# The other trace of the curvature tensor

I am using index notation here, since denoting traces in index notation is easier. Einstein summation convention assumed.

If $(M,g)$ is a Riemannian or pseudo-Riemannian manifold, and $$R^\rho_{\sigma\mu\nu}=\partial_\mu\Gamma^\rho_{\nu\sigma}-\partial_\nu\Gamma^\rho_{\mu\sigma}+\Gamma^\rho_{\mu\lambda}\Gamma^\lambda_{\nu\sigma}-\Gamma^\rho_{\nu\lambda}\Gamma^\lambda_{\mu\sigma}$$ is the Riemann curvature tensor, then the only independent trace of $R$ is the Ricci tensor $R_{\mu\nu}=R^\sigma_{\mu\sigma\nu}$, since the trace $R^\sigma_{\sigma\mu\nu}$ is zero.

If we are, on the other hand, given an arbitrary linear connection, it is necessarily a $\text{GL}(n,\mathbb{R})$-connection, and there is nothing specific to be said about the first two indices of the curvature tensor, so the tensor field $Q_{\mu\nu}=R^\sigma_{\sigma\mu\nu}$ is not necessarily zero.

What is there to be said about this tensor? What is its geometric meaning? What does it signify that for a Riemannian curvature tensor, this is zero?

I do realize that if $\nabla$ is an arbitrary $g$-compatible connection then, for and arbitrary frame $e_{a}$ (latin indices - frame indices, greek indices - coordinate indices) we have $$d^\nabla g_{ab}=0=dg_{ab}-\omega^c_{\ a}g_{cb}-\omega^c_{\ b}g_{ac},$$ so $dg_{ab}=\omega_{ba}+\omega_{ab}$, so if $e_a$ is an orthonormal frame then the connection forms are skew-symmetric, and then the curvature form $\Omega^a_{\ b}=d\omega^a_{\ b}+\omega^a_{\ c}\wedge\omega^c_{\ b}$ is also skew-symmetric. Moreover, since unlike $\omega$, $\Omega$ is gauge-covariant, this skew-symmetry is preserved even if calculated in a non-orthonormal frame.

Therefore, if $Q_{\mu\nu}$ does not vanish, then $\nabla$ cannot be metric-compatible for any metric I assume.

But I am curious about more info. Does the vanishing of $Q$ also imply that $\nabla$ is metric compatible for some metric? What else can be said about $Q$?

• The vanishing of $Q$ means that the curvature of the connection $\nabla$ on a smooth manifold $M$ takes values in ${\frak sl}(TM)$, or equivalently that $\nabla$ (at least locally) preserves a volume form. Of course, the Levi-Civita connection of a metric always locally preserves a volume form, namely the volume form of the restriction of the metric to an orientable neighborhood (endowed with either choice of orientation). An affine connection that preserves some volume form is sometimes called special, and such connections play an important role in projective differential geometry. – Travis Feb 2 '17 at 11:55
• The vanishing of $Q$ does not imply that $\nabla$ is metric-compatible: If $\nabla$ preserves a metric $g$, then its holonomy group $\textrm{Hol}_x(\nabla)$ at any point $x \in M$ is contained inside $\textrm{SO}(g_x)$, but it is possible to construct $\nabla$ for which this is not true for any $g$. – Travis Feb 2 '17 at 12:05

For any affine connection $\nabla$ on a smooth manifold, the curvature $R_{ab}{}^c{}_d$ may be uniquely decomposed as $$R_{ab}{}^c{}_d = C_{ab}{}^c{}_d + 2 \delta^c{}_{[a} {\mathsf P}_{b]d} + \beta_{ab} \delta^c{}_d \qquad (\ast)$$ for some totally tracefree $C$, called the projective Weyl tensor, and skew $\beta$; the tensor $\mathsf P$ is called the projective Schouten tensor. The First Bianchi Identity, $R_{[ab}{}^c{}_{d]} = 0$, implies that $-2 {\mathsf P}_{[ab]} = \beta_{ab}$.

