu^R(x)\nu^R(y)$ wouldn’t make sense for them anyway: not even in the case $x=y$, because the fibers of $E$ don’t have an algebra structure; even less for $x\neq y$.

Or maybe you do not interpret $\nu^R$ as a quantum field. In that case I have even less of an idea what your point is.

One thing is clear, though: If the bilinear form $M$ in my formula is symmetric, then $\mathscr{L}(\nu^R)$ vanishes for all arguments $\nu^R$. (This has nothing to do with bosonic or fermionic behavior, with commuting or anticommuting; commutators or anticommutators are not defined for the values of $\nu^R$ in the first place.)

Since you claim otherwise, you have apparently – to paraphrase Goethe – translated the situation into your own language, and henceforth it is something completely different.

Here is an exercise: Take a random hep-ph article about neutrinos (sterile or otherwise) and look at the Lagrangian that’s written down on the first few pages. Explain in simple language what kind of mathematical object the symbol $\nu$ in that formula represents. (Resist the urge to use the prefix “super”. You don’t need it.)

Posted by: Newtino on December 8, 2022 1:56 PM | Permalink

### Re: Neutrino Dark Matter

I have no idea what the source of your misunderstanding is, but the anticommutation rule you wrote down lets me suspect that you are talking about a quantum field $\nu^R$

Your suspicion is incorrect. These fields $\nu^R(x)$ anti-commute classically (even before quantization).

The physicists have no problem with that concept. Mathematicians usually do. Which is why I alluded to a formalism in which that can be made mathematically precise.

If you have no problem with classically anti-commuting fields, then we can proceed. Otherwise, you will need to reconcile yourself to a setting in which they emerge naturally.

Explain in simple language what kind of mathematical object the symbol $\nu$ in that formula represents. (Resist the urge to use the prefix “super”. You don’t need it.)

I just did. Classically, it’s an anti-commuting spinor-valued field (i.e., a section of $\Pi S$).

Crack open any textbook on QFT and this will be explained to you. If you don’t like that explanation, then read up on supermanifolds …

Posted by: Jacques Distler on December 9, 2022 8:21 AM | Permalink | PGP Sig

### Re: Neutrino Dark Matter

If the bilinear form $M$ in my formula is symmetric, then $\mathcal{L}(\nu^R)$ vanishes for all arguments $\nu^R$.

Sigh. Which clearly indicates that you have not understood a single hep-ph article.

So let’s write out that term, for a single fermion (which I will call “$\psi$” instead of $\nu^R$, because this misunderstanding has nothing to do with neutrinos).

(1)$\mathcal{L}(\psi)(x) = \operatorname{Re}(m B(\psi(x),\psi(x)))= \operatorname{Re}\bigl(m( \psi_1(x)\psi_2(x) -\psi_2(x) \psi_1(x))\bigr)$

If $\psi_\alpha(x)$ were a commuting field, then this would indeed vanish. But $\psi_\alpha$ is an anti-commuting field (and, even quantum-mechanically, $\psi_1(x)$ and $\psi_2(x)$ anti-commute.) So (1) is nonvanishing. It is, in fact, the mass term for a single Weyl fermion.

The generalization to three Weyl fermions is straightforward..

Posted by: Jacques Distler on December 9, 2022 8:37 AM | Permalink | PGP Sig

### Re: Neutrino Dark Matter

Okay, so we agree that $\nu,\nu^R,\psi$ in the present context are classical fields, and that one of us has misunderstood something very basic when reading hep-ph articles. That’s a start.

Let’s describe everything solely in language that a large majority of hep-ph people would understand. They know complex numbers, and vectors, matrices, more general tensors whose entries are complex numbers. By “spinor”, they mean a map that assigns to every spacetime point a pair (or quadruple) of complex numbers and transforms in a certain way under the universal cover of the Poincaré group.

I strongly doubt that they have ever heard of a (classical) “anti-commuting spinor-valued field”, so you will have to break that notion down into terms they understand (or do a representative poll of hep-ph authors that convinces me they already know that notion).

If an object $\psi$ is a Weyl spinor, the hep-ph folks understand that it consists of two complex numbers $\psi_1(x),\psi_2(x)$ at each spacetime point $x$ (and that the values at different spacetime points are not constrained by any equation). The three generation-components of the $\nu^R$ in my definition of $\mathscr{L}(\nu^R)$ are such Weyl spinors $\psi$, hence satisfy $\psi_1(x)\psi_2(x) -\psi_2(x)\psi_1(x) = 0$.

Now you say to me: “No no no, the $\nu^R$ you are using is a commuting field! That makes no sense! Since it describes a fermion, it must be an anticommuting field! The $\psi_1(x)\psi_2(x) -\psi_2(x)\psi_1(x)$ do not vanish!”

