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$\DeclareMathOperator{End}{\mathrm{End}}$ This question is already asked here Second Bianchi identity on tangent bundle but with no answer.

Let $M$ be a smooth manifold, and $E \to M$ a smooth vector bundle over $M$. Let $\nabla$ be a connection on $E$ (and using the same notation for any induced connections on any related bundles), with $F^\nabla$ its curvature (an $\End(E)$ valued 2-form) and $d^{\nabla}$ its exterior covariant derivative.

In the lecture notes I am reading, several equivalent characterizations of $F^{\nabla}$ are proved, probably the most useful for this question is $F^{\nabla}(X, Y)\sigma = \nabla_X(\nabla_Y \sigma) - \nabla_Y(\nabla_X \sigma) - \nabla_{[X, Y]}\sigma$ for all $X, Y \in \mathfrak X(M)$, $\sigma \in \Gamma(E)$.

Then, the (second) Bianchi identity is then shown $$ d^{\nabla}F^{\nabla} = 0 $$ (where $d^{\nabla}$ is the induced exterior covariant derivative on $\End(E)$ valued forms).

Later, specializing to the case $E = TM$ (but without choosing a metric), where now we specifically have $F^{\nabla} \in \Gamma(\Lambda^2(T^*M) \otimes T^*M \otimes TM)$, the notes say "one can compute that the Bianchi identity now can be expressed as $$(\nabla_X F)(Y, Z) + (\nabla_Y F)(Z, X) + (\nabla_Z F)(X, Y) = 0$$ for $X, Y, Z \in \mathfrak X(M)$", where we have written $$ F^{\nabla} = \frac{1}{2}F_{kij}^l e_l \otimes e^k \otimes (e^i \wedge e^j) $$ with $F^l_{kij}e_l = F^{\nabla}(e_i, e_j)e_k$, where $\{e_1, \dots, e_n\}$ a local frame for $M$ and $\{e^1, \dots, e^n\}$ its coframe.

I have tried to do this computation (just expanding definitions in terms of the characterization I gave above), but got rather lost in the sauce with all of the cyclic sums of vector fields. Is there a writeup somewhere of the direct proof of the equivalence of these identities, preferably in a relatively invariant way?

Alex Pawelko
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  • it’s easy enough (I’ll write it up more precisely later if necessary): first note that if you write locally $F=\sum_{a<b}F_{ab},dx^a\wedge dx^b$ (each $F_{ab}$ being local sections of $E$) then Bianchi’s identity $d_{\nabla}F=0$ becomes $\sum_c\sum_{a<b}\nabla_{\partial_c}(F_{ab}),dx^c\wedge dx^a\wedge dx^c$, and then rearranging this slightly, you get $\nabla_{a}(F_{bc})+\nabla_b(F_{ca})+\nabla_c(F_{ab})=0$. The only reason I go down to local coordinates is because I don’t want to deal with any Lie brackets. – peek-a-boo Jun 14 '24 at 23:46
  • Now, fix a torsion-free connection on $TM$; then you get $\sum_{\text{cyclic}}(\nabla_aF)(\partial_b,\partial_c)=0$ (in going from the $\nabla_a(F_{bc})$ to $(\nabla_aF)(\partial_b,\partial_c)$ form, you apply the Leibniz rule, and then all those extra terms will cancel out in the cyclic sum), and hence by $C^{\infty}(M)$-linearity, we get this for all vector fields $X,Y,Z$, so $\sum_{\text{cyclic}}(\nabla_XF)(Y,Z)=0$. In particular, you don’t need to expand $F$ into your $(l,k)$ ‘endomorphism indices’. – peek-a-boo Jun 14 '24 at 23:47
  • @peek-a-boo Why not post this as an answer? – Ted Shifrin Jun 15 '24 at 03:24
  • @TedShifrin I was on a lunch break when I wrote that (and I wanted to elaborate slightly in an answer :) – peek-a-boo Jun 15 '24 at 08:48

1 Answers1

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See here (and the various sublinks) for any background about the exterior covariant derivative and how to do computations with it.

