As I understand it, the holographic principle says that the maximum information in a space is limited by the surface area of a boundary of that space, and thus in principle could be encoded on that boundary, as revealed by the entropy of a black hole. I found that surprising, but maybe I shouldn't have.
Reading back through my notes on complex analysis* from long ago, I see Cauchy's integral formula tells us that if you know the value of some (complexvalued) holomorphic function around the boundary of a disk, you know the value everywhere inside the boundary, and thus you could say that all the information in the region is encoded on the boundary. For a function to be holomorphic it needs to be infinitely differentiable, but that seems like a natural outcome of physical laws as long as there are no singularities. From what little I understand of QFT, everything is a complex multivalued function, and arguably uncertainty implies the absence of singularities (at least without an event horizon shielding them).
So, for those who do know the math, why would it be surprising under QFT that the information in a region of space could be seen as encoded on the boundary? Was this really a surprising result to those in the know, or was it one of those "no one had thought of it, but it was obvious in hindsight" things? For a region of space containing no singularities in any field, the holographic principle seems like it was implied all along  or what am I missing? Are there important/irreducible places where nonholomorphic operations happen?
*Basic Complex Analysis  best textbook name since Advanced Regression.
Holomorpic vs Holograpic
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Holomorpic vs Holograpic
"In no set of physics laws do you get two cats."  doogly

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Re: Holomorpic vs Holograpic
Holomorphic functions are defined over the complex plane. Wave functions are defined over a space with real coordinates.
Cauchy's integral formula works, because being "complex differentiable" (i.e. holomorphic) is a far stronger property than being differentiable in a real space. Remember, a differentiable function over the real plane has two independent partial derivatives. You can have df/dx=a and df/dy=b for any values of a and b. But a differentiable function over the complex plane must have df/dy=i(df/dx) (assuming we use 'x' to denote the real axis and 'y' to denote the complex axis). There is only one degree of freedom, which is why you can "predict" the values of a function inside a disk from information on the boundary alone.
So no, there is no reason for a complex function defined over an Euclidean real space to obey the holographic principle. Cachy's integral formula does set a precedent, though. Perhaps something similar (even if it isn't identical) is happening in the real world. Kudos for noticing the connection.
Cauchy's integral formula works, because being "complex differentiable" (i.e. holomorphic) is a far stronger property than being differentiable in a real space. Remember, a differentiable function over the real plane has two independent partial derivatives. You can have df/dx=a and df/dy=b for any values of a and b. But a differentiable function over the complex plane must have df/dy=i(df/dx) (assuming we use 'x' to denote the real axis and 'y' to denote the complex axis). There is only one degree of freedom, which is why you can "predict" the values of a function inside a disk from information on the boundary alone.
So no, there is no reason for a complex function defined over an Euclidean real space to obey the holographic principle. Cachy's integral formula does set a precedent, though. Perhaps something similar (even if it isn't identical) is happening in the real world. Kudos for noticing the connection.
Re: Holomorpic vs Holograpic
When you move from the complex plane to R^{n}, pretty much all the same results work, you just need to replace holomorphic functions with harmonic ones.
Cauchy's theorem then is a special case of the divergence theorem when there are no r^{n1} terms in the expansion about any point within a surface. The residue theorem reduces to the divergence theorem (and knowing that the functions we're using will be sums of green's functions for Poisson's equation).
This is for real functions on R^{n}. These results should extend fairly easily, you'd just need to mess about a bit with CauchyRiemann equations a bit to work out what a harmonicholomorphic function would look like but it should certainly be doable.
Cauchy's theorem then is a special case of the divergence theorem when there are no r^{n1} terms in the expansion about any point within a surface. The residue theorem reduces to the divergence theorem (and knowing that the functions we're using will be sums of green's functions for Poisson's equation).
This is for real functions on R^{n}. These results should extend fairly easily, you'd just need to mess about a bit with CauchyRiemann equations a bit to work out what a harmonicholomorphic function would look like but it should certainly be doable.
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Re: Holomorpic vs Holograpic
True. But is there any reason to assume quantum fields are harmonic functions?
Re: Holomorpic vs Holograpic
Any reasonable function can be decomposed into the sum of harmonic functions (again, we need to let each term have up to 1 "pole", but without that all our surface integrals are 0 which is boring).
my pronouns are they
Magnanimous wrote:(fuck the macrons)

 Posts: 126
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Re: Holomorpic vs Holograpic
eSOANEM wrote:Any reasonable function can be decomposed into the sum of harmonic functions (again, we need to let each term have up to 1 "pole", but without that all our surface integrals are 0 which is boring).
It is exactly those singularities you mentioned in the parenthesis which render your main statement irrelevant.
Re: Holomorpic vs Holograpic
It's an interesting idea, but not quite as simple as you suggest. Any explanation of the holographic principle has to involve gravity in some essential way, I think. My (decidedly nonexpert) intuition for this is that with, say, EM interactions you can have screening of positive and negative charges that prevents you from knowing anything about the inside of a region from the boundary, but because gravity couples to all energy and cannot be screened it does have a holographic nature in some sense. I imagine this could be formalized into some sort of Cauchylike theorem that is satisfied by GR (and, presumably, quantum gravity) but not QFT, but I don't know exactly what this would look like offhand.
Re: Holomorpic vs Holograpic
PsiSquared wrote:eSOANEM wrote:Any reasonable function can be decomposed into the sum of harmonic functions (again, we need to let each term have up to 1 "pole", but without that all our surface integrals are 0 which is boring).
It is exactly those singularities you mentioned in the parenthesis which render your main statement irrelevant.
Huh. No.
The Cauchy integral formula equivalent would just need to subtract a constant value i.e. the sum of the strength of the sources within the region. Generally, that constant has an obvious physical interpretation, like the total charge or mass or something within the boundary so it's totally fine.
Classically (because I'm more familiar with the specifics of the maths), if you consider a black hole, the integral of the gravitational field over the event horizon will tell you the mass of the black hole, by integrating the field divided by (xa)^{2} over the surface and subtracting off the mass of the black hole, the field at coordinates a will be produced. Exactly analogously to Cauchy's Integral formula.
Edit: of course, again, complex fields will be a bit more complicated because the CauchyRiemann equations restrict the function space more than just requiring functions to be harmonic but I suspect all reasonable solutions satisfy them. Vector fields should be pretty easy though, most of the ones we care about are guaranteed to have harmonic components so we can simply carry everything through in the obvious way.
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Magnanimous wrote:(fuck the macrons)
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