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Can anyone help me and prove that in the case of the Ricci flow, the symmetries of the flow are scalings and diffeomorphisms?

Thanks for your time.

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crosspost:… – Sepideh Bakhoda May 26 '13 at 14:56
any textbook will have this proof, see for example… – Carlo Beenakker May 26 '13 at 15:31
@Carlo: Most textbooks on Ricci flow will indeed show that scalings and diffeomorphisms are symmetries of the PDE, but I am not so sure that there are many books that show the converse, namely, that any symmetry of the PDE system (once this has been properly defined) is necessarily a combination of scalings and diffeomorphisms. – Robert Bryant May 26 '13 at 15:53
Could someone say what the definition of a symmetry of the flow is? My reaction to the question was: isn't that the definition of a symmetry? – Deane Yang May 26 '13 at 18:22
@Deane: I took 'the Ricci flow' to mean the PDE system itself which one can think of as a submanifold of an appropriate jet bundle $J$ over $M\times\mathbb{R}$. Then a 'symmetry' would a self-diffeomorphism of $J$ that carries solutions to solutions (thought of via their $k$-jet graphs). @Carlo: You probably want to be more precise about what 'operations' you allow. Let $\mathcal{S}$ be the set of all Ricci-flat metrics on $\mathbb{R}^n$ and let $\sigma:\mathcal{S}\to\mathcal{S}$ be any mapping whatsoever. This is an 'operation that transforms one solution to another solution', no? – Robert Bryant May 27 '13 at 0:43

I apologise first for submitting this comment as an answer (not enough MO-cred to comment) and second for submitting a question rather than a comment:

What I can gather from the comments above is that the symmetries of the Ricci flow are completely understood: Every `symmetry' of the Ricci flow is a spatial diffeomorphism combined with scaling (and every such combination is a symmetry). My question is whether the problem is also solved for the mean curvature flow (of, say, hypersurfaces of Euclidean space)?

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