It is well-known that the curvature properties of Kähler metrics "flow downward". To give meaning to this statement, let us say that a tensor \( \alpha \) dominates a tensor \( \beta \) if any of the following implications hold: \begin{align*} \alpha > 0 &\implies \beta > 0, & \alpha &\geq 0 \implies \beta \geq 0, \\ \alpha < 0 &\implies \beta < 0, & \alpha &\leq 0 \implies \beta \leq 0. \end{align*} Strictly speaking we need to specify what it means for a tensor to be "positive" or "negative" for this to make sense. This is fairly clear for the tensors we’ll discuss here, except for the Kähler curvature tensor itself which we mean to be Griffiths positive (or negative).

If \( R \) is the curvature tensor of a Kähler metric, its various derived curvature tensors are the holomorphic sectional curvature \( H \), the Ricci tensor \( r \) and the scalar curvature \( s \). Then \( R \to H \to s \) and \( R \to r \to s \), where an arrow \( \alpha \to \beta \) means that \( \alpha \) dominates \( \beta \).

The witches of the field know that neither \( H \) nor \( r \) dominates the other. The best evidence for this seems to be indirect. We for example know that Hirzebruch surfaces (surfaces that are projective bundles over \( \mathbf{P}^1 \)) carry metrics with positive holomorphic sectional curvature, but carry no metric of positive Ricci curvature as their anticanonical bundle is not ample. There are some suggestive results in the negatively curved case though, where a compact Kähler manifold that carries a metric of negative holomorphic sectional curvature also carries a possibly different metric of negative Ricci curvature. We don’t know if these are the same metric. It would be nice if some easy local calculations said they were, but in this note we will see they don’t.

It is also folklore that all of the reverse relations in the diagram fail. Some do not hold by general considerations; for example Hirzebruch surfaces again carry a metric of positive holomorphic sectional curvature but not one of positive Griffiths curvature, for otherwise they would be \( \mathbf{P}^2 \) by Siu—​Yau’s solution of the Frankel conjecture and, lo, they are not.

We feel like it would be nice to see explicitly what doesn’t hold here. In this light reading note we work out local examples that show that the reverse of any known positivity implication fails. As other people have noted it would be nice to have compact examples of these failures but it is fairly difficult to get one’s hands on those. Suggestions are welcome.

Let \( V \) be a complex vector space of dimension \( n \), which we think of as the tangent space of a complex manifold at a given point. The curvature tensor \( R \) of a Hermitian metric on the manifold identifies with a Hermitian form \( q \) on \( V \otimes V \), defined by \[ R(x, \overline{y}, z, \overline{w}) = q(x \otimes z, \overline{y \otimes w}). \] If the metric is Kähler we get an additional symmetry \( R(x, \overline{y}, z, \overline{w}) = R(z, \overline{y}, x, \overline{w}) \) (and the ones induced by conjugating).

We write \( B(V) \) for the real vector space of Hermitian forms on \( V \). The curvature tensor of a Hermitian metric is then just a member of \( B(V \otimes V) \). We’ll call such an element an algebraic curvature tensor, and one that satisfies the symmetries of a Kähler curvature tensor an algebraic Kähler curvature tensor.

Proposition If \( R \) is an algebraic Kähler curvature tensor, then \( R \to H \to s \) and \( R \to r \to s \).

This is both well-known and mostly easily proved by taking traces in an orthonormal basis. The trickiest impliciation is to show that \( H \to s \). Berger gave the original proof by showing that \[ \frac{1}{\operatorname{Vol} S^{2n-1}} \int_{S^{2n-1}} \!\!\! H(v) \, d\sigma(v) = \frac{1}{\binom{n+1}{2}} s \] by writing \( v = \sum_{j=1}^n v_j e_j \) in an orthonormal basis \( (e_j) \), expanding the curvature tensor \( R(v, \bar v, v, \bar v) \), and integrating polynomials in \( v_j \) over the sphere. (The reader who wants to try their hand at this can see Folland’s paper on integrating polynomials over spheres for the last part.)

Let’s define \begin{align*} \mathcal{H}(V) &= \{ \text{germs of smooth Hermitian metrics on \( V \) at \( 0 \)} \}, \\ \mathcal{K}(V) &= \{ \text{germs of smooth Kähler metrics on \( V \) at \( 0 \)} \}. \end{align*} We just write \( \mathcal{H} \) if \( V \) is clear from context and will usually mean \( V = \mathbf{C}^n \). Note that \( \mathcal{K} \) and \( \mathcal{H} \) are only convex open cones in real vector spaces and not full subspaces. Evaluating the curvature tensor of a metric at \( 0 \) defines nonlinear maps and a commutative diagram \begin{align*} \mathcal{K} &\stackrel{\mathcal{R}}{\longrightarrow} B(\operatorname{Sym}^2 \mathbf{C}^n) \\ \downarrow & \qquad\qquad \downarrow \\ \mathcal{H} &\stackrel{\mathcal{R}}{\longrightarrow} B(\mathbf{C}^n \otimes \mathbf{C}^n). \end{align*} Both vertical arrows are of course injective. The horizonal arrows are neither injective nor surjective in general.

Let \( H \) be the matrix of a Hermitian metric \( h \) in a neighborhood around \( 0 \) in \( V \). The Chern connection of \( h \) is given by \( D^{1,0} = H^{-1}\partial H \) and its curvature form is \[ F = \tfrac i2 \bar\partial(H^{-1}\partial H) = - \tfrac i2 H^{-1} \bar\partial H \wedge H^{-1} \partial H - \tfrac i2 H^{-1} \partial \bar\partial H. \] The curvature tensor is then \( R(x,\overline{y}, z, \overline{w}) = h(F(x, \overline{y}) z, \overline{w}) \).

