# Characterizing Hessians among symmetric bilinear tensors

I apologize in advance if this is somewhat elementary, but:

Let $(M,g)$ be a compact Riemannian manifold. Is there a "characterization" of which symmetric bilinear tensors $B\in Sym^2(M)$ are Hessians of functions $f\colon M\to \mathbb R$?

In other words, for which $B\in Sym^2(M)$ there exists $f\colon M\to\mathbb R$ such that $B(X,Y)=Hess \ f(X,Y)$, where $Hess\ f(X,Y)=g(\nabla_X \nabla f,Y)=X(Y(f))-\mathrm df(\nabla_X Y)$?

Obviously, since $M$ is assumed compact, any continuous function will attain a min and a max, so a symmetric bilinear form that is a candidate to Hessian cannot be definite. This maybe pretty naive, but I cannot think of any other property that distinguishes Hessians among bilinear symmetric tensors in this context... For example, it doesn't seem to be the case that $B$ has to be parallel, or satisfy any other sort of similar "nice" conditions.

Note that even if we impose the extra condition that $B$ is the Hessian of a Morse function, I don't quite see if the topology of $M$ will impose any properties on $B$ via Morse theory, since $B$ does not know if it is at a (future) critical point or not, right?

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I know that the Hessian isn't exactly the differential of something, but shouldn't there be an analogue to closedness of forms for it? Then there would be topological obstructions for the implication "closed => exact" of course. – Johannes Hahn Mar 5 '13 at 17:52
@Renato I can not understand some thing in your formula $g(\nabla_X \nabla f,Y)=X(Y(f))-\mathrm df(\nabla_X Y)$, so I think there is a contradiction in this formula: fix a point $p \in M$. the left side depends only on $Y(p)$ but the right side depends on the value of $Y$ in a neighborhood of $p$ not just on $Y(p)$. Am I missing some thing? – Ali Taghavi May 20 '14 at 8:57
@AliTaghavi: Yes, you are missing something. The expression $X(Y(f))-\mathrm{d}f(\nabla_XY)$ is linear in $Y$ over the smooth functions, so its value at $p$ depends only on the value of $Y$ at $p$. You can see this by replacing $Y$ by $hY$ for some smooth function $h$; the term $X(h)Y(f)$ that is generated by the first term in the formula cancels the term $\mathrm{d}f(X(h)Y) = X(h)Y(f)$ generated by the second term. – Robert Bryant May 20 '14 at 12:26

## 3 Answers

There are local conditions, but they typically involve the curvature tensor of the underlying metric. For example, if the metric is flat, so that one can choose orthonormal coordinates $x_i$ in which the covariant derivatives are the ordinary derivatives, then the condition that a quadratic form $H = h_{ij}\ dx_idx_j$ be a Hessian is just that $$\frac{\partial h_{ij}}{\partial x_k} = \frac{\partial h_{ik}}{\partial x_j} \ .$$ This is an overdetermined system, and, if it is satisfied, there are solutions $f$ of $$\frac{\partial^2 f}{\partial x_i\partial x_j} = h_{ij}$$ and they are uniquely determined (locally) up to the addition of a function linear in the $x_i$.

In the more general case, one has local conditions in terms of the curvature of $g$, so the answer is more complicated (and more interesting). I calculate using moving frames (see below at the end for the 'Nomizu-style' interpretation), so I would explain it this way:

Let $g = {\omega_1}^2 +\cdots + {\omega_n}^2$ be the expression of $g$ in a local orthonormal coframing $\omega$. There will exist unique $1$-forms $\theta_{ij}=-\theta_{ji}$ so that (using the summation convention) $d\omega_i = -\theta_{ij}\wedge\omega_j$. (These are the first structure equations.)

If $f$ is a function defined in the domain of the coframing, one will have $$df = f_i\ \omega_i \qquad\text{and}\qquad df_i = -\theta_{ij}\ f_j + f_{ij}\ \omega_j$$ and, by definition, $\mathrm{Hess}_g(f) = f_{ij}\ \omega_i{\circ}\omega_j$.

