Friday, January 29, 2016

The Grothendieck group construction

After "snowzilla" disrupted my priorities for about a week I am falling behind in my duties and today I only have time to present a small but essential topic. So let's talk about the Grothendick group construction which appears front and center in the categorical solution to the measurement problem. 

Alexander Grothendieck

There is no easier way to show this then to see it at work in the case of the integer numbers. Basically we introduce a Cartesian pairs of natural numbers and we call the left element positive integers, and the right element negative integers like this:

\(7\equiv (7,0)\)

\(-3\equiv (0,3)\)

Then we can add and subtract the numbers in the most natural way: \(7-3=7+(-3) = (7,0)+(0,3) = (7,3)\)

But this is not what we want: we want 7-3 to be 4 and not (7,3). How to do that? 4 is (4,0) and there is no problem if (7,3) is identical with (4,0). In other words, we need to introduce an equivalence relationship where distinct pairs represent the same thing.

How can we justify (7,3)=(4,0)? One way is to observe that if we subtract the right pair from the left one we get identical answers in both slots: (7-4, 3-0) = (3,3) but we are not allowed to use subtraction because we need to rely only on the operation available from the original commutative monoid. The answer is that we need to add the elements like this:

7+0 = 3+4

In other words, the equivalence relation we seek is as follows:

The pairs a-b and c-d are equivalent \((a,b)\sim (c,d)\)

if \(a+d = b+c\). Please notice the outer-inner pattern.

But is this an equivalence relation? To prove that we need to show three properties:

  • relexivity
  • symmetry
  • transitivity
Is this reflexive: \((a,b)\sim (a,b)\)? Indeed it is because: a+b (from outer) is the same as b+a from inner.

Is this symmetric: if \((a,b)\sim (c,d)\) is \((c,d)\sim (a,b)\)? Trivial.

Is this transitive: if \((a,b)\sim (c,d)\) and \((c,d)\sim (e,f)\) do we have: \((a,b)\sim (e,f)\)?

Let's see. We have: a+d=b+c and c+f=d+e. Can we prove a+f=b+e?

If we add the first two we get: a+d+c+f = b+c+d+e or a+f + k = b+e+k where k=d+c and we need to make a tiny generalization:

\((a,b)\sim (c,d)\) if and only if there is a k such that a+d+k = b+c+k.

So what does this have to do with quantum mechanics? It will turn out that those a,b,c,d numbers will be the dimensions of the Hilbert spaces involved in the measurement problem. Also if in the case of integers we have the Cartesian pair: (positive number, negative number) in the case of quantum mechanics we have the Cartesian pair: (Quantum system, Observer).

No more hokey pokey endless philosophical debates about the role of the observer in quantum mechanics which devolve into arguments about consciousness, but a sharp (and unique, and natural) mathematical construction which will allow us to bring about rigorous proofs. Please stay tuned.

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