## View abstract

### Session S23 - Group actions in Differential Geometry

Wednesday, July 14, 12:20 ~ 12:50 UTC-3

## Prescribing Ricci curvature on a product of spheres

### Anusha Krishnan

#### Syracuse University, USA   -   This email address is being protected from spambots. You need JavaScript enabled to view it. document.getElementById('cloak2f26a7e4212d871c5d099c71d068950a').innerHTML = ''; var prefix = '&#109;a' + 'i&#108;' + '&#116;o'; var path = 'hr' + 'ef' + '='; var addy2f26a7e4212d871c5d099c71d068950a = '&#97;kr&#105;sh03' + '&#64;'; addy2f26a7e4212d871c5d099c71d068950a = addy2f26a7e4212d871c5d099c71d068950a + 'syr' + '&#46;' + '&#101;d&#117;'; var addy_text2f26a7e4212d871c5d099c71d068950a = '&#97;kr&#105;sh03' + '&#64;' + 'syr' + '&#46;' + '&#101;d&#117;';document.getElementById('cloak2f26a7e4212d871c5d099c71d068950a').innerHTML += '<a ' + path + '\'' + prefix + ':' + addy2f26a7e4212d871c5d099c71d068950a + '\'>'+addy_text2f26a7e4212d871c5d099c71d068950a+'<\/a>';

Given a symmetric 2-tensor T on a manifold M, does there exist a Riemannian metric g such that Ric(g) = T? I will discuss some classical results as well as some recent work in the presence of symmetry.

Joint work with TImothy Buttsworth (University of Queensland, Australia).