We can pick coordinates such that \( H = I_n \) at \( 0 \). If \( h \) is Kähler we can further get \( \partial H = 0 \) at \( 0 \). We will arrange this in our examples. We now restrict to \( V = \mathbf{C}^2 \).

Proposition. The image of \( \mathcal{K}(\mathbf{C}^2) \) in \( B(\mathbf{C}^2 \otimes \mathbf{C}^2) \) is the set of matrices of the form \[ \begin{pmatrix} x & a & a & c \\ \bar a & z & z & b \\ \bar a & z & z & b \\ \bar c & \bar b & \bar b & y \end{pmatrix} \] where \( x, y, z \in \mathbf{R} \), \( a, b, c \in \mathbf{C} \).

Proof: If \( h \) is a Hermitian metric on a neighborhood of \( 0 \) in \( \mathbf{C}^2 \) we can write it in matrix form as \[ H = \begin{pmatrix} a & c \\ \overline{c} & b \end{pmatrix}. \] After a change of coordinates, we can assume that \( H(0) = I_2 \). Assuming \( h \) is Kähler we can also arrange that \( \partial H(0) = 0 \). Then the matrix of the curvature tensor in the basis \( (\partial / \partial z \otimes \partial / \partial z, \partial / \partial z \otimes \partial / \partial w, \partial / \partial w \otimes \partial / \partial z, \partial / \partial w \otimes \partial / \partial w) \) is \[ R = \begin{pmatrix} a_{z \overline{z}} & \overline{c}_{z \overline{z}} & a_{z \overline{w}} & \overline{c}_{z \overline{w}} \\ c_{z \overline{z}} & b_{z \overline{z}} & c_{z \overline{w}} & b_{z \overline{w}} \\ a_{w \overline{z}} & \overline{c}_{w \overline{z}} & a_{w \overline{w}} & \overline{c}_{w \overline{w}} \\ c_{w \overline{z}} & b_{w \overline{z}} & c_{w \overline{w}} & b_{w \overline{w}} \end{pmatrix}. \] We’ve written, for example, \( a_{z \overline{w}} \) instead of \( \frac{\partial^2 a}{\partial z \partial \overline{w}} \). The metric \( h \) is Kähler, so we have \[ a_w = c_{z}, \quad a_{\overline{w}} = \overline{c}_{\overline{z}}, \quad b_z = \overline{c}_w, \quad b_{\overline{z}} = c_{\overline{w}}, \] and thus \[ c_{z \overline{z}} = a_{w \overline{z}}, \quad \overline{c}_{w \overline{w}} = b_{z \overline{w}}, \quad c_{z \overline{w}} = a_{w \overline{w}}, \quad \overline{c}_{w \overline{z}} = b_{z \overline{z}} \] Then we get \[ R = \begin{pmatrix} a_{z \overline{z}} & a_{z \overline{w}} & a_{z \overline{w}} & a_{w \overline{w}} \\ a_{w \overline{z}} & a_{w \overline{w}} & a_{w \overline{w}} & b_{z \overline{w}} \\ a_{w \overline{z}} & a_{w \overline{w}} & a_{w \overline{w}} & b_{z \overline{w}} \\ a_{w \overline{w}} & b_{w \overline{z}} & b_{w \overline{z}} & b_{w \overline{w}} \end{pmatrix}, \] where some simplifications happen because \( a \) is real so \( a_{w \overline{z}} = a_{z \overline{w}} \) and we can ping-pong to \( b_{z \overline{z}} = a_{w \overline{w}} \). This is exactly of the form we announced.

We will conclude the proof by taking the above Hermitian matrix as given and constructing a germ of a Kähler metric whose curvature tensor agrees with it at \( 0 \). Suppose we let \[ f(z,w) = a_{z \overline{z}} z \bar z + a_{z \overline{w}} z \overline{w} + a_{w \overline{z}} w \overline{z} + a_{w \overline{w}} w \bar w. \] Then \( f(0) = 0 \) and \( df(0) = 0 \) and \[ \operatorname{Hess} f(0) = \begin{pmatrix} a_{z \overline{z}} & a_{z \overline{w}} \\ a_{w \overline{z}} & a_{w \overline{w}} \end{pmatrix}. \] We also let \[ g(z,w) = a_{w \overline{w}} z \bar z + b_{z \overline{w}} z \overline{w} + b_{w \overline{z}} w \overline{z} + b_{w \overline{w}} w \bar w \] and \[ h(z,w) = h_1 z \bar z + h_2 z \overline{w} + h_3 w \overline{z} + h_4 w \bar w. \] We want \[ a_{w \overline{z}} \overline{z} + a_{w \overline{w}} \bar w = f_w = h_z = h_1 \overline{z} + h_2 \overline{w}, \] so we set \( h_1 = a_{w \overline{z}} \) and \( h_2 = a_{w \overline{w}} \). We also want \[ a_{w \overline{w}} \overline{z} + b_{z \overline{w}} \overline{w} = g_z = \overline{h}_w = \overline{h}_3 \bar z + \overline{h}_4 \overline{w} \] so \( h_3 = \overline{h_2} \) and \( h_4 = b_{z \overline{w}} \). The germ of the Kähler metric \[ h(z,w) = \begin{pmatrix} 1 + f(z,w) & h(z,w) \\ \overline{h(z,w)} & 1 + g(z,w) \end{pmatrix} \] then has the curvature tensor \( R \) above.