To determine whether a given symmetric quadratic form $H = h_{ij}\ \omega_i{\circ}\omega_j$ can be written as $H = \mathrm{Hess}_g(f)$ for some $f$, one computes the exterior derivative of the equations $df_i = -\theta_{ij}\ f_j + h_{ij}\ \omega_j$ and finds that one must have $$h_{ikl}-h_{ilk} = -R_{ijkl}\ f_j\ , \tag{1}$$ where $dh_{ij} = -\theta_{ik}\ h_{kj} + \theta_{kj}\ h_{ik} + h_{ijk}\ \omega_k$ and the $R_{ijkl}=-R_{ijlk}$ are the components of the Riemann curvature tensor in this coframing, as defined by the second structure equations $$d\theta_{ij} = -\theta_{ik}\wedge\theta_{kj} + \tfrac12\ R_{ijkl}\ \omega_k\wedge\omega_l\ .$$

The system (1) gives necessary conditions for $f$ to exist, since, in most cases, it completely determines the only possible functions $f_i$ that could be its derivatives in this coframing.

It may be worth pausing to interpret (1) as a global equation. In Nomizu-style notation, one has $\mathrm{Hess}_g(f) = (\nabla\nabla f)^{\flat\flat}$, but I am going to ignore the distinction between $T=TM$ and $T^\ast=T^*M$ since we have a metric $g$ that is $\nabla$-parallel and just write $\mathrm{Hess}_g(f) = \nabla\nabla f$. Applying the Bianchi identities, one has $$\sigma\bigl(\nabla\bigl(\mathrm{Hess}_g(f)\bigr)\bigr) = \mathsf{R}(\nabla f),$$ where $\sigma:\mathsf{S}^2(T^\ast)\otimes T^\ast\to T^\ast\otimes\Lambda^2(T^\ast)$ is the canonical skew-symmetrization operator, and $\mathsf{R}:T\to T^\ast\otimes\Lambda^2(T^\ast)$ is the usual mapping induced by the curvature operator. In particular, if $H = \mathrm{Hess}_g(f)$, then one must have $$\sigma\bigl(\nabla H\bigr) = \mathsf{R}(\nabla f)\tag{1'}$$ which, in my moving frames notation, is the system (1).

Now, the image of $\sigma$ is, as usual, the kernel of the further skew-symmetrization map $\sigma: T^\ast\otimes\Lambda^2(T^\ast)\to \Lambda^3(T^\ast)$ and, fortunately, the first Bianchi identity implies that $\mathsf{R}$ also takes values in this kernel, which has dimension $n\cdot \tfrac12 n(n{-}1) - \tfrac16 n(n{-}1)(n{-}2)= \tfrac13 n(n^2{-}1)$.

We already know what happens when $\mathsf{R}\equiv0$, namely, the first order system of equations $\sigma\bigl(\nabla H\bigr)=0$ are necessary and sufficient that $H$ be locally expressible as a Hessian.

However, the 'generic' situation is that $\mathsf{R}$ is injective, so let's consider that case. (I'll say a bit more about what happens in the intermediate cases at the end.) When $\mathsf{R}$ is injective, let $\mathsf{R}(T)\subset T^\ast\otimes\Lambda^2(T^\ast)$ be the image subbundle of rank $n$. The equations $(1')$ then say that $H$ must satisfy the system of $\tfrac13 n(n^2{-}1) - n = \tfrac13 n(n^2{-}4)$ first order equations $$\sigma(\nabla H) \equiv 0\ \text{modulo}\ \mathsf{R}(T).\tag{1''}$$ Assuming that these equations do hold, then there is a unique vector field $F(H)$ on $M$ such that $$\sigma(\nabla H) = \mathsf{R}\bigl(F(H)\bigr),$$ and this $F(H)$, which is a linear, first order differential expression in $H$ (whose coefficients involve the curvature of $g$), is the only possible candidate for $\nabla f$. However, in order for this to solve our problem, it must satisfy $$\nabla \bigl(F(H)\bigr) - H = 0,\tag{2}$$ and this is a system of $n^2$ first-order equations on $F(H)$, which is, of course, a system of $n^2$ second-order equations on $H$. If $H$ does satisfy this system, then, because $\nabla\bigl(F(H)\bigr)=H$ is symmetric, it will follow that $F(H)$ is (locally) a gradient vector field of some function $f$, uniquely determined up to an additive constant.

For example, when $n=2$ and the Gauss curvature of $g$ is nowhere zero, the equations $(1'')$ are trivial since $\mathsf{R}(T) = T^*\otimes\Lambda^2(T^\ast)$. Thus, the result of this analysis is that there exists a linear, second order differential operator $\mathsf{S}_g$ from symmetric quadratic forms to quadratic forms, namely $\mathsf{S}_g(H) = \nabla\bigl(F(H)\bigr)-H$, such that a symmetric quadratic form $H$ is (locally) a Hessian with respect to $g$ if and only if $\mathsf{S}_g(H)=0$. Moreover, when the first deRham cohomology group of $M$ vanishes and $\mathsf{S}_g(H)=0$, there will exist a global function $f$ such that $\mathrm{Hess}_g(f)=H$, and this $f$ is unique up to an additive constant.

