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Stream: learning: reading & references

Topic: Mastering Bott Periodicity

Andrius Kulikauskas (Apr 10 2024 at 10:41):

In this thread, I want to work on mastering Bott periodicity along with others who may be interested, and certainly with the help of those who can teach us. I have made a video about my own interests, which are metaphysical, and have written out the transcript with slides: Bott Periodicity Models Consciousness? Preliminary Exploration.

Bott Periodicity is related to category theory on many levels and in many variants so I hope this is a fine place to work on this! Please jump in with your related interests, comments, insights, questions, suggestions, corrections, ideas...

Andrius Kulikauskas (Apr 10 2024 at 10:52):

I myself want to have intuition on how real Bott periodicity functions as a clock with periodicity 8. My PhD is in algebraic combinatorics so I am most inclined to Clifford algebra calculations. I would like to also understand in terms of Lie theory and symmetric spaces. For me, the homotopy theory and K-theory are optional, but I have studied some of Hatcher's Algebraic Topology and ultimately that would surely add insight.

Metaphysically, my goal is to understand CPT symmetry in this context. Charge conjugation, parity, time reversal are metaphysical concepts that apparently pop out of the math. I think they relate to what I call "divisions of everything", the twosome, threesome, foursome, which I talk about in my presentation Time and Space as Representations of Decision-Making. I talk about an eight-cycle of divisions of everything, proceeding from the nullsome to the sevensome, with the eightsome (the logical square with four corners and four sides) collapsing into the nullsome (for example, if all are known and all are unknown, then the system is empty).

Charge conjugation has to do with (not) distinguishing particles (what is) and holes (what is not), thus existence. Parity has to do with participation, whether our rules apply in a mirror world, and in particular, whether a learning three-cycle (taking a stand, following through, reflecting) functions in the opposite direction. Time reversal has to do with knowledge, whether there is change, whereby a concept like How makes sense. You may disagree but all the better, as this is on my mind, and your views may surely help.

Andrius Kulikauskas (Apr 10 2024 at 11:07):

My current plan of study is:

1) Understand how to calculate the Clifford algebras $C_{0,k}$ (with generators squaring to $-1$) as matrix algebras. I think I need to use the homomorphisms for the recurrence relations.

• $\textrm{Cl}_{p+2,q}(\mathbf{R}) = \mathrm{M}_2(\mathbf{R})\otimes \textrm{Cl}_{q,p}(\mathbf{R})$
• $\textrm{Cl}_{p+1,q+1}(\mathbf{R}) = \mathrm{M}_2(\mathbf{R})\otimes \textrm{Cl}_{p,q}(\mathbf{R})$
• $\textrm{Cl}_{p,q+2}(\mathbf{R}) = \mathbf{H}\otimes \textrm{Cl}_{q,p}(\mathbf{R})$

2) Understand what are the representations (and also the $\mathbb{Z}_2$-graded representations) of the Clifford algebras. Calculate the groups $A_k=M(C_{0,k})/i^*(M(C_{0,k+1}))$ as in Dale Husemoller's book Fibre Bundles and Bott, Attiyah, Shapiro's paper Clifford Modules.

3) Do the calculations of the Lie group embedding described in the very helpful paper Michael Stone, Ching-Kai Chiu, Abhishek Roy. Symmetries, Dimensions, and Topological Insulators: the mechanism behind the face of the Bott clock. which is like a tutorial for physicists.

4) Understand in that paper how they talk about CPT symmetry and how it relates to Bott periodicity.

5) Understand how CPT symmetry relates to random matrix ensembles.

6) From there it would be great to study @John Baez 's video on Symmetric Spaces in terms of a free-forgetful adjunction.

7) I also want to understand the loop space - suspension adjunction.

Andrius Kulikauskas (Apr 13 2024 at 19:50):

I am working on my Step 1) above. I am studying José Figueroa-O'Farrill's lecture notes for his Projective Geometry course on Spin Geometry. His Lecture 2 is about Clifford algebras: classification. In Theorem 2.3, he states the three recurrence relations and he describes the homomorphism with which he proves $CL(n+2,0)\cong Cl(0,n)\otimes Cl(2,0)$ and then I will need to work out the other two myself. His notation is that the left entry counts the generators that square to $-1$, and the right entry counts the generators that square to $+1$. Just now I am realizing that here he is using the ungraded tensor product, and I think I could figure out what he is doing with that. Whereas at the beginning of the section he introduced the $\mathbb{Z}_2$-graded tensor product, which I am wrestling with, and which will be important later. That is why today I thought I should I start with his Lecture 1 on Clifford algebras: basic notions.

Andrius Kulikauskas (Apr 13 2024 at 20:03):

Lecture 1 was helpful for me to work out some basic points. In the relationship between the symmetric bilinear form and the quadratic form, I came to realize that the "symmetric" part is essential, so that $\frac{1}{2}(B(x,y)+B(y,x))=B(x,y)$. I wrote out some of the diagrams to work out that, for example, why the category of quadratic vector spaces has an initial object (zero vector space with zero quadratic form) but not a terminal object (the unique map needs to respect norms! so the terminal object can't be too small, and if it's too big, then multiple maps will be possible). Nor are there products (I think because you can try to pair generators but then you can't pair their scalars, unfortunately).

Andrius Kulikauskas (Apr 13 2024 at 20:25):

Meanwhile, I watched a couple of very good video lectures by Greg Moore, Quantum Symmetries and the 10-Fold Ways, Part 1 and Part 2. He is very concrete which helps me greatly because I am an algebraic combinatorialist. I was glad to see his concrete form for Morita equivalence, which I will need to find. I will surely relisten to the lectures with my full attention. In Part 1, he mentioned Wigner's theorem and it's connection to CPT symmetry, which is crucial for me. And it means that this is all connected to the foundations of quantum mechanics, precisely what my friend John Harland is teaching me, so that I could discuss with him his research program, which we talked about in this video, Extradynamical Evolutionary Foundation for Physics.

