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- Hacker News
- What is the (co)homology of this space?
- That of the underlying, hypothetical universal brain topology?by chombier
- Cyberphrenology. In any two random graphs, you'll find an isomorphic graph which is can be up to log of the size of the graphs.
And if the LLM has been trained up to the limit of what data it can hold, it is going to be random. Proof below if it isn't obvious.
The entire effort of all people who are trying to understand how LLMs work, how they represent their data, its all bound to fail.
Proof: a LLM is a very good approximation of the Solomonov/Levin/Kolmogorov universal probability function on tokens. As such, it will be random--pure white noise--because if you found any patterns in there, you could exploit the regularity and come up with a smaller set of weights for the same LLM.
There are no patterns there to be found. They have all been factored out by training the neural net until it couldn't learn any more.
by rhelz - The weights aren’t compressed. So there are interpretable redundancies in practice.by canjobear
- /a smaller set of weights for the same LLM./
Distillation is alive and well... Earlier work on model printing also found that it's pretty easy to find smaller sets of parameters which can replicate the behavior of the entire network with pretty good fidelity.
Large parameter counts give space to explore, and give routes out of what would be local minima in a lower dimensional space.
In other words, there's no guarantee that any given trained model is a minimal representation of its training set.
by sdenton4 - I’ve never liked that this was called “the platonic representation hypothesis”. Lots of weird baggage attached and seems like a waste of a good name.by stephantul
- "We believe these representations are not serious, they're just really good friends."by pksebben
- Let's assume that monotonocity of pair-wise distances are preserved.
Without knowing the details of how the paper solved the problem, my first attempt would be to find the diametrically distant pair of points in the two different embeddings and assume that the pair is the same pair. Then find the next distant pairs and so on.
After sufficiently many such pairs have been found, or better still, the largest d-simplex is found, find that scaled rigid body transformation that makes the corresponding pairs coincide. Proceeding this way ought to be less work than solving a generic graph isomorphism problem.
by srean - I think a less stringent, but still workable assumption is that for very similair objects, their distances will be small. This is much easier to accomplish than agreement across all pairs.by robrenaud
- I'm not as heavy on the maths stuff involved in this as other people commenting appear to be.
But the idea makes sense, of course there is still recoverable data in embeddings, that's the point. Though as I constantly find the more you try to squeeze into an n bit vector the more watered down everything gets.
I suppose a latent space could be encrypted/mapped in some way to resolve that, but how many people are exposing their vectors in the first place?
by fennecfoxy - The point of the paper isn't that embeddings contain information, it's that even if you don't know what model generated a set of embeddings you can still recover information from the geometry of the point cloud itself.
The fact that this is possible also adds some pretty strong restriction to the set of possible maps you could use to remove that information. No linear map will work since all embedding spaces are ~an orthonormal matrix apart, so some form of encryption is necessary. This wasn't known until very recently.
- I am not familiar with the standards of publishing in machine learning, but as someone trained in a mathematics background, this paper seems relatively light on details and heavy on exposition. Is that typical? Is this a really novel idea? Not trying to be snarky, just trying to understand how meaningful this is.by ironSkillet
- You are not wrong. But this has by no means proven its up to the standard of being publishable in a machine learning journal. Its on arXiv.org, which, lets face it, at the end of the day is a vanity press.by rhelz
- This isn’t math. The exposition IS the details.by adastra22
- It isn't a maths paper, so the conventions are different.by Nail2680
- They link to their code on GitHub - see footnote 2 on page 2. I don't see it linked anywhere else, which makes it easy to miss. https://github.com/rjha18/vec2vec/by wging
- It's cool that it proves that a bunch of vectorized outputs from an unknown embedder on an unknown dataset is in no way private, because of this ability to reverse engineer the embedder.
I talked to the author at his poster session at neurips and was able to get the gist, though I had read a lot about the platonic representation hypothesis, and this was one of my top 10 favorite papers in the conference.
by robrenaud - Dupe: https://news.ycombinator.com/item?id=44054425
Note this is version 4 of the paper and the original post was version 1 (I think?)
OpenReview (for NeurIPS) for the curious: https://openreview.net/forum?id=jiCLUPq5xv
by nickledave - One way to pose/(think about) the problem is that there are two finite metric spaces linked by an unknown odometry (damn you autocorrect). The problem is to recover that unknown isometry.
This, like graph isometry, can be very computationally intensive in the worst case. However, heuristics to aid matching one vertex on one graph to another vertex on another graph using local, semilocal structural signatures can be very effective on particular cases.
One can of course argue that the spaces are not designed as metric spaces. Even if true, these might be metrizable topological spaces.
More generally, if these are indeed non-metric spaces one can still pose it as finding the unknown isomorphism between two poset spaces.
In my other comment I was using the property of maximal chains -- Identify the longest chains in both posets. The isomorphism must map the longest chain in Poset 1 directly to a longest chain in Poset 2, preserving the exact linear order.
by srean - I've been working on research related to this in the context of diferent LLMs, and I can tell you that similarity != executability.
While you can make embeddings across different LLMs similar (i.e. universally looking) the few percentage of R2 that you are missing in translating the universal representation into a native representation are precisely those that make the hidden states executable in the LLM (which makes them useful).
They do this with simple embedding models because there the purpose of the embeddings is to measure similarity, but if you would like to use this principle to turn latent representations of one LLM into latent representations that are understandable/executably by a different LLM, you will fail.
by dankai