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A note on negative isotropic curvature. (English) Zbl 1073.53050
Let $$(M, g)$$ be a Riemannian manifold, $$\dim M\geq 4$$, and denote by $$\langle\;,\;\rangle$$ the complex bilinear extensions of the inner products on $$TM$$ to the complexification $$TM\otimes\mathbb{C}$$. A subspace $$W$$ of $$T_pM\otimes\mathbb{C}$$ is isotropic if $$\langle w, w\rangle= 0$$, for all $$w\in W$$. The metric $$g$$ has positive (negative) isotropic curvature if $$\langle R(u\wedge v),\overline u\wedge\overline v)> 0$$ $$(< 0)$$, whenever $$u$$, $$v$$ span an isotropic 2-plane, $$R: \Lambda^2 TM\otimes\mathbb{C}\to \Lambda^2 TM\otimes\mathbb{C}$$ denoting the complex linear extension of the curvature operator $$R$$. Manifolds with curvature operators of a fixed sign also have isotropic curvature of the same sign but the relationship between the sign of isotropic curvature and the classical curvatures is not clear.
The author proves that any smooth orientable closed 4-manifold admits a metric with negative isotropic curvature.
In fact, a closed 4-manifold $$M$$ has negative isotropic curvature if and only if $${s\over 6}I- W< 0$$, where $$s$$ is the scalar curvature and $$I$$, $$W$$ are the identity and the Weyl curvature operators on $$\Lambda^2TM$$, respectively. Then, putting $$\sigma_g= {s\over 6}+ |W|$$, one considers the functional $$F$$ on the space of metrics on $$M$$ given by: $$F(g)= \int_m \sigma_g \,dv_g$$. The author explicitely constructs a metric $$g$$ with $$F(g)< 0$$, gluing in a product of a circle with a hyperbolic 3-manifold of finite volume. This entails the existence of a metric $$\overline g$$ in the conformal class of $$g$$ which has negative isotropic curvature.

##### MSC:
 53C21 Methods of global Riemannian geometry, including PDE methods; curvature restrictions
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