Andrius Kulikauskas (Apr 15 2024 at 18:39):

I watched a couple of videos by Tobias Osborne about Wigner's theorem. He talks about it in the first video of his series on Symmetries and Quantum Mechanics And he gives a proof of it in the last video of his series on Advanced Quantum Theory The point for me is that, vaguely speaking, symmetric transformations of ray space are either linear unitary operators or antilinear antiunitary operators. So that gives some context for CPT symmetry where charge conjugation C is unitary, and parity P is unitary, and time reversal T is antiunitary.

Andrius Kulikauskas (Apr 19 2024 at 21:43):

I have been learning how to express Clifford algebras as matrix algebras. I have a PhD in Math, but still, I find it hard going, taking hours to figure out things that are surely trivial to others. I envy those who can just walk down a corridor and ask a colleague for a quick explanation. Which is part of the reason I write here.

Some of the challenges are notation. For example, in José Figueroa-O'Farrill's notes, in his Chapter 2, $Cl(1,0)=\mathbb{C}$, which is to say, the left generators square to $-1$ and the right generators square to $+1$. But in his Chapter 1, when he writes $n=r+s+t$, it is the s that counts the generators that square to $+1$, and only then comes the t that counts the generators that square to $-1$, and so he writes of the Clifford algebra $Cl(s,t)$, which is the opposite as in Chapter 2. In Theorem 2.3 on the recursion relations, he writes of a quadratic vector space that is isomorphic to $\mathbb{R}^{s,t}$. I got stumped and after many hours I started to realize, for example, the subtle distinctions between the quadratic vector space and the Clifford algebra we seek. Or the input variables and the output variables in the homomorphism. I was ultimately stumped by his expression $-Q(v)$.

Andrius Kulikauskas (Apr 19 2024 at 21:46):

After several hours I switched to Husemoller's book and his Theorem 5.6 in Chapter 12. He clearly distinguishes between the $e_1,\dots e_k$ that generate $C_k$ and square to $-1$ and the $e'_1,\dots e'_k$ that generate $C'_k$ and square to $+1$. It all made much more sense.

Andrius Kulikauskas (Apr 19 2024 at 21:52):

Then I read Husemoller's Proposition 6.4 because I am trying to understand the irreducible modules over $F(n)$ and $F(n)\oplus F(n)$. I think in terms of matrices and representations (of groups!) and am just starting to think in terms of algebra representations. I have never got into modules and ring theory. I became confused what $F(n)$ meant, whether a column or a matrix. Then I read on Math Stack Exchange about simple $M_n(D)$ modules. The third answer helped me. I was confused to think that each column is a module. But finally it dawned on me that they were all isomorphic, and from that point of view, there was only one irreducible module.

Andrius Kulikauskas (Apr 19 2024 at 22:08):

Another thing that was curious to learn from Wikipedia was that direct sum means different things for groups, for rings, for algebras, and so I had to stop and wonder what it meant in my case, for example, $\mathbb{R}\oplus\mathbb{R}$ or $\mathbb{H}\oplus\mathbb{H}$. In particular, the article on Direct Sum of Modules explained that in the case of algebras such as $\mathbb{R}\oplus\mathbb{R}$, a direct sum means what category theory calls a product.

Andrius Kulikauskas (Apr 19 2024 at 22:13):

Since then I have been working to variously understand $\mathbb{R}\oplus\mathbb{R}$ as relevant for me. Here are my notes on these split-complex numbers. After many hours, I am sorting out the relationship between $Cl_{1,0}\cong \mathbb{R}\oplus\mathbb{R}$, where the Clifford algebra has identity $1$ and the direct sum has identity $(1,1)$. For quite some time I was wondering why the Clifford algebra yielded one isomorphic module, whereas the direct sum yielded two modules for either half. Then I realized that the one must not be simple, it must break down into two, like those two. These are the simple things I think about!

But I have figured out how to write out their two irreducible modules and how they come together for one irreducible modules. I will write that out and make a Math4Wisdom video.

Andrius Kulikauskas (Apr 19 2024 at 22:15):

My question, if anybody would help me here, is how do I know what the simple modules are, in this case? What is the argument or technique to establish that? I suppose it is straightforward and I may well study the module theory from scratch. But I welcome your help to save me these hours, at least for now!

John Baez (Apr 20 2024 at 18:13):

The two simple modules of $\mathbb{R} \oplus \mathbb{R}$ are these.

1) In the first, $\mathbb{R} \oplus \mathbb{R}$ acts on $\mathbb{R}$ as follows:

$(a,b)(c) = ac$

2) In the second, $\mathbb{R} \oplus \mathbb{R}$ acts on $\mathbb{R}$ as follows:

$(a,b)(c) = bc$

Andrius Kulikauskas (Apr 20 2024 at 20:01):

@John_Baez Thank you! I appreciate this very much.

This shows existence. Now I will think how to show that there are no other simple modules M. I have an idea how to do that by using these two simple modules, and creating maps with them, and showing that any other module, if it was different, would have such a submodule, thus not be simple.

I am making progress on my notes for my video on Split complex numbers.

