In the notes I took as an undergraduate (Dr. Stewart's 8th lecture on General Relativity (M16), probably 1985) some properties of the Riemann tensor are given. For these, take your co-variant differential operator, D, which annihilates the metric, (G: g |T), and extract the linear map as which antiSymm([G,G])&on;D&on;D acts on the gradient bundle, G, of your smooth manifold. This is a linear map from G to G&tensor;G&tensor;G and contracting it with the metric delivers a tensor, R = Riemann(D)·g, of rank G&tensor;G&tensor;G&tensor;G, antisymmetric in its first two factors by construction. The quoted identities claim (once translated into my terms):
Note that several of these properties are shared with (d^A)×(d^A) for any covector field A, e.g. the electro-magnetic covector potential.
In some later notes (Dr. Garry Gibbons, 15th lecture, my part III year) I see the Riemann tensor introduced as (in my terms) a [G,G,G,T]-tensor for which
This second course gives these three identities for any differential operator's Riemann tensor; it gives the τ[2,3,0,1] symmetry and the τ[1,0,2,3] anti-symmetry of R·g as arising for the differential operator which annihilates g. It also gives the following for the Ricci tensor, Ricci = τ[0,*,1,*](R):
and duly defines the Einstein tensor to be D's argument in this last equation.
The given derivations rely on use of normal co-ordinates at each point – i.e. such co-ordinates as render the Christoffel symbols zero at that point. I'm interested in seeing how far similar results can be obtained for general Leibniz operators. For these purposes, D (of rank G) is replaced by a tensor operator Q, of rank R, with torsion(Q) defined as antiSymm([R,R])&on;Q&on;Q, which can be shown to act as a linear map on each rank of the tensor bundle; then Riemann(Q) is (: torsion(Q) :R) and Ricci(Q) is obtained by taking its trace in analogous manner to D's case. The case where Q annihilates some linear (R: g |dual(R)) is considered as a special case of particular significance, analogous to D's annihilation of the metric.
Below, I mess around in search of Bianchi's identities for the general case. The antisymmetry under τ[1,0,2,3] arises trivially from the definition of Riemann(Q). Where possible, I do this using Riemann(Q) without contracting it with g.
That antisymmetrising over T's first three factors yields zero arises with slightly more effort, on condition of there being some basis, b, of rank(Q) for which Q&on;b's outputs are all symmetric, as a consequence of the corresponding truth for Riemann(Q), without any g-contraction (or, indeed, any reference to there being a g which Q annihilates).
I first show that cycling among the first three terms and summing the results delivers zero: from this, we have that the triply-symmetric and triply-antisymmetric parts of the Riemann tensor are both zero.
Thus, if we can ever find (in some neighbourhood of each point of M) a local basis, b, of Q's rank for which each value taken by Q&on;b is symmetric (throughout the neighbourhood), then the above is zero (since the three terms in the above sum are then, variously, (j,k)-, (i,j)- and (i,k)-symmetric whence their summed product with their wholly antisymmetric multiplier is zero). (For the case of a torsion-free differential operator, such a basis is readily generated from a chart.) In such a case, antisymmetrising over the first three factors of Riemann(Q) also yields zero.
When we contract Riemann(Q) with a [R,R]-rank tensor, g, which Q (of rank R) annihilates, we get a [R,R,R,R]-rank tensor Riemann(Q)·g which is [1,0,2,3]-antisymmetric, [0,1,3,2]-antisymmetric and annihilated by (τ[0,1,2] +τ[1,2,0] +τ[2,0,1]). So consider any [R,R,R,R]-rank tensor, H, annihilated by this last combination and anti-symmetric under [0,1,3,2] – i.e. with the same symmetries, but not bothering to assume the [1,0,2,3]-assymmetry – and infer:
which we can then substitute into itself to get
which we can re-arrange as
whence we infer that (τ[0,1,2,3]+τ[1,0,3,2])(H) is [2,3,0,1]-symmetric. Now, in the case of H = Riemann(Q).g, which is also [1,0,3,2]-symmetric, due to being antisymmetric under both [1,0,2,3] and [0,1,3,2], (τ[0,1,2,3]+τ[1,0,3,2])(H) is just 2.H; so its' being [2,3,0,1]-symmetric implies that H is too: thus Riemann(Q)·g is [2,3,0,1]-symmetric.
If rank(Q) is R and Z = antiSymm([R,R]), we can construe Riemann(Q)·g as a symmetric rank [Z,Z] tensor; if N = antiSymm([dual(R),dual(R)]), which is dual(Z), this can equally be construed as a linear (Z:|N), i.e. as a metric on N. (It can, indeed, equally be construed as a metric on dual(R)&tensor;dual(R), rather than on N, but it annihilates the symmetric part of such tensors, so its interesting action is on N.) Thus general relativity's Riemann tensor is a metric on two-dimensional area elements, for example.
Aside from the seductive option of looking for a rank-Z Leibniz operator which annihilates Riemann(Q)·g, thereby construing it as g's equivalent in a fresh layer of analogous analysis, what does this get us ? Well, first off, it implies that we can express Riemann(Q)·g as a sum(: k(i).z(i)×z(i) ←i |n) for some (Z: z |n) and ({scalars}: k |n) with n no bigger than Z's dimension. (Furthermore, k can be ({−1,+1}: k |n) if we want.) For general relativity, dim(R) is 4, so dim(Z) is 6 and we need at most six terms in this sum.
At the same time, we have antiSymm({R}:|dim(R)) one-dimensional, as is antiSymm({dual(R)}:|dim(R)) and a linear map induced from g between these, det(g) = (: wedge(g&on;u) ← wedge(dual(R):u|dim(R)) :) implying the existence of an antiSymm({R}:|dim(R)) tensor, μ, for which det(g) is ±μ×μ – this μ generalises the measure, which mediates integration, when g is the metric of space-time. We can construe μ as a linear map (antiSymm({R}:|a): μ |antiSymm({dual(R)}:|c)) whenever a+c is dim(R). In our favourite example, dim(R) is 4 and this allows us the special case (Z: μ |N), just like Riemann(Q)·g, though antisymmetric rather than symmetric. We're also guaranteed that this reading of μ is invertible, i.e. an isomorphism between Z and N, which is more than we can be sure of for Riemann(Q)·g.
A metric on area elements should let us integrate up the analogue, for
2-surfaces, of the lengths of lines. Presumably this would amount to
determining (something like) the mass enclosed by the 2-surface, in a way that
allows any 3-surface spanning the 2-surface as the 3-dimensional space-like
volume enclosed
– much as, in three dimensions, integrating curl
of a vector field over a 2-surface spanning a loop yields the vector field's
integral round the loop, independent of which spanning 2-surface you
chose.
Note that the natural
thing to integrate along a line is a 1-form;
if the 1-form is a gradient, the integral will only depend on the
end-points. This is analogous to integrating a 2-form over an area and the
result only depending on the bounding curve when the 2-form is a curl,
i.e. the result of applying d^ to some 1-form. A length, however, is arrived
at by integrating the square root of the result of applying a metric to the
(tensor) square of displacement along the line. What the Riemann tensor
offers for areas is analogous to this, not to the natural
integration
of a 2-form.
What is the action of Q on the Ricci tensor ? (now using R = rank(Q))
but antiSym([R,R,R])&on;Q&on;(:Q:R) = 0 in the cases of interest, and Q annihilates all τ operators,
Written by Eddy.