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wikipedia doesn't say, nor my Berger Panorama book (but I might google Levi-Civita to get rid of one level of brackets) and the library is far (actually not, but it has German Schließungszeiten and I got no key). (I guess the differential Bianchi identity is not due to Bianchi? So who did which?)

To clarify on the "vice versa": according to my Swiss cheese memory, there might be something completely different actually due to Bianchi. Some book was telling this quite nice...

Said MO answer is here

edited Sep 23 2010 at 2:30

asked Sep 23 2010 at 2:06

Martin Gisser
418●2●10

	Meanwhile I've found this paper: Ph. Delanoe, On bianchi identities, Rendiconti del Circolo Matematico di Palermo Volume 51 (2002), 237-248. springerlink.com/content/k7qlt4l33568j70q According to the introduction (if I parse it right), the first "algebraic" Bianchi identity is due to Riemann and Christoffel. The second "differential" is due to Bianchi. (Delanoe extends Kazdan's "another proof" to arbitrary vector bundles. Alas he does not mention M. Schlicht's extension to arbitrary principal bundles.) – Martin Gisser Oct 18 2010 at 0:48
	Actually, your « Swiss cheese memory » seems to be a misnomer, as was pointed out by Tom Leinster in his book « Higher Operads, Higher Categories » (he acknowledges the help of Paul-André Melliès for that). See bottom of p. 63 of arxiv.org/pdf/math/0305049. – Jonathan Chiche Dec 11 2010 at 13:40

Andrey Rekalo · Answer 1 · 2010-12-12 08:07:45Z UTC

L.P. Eisenhart claims in his Riemannian Geometry (Princeton University Press, 1926, p. 82) that Bianchi was the first who discovered the algebraic identity in 1902.

It is also indicated in Italian Mathematics Between the Two World Wars by Guerraggio and Nastasi (Birkhäuser Verlag, 2006, p. 15) that Bianchi obtained both of the identities which were published in his paper "Sui simboli di Riemann a quattro indici e sulla curvatura di Riemann", Rend. Acc. Lincei, 11 (1902), pp. 3–7.

Kallikanzarid · Answer 2 · 2010-10-16 10:01:26Z UTC

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The first one is due to Ricci, and the second one is due to Bianchi.

answered Oct 16 2010 at 10:01

Kallikanzarid
134●7

Cristi Stoica · Answer 3 · 2010-10-16 08:06:43Z UTC

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Wikipedia saids that the algebraic one at least is due to Ricci:

Curvature of Riemannian manifolds - Symmetries and identities

answered Oct 16 2010 at 8:06

Cristi Stoica
1,848●13●24

Which Bianchi identity is due to Bianchi (or not, since it might be due to Ricci (according to Levi-Civita (according to MO))) or vice versa?

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