I still want to think in what sense this all carries over for $\mathbb{H}\oplus\mathbb{H}$. And then I can think about the $\mathbb{Z}_2$ graded representations and try to calculate the groups $0$ and $\mathbb{Z}$ that show up in Bott periodicity. That is why I care about this $\mathbb{R}\oplus\mathbb{R}$. Likewise, I can think how this carries over to $M_8(\mathbb{R})\oplus M_8(\mathbb{R})$.

Andrius Kulikauskas (May 07 2024 at 15:57):

I am growing in my understanding of Bott periodicity in its various forms. I am especially interested in the embeddings of Lie groups

$O(16)\supset U(8)\supset Sp(4)\supset Sp(2)\times Sp(2)\supset Sp(2)\supset U(2)\supset O(2)\supset O(1)\times O(1)\supset O(1)$

as detailed in the paper by Michael Stone, Ching-Kai Chiu, Abhishek Roy. Symmetries, Dimensions, and Topological Insulators: the mechanism behind the face of the Bott clock. They make it very concrete.

What impresses me is how they describe this embedding by applying, one by one, any set of mutually anticommuting orthogonal complex structures $J_1, J_2, J_3,\dots$ This gave me the idea that I've been looking for to explain how Bott periodicity might model the eight-cycle of conceptual frameworks which I have documented, however tentatively.

My idea is to consider how to go back up from an embedded Lie group to the starting point, $O(16)$. How does one assemble the orthogonal complex structures to fill that gap? Each orthogonal complex structure would be a perspective and their assembly would be the cognitive framework which describes the available shifts in perspective. I think of these frameworks as "divisions of everything". The mutual anticommuting I think implies, in a sense, that there is no overlap, just as there isn't in a framework where the perspectives divide up everything.

The orthogonal complex structures are symmetry-breakings and so considering the reverse direction is removing structure and restoring symmetry. Perhaps that can be phrased as an adjunction.

I am now investigating this concretely. To go from $U(8)$ to $O(16)$ there is basically only one way and it is trivial. So that is what I call the onesome, the division of everything into a single perspective.

Next, to go from $Sp(4)$ to $O(16)$, then we take two steps, adding an orthogonal complex structure and then another. I suspect that is straightforward enough. And it would match what I call the twosome, the division of everything into two perspectives, where one perspective ("opposites coexist" as in free will) is followed by another perspective ("all is the same" as with fate).

What I am hoping is that it gets interesting going from $Sp(2)\times Sp(2)$ to $O(16)$. Suppose there were three ways to proceed, I imagine, starting with the left $Sp(2)$ or the right one or perhaps a diagonal one. Then suppose that all three ended up getting used, A, B, C, and suppose it went in a particular order, A to B to C to A. Then I would get the learning three-cycle ("take a stand", "follow through", "reflect") that I am looking for, where you can start at any point and then proceed cyclically. Similarly, I have this fantastic hope that in the further cases, the framework I seek to model do assemble themselves as we restore the original symmetry of $O(16)$ or $O(16r)$ or $O(\infty)$ if you prefer.

This may well fail but I am inspired to think that I am arriving at the mathematical concepts with which I might best model what I am trying to say about perspectives and frameworks. And if what I am saying is coherent, then there will be a mathematical model. Or otherwise I may understand, where are my observations failing me, as regards these frameworks.

Andrius Kulikauskas (May 30 2024 at 12:08):

I have been studying the helpful paper by Michael Stone, Ching-Kai Chiu, Abhishek Roy. Symmetries, Dimensions, and Topological Insulators: the mechanism behind the face of the Bott clock., notably the section on the Lie group embeddings. More and more, it seems like it can model what I want.

Given the orthogonal group $O(16)$ and mutually anticommuting linear complex structures $J_1,J_2,\dots J_k$, they are considering the subgroup which commutes with all of the $J_i$. A general theme is that $J_1$ commutes with rotations but not with rotoreflections. Thus it preserves rotations but rejects rotoreflections, which is to say, rotations combined with a reflection. For me, the mathematical reflection models a mental reflection.

In my understanding, a single $J_1$ models a perspective. Two mutually anticommuting linear complex structures define a quaternionic structure, where $J_2$ is antilinear, thus models reflection. Together they model a perspective on a perspective.

Three mutually anticommuting linear complex structures define a split-biquaternionic structure, which relates two quaternionic structures V_+ and V_-, so to speak, two minds. The operator $K=J_1J_2J_3$ acts as $Kv=v$ on $V_+$ and $Kv=-v$ on $V_-$. For me, $J_1,J_2,J_3$ model a perspective upon a perspective on a perspective.

In this way, I can distinguish three minds:

• the Unconscious, adding a single perspective $J_1$
• the Conscious, adding a perspective $J_2$ on a perspective $J_1$
• Consciousness, adding a perspective $J_3$ upon a perspective $J_2$ on a perspective $J_1$

I have a lot more to understand in this section of the paper. I am trying to understand the entire chain of embeddings.

Todd Trimble (May 30 2024 at 12:52):

Somewhere there must be a community of people who would be passionately interested in following your speculations on Philosophy of Mind, although I'm not confident you will find them here.

I'm only going to say this quickly, because I'd guess you'd find it interesting, but I plan not to continue the discussion past this point (because I will quickly get out of my depth in a way I'd be uncomfortable discussing in public).

Todd Trimble (May 30 2024 at 12:52):

You may know that Wolfgang Pauli was in analysis with a Jungian psychologist, in the early 1930's (with Jung keeping a close watch over the case); he was in some mental distress before the analysis and may have thought he was going insane. It's probably not as well known that the analysis of Pauli's dreams, which Pauli worked very hard on with his analyst, became essentially part II of Jung's Psychology and Alchemy. The course of the analysis more or less concluded with a spontaneous vision (or perhaps it was a very vivid dream) Pauli had, of a kind of 4-dimensional "World Clock", which Pauli in his conscious reflections subsequently experienced as a kind of healing synthesis or culmination of the analysis. I imagine that for Pauli, this was connected in some way with quaternions and Pauli matrices and so forth and so on.