Well, I meant indeed that the three generation-components of my $\nu^R$ are Weyl spinors $\psi$ in the sense above, which you regard as examples of “commuting fields”. And yes, I claim that is exactly what the hep-ph people mean when they write down fermion Lagrangians like mine: their $\psi_1(x),\psi_2(x)$ are plain old boring complex numbers and are multiplied as such.

Before we can figure out what the hep-ph people really mean, you will have to remind them (and explain to me), in the elementary terms above, what objects the anticommuting $\psi_i(x)$ in your formula (1) are. How can I represent them on a computer, how do I tell the computer how to multiply them (to make sense of your formula (1))? (So that, at least in principle, the hep-lat folks could compute predictions with them that can be compared to experimental data.)

Posted by: Newtino on December 9, 2022 11:14 PM | Permalink

### Re: Neutrino Dark Matter

This has gotten silly.

I strongly doubt that they have ever heard of a (classical) “anti-commuting spinor-valued field” …

I’m sorry but I guarantee that everyone on hep-ph is intimately familiar with anti-commuting spinor-valued fields. They are the bread-and-butter of every working high energy physicist.

You really need to crack open any textbook on quantum field theory. Five minutes reading (less than you probably spent composing that comment) would convince you that you are wrong.

Come back when you understand this basic point.

In fact, you had best convince yourself that neither the kinetic term nor the mass term for a Weyl fermion would make sense if they were commuting spinor-valued fields.

you will have to remind them (and explain to me), in the elementary terms above, what objects the anticommuting $\psi_i(x)$ in your formula (1) are.

I told you how the physicists understand them and I alluded to how mathematicians (at least those who are serious about understanding what the physicists mean) understand them.

If you don’t like what the physicists do, you can read Leites for how to do it “right”.

Posted by: Jacques Distler on December 9, 2022 11:49 PM | Permalink | PGP Sig

### Re: Neutrino Dark Matter

“This has gotten silly.” Another point on which we agree.

“I guarantee that everyone on hep-ph is intimately familiar with anti-commuting spinor-valued fields.” (Emphasis mine.) I wonder why I’m not impressed by that guarantee.

“I told you how the physicists understand them […]” No you didn’t. You just claim again and again, “Everyone knows that!” You have to tell me how to implement your $\psi_i(x)$ objects on a computer – what type of data represents them – and how to write the multiplication subroutine. I could and would explain how complex numbers (my $\psi_i(x)$ objects) can be implemented on a computer and be multiplied, if someone needs that basic information.

I’m a mathematician, my philosophy is clear and simple: When someone is in case of disagreement not willing to spell out his definitions, he should not be taken seriously. Further communication with him is then a waste of time.

Let’s call it a day.

Posted by: Newtino on December 10, 2022 12:57 AM | Permalink

### Re: Neutrino Dark Matter

“I guarantee that everyone on hep-ph is intimately familiar with anti-commuting spinor-valued fields.” (Emphasis mine.) I wonder why I’m not impressed by that guarantee.

Yes, indeed that is puzzling.

I’m a high energy theoretical physicist. This is one of the most elementary facts in my field. That you would claim to have even a passing familiarity with high energy physics and be oblivious to — nay, deny — even the most basic facts about the subject, even when corrected by someone working in the field ….

Puzzling.

You have to tell me how to implement your $\psi_i(x)$ objects on a computer – what type of data represents them – and how to write the multiplication subroutine.

Seriously?

There are computer packages that handle arbitrary noncommutative algebras. $\mathbb{Z}_2$ graded-commutative algebras are a piece of cake All you need to do is keep track of signs $\psi_i(x)\psi_j(y)= -\psi_j(y)\psi_i(x)$. In fact, for the purpose at hand, we don’t even need these at non-coincident points, it suffices that $\psi_i(x)\psi_j(x)= -\psi_j(x)\psi_i(x)$.

I’m a mathematician

Apparently one who can’t be bothered to find out what physicists actually do, before berating them for doing it wrong …

Let’s call it a day.

Yep. There’s clearly no possibility you will actually open up a QFT textbook and realize that you are wrong. So there’s really nothing useful that can happen here.

Posted by: Jacques Distler on December 10, 2022 2:21 AM | Permalink | PGP Sig

### Re: Neutrino Dark Matter

$\mathbb{Z}_2$ graded-commutative algebras are a piece of cake.” Indeed! You seem to have incredible difficulty spelling out which one you’re using, and how!

I assume you’re using a complex algebra $A$ here. What is its even part $A_0$ (dimension, please), what is its odd part $A_1$ (dimension, please), what is the multiplication map (explicit formula, please)?

I assume your objects live in the odd part. But which objects precisely? Your $\psi(x)$? Your components $\psi_i(x)$? Again, how is the multiplication of components in your formula (1) defined?