When $n>2$, the equations $(1'')$ are never trivial, so the condition that $H$ be a Hessian with respect to $g$ involves first order conditions and second order conditions. What is interesting is that, when $n$ is sufficiently large and $\mathsf{R}$ is sufficiently 'generic', it appears (I haven't checked for sure) that $(1'')$ can actually imply $(2)$, so that the conditions become first order again. (This is not true when $n=3$, though.)

Finally, if $\mathsf{R}$ is not injective, one may have to go to higher order derivatives of $H$ to determine whether it is a Hessian. This is an interesting case, but I don't have time to go into it right now.

Remark: Of course, the metric $g$ is not really essential in this story, since it really depends more on $\nabla$, than on $g$. The equation $\nabla(df) = H$ makes sense without any metric and is a reasonable equation, as long as $\nabla$ is a torsion-free connection on $M$. Thus, one could ask for characterizations of those symmetric quadratic differentials $H$ that can be written in the form $H = \nabla(df)$ where $\nabla$ is a given torsion-free connection on $M$. The answer in this case would have essentially the same form as the answer above since, if one carries out the computations carefully, one sees that the metric $g$ plays essentially no rôle in the problem.

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Nice answer! I guess that the translation to my language is that the local integrability of the Lagrangian plane field involves the curvature of the Ehresmann connection. – alvarezpaiva Mar 5 '13 at 18:09

Given $f : N \to \mathbb R$, there is its derivative $Df : TN \to \mathbb R$, which has its dual gradient $\nabla f: N \to TN$, which you can covariantly differentiate $c \circ D\nabla f : TN \to TN$, and this map is dual to the Hessian $Hf : TN \oplus TN \to \mathbb R$. From this perspective there's two things to determine. Here $c : T^2 N \to TN$ is the connection, in the Ehresmann formalism.

1) Your Hessian is adjoint to a bundle map $TN \to TN$, is this the covariant derivative of a vector field on $N$?

2) If the answer to (1) is yes, among the solution vector fields from (1) is there a vector field which is the gradient of a real-valued function on $N$?

(2) Has the traditional cohomological answer so I'll focus on (1).

A bundle map $TN \to TN$ is the covariant derivative of a vector field $N \to TN$ means that you could write the vector field as a type of holonomy integral, by adding to your parallel transport an integral of the bundle map $TN \to TN$. So there will be a local triviality condition on this holonomy integral, as well as a global triviality condition.

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This is more of a comment/thinking aloud than an answer, but perhaps the symplectic viewpoint helps:

Think of the Hessian with respect to a Riemannian metric as follows: through the Legendre transfom, the Levi-Civita connection can be seen as a splitting of the tangent bundle of the cotangent bundle of $M$ into vertical and horizontal bundles: $T(T^*M) = V(T^*M) \oplus H(T^*M)$. At each point $p_x \in T^*M$, the subspace $V_{p_x}(T^*M) = V_{p_x}$ and $H_{p_x}(T^*M) = H_{p_x}$ are Lagrangian subspaces.

If $f$ is a function on $M$, the graph of its differential is a Lagrangian submanifold $L \subset T^*M$. The tangent space of $L$ at the point $df(x)$ is a Lagrangian subspace of $T_{df(x)} T^*M$ which is transverse to the vertical space $V_{df(x)}$ and, therefore, we can think of $T_{df(x)}L$ as the graph of a linear transformation from $H_{df(x)}$ to $V_{df(x)}$. Modulo some basic identifications, this linear map is the Hessian of $f$ at $x$.

A nice thing about this description is that it works on Finsler spaces since the splitting $T(T^*M) = V(T^*M) \oplus H(T^*M)$ just comes from the linearlization of the geodesic flow and the convexity of the Hamiltonian.

If you start with a Hessian, this description suggests the following procedure to construct a function $f$: given a point $p_x \in T^*M$, use the Hessian of $f$ at $x$, the identification of $V_{p_x}$ and $H_{p_x}$ with $T_xM$ and the decomposition $T_{p_x}T^*M$ to construct a Lagrangian subspace in $T_{p_x}T^*M$. One ends up with a field of Lagrangian subspaces in $TT^*M$. We have to find an integral manifold $L$. Such a manifold will be Lagrangian by construction and, given that the constructed Lagrangian subspaces are transverse to the vertical subspaces, it will also project in a locally-diffeomorphic way onto $M$. It's a good candidate for the graph of $df$.

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