Andrius Kulikauskas (May 30 2024 at 13:42):

@Todd Trimble Thank you. I knew nothing about this. I will share this with my group, keep it in mind and try to learn more.

I found the book Atom and archetype : the Pauli/Jung letters, 1932-1958 on Archive.org Thank you again.

Andrius Kulikauskas (Jun 11 2024 at 20:43):

I feel like I am making progress, learning to think in terms of products of mutually anticommuting linear complex structures such as $J_1J_2$ and $J_1J_2J_3$ which are organizing the structures. I am studying with my friend John Harland, who has a PhD in functional analysis and a passion for quantum physics. I uploaded my video presentation to him, "Mental Perspectives as Linear Complex Structures. Bott Periodicity Update." https://www.youtube.com/watch?v=9xucCcfj_lY

The concreteness of this approach and my long term exposure is helping me feel more at home with the many flavors of Bott periodicity, which I am presently relating and drawing from. I now see that Max Karoubi connects these flavors with his paper "Bott periodicity in topological, algebraic and Hermitian K-theory". Category theory plays an important role here, it seems, towards presenting a general theory of periodicity, as in Karoubi's conclusion, noting work by Morel, Voevodsky, Quillen and Hornbostel.

Morel and Voevodsky [35] have proved that algebraic K-theory is representable by an infinite Grassmannian in the unstable motivic homotopy category. Moreover, Voevodsky [50] has shown that this, together with Quillen’s computation of the K-theory of the projective line, implies that algebraic K-theory is representable in the stable motivic homotopy category by a motivic (2, 1)-periodic Ω-spectrum.

Andrius Kulikauskas (Jul 13 2024 at 11:29):

@Todd Trimble I have revisited your helpful notes on super division algebras. I am understanding much more of it, perhaps everything after the category theory preamble :smile: Thank you for publishing these notes.

In particular, the three minds that I seek to model can be identified with different aspects of a super division algebra $A+Ae$

• the even element $A$ is the Unconscious, unreflected experience
• the odd element $Ae$ is the Conscious, unreflected experience reflected by $e
• the automorphism $a\rightarrow a'$ defined by $ae=ea'$ indicates Consciousness, which relates the other two minds

It's interesting to think about the automorphism. It means that there is a map from $a\rightarrow ae=ea'$. But also it means that the distinction of $a$ and $a'$ (opposites coexist, as with free will) is identified with a nondistinction ($ae=ea'$) (all is the same, as with fate). Anyways, it is very meaningful for me that every one of the Clifford algebras can be thought of as split into even and odd parts in this meaningful way. In the cases of $\mathbb{R}$ and $\mathbb{H}$ (up to Morita equivalence of superalgebras) the odd part is zero, if I understand correctly.

I am just realizing that here it is essential to consider Morita equivalence of superalgebras (rather than just algebras). For in the case of algebras we have $Cl(0,2)\cong H \cong_{Morita} Cl(0,4)$ which had made me think that there must be something more to it. So the emphasis on superalgebras fits very well with my attempt to model these three minds.

The part about the repeated tensoring of $Cl(0,1)$ is very interesting for me to think about, especially because of how it relates to applying linear complex structure $J$, which yields the symmetric spaces. It's very interesting how the quaternions get expressed in ever new ways, first as $Cl(0,2)$ using both even and odd parts, then as the even part of $Cl(0,3)$, then identifying that even part with the whole (per Morita equivalence) because the odd part is basically the real $2x2$ matrices which are Morita equivalent to $R$. So I need to study this concretely but I am very interested.

Todd Trimble (Jul 15 2024 at 16:00):

Glad you found them helpful -- to me, now, they read a little terse. But they hold a special place for me, as a reminder of how people like @John Baez have helped me. The note was a reaction to a problem he floated before me, by way of his striking up a mathematical conversation, around the time that I was considering returning to mathematics after some years of being away, thinking I was done with it. The encouragement I got from John, and from Eugenia Cheng and Peter May when I reached out to them around the same time, were instrumental in bringing me back.

And actually, it was John who published that note on his website! All I did there was write him an email. So the thanks really go to him.

John Baez (Jul 15 2024 at 18:46):

Thanks, Todd! I'm sure glad you're "back".

Andrius Kulikauskas (Jul 24 2024 at 19:18):

@Todd Trimble Thank you for recounting how @John Baez, Eugenia Cheng and Peter May encouraged you and embraced you. I find that heartening. Thank you also to John for publishing your email!

Todd, I found more of your ideas that John published and also your page of ideas at nLab.

I am curious, why and how did you get interested in super division algebras and Bott periodicity?

Todd Trimble (Jul 25 2024 at 02:26):

I don't know exactly when as a graduate student at Rutgers I was first exposed to Bott periodicity, but it was sometime then that I did a first reading of Milnor's Morse Theory, which was definitely inspiring. I was exposed to a certain amount of algebraic topology anyway as a grad student, being around people like Shaneson and Landweber and Weibel, so naturally it would have been "in the air". The business with super division algebras was due to John's conversation opener, and the fact that I've been known to enjoy a challenge on occasion. It just seemed like an inviting problem at the time. I have a history of getting temporarily involved with some topic but not particularly following up on it over the years.

Andrius Kulikauskas (Jul 31 2024 at 20:13):

@Todd Trimble Thank you for that context.

