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The Training Example Lie Bracket

pbement.com
33 points·by pb1729·3 months ago·17 comments

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pb1729
·3 months ago·discuss
Yeah, this is a good point. IIRC, I wasn't able to get the network to train very well at all with standard SGD. I don't think I thought to try Adam with β1 = 0, I will try it (& recompute brackets) if I get some time.

If we have built up a momentum M, then the two orderings are:

M' = M + εv1

θ' = θ + M' = θ + M + εv1

M'' = M' + εv2(θ') = M + εv1 + ε(v2 + (M + εv1)⋅∇v2)

M' = M + εv2

θ' = θ + M' = θ + M + εv2

M'' = M' + εv1(θ') = M + εv2 + ε(v1 + (M + εv2)⋅∇v1)

Then the resulting difference in momenta M'' is:

ε^2*[v1, v2] + ε(M⋅∇)(v2 - v1)

So there is an extra term which is not actually a Lie bracket itself. I think the bracket can still be informative on its own, but it's definitely no longer the sole component of what happens when order is swapped.

One other inconsistency that is a little less bad is BatchNorm. Since it needs a whole batch to work, and we're just comparing individual examples, I computed the Lie brackets with the BatchNorm layers in eval mode, not train mode.

I don't know if there is any relevance of this to Muon, even if so, it would likely be very messy to compute.
pb1729
·3 months ago·discuss
If I understand your question correctly, the answer is that not only are the Lie brackets non-commutative, they're anti commutative (swapping the order negates the bracket). But this ironically means they end up having the same RMS, because the squaring part of the RMS gets rid of the sign.