# weakest topology so that space of riemannian metrics is open in space of symmetric tensors

Let $M$ be a smooth n-dim manifold. Then on any local chart we can write a symmetric tensor as a symmetric $n \times n$ matrix with the entries being smooth functions on $\mathbb R^n$. The space of Riemannian metrics is the space of matrices that are invertible. What is the weakest endowment of a topological vector space structure on $\Gamma(M,T_{2,0}M)$ that is weaker than te $C^\infty(M,T_{2,0}M)$ topology, that makes the space of Riemannian metrics open in the space of symmetric tensors(given the subspace topology)?

• You probably want to specify a bit more on the topology, for example that it turns $T^*M$ into a topological vector space. Otherwise I would also not see how a topology on $T^*M$ induces one on the tensor of $T*M$ with itself. The weakest topology on the symmetric tensors that makes the Riemannian metrics open is the topology with 4 open sets: The symmetric tensors, the empty set, the Riemannian metrics and its complement. – Thomas Rot Oct 19 '16 at 11:07
• Thanks. I edited the question. – user062295 Oct 19 '16 at 11:49

## 1 Answer

A good candidate for such a topology is the natural analog of the topology of the product topology i.e. the topology of point-wise convergence. This means that for each $x\in M$, you take the evaluation map $\Gamma(M,T_{2,0}M)\to S^2T^*_xM$ (where the target is just the fiber over $x$ of $T_{2,0}M$). Then take the initial topology with respect to these maps. Hence a basis for the open subsets in $\Gamma(M,T_{2,0}M)$ would be of the form $\{s\in\Gamma(M,T_{2,0}M):s(x_i)\in U_i \forall i=1,\dots,k\}$ where $x_i\in M$ are points and for each $i$, $U_i$ is an open subset in $S^2T^*_{x_i}M$. I think it is rather easy to see that this defines a topological vector space structure and that Riemannian metrics form an open subset. I am not sure whether it indeed is the weakest such topology.