I am making progress from various directions. I feel encouraged that this is relating to the "three minds" I am trying to model. I have purchased the domain TheoryTranslator.com and there I have a table of 33 examples so far, how various thinkers refer to these three levels of awareness. Your exposition of the super division algebras is very interesting to me because it suggests a pair of minds, where the second is a reflection of the first.

The paper by Stone, Chiu, Roy got me thinking of a process whereby the application of a linear complex structure $J_k$ discards the "reflected" knowledge from the "unreflected" raw experience. I submitted to the Models of Consciousness Conference an abstract for a talk Modeling Introspected Contexts With Mutually Anticommuting Linear Complex Structures but that was rejected. But I am hopeful that if I work on this for a year then I can get the model I want and present it as a paper.

I found very helpful J.-H. Eschenburg's Lecture Notes on Symmetric Spaces and I will go back to that. Currently, I am studying Gregory Moore's paper Quantum Symmetries and K-Theory. I am trying to understand how he connects the C, P, T symmetries with the super division algebras and also how that relates to the application of the linear complex structures and the resulting symmetric spaces.

I am very happy that my friend John Harland, who has a PhD in functional analysis, is meeting with me twice a week by zoom to work on this. He has been studying the embeddings $O(16r)\supset U(8r)\supset Sp(4r)$. He has talked to me about the geometry of preserving the real, complex and quaternionic norms, and also of complex multiplication and quaternionic multiplication. This Math StackExchange post was helpful. It's great to have a friend to work with.

Andrius Kulikauskas (Nov 01 2024 at 19:17):

@Todd Trimble at Rutgers did you know Gregory Moore? These last few months I have been studying his very helpful paper Quantum Symmetries and K-Theory, notably section 2.6 Realizing the 10 classes using the CT groups. I got stuck trying to calculate the $(\phi ,\chi )$ representations of these $CT$ groups. I was confused how the complex vector spaces related with the real vector spaces. But I found his paper Quantum Symmetry and Compatible Hamiltonians and am studying 13.3 Real Clifford algebras and Clifford modules of low dimension. I got stuck again with calculating the graded tensor product at (13.125) and (13.131) as to where the fact that $e_i^2=1$ and $e_i^2=-1$ comes in. Today I found his Linear Algebra User's Manual with section 23. $\mathbb{Z}_2$ -graded, or super-, linear algebra.

I will write out my question and maybe somebody can help. In any event, it is helpful for me to write it out.

Andrius Kulikauskas (Nov 01 2024 at 19:58):

Consider the real Clifford algebra $Cl_{+1}$ with a single generator $e$ for which $e^2=1$. As a graded algebra, it has a unique irreducible representation $\tilde{\eta}=\mathbb{R}^{1|1}$, which is to say, one real dimension even, one real dimension odd, where the generator $e$ is mapped

$\rho (e)=\begin{pmatrix}0 & 1 \\1 & 0 \\ \end{pmatrix}$

So far, so good. Now consider the real Clifford algebra $Cl_{+2}$ with two generators $e_1, e_2$ where $e_1^2=1, e_2^2=1$. As a graded algebra, we want to construct the graded tensor product $\tilde{\eta}^2:=\tilde{\eta}\hat{\otimes}\tilde{\eta}$ where the vector space is $\mathbb{R}^{2|2}$. For the first factor, we set even and odd basis elements $v_0,v_1$ with $\rho(e)v_0=v_1, \rho(e)v_1=v_0$ and similarly for the second factor we set even and odd basis elements $w_0,w_1$ with $\rho(e)w_0=w_1, \rho(e)w_1=w_0$.

We know (12.23) that the graded tensor product includes a possible sign. If $T:A\rightarrow B$ and $T':A'\rightarrow B'$ are linear operators on super vector spaces then we can define the $\mathbb{Z}_2$ graded tensor product

$(T\otimes T')(v\otimes v')=(-1)^{\mathrm{deg}(T')\mathrm{deg}(v)}T(v)\otimes T'(v')$
where only the $T'$ and $v$ are involved in calculating the sign.

Andrius Kulikauskas (Nov 01 2024 at 20:00):

We define $\rho(e_1)=\rho(e)\hat{\otimes}1$. I try to calculate that. I note that $T'=1$ is even so $\mathrm{deg}(T')=0$ and $\mathrm{deg}(v_0)=0$ but $\mathrm{deg}(v_1)=1$. So I get:
$\rho(e)\hat{\otimes}1(v_0\hat{\otimes}w_0)=+v_1\hat{\otimes}w_0$
$\rho(e)\hat{\otimes}1(v_1\hat{\otimes}w_1)=-v_0\hat{\otimes}w_1$
$\rho(e)\hat{\otimes}1(v_0\hat{\otimes}w_1)=+v_1\hat{\otimes}w_1$
$\rho(e)\hat{\otimes}1(v_1\hat{\otimes}w_0)=-v_0\hat{\otimes}w_0$

We list the basis in this order: $v_0\hat{\otimes}w_0, v_1\hat{\otimes}w_1, v_0\hat{\otimes}w_1, v_1\hat{\otimes}w_0$

My calculation gives the matrix

$\begin{pmatrix}0& 0 & 0 & 1 \\ 0& 0 & -1 & 0 \\ 0& 1 & 0 & 0 \\ -1& 0 & 0 & 0 \\ \end{pmatrix}$

but the matrix given in the paper is

$\begin{pmatrix}0& 0 & 0 & 1 \\ 0& 0 & 1 & 0 \\ 0& 1 & 0 & 0 \\ 1& 0 & 0 & 0 \\ \end{pmatrix}$

What have I done wrong?

