# Arcadian Functor

occasional meanderings in physics' brave new world

Name:
Location: New Zealand

Marni D. Sheppeard

## Sunday, December 10, 2006

### M Theory Lesson 6

In the Higgs poll over at PF a whopping 50% of people are voting for no Higgs found in the next three years. Well, back to M Theory again.

Points take many guises. Eventually we will have the option of throwing them away altogether, using topos theory. After all, the question that really needs answering is, what is an observable? But things get pretty interesting before then. In Brannen's Clifford algebra one hunts out idempotents, which satisfy the simple projector relation tt=t. Similarly, one might use points in projective (twistor) space, described by the Jordan algebra of Michael Rios.

Are there other ways in which points are naturally associated to operators satisfying tt=t? Why yes, using categories in a fairly simple way, as follows. One can think of the objects of a category as the identity arrows on those objects. For example, a set being a discrete category which only has objects, only has identity arrows. Now let t be the target map, sending an arrow f to its target identity arrow tf. Then clearly tt=t. Similarly, one can discuss a source map s.

For those who are really keen on playing, remember that a topology t in the form of an arrow from Omega to Omega in an elementary topos also obeys the relation tt=t.

#### 10 Comments:

L. Riofrio said...

Quite a poll, it shows that people have lost faith in the "God Particle" Higgs. All this and LHC hasn't even been turned on. Your posts do give one faith in M-theory.

December 10, 2006 1:13 PM
Mahndisa S. Rigmaiden said...

12 10 06

Gee, when we learned that 1^2=1 and 0^2=0, that was just the tip of the iceberg wasn't it? Now I am upset with my elementary school math instructors for not explaining the deeper implications sooner;)

What is an operator that looks sorta like the roots of unity equation? U^N=U is that an idempotent operator?
hhehehehehehe

What about parity operators acting on an even wavefunction in 1-D=> pp|psi+>= |psi+>?

Oh I see that for even wavefunctions, pp|psi+>=|psi+>=p|psi+>
Because p|psi+>=|psi+>

Or to be more precise, does the + eigenvector of the parity operator represent the idempotent concept? Please explain the error in my thinking here...Thanks:)

Hmmm, it seems quite interesting that observables are correlated to idempotents in these frameworks. But from a gut feeling, when I see something like UU=U or even U^N=U, I think of trancendental equations e.g. free end boundary condition on a string. You end up getting results that must be taken as the intersection between quantity kn and tan(kn) or cot(kn), I hope I am not rambling here. But what I am hinting at is that that trancendental equation defines one quantity in terms of itself,which is what I see with idempotents.

Sorry for rambling, and I hope you will correct me if you find error in my logic.
Warmest Regards:)

December 11, 2006 2:17 PM
Mahndisa S. Rigmaiden said...

12 10 06

Gee Kea, you really got me to thinkin'!!!! I really like your category theoretic explanation of the last sentence. Sorry I forgot to mention that.

December 11, 2006 2:22 PM
Kea said...

Now I am upset with my elementary school math instructors

I know how you feel! We do end up with 'roots of unity' too, but Carl is way ahead of me on that one. I guess one can think of the p^n = 1 relation as an exponentiated nilpotency equation. Actually, there is a HARD mystery in Operad Theory related to your thoughts: marked points on the edge of a cylinder containing discs gives an important example of an operad. But the points can be rotated before gluing cylinders. The mystery is that nobody seems to understand the categorical structure behind this important (geometric) operad - the rotation is like a permutation of sources and targets. Now there's a problem for you, Mahndisa. This is something we need to understand.

December 11, 2006 2:40 PM
Kea said...

Re parity: still thinking about that.

December 11, 2006 2:41 PM
Kea said...

Sorry, I should have mentioned that there are half discs along the edges, which define boundary points.

December 11, 2006 2:51 PM
mitchell porter said...

elementary school math instructors

Curriculum reform now! Bring back Mathematics Made Difficult!

December 11, 2006 3:26 PM
Kea said...

Hi Mitchell

Nice to see you here! Yeah, let's teach the kids diagrammatic calculi. They like drawing, right?

December 11, 2006 3:44 PM
Kea said...

Heh, Carl, you tipped the no Higgs result over the 50% mark. You can't call yourself a minority any more.

December 11, 2006 3:47 PM
Mahndisa S. Rigmaiden said...

12 10 06

Hello Kea:
Thanks for your response. The combinatorial approach should be explored further to answer the question you pose about the boundary points on the cylinder. And what a challenge. I will think about that.

Regarding parity operations, if a wavefunction is even under the parity operation (restricting discussion to one dimension for simplicity) then we will always have:p|psi(x)>=+1|psi(x)>=|psi(x)> We have the ket multiplied by the identity element;) Therefore, we can take any power of the parity operator, operate on an even wavefunction and get that wavefunction back again. So this would be an nth order polynomial equation and an idempotent, and since P is unitary we can construct it as an exponential, so I see where the exponentiated nilpotency equation you evoked earlier comes in.

What is so interesting is that the P^2|arbitrary psi>=+1|psi> regardless of the evenness or oddness of |psi>. However, the same cannot be said of P^n acting on an arbitrary function. I know that for an even function the result makes sense, but for an odd function there is a phase factor that I am omitting;)

December 11, 2006 7:10 PM