- #26

samalkhaiat

Science Advisor

- 1,758

- 1,108

Hopefully Sam will answer but in the meantime this is my understanding.

You have to use analytic continuation, that is, impose dependence on Riemann zeta function, in this particular case for s=1, and the particular series 1+2+3+.... , in other bases you get other convergences. This comes at the cost of losing basis-independence of the trace i.e. incompatibility with linearity that Sam found bizarre above.

I see no question to answer, so I just make the following remarks for added clarification:

1) Divergent series are not “the invention of the devil”. Some divergent series can be summed rigorously. For example, the “usual sum” of [itex]1 - 2 + 3 - 4 + \cdots[/itex] diverges, but we can rigorously show that its Abel sum exists and is equal to [itex]1/4[/itex]. Other divergent series need to be regularized: by introducing suitable cutoff function, one can justify the use of the Euler-Maclaurin formula to them.

2) The Euler sum [itex]\sum_{1}^{\infty} k = - 1 / 12[/itex]:

A)

**In physics**, it has passed the test:

(i) The whole derivation of the Casimir force can be summarized by the replacement [itex]\sum^{\infty} k \to ( - 1 / 12 )[/itex].

(ii) In bosonic string theory, the mass of particles in the [itex]J[/itex] excitation is given by [tex]\alpha^{ ’ } m^{2} = J + \frac{D - 2}{2} \sum_{n = 1}^{\infty} n .[/tex] It is also known that the [itex]J = 1[/itex] excitations are massless. Thus, by using the Euler sum in the relation above, we conclude that bosonic string theory is consistent only in space-time of dimension [itex]D = 26[/itex].

B)

**In mathematics**, “all roads take you to Rome”:

(i)

**Zeta function**regularization leads to [tex]\zeta ( - 1 ) = - \frac{1}{12} .[/tex]

(ii)

**The Heat Kernel**Regularization gives you [tex]\lim_{r \to 0} \left( \frac{e^{r}}{( e^{r} - 1)^{2}} - \frac{1}{r^{2}} \right) = \lim_{r \to 0} \left( - \frac{1}{12} + \mathcal{O} ( r^{2} ) \right) = - 1 / 12 .[/tex]

(iii)

**Ramanujan’s summation**method gives us (in terms of Bernoulli’s numbers) [tex]C = - \frac{B_{1}}{1!} f ( 0 ) - \frac{B_{2}}{2!} f^{ ’ } ( 0 ), [/tex] which for our case, [itex]f(x) = x[/itex], leads to [itex]( - 1 /12)[/itex].

3) There is no such thing as “free lunch” : all regularization methods are either

**not linear or not stable**.

4) The main point of my previous post is to

**highlight**point (3): The trace of infinite-dimensional matrices is neither rigorous nor nice: (i) the very notion of diagonal elements lose its meaning. This is why I wrote “diagonal elements”. (ii) the defining properties of the trace in finite dimension, i.e. linearity and invariance (or cyclicity), can not both be maintained in infinite-dimension. This is what happened when we “traced” the number operator [itex]A^{\dagger} A[/itex]: the trace was invariant (cyclic) but not linear.

Sam