Andrius Kulikauskas (Nov 01 2024 at 20:11):

I think I figured it out. When I calculate the sign, I need to take the product $\textrm{deg}(T')\textrm{deg}(v)$. So that is $0$ in three out of four logical cases. It is only nonzero if both $\textrm{deg}(T')$ and $\textrm{deg}(v)$ are odd. But that can't happen because $\textrm{deg}(T')=0$.

I am glad I asked and wrote this out. Thank you!

Todd Trimble (Nov 06 2024 at 21:49):

at Rutgers did you know Gregory Moore?

Sorry, I was gone all last week with very limited internet access, so just getting back to things now. No, I didn't know him then nor do I now, but I've seen his name occasionally.

The term "CT group" is not familiar to me, so I'm glad you figured out your problem on your own!

John Baez (Nov 07 2024 at 01:52):

Gregory Moore and I took the same QFT class when I was a grad student at MIT.

Andrius Kulikauskas (Nov 07 2024 at 10:02):

@Todd Trimble @John Baez I am very glad to hear from you! I wrote to Gregory Moore on October 25th but I didn't hear from him. In particular, I asked: Please, could you advise, is there a community where I could participate with my questions so that I could master the details that I care about?

For now, I will share my questions and efforts here at this Zulipchat, and if I am successful, then in the context of Bott periodicity, this may lead to insight about Morita equivalence, which is to say, category theory.

At Math Stack Exchange, I ended up answering my own question, What are the group extensions of $\mathbb{Z}_2$ by $U(1)$? This is important for defining $CT$-groups, which in this case are $\textrm{Pin}_+(2)$ and $\textrm{Pin}_-(2)$. But ultimately, for the understanding that I seek, this may mostly be irrelevant.

What it boils down to, I think, is that $a^4=1$ has two solutions for $a^2$, namely $a^2=1$ and $a^2=-1$. So this yields two models: $\mathbb{Z}_2$ and $\mathbb{Z}_4$. In modeling cognition, $\mathbb{Z}_2$ has two states - the world (the identity) and the unconscious (raw, direct experience). Whereas $\mathbb{Z}_4$ has an additional two intervening states that observe (consciously) the movement from world to unconscious (as perception) and from unconscious to world (as action). Which is to say, an operator $T^2=1$ models unconscious raw experience and an operator $T^2=-1$ models conscious reflected experience.

@Todd Trimble I have started a Theory Translator where, so far, I have presented 179 examples of this distinction between raw experience (answers) and reflected experience (questions) and how they are related (investigations). This includes a reference to your note on classifying real super division algebras. Another example that comes up in quantum physics concerns states (answers), observables (questions) and measurements (investigations).

Anyways, this is what I think I understand. I will write about what I don't understand but would like to.

Andrius Kulikauskas (Nov 07 2024 at 10:32):

I need to be able to calculate the $(\phi, \chi)$ representations of a bigraded group $G$ which Gregory Moore defines in his paper Quantum Symmetries and K-Theory, Section 2.1 Gapped systems and the notion of phases, Definition 2.3

Suppose $G$ is a bigraded group, that is, it has a homomorphism $G \rightarrow \mathbb{Z}_2 × \mathbb{Z}_2$ or, what is the same thing, a pair of homomorphisms $(\phi, \chi)$ from $G$ to $\mathbb{Z}_2$ . Then we define a $(\phi, \chi)$-representation of $G$ to be a complex $\mathbb{Z}_2$ -graded vector space $V = V_0\oplus V_1$ and a
homomorphism $ρ : G → \textrm{End}(V_\mathbb{R} )$ such that $\rho (g)$ is $\mathbb{C}$ − linear if $\phi (g) = +1$, $\rho (g)$ is $\mathbb{C}$ − anti-linear if $φ(g) = -1$, $\rho (g)$ is even if $\chi(g) = +1$, $\rho (g)$ is odd if $\chi(g) = -1$.

Which is to say, every matrix $\rho(g)$ is either even linear, odd linear, even anti-linear or odd anti-linear, and that is controlled by the two binary switches $\phi(g)$ and $\chi(g)$.

Andrius Kulikauskas (Nov 07 2024 at 12:09):

What is $V_\mathbb{R}$ ? It is defined in section 1.6.2 Real structure on a complex vector space.

Suppose $V$ is a complex vector space. Then a real structure on $V$ is an antilinear map $C : V \rightarrow V$ such that $C^2 = +1$.

If $C$ is a real structure on a complex vector space $V$ then we can define real vectors to be those such that $C(v) = v$. Let us call the set of such real vectors $V_+$ . This set is a real vector space, but it is not a complex vector space, because $C$ is antilinear. Indeed, if $C(v) = +v$ then $C(iv) = −iv$. If we let $V_−$ be the imaginary vectors, for which $C(v) = −v$ then we claim $V_\mathbb{R} = V_+ \oplus V_−$. Multiplication by $i$ defines an isomorphism of real vector spaces: $V_+ \cong V_−$. Thus we have $\textrm{dim}_\mathbb{R} V_+ = \textrm{dim}_\mathbb{C} V$.

Andrius Kulikauskas (Nov 07 2024 at 12:34):

I am lost untangling this. I wonder whether $V^0=V_+$ and$$V^1=V_-$$ ?

According to section 2.2.1 Super vector spaces:

A $\mathbb{Z}_2$ -graded vector space over a field $κ$ is a vector space over $κ$ which, moreover, is written as a direct sum $V = V^0 \oplus V^1$. The vector spaces $V^0 , V^1$ are called the even and the odd subspaces, respectively. We may think of these as eigenspaces of a “parity operator” $P_V$ which satisfies $P_V^2 = 1$ and is $+1$ on $V^0$ and $−1$ on $V^1$ . If $V^0$ and $V^1$ are finite dimensional, of dimensions $m, n$ respectively we say the super-vector space has graded dimension or superdimension $(m|n)$.

