In General Relativity, one uses the Riemann Tensor in its coordinate form $R_{abcd}$, and proves the Second Bianchi Identity-

$R_{abcd;e} + R_{abde;c} + R_{abec;d} = 0$

It is said that this identity amounts, in the Riemmanian geometry sense, to the saying that, loosely, "the boundary of the bounday is null". So my questions are

Question 1: What would be a more rigorous statement of "the boundary of the boundary is null" and how does the above identity is equivalent ot it?

Question 2: Is there an analog to this geometrical intuition in general relativity?

I assume both questions might be answered with a reference, which is even better.

  • 2
    $\begingroup$ If I recall, the 2nd Bianchi identity is just a form of the Jacobi identity. You take the covariant derivative of the Riemann curvature tensor, use the fact that you can express this in terms of covariant derivatives of "evaluations" of the tensor, and then the Jacobi identity is staring you in the face. So whatever intuition you bring to the Jacobi identity comes forward to the Bianchi identity. There are "boundary of boundary" interpretations although I think of those more in the spirit of Taylor expansions. $\endgroup$ – Ryan Budney May 10 '16 at 5:48
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    $\begingroup$ Any connection $\nabla$ on a vector bundle $E$ induces an exterior derivative on $E$-valued forms. Unlike the usual exterior derivative, $(d^\nabla)^2\neq 0$. (One can identify $(d^\nabla)^2$ with the curvature.) The Bianchi identity that you quote essentially states $(d^\nabla)^3=0$: see sec. 3.3.5 of www3.nd.edu/~lnicolae/Lectures.pdf $\endgroup$ – Liviu Nicolaescu May 10 '16 at 9:48
  • $\begingroup$ To complete @LiviuNicolaescu's comment, it may help to be reminded of the connection between the boundary operator $\partial$ and exterior differentiation. See e.g. mathoverflow.net/questions/46252/… or mathoverflow.net/questions/60059/… $\endgroup$ – Willie Wong May 10 '16 at 13:20

From The "Foreword to Feynman Lectures on Gravitation" by John Preskill and Kip S. Thorne:

In §9.3, Feynman comments that he knows no geometrical interpretation of the Bianchi identity, and he sketches how one might be found. The geometrical interpretation that he envisions was actually implicit in 1928 work of the French mathematician Elie Cartan [Cart 28]; however, it was largely unknown to physicists, even professional relativists, in 1962, and it was couched in the language of differential forms, which Feynman did not speak. Cartan’s interpretation, that “the boundary of a boundary is zero,” was finally excavated from Cartan’s ideas by Charles Misner and John Wheeler in 1971, and has since been made widely accessible by them; see, e.g., chapter 15 of [MTW 73] at the technical level, and chapter 7 of [Whee 90] at the popular level.

[cart 28] is Cartan's 1928 lectures in French about the geometry of Riemannian manifolds. Extended English version can be found in the Cartan's book "Geometry of Riemannian Spaces": http://www.amazon.com/Geometry-Riemannian-Spaces-Lie-Groups/dp/0915692341

[MTW 73] is Misner, Thorne, and Wheeler's classic book "Gravitation": http://www.amazon.com/Gravitation-Charles-W-Misner/dp/0716703440

[Whee 90] is Wheeler's book "A Journey into Gravity and Spacetime": http://www.amazon.com/Journey-Gravity-Spacetime-Scientific-American/dp/0716750163

"Feynman Lectures on Gravitation" can be found here: http://hixgrid.de/pg/file/read/3511/feynman-lectures-on-gravitation-frontiers-in-physics

The exposition in MTW is actually quite short:

  1. Take a really small cube. Each of the three terms in the second Bianchi identity correspond to the difference between opposing faces of the effect of parallel transport of a vector around the boundary of the face. It is easy to see that if you sum over all three pairs of opposing faces, the paths cancel and so the net sum should be zero.

MTW Figure

  1. That the "paths cancel" in the preceding point is precisely the statement that "the boundary of a boundary vanishes" (the boundary of a cube is its faces, and the boundary of that are the edges).
  • $\begingroup$ I intuitively understand 1 ,at least for flat $R$. As for 2 - we don't show that the edges vanish. How do you connect 1 and 2? $\endgroup$ – Amir Sagiv May 12 '16 at 19:19
  • $\begingroup$ Each edge belongs to two different faces and thus enters in the definition of boundary twice with opposite signs. For motivation why the boundary of the cube is defined as an alternating sum, see, for example, books.google.ru/… $\endgroup$ – Zurab Silagadze May 13 '16 at 5:22

The second Bianchi identity can also be viewed as a consequence of the diffeomorphism invariance of the Riemann curvature tensor, i.e. the fact that $\operatorname{Rm}_{\varphi^*g} = \varphi^* \operatorname{Rm}_g$ for any metric $g$ and diffeomorphism $\varphi$. Differentiating with respect to $t$ the equation $\operatorname{Rm}_{\varphi_t^*g} = \varphi_t^* \operatorname{Rm}_g$ , where $\{\varphi_t\}$ is a one-parameter family of diffeomorphisms, yields a proof of the first and second Bianchi identities.

This is discussed in (for instance) Chapter 5.3.1 of The Ricci Flow in Riemannian Geometry: A Complete Proof of the Differentiable 1/4-Pinching Sphere Theorem by Ben Andrews and Christopher Hopper.

In terms of a relativistic interpretation, this is related to the idea of general covariance, which requires the laws of physics to be diffeomorphism invariant.

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    $\begingroup$ I believe the first mathematician to observe this was Dennis DeTurck. This led to what is now known as the DeTurck trick, which was used originally for the prescribed Ricci problem but later for the Ricci flow. $\endgroup$ – Deane Yang May 12 '16 at 4:36
  • $\begingroup$ Graham, I'm not familiar with the notation ${\rm Rm}_g$, could you refer me or define it? $\endgroup$ – Amir Sagiv May 12 '16 at 19:21
  • $\begingroup$ @AmirSagiv $\operatorname{Rm}_g$ is the Riemann tensor computed from the metric $g$, while $\operatorname{Rm}_{\varphi^*g}$ is the Riemann tensor for the pulled-back metric tensor $\varphi^* g$. $\endgroup$ – Graham Cox May 15 '16 at 13:15
  • $\begingroup$ I'm a bit confused. GR is invariant under all diffeomorphisms? In what way? $\endgroup$ – Amir Sagiv May 17 '16 at 6:00
  • $\begingroup$ @AmirSagiv In the sense that any predictions made by the theory should not depend on a particular choice of coordinates. $\endgroup$ – Graham Cox May 23 '16 at 18:58

Paul Gauduchon, "Calabi's extremal K\"ahler metrics: An elementary introduction", see Section 1.18 at page 44.


Look at A visual introduction to Riemannian curvatures by Yann Olivier. It provides nice geometrical intuitions for Riemannian quantities such as sectional curvature or Ricci curvature and also a visual interpretation of the Bianchi identity (page 7).

  • 2
    $\begingroup$ For the record, on p.7, the linked PDF file talks about the first or algebraic Bianchi identity $R_{[abc]d} =0$. $\endgroup$ – Igor Khavkine May 11 '16 at 11:49

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