Posted by: Newtino on December 10, 2022 3:07 AM | Permalink

### Re: Neutrino Dark Matter

As a mathematician: implicitly, I believe Jacques Distler is assuming that $\psi$ lives in an algebra bundle constructed over spacetime M whereby the local fibre has (at least) $2^floor{\frac{\dim M}{2}}$ anti-commuting generators, or twice that many if you wish to assume \emph{a priori} independent conjugate spinors. Each $\psi_a(x)$ is a coefficient of one of these generators.

I may be wrong, and if so, I welcome correction.

Posted by: SOMC on December 10, 2022 3:32 AM | Permalink

### Re: Neutrino Dark Matter

Thank you, that makes some sense. In the case $\dim(M)=4$ it produces $4$ generators, though. Since we have only $\psi_1,\psi_2$, we must bring in a representation-theoretic splitting…

Posted by: Newtino on December 17, 2022 3:55 PM | Permalink

### Re: Neutrino Dark Matter

I deleted a comment where one participant had become quite rude. Admittedly, they were goaded by rudeness on the other side. But I have to impose a cutoff somewhere as people descend the scale of decreasing politeness. I should have stopped y’all earlier, but I was traveling around, not reading all my emails.

I don’t completely understand what it is about particle physics that makes people so rude. People don’t seem to act this way when it comes to etale cohomology or other equally esoteric subjects. It’s possible to ask questions about the role of supercommutative algebras in the Standard Model Lagrangian, and answer them, while remaining perfectly polite and cheerful. It’s a fun subject, after all!

Posted by: John Baez on December 11, 2022 1:48 PM | Permalink

### Re: Neutrino Dark Matter

I had seen Jacques Distler’s now deleted comment a week ago, but was then quite busy for a few days. I had planned to reply today.

To the extent that I misbehaved, I apologize and will try to do better.

Unfortunately, the deleted comment contained not only rude remarks but also Distler’s first substantial — but in my opinion still insufficient — steps toward a definition of the Lagrangian he wrote down. I would have liked to ask follow-up questions, to hopefully get to the bottom of things eventually. But the Internet Archive does not have a snapshot of his comment, so I guess bilateral rudeness killed the last chance of progress in that regard.

Posted by: Newtino on December 17, 2022 3:58 PM | Permalink

### Re: Neutrino Dark Matter

Do I understand correctly that John is interested in a Dirac-only right-handed neutrino? I had thought that a Dirac-only right-handed neutrino “merely” acts as the other half of a Dirac neutrino field. Hence the left and right chiralities oscillate into one another with the same mass. So a Dirac-only right-handed neutrino couldn’t possibly be heavy. Am I misunderstanding?

Posted by: Luther Rinehart on December 11, 2022 1:14 AM | Permalink

### Re: Neutrino Dark Matter

You may be exactly right. I wish there were a really nice explanation of conventions concerning the CKM and PMNS matrices, the various main options for neutrino masses, and so on. As it is, whenever I think about this stuff, about once a decade, I need to refresh my knowledge. I find a lot of the explanations fragmentary and confusing.

Posted by: John Baez on December 11, 2022 2:28 PM | Permalink

### Re: Neutrino Dark Matter

“I had thought that a Dirac-only right-handed neutrino “merely” acts as the other half of a Dirac neutrino field.”

Maybe you know all this already, but just to make sure: The left-handed half is not a field of the Standard Model. The Dirac-only right-handed neutrino field in John Baez’s context (as I understand it) is indeed nothing but one half of an ordinary Dirac spinor field (or rather, three copies of such a field, corresponding to the three particle generations). Physically, it interacts only via gravitation and Yukawa-type terms.

The left-handed neutrino field in the Standard Model is a more complicated object, because it interacts also via the weak force. Let $G$ be the Standard Model gauge group. (Usually one takes $G = SU(3)\times SU(2)\times \text{U}(1)$, but using $G = \text{S}(\text{U}(2)\times \text{U}(3))$ is more elegant once we spell out the details, which I won’t do here.) Mathematically, the bundle the left-handed neutrino field lives in is constructed with a representation not of the group $\Spin^\circ(3,1) = \SL(2,\mathbb{C})$ but of the group $\SL(2,\mathbb{C})\times G$. (Of course the bundle for the right-handed neutrino field can also be regarded as being constructed with a representation of $\SL(2,\mathbb{C})\times G$, but that is a different representation where the $G$ component acts trivially.)

Then there is the further complication that in this model, we have two types of “mass-generating” terms in the Lagrangian: the ones coming from Yukawa-type terms (which involve the Higgs field), and additional terms which are just quadratic in the right-handed neutrino fields. In this situation, we should first clarify what precisely we mean by “mass”…

Posted by: Newtino on December 17, 2022 3:54 PM | Permalink