So here I feel confused. It seems that $V^0, V^1$ are complex vector spaces whereas $V_+$ and $V_-$ are real vector spaces. So they can't be equal. I appreciate help sorting this out.

Andrius Kulikauskas (Nov 08 2024 at 10:00):

Gregory Moore has a later book, Linear Algebra User's Manual, where he explains in section 4.3 what $V_\mathbb{R}$ is:

any vector space over $κ = \mathbb{C}$ is, a fortiori also a vector space over $κ = \mathbb{R}$. Let us call it $V_\mathbb{R}$. It is the same set, but now the vector space structure on this Abelian group is just defined by the action of real scalars. Then we will see that: $\textrm{dim}_\mathbb{R} V = 2\;\textrm{dim}_\mathbb{C} V$

Then in section 9.2 he explains that a real structure on a complex vector space produces a real vector space $V_+$ of half the real dimension of $\textrm{dim}_\mathbb{R} V$ and, in fact, $\textrm{dim}_\mathbb{R} V=V_+\oplus V_-$.

Now I need to understand what is meant by $\textrm{End}V_\mathbb{R}$. This goes back to my question of whether $V_+=V^0, V_-=V^1$?

David Michael Roberts (Nov 08 2024 at 13:02):

$\dim_\mathbb{R} V = V_+\oplus V_-$ is a type error: the LHS is a number, the RHS is a vector space. Perhaps you meant $V_\mathbb{R} = V_+\oplus V_-$?

$V_{\mathbb{R}}$ is literally the same set as $V$, but where you forget that you can multiply by anything other than real numbers. If everything is finite-dimensional, then picking a basis so that $V\simeq \mathbb{C}^n$, then $V_\mathbb{R} = (\mathbb{R}^2)^n$. A real structure on a complex vector space is like singling out the first copy of $\mathbb{R}$ in all those $$\mathbb{R}^2$$s, and collecting all those into the $V_+$. But without picking a basis first.

Endomorphisms of $V_\mathbb{R}$ are real-linear endomorphisms, which means that they only have to respect scalar multiplications by real numbers, not by all complex numbers, so there are few conditions, which lead to more maps satisfying those conditions.

Todd Trimble (Nov 08 2024 at 14:02):

David, you made the same type error (second paragraph at the beginning) ;-)

Andrius Kulikauskas (Nov 08 2024 at 16:24):

@David Michael Roberts (The Higher Geometer!) Thank you for your helpful explanations. Thank you, also, for spotting my error, which I have corrected. I will share more about the $(\phi, \chi)$-representations that I am trying to understand. Hi @Todd Trimble

Andrius Kulikauskas (Nov 08 2024 at 19:38):

@David Michael Roberts @Todd Trimble I am trying to understand this theorem in Section 2.6 of Gregory Moore's Quantum Symmetries and K-Theory.

CTGroups-Figure11.png

Theorem There is a one-one correspondence, given in the table above, between the ten CT groups and the ten real super-division algebras (equivalently, the 10 Morita classes of the real and complex Clifford algebras) such that there is an equivalence of categories between the (φ, χ)-representations of the CT group and the graded representations of the corresponding Clifford algebra.

Basically, we're associating $\mathbb{Z}_2\times\mathbb{Z}_2$-graded group representations with $\mathbb{Z}_2$-graded Clifford algebra representations. I want to understand the details but I am still struggling to understand the definitions. Anyways, let me describe the simplest case.

Consider the group extension $1\rightarrow U(1)\rightarrow G\rightarrow U\rightarrow 1$ when $U=1$, the trivial group. Then $G\cong U(1)$, the circle group, is the $CT$ group. Moore writes:

First, consider the subgroup $U = \{1\}$. A $(φ, χ)$ representation $W$ is simply a $\mathbb{Z}_2$ -graded complex vector space, so $V = W$ is a graded $Cℓ_0$ -module.

What apparently he is saying is that the representation space is given ($\mathbb{Z}_2$ -graded complex vector space $V=V^0\oplus V^1$) and the representation $\rho :G\rightarrow\textrm{End}(V_\mathbb{R})$ is obvious. But I can't figure out what it is!

According to Circle group: Representations, the irreducible real representations of the circle group are the trivial representation and also $\phi_n, n\in\mathbb{Z}, n> 0$ given by

$\rho_n(e^{i\theta})=\begin{pmatrix} \cos n\theta & -\sin n\theta \\ \sin n\theta & \cos n\theta \\ \end{pmatrix} = \begin{pmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \\ \end{pmatrix}^n$

These matrices are all complex linear. Now, if they had to be either even or odd, then by the grading they would all be prohibited, for they are neither even nor odd. We would be left with the $1$-dimensional trivial representation $\rho(g)=(1)$ for all $g\in G$. So I suppose that is what must be meant. This means that $V_\mathbb{R}$ should be understood as $V_\mathbb{R}=V_+\oplus V_-$ where $V_+=V^0_\mathbb{R}$ and $V_-=V^1_\mathbb{R}$.

