I met with the following problem. Consider real manifold $M^{2n}$ with operator field $R$ (that is the tensor field of type $(1,1)$). We are to find a symmetric connection $\Gamma^k_{ij}$ such that tensor $\nabla R$ is symmetric in lower indices.

We write an equation $\nabla R^k_{i,j} - \nabla R^k_{j,i} = 0$. Expanding it and using symmetry we get $$ \frac{\partial R^k_i}{\partial x^j} - \frac{\partial R^k_j}{\partial x^i} - \Gamma^k_{a b} T^{a b}_{ij} = 0, $$ where $T^{ab}_{ij} = \delta^a_i \otimes R^b_j - R^a_i \otimes \delta^b_j$. This tensor can be viewed as a map from $V \otimes V$ to itself. It is easy to check that this map has no kernel if $R$ has no real eigenvalues. Ok, we put this restriction on $R$ and get basically almost complex structure. Then we invert $T^{ab}_{ij}$ and get the needed connection.

Is it correct? Do I define the connection correctly?

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    $\begingroup$ I find tensorial notation confusing. You want $(\nabla R)(X,Y)=(\nabla R)(Y,X)$, or in expanded version $\nabla_X(R(Y))-R(\nabla_XY)=\nabla_Y(R(X))-R(\nabla_YX)$. I don't see any terms containing $\Gamma_{ij}^?$ or $\Gamma_{ji}^?$, so you are probably missing them. Or do you mean "torsion free connection" when you say "symmetric"? - in that case, your formula is probably ok. $\endgroup$ Dec 30, 2015 at 15:33

1 Answer 1


N.B.: I'm fixing my answer, which was off for two reasons: First, I didn't correctly interpret the OP's notation. (Thanks, Sebastian, for pointing that out!) Second, I didn't check the case when the Jordan normal form of $R$ has blocks of size $2$ or more (i.e., multiple eigenvalues but not multiple eigenvectors), and my original answer did not clearly treat this case.

By the way, even and odd dimensions have nothing to do with this problem.

In local coordinates $x^i$ and with connection coefficients $\Gamma^i_{jk}=\Gamma^i_{kj}$ (since the connection is assumed to be torsion-free (aka symmetric)), the formula for covariant derivative of a $(1,1)$-tensor is $$ \nabla\left(R^i_j\ \frac{\partial\ \ }{\partial x^i}\otimes \mathrm{d}x^j\right) = \left(\frac{\partial R^i_j}{\partial x^k} + R^\ell_j\Gamma^i_{\ell k}- R^i_{\ell}\Gamma^\ell_{jk}\right)\ \frac{\partial\ \ }{\partial x^i}\otimes \mathrm{d}x^j\otimes \mathrm{d}x^k. $$ Thus, you want to solve these equations: $$ R^\ell_j\Gamma^i_{\ell k}- R^\ell_k\Gamma^i_{\ell j} = \frac{\partial R^i_k}{\partial x^j} - \frac{\partial R^i_j}{\partial x^k} \qquad\text{and}\qquad \Gamma^i_{jk}=\Gamma^i_{kj}\,. $$

Note that the equations with upper index $i$ and upper index $i'\not=i$ do not interact, so what you really want to know is the conditions on a linear transformation $R = (R^i_j)$ in order that the equations $$ R^\ell_j S_{\ell k}- R^\ell_k S_{\ell j} = A_{jk} $$ be solvable for a symmetric form $S = (S_{jk})=(S_{kj})$ for any given anti-symmetric form $A = (A_{jk})= (-A_{kj})$. I.e., in matrix form, one wants to solve the equation $$ RS - SR^T = A $$ for a symmetric matrix $S$ given any anti-symmetric matrix $A$, so it's a question of when this linear map from symmetric matrices (a vector space of dimension $\tfrac12n(n{+}1)$) to anti-symmetric matrices (a vector space of dimension $\tfrac12n(n{-}1)$) is surjective. (For dimension reasons, it could never be an isomorphism.)

This problem is equivariant with respect to the action of $\mathrm{GL}(n,\mathbb{R})$, i.e., $$ aRa^{-1} (aSa^T) - (aSa^T)(aRa^{-1})^T = a(RS-SR^T)a^T, $$ so it suffices to put $R$ in Jordan normal form and check surjectivity of this linear map in this case.

What one finds is that the map is surjective if and only if there do not exist two distinct Jordan blocks belonging to the same eigenvalue. In other words, $R$ should not have two linearly independent eigenvectors belonging to the same eigenvalue. It's OK for $R$ to have generalized eigenspaces of dimension greater than $1$, but not a true eigenspace of dimension greater than $1$. An equivalent way to express this condition is that the minimal polynomial of $R$ have degree $n$. As expected, this is an open condition on $R$, since it asserts that the $n$ matrices $I,R,\ldots, R^{n-1}$ be linearly independent. Whether the eigenvalues be real or not is immaterial.

Note that, even when $R$ satisfies this condition, the solution $S$ is not unique; there is an $n$-parameter family of solutions. (Which leads, at each point of $M$, to an $n^2$ parameter family of solutions to the original problem.)

  • $\begingroup$ Thank you very much, for taking time to answer my question. When I was talking about the complex eigenvalues I meant the sufficient, not necessary conditions. You showed the iff-condition, which is really awesome. The thing is that in terms of torsionless connection $\nabla$ the Nijenhuis tensor of the operator field can be rewritten as $$ N_R (X, Y) = (\nabla R) (RX, Y) +(\nabla R) (X, RY) - (\nabla R) (RY, X) - (\nabla R) (Y, RX). $$ This means that if connections do exist the Nijenhuis tensor vanishes $\endgroup$ Jan 1, 2016 at 12:03
  • $\begingroup$ Hmmm.... This is hard to believe, because the above condition that guarantees the exisence of a torsion-free connection that makes $\nabla R$ symmetric is an open condition on $R$. If your formula were correct, then this would imply that the Nijenhius tensor of a $(1,1)$-tensor vanishes generically, and this isn't true. $\endgroup$ Jan 1, 2016 at 20:48
  • $\begingroup$ I know! But look at the formula $\endgroup$ Jan 1, 2016 at 21:24

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