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I Number of independent components of the Riemann tensor

  1. Mar 27, 2017 #1
    I've thought of a new way (at least I never read it anywhere) of counting the independent components of the Riemann tensor, but I am not sure whether my arguments are valid, so I would like to ask whether my argument is sound or total bonkers.
    The Riemann tensor gives the deviation of a vector A when transported around a loop with vectors u and v:
    [tex] \delta A^\alpha = -R^\alpha_{\beta\mu\nu} A^\beta u^\mu v^\nu[/tex]

    A parallel transport is only non-trivial if u and v are different (since u=v just means going forward and backward along a line). This also shows that the tensor is antisymmetric in the final two indices because reversing the vectors means travelling along the loop in the opposite direction. There are six possible planes that can be spanned by u and v (01, 02, ..., 23).

    If I fix the plane, what the Riemann tensor does is create an infinitesimal rotation of the initial vector A, resulting in the deviation. An infinitesimal Rotation is antisymmetric (so we get antisymmetry in the first two indices), in 4 D it has 6 independent components.

    This leaves me with 6x6=36 independent components.

    To get rid of the remaining 16, I use the argument presented in this paper by Ollivier
    see fig. 5 and 6.
    What this shows is that one can understand the first Bianchi identity geometrically by constructing a cube with parallel transport of vectors and noting that the endpoints of different paths will form a triangle.

    In normal derivations of the number of independent components, the Bianchi identity can only eliminate one of them because the symmetry [tex]R_{\alpha\beta\mu\nu}=R_{\mu\nu\alpha\beta}[/tex] has already been established. Since I did not do this, I can exploit the Bianchi identity using the cube construction from the Ollivier-paper. There are four distinct cubes in 4D, each of them yields one Bianchi-identity which fixes four components up to a total of 16.
    Thus, 16 components of the Riemann tensor have been fixed this way, leaving 20 as it should be.

    I realize that the argument has the big disadvantages of (i) being a bit heuristic and (ii) not establishing the symmetry between the first and last two indices. Nevertheless, I find it interesting - if it is correct.

    I'd be very grateful if one of the experts could tell me whether this argument actually works out or where I made a mistake.
  2. jcsd
  3. Mar 28, 2017 #2


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    I can vouch for the first part of your analysis. There's a description of a similar technique in MTW's text "Gravitation". https://www.amazon.com/Gravitation-Charles-W-Misner/dp/0716703440 I believe it's also related to something that's known as the "Bel decomposition", but I don't have a good reference on that in English, alas. Wiki has a stub article <<here>>, but the references it cites are all not in English (and I haven't read them). MTW goes through the analysis without ever using the term "Bel decomposistion".

    The second part of your analysis I'm not familiar with, and I don't have the time to read your paper at the moment, though it looks interesting. But I can give a fast run-through of a different analysis I've seen, though it won't be very detailed.

    To go a step further, go from viewing the Riemenn as a 6x6 matrix to four 3x3 matrices, by dividing the six pairs of vectors (bivectors) that you've already noticed into those that contain time as part of the bi-vector, and those that don't. To do this requires, as the wiki stub mentions, singling out which vector represents "time". Wiki mentions the formal way of doing this, which is to define a unit vector field that points in the "time" direction, which is called a "timelike-congruence". I should add that I usually envision said time-like vector field as having zero vorticity - I'm not quite sure if this essential or just laziness on my part.

    If one chooses a specific local orthonormal basis, the choice is obvious by looking at the signs of the diagonal metric - one component will have a negative sign, that will be the vector that is "time". Of course, there are many ways we can choose a local orthonormal basis.

    Let's call the set of three bi-vectors that contain time "T", and the set of three bi-vectors that don't contain time S. Then we can divide our 36 components, arranged in a 6x6 matrix, as a set of four 3x3 matrices (which also has 36 components). We name these 3x3 matrices TT, TS, ST, and SS, T representing a bivector that contains time, and S representing a bivector that doesn't.

    Next we note that TS and ST turn out to be the same, by the Bianchii identies. This leaves us with three 3x3 matrices - we're down to 27 components.

    The next observation is that the TT matrix must be symmetric (in the wiki stub article, and in MTW, it's given a name, the electirc part of the Riemann). Similarly, the SS matrix is also symmetric. (This is also named, the topogravitic part).

    The ST matrix doesn't have any symmetries. It's referred to as the "magnetic part". The two symmemtrical matrices each have 6 unique components, and the assymetrical matrix has 9. So we're at 6+6+9=21 unique nozero components.

    The 21 components are not all independent, though - it turns out that there is a constraint equation from the Bianchi identies. So the number of degrees of freedom drops from 21 down to 20, because of this constraint. So there are 21 unique non-zero components in the Riemann (down from 256), which we can organize into three 3x3 matrices which we can give names to. And there's one constraint equation that links these 21 components, so they're not all independent.
    Last edited by a moderator: May 8, 2017
  4. Mar 29, 2017 #3
    Thanks a lot for the detailed answer. Glad to hear that my intuition for the first part (which I came up with myself) is actually well-known.

    I'll have to look through the MTW again (it's a great book, but finding things is sometimes a bit painful...) to make sure I get all this.
  5. Mar 29, 2017 #4


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    I found some discussion of number of components on pg 326. On pg 360, there was a discussion of "matrix display of the Riemann", exercise 14.14. I thought there was more on the topic than an exercise - and perhaps there is - it really is hard to find things in the book :(
  6. Mar 30, 2017 #5
    Thanks for taking the pains to dig through the MTW. I re-checked this and the MTW indeed uses the "1st Bianchi identity" (not using this name though) to establish the symmetry upon switching the first and last two indices. So my argument that these can be used to reduce the number from 36 to 20 seems to be valid.
  7. Mar 30, 2017 #6


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    The argument from symmetries seems to be the cleanest to me - and in exercise 14.14, MTW justifies the matrix display of the Riemann (which I see as basically being equivalent to your proposal) solely in terms of the symmetries.

    MTW also mention that some symmetries are derived from the existence of a metric, something that I didn't really recall. Following up a bit on the topic of metric-induced symmetries, I found http://www.physicspages.com/2014/04/05/riemann-tensor-symmetries/. Their approach appealed to me more than MTW's treatment (though MTW has better provenance than a web page, of course). Some details of putting it all togheter again are still fuzzy to me, such as why MTW used ##R^{ij}{}_{kl}## vs the web-page use of ##R_{ijkl}##.

    I still like the decomposition idea, though, it gives some physical insight. One winds up decomposing the RIemann into two 3x3 symmetric matrices, and one 3x3 trace-free matrix (plus the transpose of said trace-free matrix). This gives the number 20 by fairly simple counting arguments. The underlying justification for the breakdown basically rests on the symmetries of the Riemann, though.
  8. Apr 1, 2017 #7
    Yes, the decomposition idea is also nice. I'm quite happy that something that I so far only saw as an invlved mathematical argument can actually be interpreted intuitively; even in two slightly different ways.
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