Fix a local coordinate chart $(U,x=(x^1,\dots, x^n))$ on $M$, and expand $F=\frac{1}{2!}F_{ab}\,dx^a\wedge dx^b$ (I have the extra $\frac{1}{2!}$ because I didn’t restrict to $a<b$), with each $F_{ab}=F\left(\frac{\partial}{\partial x^a}\wedge \frac{\partial}{\partial x^b}\right)$ being a local section of $\text{End}(E)$. Then, Bianchi’s identity in local coordinates becomes \begin{align} 0&=d_{\nabla}F\\ &=\frac{1}{2!}d_{\nabla}(F_{ab})\wedge dx^a\wedge dx^b\\ &=\frac{1}{2!}\nabla_{c}(F_{ab})\,dx^c\wedge dx^a\wedge dx^b\tag{!}\\ &=\sum_{a,b,c}\left[\nabla_a(F_{bc})+\nabla_b(F_{ca})+\nabla_{c}(F_{ab})\right]\,dx^a\otimes dx^b\otimes dx^c, \end{align} where the last step is some simple enough algebra with permuting stuff. Since the collection of all $dx^a\otimes dx^b\otimes dx^c$ forms a basis for $(0,3)$ tensor fields it follows the coefficient vanishes identically (granted, here we’re talking about bundle-valued objects, but still, the argument goes through). Note that I could not simply conclude from $(!)$ that $\nabla_{c}(F_{ab})=0$ because the sum in that step is over all $c,a,b$, note just those for which $c<a<b$ (which is important in order for the collection of $dx^c\wedge dx^a\wedge dx^b$ to form a basis). So, the upshot of all this is that for all $a,b,c\in\{1,\dots, n\}$, \begin{align} \nabla_a(F_{bc})+\nabla_b(F_{ca})+\nabla_{c}(F_{ab})&=0.\tag{$*$} \end{align} Note that algebraically speaking, this is essentially the same conclusion you’d arrive at if you had a usual scalar-valued 2-form $F$ and you wrote out $dF=0$ explicitly in coordinates: $\frac{\partial F_{bc}}{\partial x^a}+\frac{\partial F_{ca}}{\partial x^b}+\frac{\partial F_{ab}}{\partial x^c}=0$. In this sense, the bundle-valued expression for Bianchi’s identity is not any more complicated.

Next, fix a torsion-free connection $\nabla$ on $TM$, and denote all induced connections by the same symbol $\nabla$. Then, essentially by definition, for all (local) vector fields $X,Y$ on $M$ we have (as an equality of (local) sections of $\text{End}(E)$) \begin{align} \nabla_X\left(F(Y,Z)\right)&=(\nabla_XF)(Y,Z)+F(\nabla_XY,Z)+F(Y,\nabla_XZ),\tag{$**$} \end{align} where the $\nabla_X$’s act respectively on (local) sections of $\text{End}(E),\text{Hom}(\bigwedge^2(TM);\text{End}(E)),TM,TM$. Now, keeping in mind that the connection on $TM$ being torsion-free (and the fact that coordinate vector fields have vanishing Lie bracket) implies that $\nabla_{\partial_a}(\partial_b)=\nabla_{\partial_b}(\partial_a)$, and so applying $(**)$ with coordinate vector fields and plugging it into $(*)$ gives (after a total of 6 terms cancelling out) \begin{align} (\nabla_{\partial_a}F)(\partial_b,\partial_c)+ (\nabla_{\partial_b}F)(\partial_c,\partial_a)+ (\nabla_{\partial_c}F)(\partial_a,\partial_b) &=0. \end{align} Now, by $C^{\infty}(M)$-linearity, we get that for all vector fields (or simply tangent vectors) $X,Y,Z$, \begin{align} (\nabla_XF)(Y,Z)+ (\nabla_YF)(Z,X)+ (\nabla_ZF)(X,Y) &= 0.\tag{$***$} \end{align} This shows that $d_{\nabla}F=0$ implies $(***)$ for a torsion-free connection on $TM$. Undoing these steps shows the converse, hence these are equivalent expressions of Bianchi’s identity.

Now of course you can specialize to $E=TM$ and fix a single torsion-free connection and the entire argument above goes through. And finally if you have a semi-Riemannian manifold $(M,g)$ and you equip it with the Levi-Civita connection, then you have $(***)$ as one version of Bianchi’s identity. People often consider not just the curvature as an endomorphism-valued 2-form (or equivalently, the corresponding $(1,3)$ tensor field), but the corresponding $(0,4)$ tensor field $R$ defined by metric duality. Then, $\nabla R$ will be a $(0,5)$ tensor-field, and one has the version of Bianchi identity (e.g Lee’s Introduction to Riemannian Geometry, Proposition 7.13) \begin{align} (\nabla R)(V,W,X,Y,Z)+ (\nabla R)(V,W,Y,Z,X)+ (\nabla R)(V,W,Z,X,Y) &=0. \end{align} This follows essentially trivially from above because the metric isomorphisms commute with the covariant derivative (by metric-compatibility); also notice that the vector fields $V,W$ remain untouched (these correspond to the endomorphism nature of the curvature originally).

peek-a-boo
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