Now, taking the trace over ${}^c{}_d$ gives $$Q_{ab} := R_{ab}{}^c{}_c = -2 {\mathsf P}_{[ab]} + n \beta_{ab} = (n + 1) \beta_{ab}.$$ Then, taking the trace of $(\ast)$ over ${}^c{}_a$ implies that $Q_{ab} = -2 R_{[ab]}$, so $Q$ is, up to a constant multiple, the skew part of the Ricci curvature of $\nabla$.

A more concrete geometric interpretation is this: Computing from $(\ast)$ gives that the curvature of the connection $\nabla$ induces on the anticanonical bundle $\Lambda^n TM$ is $Q$, or equivalently that the curvature of the connection induced on the canonical bundle $\Lambda^n T^*M$ is $-Q$. Thus, this bundle locally admits parallel sections, that is, $\nabla$ (locally) preserves a volume form on $M$, iff $Q = 0$, corresponding to the fact that (by definition) $Q = 0$ iff $R$ takes values in ${\frak sl}(TM)$. In particular, the Levi-Civita connection of any metric $g$ preserves the volume form of the restriction of that metric to any open orientable subspace (endowed with either choice of orientation), so $Q = 0$ for a Levi-Civita connection, or, like you say, for any metric connection.

Expanding the Second Bianchi Identity, $\nabla_{[e} R_{ab]}{}^c{}_d$, using $(\ast)$ implies that $dQ = 0$, so $Q$ defines a second cohomology class $[Q] \in H^2(M)$. If $\nabla$ is torsion-free, any connection projectively equivalent to $\nabla$, that is, sharing the same (unparameterized) geodesics as $\nabla$, has the form $$\hat\nabla_a \xi^b = \nabla_a \xi^b + \Upsilon_a \xi^b + \Upsilon_c \xi^c \delta^b{}_a$$ for some $\Upsilon \in \Gamma(T^*M)$ (and any choice of $\Upsilon$ gives projectively equivalent connections). The corresponding tensors $Q, \hat Q$ are related by $\hat Q = Q + 2 (n + 1) d\Upsilon$, and in particular, they differ by an exact form. Thus, the cohomology class $[Q] = [\hat Q]$ is actually an invariant of the projective structure---that is, the equivalence class of projective equivalent connections---that $\nabla$ defines. On the other hand, this transformation rule for $Q$ shows that locally $\nabla$ is projectively equivalent to one with $Q = 0$ (such connections are sometimes called special). So, in the setting of local projective differential geometry, we may as well just work with special connections, which enjoy the convenient feature that ${\mathsf P}_{ab}$ and $R_{ab}$ are symmetric.

This formulation can be found, by the way, in $\S$3 of the following reference:

T. Bailey, M. Eastwood, A.R. Gover, "Thomas' structure bundle for conformal, projective and related structures" Rocky Mountain J. Math, 24 (1994), 1191-1217.

It is not true that vanishing of $Q$ implies that $\nabla$ is a Levi-Civita connection, or even that it is projectively equivalent to one. Naively one should expect as much: Vanishing of $Q$ implies that the (local) holonomy group of $\nabla$ based at any point $x$ is contained in $\textrm{SL}(T_x M)$, but if $\nabla$ is the Levi-Civita connection of a metric $g$, the holonomy group must be contained in the much smaller group $\textrm{SO}(g_x)$.

A simple example is the connection $\nabla$ on $\Bbb R^3$ whose nonzero Christoffel symbols are specified (in the canonical coordinates $(x^a)$) by $\Gamma_{21}^3 = \Gamma_{31}^2 = x^2$. (The projective structure this connection defines is the so-called Egorov projective structure, which is interesting for other reasons, too.) The nonzero components of curvature are specified by $R_{23}{}^1{}_2 = -R_{32}{}^1{}_2 = -1$, so $\nabla$ is special, but $\S$2.3 of the below reference shows that it is not a Levi-Civita connection, nor even projectively equivalent to one.

M. Dunajski, M. Eastwood, "Metrisability of three-dimensional path geometries", European J. Math (2015), 809-834.