3 ms·
Is the "group" in renormalization group the same "group" in group theory?
by reader5000 11y ago
Is the "group" in renormalization group the same "group" in group theory?
- oneloop 11y agoAlmost. The name "group" in renormalization group was inspired by the groups in group theory, but in reality the renormalization "group" isn't a group but a semi-group. A semi-group satisfies the same axioms as a group except for the existence of inverse. And that is the mathematical reason why you can describe big things in terms of smaller things, but you can't describe small things in terms of bigger things: the renormalization (semi-) group flows from the ultraviolet to the infrared, but not the other way around :-)
- ximeng 11y agooneloop, your comment looks helpful but you are hell-banned, you may want to email HN to be reinstated. -- Edit - he's been reinstated.
- chmartin 11y agowhat does hell-banned mean?
- ximeng 11y agoComments are marked as dead (can only be seen by people with showdead set in their profile), but appear visible to the user so that they may not realise why nobody responds to them or upvotes their comments. Normally a punishment for bad behaviour or poor commenting, but seems inappropriate in this case.
- selimthegrim 11y agoHis comment below about a semigroup is absolutely correct, if that helps his case any. Qualifications: I studied RG in classes at Santa Barbara
- chmartin 11y agoKinda. The 'group' refers to a group scale and/or conformal transformations. In the context of Deep Learning, the 'scale' transform is akin to adding layers.
- eeperson 11y agoYes, it is the same as the "group" in group theory. However, I the name is a misnomer since I believe it is technically a semigroup because the binary operator is not invertable.
- reader5000 11y agoWell, if say the elements of the permutation group are permutation functions/mappings and the operator is function composition, in a renormalization group the elements would be "renormalization functions"? and the operator would be function composition?