Given that we have the $\mathbb{Z}_2$-graded complex vector space $V_\mathbb{R}=V_+\oplus V_-$, we have two irreducible $(\phi,\chi)$-representations of $G$, namely:

$\begin{pmatrix} 1 & 0 \\ 0 & 0 \\ \end{pmatrix}$ and $\begin{pmatrix} 0 & 0 \\ 0 & 1 \\ \end{pmatrix}$

This accords with the graded representations of the complex Clifford algebra $\mathbb{C}\ell_0$ he describes in Quantum Symmetries and Compatible Hamiltonians in Section 13.2.1 Structure of the (graded and ungraded) algebras and modules. He writes:

Of course $\mathbb{C}\ell_0\cong\mathbb{C}$ is purely even. Nevertheless, as a superalgebra it has two inequivalent irreducible graded modules $M_0^+ \cong C^{1|0}$ and $M_0^−\cong C^{0|1}$.

Super-modules are defined in QS&CH Section 12.4 Modules over superalgebras.

A super-module M over a super-algebra A (where A is itself a superalgebra over a field κ) is a supervector space M over κ together with a κ-linear map A × M → M defining a left-action or a right-action.

So the map is multiplication of complex numbers as in $\mathbb{C}\times C^{1|0}\rightarrow C^{1|0}$

So I suspect I am on track. Going back, in the opposite direction, he says in QS&CH, Section 16. Realizing the 10 classes using the CT groups

For $\mathbb{C}\ell_0$ we take $W = V$

So the two situations are very similar. But for the group we have a trivial representation, whereas for the Clifford algebra, we have multiplication by $c\in\mathbb{C}$. At least, that's what I think.

Andrius Kulikauskas (Nov 08 2024 at 20:08):

My intuition, which I doubted for a while, but now return to, is that this is all about breaking a $2\times 2$ matrix into its 4 components.

$\begin{pmatrix} 1 & 0 \\ 0 & 1\\ \end{pmatrix}$ is even complex linear, as with identity

$\begin{pmatrix} 1 & 0 \\ 0 & -1\\ \end{pmatrix}$ is even complex antilinear, as with time reversal $\bar{T}$

$\begin{pmatrix} 0 & 1 \\ 1 & 0 \\ \end{pmatrix}$ is odd complex antilinear, as with charge conjugation $\bar{C}$

$\begin{pmatrix} 0 & -1 \\ 1 & 0\\ \end{pmatrix}$ is odd complex linear as with the product, parity $\bar{S}=\bar{C}\bar{T}$

If this is on track, then it starts to provide the metaphysical intuition that I am seeking. Time reversal fixes position and flips momentum, which is to say, it distinguishes an absolute frame (for position) and a relative frame (for momentum). Then charge conjugation swaps absolute frame and relative frame, perhaps thereby defining a perspective. My goal is to get metaphysical intuition on these perspectives and then relate that back to the construction described by Stone, Chiu, Roy in terms of Lie group embeddings and $0,1,2,\cdots 7$ mutually anticommuting linear complex structures. I think the later may encode the "divisions of everything" that I have documented, the ways of carving up mental space into perspectives, much like a chain complex or exact sequence carve up conceptual space with homomorphisms.

David Michael Roberts (Nov 09 2024 at 02:58):

Todd Trimble said:

David, you made the same type error (second paragraph at the beginning) ;-)

That's what comes of doing cut-and paste to avoid typing maths... :-/

John Baez (Nov 09 2024 at 16:09):

While we're picking on David's typography I'll point out that $$\mathbb{R}^2$$s failed to render properly because the 's' was right next to the dollar sign. This is an annoying bug/feature in the TeX rendering here: you can put a dash or comma or period next to dollar sign and the TeX works fine, like this: $\mathbb{R}^2$, but a letter will kill it. So you have to put in a little space and write $\mathbb{R}^2$ s.

Andrius Kulikauskas (Nov 15 2024 at 21:32):

I have published a YouTube video of a presentation that I gave to John Harland and Thomas Gajdosik about my research these last few months: Bott Periodicity for Clifford Algebra Maniacs.

Here are my slides.

Andrius Kulikauskas (Nov 15 2024 at 21:35):

@Todd Trimble I talk about your classification of super division algebras. I include a slide where I relate that to the Chomsky hierarchy of automata.
BC11-ChomskyHierarchy.png

Andrius Kulikauskas (Nov 23 2024 at 16:32):

Fabrice Pfaff has created a wonderful online tool for exploring Clifford algebras and their matrix representations. He's also published a video where he explains it.

Andrius Kulikauskas (Nov 30 2024 at 21:36):

Today I saw a great exposition of the Krebs cycle (the citric acid cycle) which fuels all living cells. The elegant picture they drew presents it as an eight-cycle of steps. That fits with my hunch that it is organized by Bott periodicity. The idea would be that Bott periodicity is a template for order which, over a couple of billion years, brought order to organic chemistry in the form of metabolic patterns and ultimately living cells. I think this is not so far fetched given that Bott periodicity cycles through various ensembles of random matrices.

Andrius Kulikauskas (Nov 30 2024 at 21:40):

Nanorooms. Krebs Cycle. Schematic
Nanorooms. Krebs Cycle. Detailed

Morgan Rogers (he/him) (Dec 02 2024 at 17:10):

Nope, the Krebs cycle doesn't seem related to Bott periodicity to me, besides the superficial fact that they both realise the cyclic group of order 8. I've tried to tell you this before less patiently @Andrius Kulikauskas : numerology isn't mathematically or scientifically watertight. Just because two things have the same number of elements or even have the same group structure doesn't imply any deeper relationship between them.

Andrius Kulikauskas (Dec 02 2024 at 17:43):

@Morgan Rogers (he/him) Thank you for alerting me. I have deleted my posts on the Krebs cycle.

Morgan Rogers (he/him) (Dec 03 2024 at 09:26):

I appreciate it (although the Krebs cycle is an interesting subject, it was off topic). Fortunately there is plenty more to explore in the world of Clifford algebras :innocent: