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Suppose we have a curvature-like tensor $R\in \wedge^2 T^*_p M \otimes T^*_p M \otimes T_p M$ on a manifold $M$, that is $R(X,Y)Z = - R(Y,X)Z$. How does one determine whether or not this is a curvature tensor for some torsion-free affine connection?

One obvious condition is that the first Bianchi identity $R(X,Y)Z + R(Y,Z)X + R(Z,Y)X = 0$ must be satisfied. Are there other conditions that can be checked?

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  • $\begingroup$ You may consider the references eudml.org/doc/74779, eudml.org/doc/82291, and sciencedirect.com/science/article/pii/S0021782409001317, and then look through the references therein. $\endgroup$
    – mdg
    Oct 16, 2015 at 20:19
  • $\begingroup$ @mdg: However, the references that you give do not address the problem of solving for a torsion-free affine connection, and this changes the character of the problem completely. It is not true in dimensions $3$ and $4$ (which is what those articles consider, but without the torsion-free condition, which doesn't make sense in their general context), that every curvature-like tensor on the tangent bundle that satisfies the first Bianchi identity is the curvature of a torsion-free affine connection, as my sketch below indicates. $\endgroup$ Oct 16, 2015 at 21:40
  • $\begingroup$ @Robert Bryant Are the DeTurck-Talvacchia papers, and other papers in the references, specific to dimensions 3 & 4? Are you saying to just dismiss the papers altogether? I think there may be some useful ideas in them... $\endgroup$
    – mdg
    Oct 16, 2015 at 22:42
  • $\begingroup$ @mdg: No, I'm not suggesting that they should be dismissed, it's just that the problems they treat won't have much to say about this particular problem. They do consider general dimensions at the beginning, but the bulk of those two papers are about the specific dimensions $3$ and $4$, where the problems can be cast as determined problems. Many of the references treat (other) problems in general dimension, but I can't think of anything in most of those papers (other than the general papers on exterior differential systems and/or formal integrability) that would apply to this specific problem. $\endgroup$ Oct 17, 2015 at 1:14

1 Answer 1

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Yes, there are typically many further conditions. In dimension $n$, the space of curvature-like tensors that satisfy the first Bianchi identity are the sections of a bundle of rank $\tfrac13n^2(n^2{-}1)$, while the space of torsion-free connections is the space of sections of an affine bundle of rank $\tfrac12n^2(n{+}1)$, so, when $n>2$, the partial differential equation for the connection $\theta$ given the curvature-like tensor $\Omega$, i.e., $\mathrm{d}\theta + \theta\wedge\theta = \Omega$, is typically over-determined and has no solutions. [Added comment: For a discussion about why over-determined systems (such as this one when $n>2$) generally have no solution, see Rigorous justification that overdetermined systems do not have a solution.] A trivial example of such an extra condition (when $n>2$) is that one must have $\mathrm{d}\bigl(\mathrm{tr}(\Omega)\bigr) = \mathrm{d}\bigl(\mathrm{tr}(\mathrm{d}\theta)\bigr) = 0$, but there are many more.

In fact, right away, one can see that the second Bianchi identity, i.e., $\mathrm{d}\Omega = \Omega\wedge\theta - \theta\wedge\Omega$ constitutes a number of inhomogeneous linear algebraic equations on $\theta$, and, when $n$ is sufficiently large ($n>3$ will do), these equations typically will have no solution, so such $\Omega$, even though they satisfy first Bianchi, cannot be the curvature of any connection, much less a torsion-free one.

Addendum: In my original answer, I wrote "This problem has been well-studied in the intermediate cases when $n$ is not too large (so that there are some interesting PDE problems to discuss).", but when the OP asked for references, I searched for a little while but couldn't find any that treated this particular problem. I am sure that I have seen this problem treated somewhere, but I couldn't turn anything up. Sorry about that. However, I can sketch the analysis via exterior differential systems, as it is straightforward. Perhaps this will be useful to the OP. Here is how it goes:

To understand the local solvability, suppose that a curvature-like tensor $R$ satisfying the first Bianchi identity has been specified on $M^n$. Choose a coframing $\eta = (\eta^i):TU\to \mathbb{R}^n$ on an open set $U\subset M$. (One could restrict to the case of a coordinate coframing, i.e., take $\eta^i = \mathrm{d}x^i$ for some local coordinate system $x = (x^i):U\to\mathbb{R}^n$, but the extra freedom of using a general coframing is sometimes useful.) Then $R$ will be represented by an $n$-by-$n$ matrix $\Omega = (\Omega^i_j)$ of $2$-forms on $U$, say $\Omega^i_j = \tfrac12 R^i_{jkl}\,\eta^k\wedge\eta^l$, where $R^i_{jkl}=-R^i_{ilk}$. The condition that $R$ satisfy the first Bianchi identity is simply that $\Omega\wedge\eta = 0$, i.e., that $R^i_{jkl}+R^i_{klj}+R^i_{ljk}=0$. (Note that the $R^i_{jkl}$ are $\tfrac13n^2(n^2{-}1)$ in number.) By Cartan's Lemma, the equations $\Omega\wedge\eta = 0$ imply that there exist $1$-forms $\rho^i_{jk}=\rho^i_{kj}$ on $U$ such that $\Omega^i_j = \rho^i_{jk}\wedge\eta^k$.

A torsion-free connection on $U$ will be represented, relative to the coframing $\eta$, by an $n$-by-$n$ matrix $\theta = (\theta^i_j)$ of $1$-forms on $U$ that satisfies the first structure equation $\mathrm{d}\eta = -\theta\wedge\eta$. We want to know when it is possible to choose $\theta$ so that it satisfies the second structure equation $\mathrm{d}\theta + \theta\wedge\theta = \Omega$.

Now, if $\mathrm{d}\eta^i = -\tfrac12 T^i_{jk}\eta^j\wedge\eta^k$, then we must have $$ \theta^i_j = (T^i_{jk} + p^i_{jk})\eta^k $$ for some (as yet unknown) functions $p^i_{jk}=p^i_{kj}$. (Note that the $p^i_{jk}$ are $N=\tfrac12n^2(n{+}1)$ in number.) Let us regard the $p^i_{jk}$ as fiber coordinates on the (trivial) bundle $V = U\times \mathbb{R}^N$ over $V$. We now work on $V$, where we have $\mathrm{d}\eta = -\theta\wedge\eta$.

Differentiating the first structure equation yields that the matrix $\Theta = \mathrm{d}\theta + \theta \wedge\theta $ satisfies $\Theta\wedge\eta = 0$, so it follows (again, by Cartan's Lemma) that there exist $$ \pi^i_{jk}=\pi^i_{kj} = \mathrm{d}p^i_{jk} + (\text{terms in $\eta$}). $$ so that $\Theta^i_j = \pi^i_{jk}\wedge\eta^k$.

Thus, we can write $$ \Upsilon = \mathrm{d}\theta+\theta\wedge\theta - \Omega = \bigl((\pi^i_{jk}-\rho^i_{jk})\wedge\eta^k\bigr) = \bigl(\tilde\pi^i_{jk}\wedge\eta^k\bigr), $$ and we see that the algebraic ideal generated by the components of $\Upsilon$ is generated by the $2$-forms $\Upsilon^i_j = \tilde\pi^i_{jk}\wedge\eta^k$. A solution to our problem will be a section $u:U\to V = U\times \mathbb{R}^n$, that pulls back $\Upsilon$ to be zero.

Now, when $n=2$, the algebraic ideal suffices because we are only looking for a $2$-dimensional integral manifold, and the exterior derivative of $\Upsilon$ is a $3$-form, which will necessarily vanish when pulled back via any section $u$. Now, it is easy to see, from the above description that, in this case, the Cartan characters of the system are $(s_1,s_2) = (4,2)$ and the space of integral elements at each point has dimension $S = 8 = s_1 + 2s_2$, so the system is involutive, and local solutions exist and depend on $s_2 = 2$ functions of $2$ variables (at least in the analytic case). (It is not hard to show that, in fact, local solutions always exist, even in the smooth case, when $n=2$, but let me skip over that discussion now. Basically, one can impose two conditions, such as $p^i_{jj}=0$, and then the restricted system becomes elliptic.)

