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I'm looking at a paper of J. A. Thorpe ("Some Remarks on the Gauss-Bonnet Integral"). In the paper he defines "higher-order" notions of curvature. One thinks of the usual curvature tensor as $R\in \Lambda^2T^*M\otimes\Lambda^2T^*M$ and then we take simultaneous wedge products of $R$ with itself $k$ times in each of these tensor component to get $R^k\in\Lambda^{2k}T^*M\otimes\Lambda^{2k}T^*M$. Thorpe calls this (up to a positive multiple which won't concern us) $R_p$, where $p=2k$. It has the additional property of being symmetric in the transpose of the $2k$-forms over the tensor product, which just follows from the analogous symmetry of the ordinary Riemann curvature.

Thorpe then proves a lemma about this $R_p$ which he calls the (first) Bianchi identity for $R_p$:

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For $p=2$ i.e. when we're just dealing with the ordinary curvature tensor this is just the familiar first Bianchi identity (by applying the typical curvature symmetries), and by looking carefully at the definition this general one just follows from this special case. All fine and good so far.

But then Thorpe says the following:

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This has been seriously bothering me over the past 24 hours; I'm a slow thinker when it comes to combinatorics but I don't see how one can reduce the full antisymmetrization to just these $p+1$ terms. The full antisymmetrization has $(2p)!$ terms and the alternating property of $R_p$ in the first and last $p$ variables should allow one to cut this number down by a factor of $p!p!$; additionally the transpose symmetry should allow to halve that, leaving $\frac{1}{2}\binom{2p}{p}$ distinct kinds of terms to reduce to. (This seems to agree in the case when $p=2$ and we're working with the ordinary curvature. There are $24$ terms if we full antisymmetrize the Riemann tensor and $\frac{1}{2}\binom{4}{2}=3$, which is the number of terms in the typical cyclic sum statement of the Bianchi identity.)

I would just think that maybe this is wrong and not a big deal but Kulkarni has a paper (On the Bianchi Identities) with a very similar statement of the Bianchi identity in the formula below (apologies, but I won't bother to try to elaborate on the notation here):

enter image description here

Can anyone clarify what is going on here? Is it correct that Thorpe's statement (5') follows from his statement (5) and I just can't see it?

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  • $\begingroup$ Cited by 110 -- perhaps one of them has taken the trouble to fill in the gap? $\endgroup$ – darij grinberg May 23 '18 at 10:21
  • $\begingroup$ My impression is that the equality (5'), combined with the antisymmetry of $R_p$ in the first $p$ arguments and in the last $p$ arguments, should show that $R_p$ factors through some Schur functor (probably the one corresponding to the two-column partition whose both columns have size $p$). This should at least yield (5). Now, conversely, to get (5') from (5), I don't think the given data suffices; probably the construction of $R_p$ as the $k$-th power of $R$ matters. $\endgroup$ – darij grinberg May 23 '18 at 10:25
  • $\begingroup$ @darijgrinberg Def (5') implies (5). Not hard to see that up to a multiple the LHS of (5') is the same as antisym in the first $p+1$ arguments. If $R^k$ vanishes under a restricted antisym like this it will vanish under a full antisym. I agree with you that the data is insufficient to go from (5) to (5') on an a priori level. It is true that (5') holds for $R^k$ (see Kulkarni's paper ref'd above), but it requires independent proof. What is true is that antisym over $2p-1$ indices is a priori equivalent to full antisym. I suspect Thorpe just got overzealous stating such a property. $\endgroup$ – Brian Klatt May 23 '18 at 18:37

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