The holographic principle says that the maximum amount of information that can be contained in a volume (bulk) equals the amount of information that can be contained on the surface (boundary) of that bulk. On first thought, this seems absurd: Common sense tells us that there is so much more matter comprising a bulk than comprising its surface. But here I give a simple argument why the boundary is the limiting factor. The argument relies on several assumptions.

1. Space is discrete
2. There is a smallest volume of space
3. Space is a cubic tiling at that smallest scale
4. those smallest volumes (cubes) are binary-valued
5. Time is discrete
6. messages (signals) can only be sent locally, i.e., to an immediately neighboring cube.

So here’s the argument.

The apparent (emergent) universe we inhabit has three macroscopic spatial dimensions.  The theoretical smallest length (distance) is one Planck length.  So let’s assume that space is a 3D cubic tiling of volumes, i.e., cubes, one Planck length on a side, as in Fig. 1. I call them “planckons“. But note that despite the ending “ons”, these are not particles like those of the Standard Model (SM). In particular, they do not move: they are the units of space itself. Because they are so small and because they can’t have any internal structure, let’s assume that they are binary.  Either matter exists in a cube or it’s empty, i.e., either a planckon exists (“1”), or it does not (“0”).  Planckons have no other properties, e.g., spin, charge.  Functionally, they are just bits.

Now consider a cube of space, C, e.g., 8 Planck lengths on a side, as in Fig. 2.  C is a bulk.  C contains 512 planckons.  Since they are binary, there are 2⁵¹² possible states of C.  So, C can contain 512 bits of information. Now consider a layer of planckons that just wraps C. It consists of 6 x 64 = 384 planckons.  That 1-Planck-cube-thick layer is the boundary, B, between C and the rest of the universe, U. It constitutes a communication channel between C and U. But B consists of only 384 planckons, so it has only 2³⁸⁴ states and so B can contain at most 384 bits.  Now think of passing a message from C into U.  All possible messages from C must ultimately be one of the 2³⁸⁴ possible states of B.  So, no matter which of the 2⁵¹² possible states of C exist, any message sent out of C must be mapped to one of the 2³⁸⁴ states of B.  So, there is a many-to-1 mapping from states of C to states of B. Thus, C is limited to containing at most 384 bits of information.  Also note that all possible messages from U to C have that same constraint. So there is also a vastly larger many-to-1 mapping from the states of U to the states of B. So, no matter how much information is in U, it can never store more than 384 bits in C.

What about multi-time step messages?

You might ask, why can’t we consider messages that take two (or more) time steps to send across the boundary (channel)? For the case of 2-time-step-long messages, there would 2³⁸⁴ x 2³⁸⁴ = 2⁷⁶⁸ possible messages, thus 768 bits. In this case, since 768 > 512, my argument would fail: the number of bits that can be stored in the bulk would not be upper-bounded by the amount of information storable in the boundary. But consider the case of passing a 2-time-step message from C through B to U. In this case, the order of the two steps is essential. Thus, there would have to be some physical reification of that order information stored in the bulk as well. But we’ve already used up all 512 bits (planckons) of the bulk in counting up the number of possible states of the bulk. That leaves no physical substrate to encode the sequential order of the two steps of the message.

I suppose we could consider allocating some fraction of the bulk’s 512 bits to representing sequential order information. Perhaps we would find some partitioning of the 512 bits into state-representing bits and order-representing bits and an accompanying explanation of how the order information would be encoded/enforced in the passing of messages, which would allow more information than 384 bits to be stored in the bulk. But Bekenstein, Hawking, ‘t Hooft, and others have shown that the amount of the information storable in a black hole (the densest possible bulk) is a 1/4 the area of its surface in Planck units, which is already 4 times lower than my simple argument would allow. This suggests finding such a partitioning/explanation would be unlikely.