However, when $n>2$, the ideal generated by the components of $\Upsilon$ is not differentially closed. In fact, one has $$ \mathrm{d}\Upsilon = \mathrm{d}(\Theta - \Omega) = \Theta\wedge\theta-\theta\wedge\Theta - \mathrm{d}\Omega = (\Omega\wedge\theta-\theta\wedge\Omega - \mathrm{d}\Omega) + (\Upsilon\wedge\theta-\theta\wedge\Upsilon), $$ so, to get a differentially closed ideal, one must add the components of the $3$-form $\Psi = (\Omega\wedge\theta-\theta\wedge\Omega - \mathrm{d}\Omega)$. Note that the $3$-form $\Psi$ does not involve any derivatives of $\theta$, and, in fact, is linear in the 'unknowns' $p^i_{jk}$. In fact, $$ \Psi^i_j = \tfrac12\bigl(R^i_{qkl}p^q_{jm}-R^q_{jkl}p^i_{qm} - S^i_{jklm} \bigr)\,\eta^k\wedge\eta^l\wedge\eta^m $$ for some functions $S^i_{jklm}$ on $U$, so it follows that the graph of any solution to the problem must lie in the affine `subbundle' of $V = U\times\mathbb{R}^N$ defined by the vanishing of these coefficients. Most of the time, when $n>3$, this is more equations than unknowns $p^i_{jk}$, and this locus will be empty. It is a nontrivial set of conditions on $R$ (and its derivative) that this locus be nonempty. (Of course, a (small) part of these conditions is that $\mathrm{d}\bigl(\mathrm{tr}(\Omega)\bigr) = 0$.)

When $n=3$, things are a little more interesting. Typically, as long as $\mathrm{d}\bigl(\mathrm{tr}(\Omega)\bigr) = 0$ and the tensor $R$ is 'algebraically generic' in the appropriate sense, the equations above will define an affine bundle $W\subset V$ of rank $9$ over $U$, and generically, the Cartan characters will be $(s_1,s_2,s_3) = (9,0,0)$. However, there will be nontrivial torsion, and the symbol will not be involutive, so you will get more obstructions at the next level. A detailed analysis is somewhat messy, but there are interesting special cases in which the curvature takes values in a simple subalgebra, such as ${\frak{so}}(3)$ or ${\frak{so}}(2,1)$.

Addendum 2: The OP asked for 'conditions that can be checked', and the above discussion does not really address that question. I had one further thought about this in the dimensions above $n=3$ that I thought that OP might find useful, so here it is:

When $n\ge 4$, there is an algorithmic way to proceed that works in the 'generic' case. What I mean by 'generic' is this: Say that a curvature-like tensor $R$ that satisfies the first Bianchi identity is algebraically generic if the kernel of the mapping $L(\phi) = \Omega\wedge\phi - \phi\wedge\Omega$, from $1$-forms with values in $n$-by-$n$ matrices to $3$-forms with values in $n$-by-$n$ matrices, consists of the elements of the form $\phi = \alpha\,I_n$, where $\alpha$ is a $1$-form (such elements are always in the kernel of $L$). It is not hard to show that, when $n\ge 4$, the generic curvature-like tensor $R$ that satisfies the first Bianchi identity is algebraically generic.

Suppose that one is given an algebraically generic $R$ and one wants to check whether it is the curvature of a torsion-free connection. Here are the steps:

  1. Test whether $\mathrm{d}\Omega$ is in the image of $L$, i.e., whether there exists a traceless matrix with $1$-form entries $\phi$ such that $\mathrm{d}\Omega = L(\phi) = \Omega\wedge\phi-\phi\wedge\Omega$. Because of the algebraically generic hypothesis, if $\phi$ exists (which is a matter of linear algebra), it will be unique. If no such $\phi$ exists, then there is no solution.

  2. Supposing that $\phi$ exists, consider the $2$-form $\mathrm{d}\eta + \phi\wedge\eta$. Either this can be written in the form $-\alpha\wedge\eta$ where $\alpha$ is a (scalar-valued) $1$-form, or it cannot. (When it can be written in this form, $\alpha$ is necessarily unique.) If it can, then $\theta = \phi + \alpha\,I_n$ is a torsion-free connection form that satisfies $\mathrm{d}\Omega = \Omega\wedge\theta-\theta\wedge\Omega$ (and it is the only such torsion-free connection form). If it cannot, there is no solution.

  3. Finally, check whether $\mathrm{d}\theta + \theta\wedge\theta = \Omega$. If this equation holds, then you have found the (unique) torsion-free connection that solves the problem. If this $\theta$ doesn't work, then there is no solution.

Thus, you can usually check whether your problem has a solution in dimensions greater than $3$. The only interesting cases that remain are those in dimension $3$ (where the problem can be subtle, since $L$ always has a large kernel) or algebraically special curvature-like tensors in higher dimensions.

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    $\begingroup$ Robert, thanks for your answer. Could you give some references for the work done on the intermediate cases? $\endgroup$ Oct 16, 2015 at 18:33

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