The n-Category Café

thanks, sorry, my bad. then there’s Weil and Weyl…

but… could $\pi_0 O(1,3)$ interact in interesting ways with representation theory graded somehow by this ten-element monoid?

It certainly interacts in some way!
$$\pi_0 O(1,3) \cong \mathbb{Z}\!/\!2 \times \mathbb{Z}\!/\!2$$ is generated by time reversal T and space reversal, or ‘parity’ P. Physicists like to think of it as a subgroup of $\mathbb{Z}\!/\!2 \times \mathbb{Z}\!/\!2 \times \mathbb{Z}\!/\!2$ where the third generator, C, switches particles and antiparticles.

In physics this larger group often acts on a Hilbert space $\mathbf{H}$ in such a way that P acts as a unitary operator while C and T act as antiunitary operators. Sometimes $\mathbf{H}$ is $\mathbb{Z}\!/\! 2$-graded in such a way that ‘particles’ are even and ‘antiparticles’ are odd. Then C acts as an odd antiunitary operator while T acts as an even antiunitary operator.

In my talk slides I discuss a more general situation where we have an irreducible representation $\rho$ of a group on a $\mathbb{Z}\!/\! 2$-graded Hilbert space. This superficially has little to do with the story I’ve just been telling: yes there’s $\mathbb{Z}\!/\! 2$-graded Hilbert space, and a group, but we don’t necessarily have $O(3,1)$ around. Nonetheless, for any such representation, we have:

The Tenfold Way. The irreducible unitary representations $\rho$ of a $\Z/2$-graded group $G$ on a super Hilbert space $\mathbf{H}$ come in 10 types, based on the commutant: the real-linear operators that commute with $\rho(g)$ for all $G$.

In 9 of these types the commutant contains:

• an even antiunitary $T$ with either $T^2 = 1$, $T^2 = -1$, or no such $T$

and

• an odd antiunitary $C$ with either $C^2 = 1$,$C^2 = -1$, or no such $C$.

In the 10th type the commutant contains:

• no such $T$ or $C$, but an odd unitary $S$; we may assume $S^2 = 1$.

In short, even when $T$ and $C$ aren’t put in via the Lorentz group O(1,3), they magically appear (or not) on their own, having the properties we expect in the O(1,3) case!

Thanks!

I think someone has already looked into Jordan superalgebras.

Someone has even started an nLab page on them: Jordan superalgebra, where we learn that over an algebraically closed field of characteristic $0$

The only exceptional finite-dimensional example is the 10-dimensional Jordan superalgebra $K_10$.

That someone was David Corfield himself two and a half years ago:

https://nforum.ncatlab.org/discussion/11491/

Dear professor John Baez,

I’ve watched your lectures on the ten fold way organized by Cohl Furey, but I am really curious about the missing Octonions.

The sequence stops at H+H (wikipedia says these are split biquaternions instead of split quaternions), instead of O.

But if you keep track of the real minimal representation, from week 104:

“C0 is 1; C1 is 2; C2 is 4; C3 is 4; C4 is 8; C5 is 8; C6 is 8; C7 is 8; C8 is 16;

The sphere S0 admits 0 linearly independent vector fields. The sphere S1 admits 1 linearly independent vector fields. The sphere S3 admits 3 linearly independent vector fields. The sphere S7 admits 7 linearly independent vector fields. ” On https://math.ucr.edu/home/baez/octonions/node9.html

you said

“On First of all, as a smooth manifold $\KP^1$ is just a sphere with dimension equal to that of $\K$”

I can notice that there is enough space for you to make a 8 fold way to get Octonions somewhere by successive embedding of hopf fibrations, by picking up spheres related to protective algebras around the 8 path if you defined. I am not sure if that is possible.