Talk:Partition of unity

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This should link to gluing axiom. Also, a more general definition could be given (in terms of sheaf morphisms). Finally, I think there could be some explanation why one usually insists on considering finite subcovers. — MFH:Talk 23:56, 30 March 2006 (UTC)Reply

Get started! -lethe ^{talk +} 00:22, 31 March 2006 (UTC)Reply

The text says analytic partitions of unit do not exist - this should be better explained! I think what does not exist are partitions with functions that are zero inside some open sets. Albmont 12:09, 11 April 2007 (UTC)Reply

Well, every point has a neighborhood where all but finitely many of functions are identically zero. If the functions are analytic (and the space is connected), these functions are then zero everywhere and we might as well assume we have finitely many functions overall. Similary, subordinating an analytic partition to an open cover doesn't make much sense either. So while analytic partitions may exist in some simple cases, the whole idea doesn't make much sense and e.g. doesn't allow one to prove results for (complex) holomorphic functions similar to results about smooth functions. This can definitely be added to the article as well as some discussion of differential geometry-related issues. I don't remember now, but there are some classical facts (about smooth manifolds) derived using the partition-of-unity technique, and such proofs are usually insightful and elegant, so worth adding to the article. Commentor (talk) 05:58, 9 March 2008 (UTC)Reply

For instance, it is used in the proof of the Mayer-Vietoris_theorem. There we have ${\displaystyle M=U\cup V}$ and we want to represent a smooth function (more generally, differential form) on ${\displaystyle U\cap V}$ as a difference of a function (or form) on U and another one on V, to complete the proof that the sequence is exact. To do that one finds a partition of unity ${\displaystyle \rho _{U}}$, ${\displaystyle \rho _{V}}$ subordinate to the cover ${\displaystyle \{U,V\}}$ and represents f as ${\displaystyle \rho _{V}}$ ${\displaystyle f-(-\rho _{U}}$ ${\displaystyle f)}$, the former being a function on U and the latter on V (not vice versa!) Commentor (talk) 06:20, 9 March 2008 (UTC)Reply

I second the comment about analytic functions; you have not made it clear what constraints are on the functions of the partition. E.G. the set of functions f=1 (just the one function) satisfies the criterion of analytic (it's a polynomial) and the set has only a finite number,1, of elements which is non-zero. User:YouRang 21:14 July25,2008 (UTC) —Preceding unsigned comment added by YouRang? (talk • contribs)

Why is the example figure mentioning a circle? This confuses people more then it should: The four functions are simply a partition of unity. To first tell us that this was starting out as a partition of the circle (line) and then was rolled out as a function to end up as a partition of unity, it needlessly uses a context that is no longer valid or important. MWinckler (talk) 13:05, 30 June 2021 (UTC)Reply