How can we show using coordinate transformations that every smooth manifold $M$ has at least one Riemannian Metric?
This is one of the problems in my past finals on Calculus on Manifolds, and it seems very puzzling to me as I do not even know where to start. Can anyone at least give me some hints or some start so I can try to finish it?
My Solution: Using the hints and applying the idea of partitions of unity, first let us define the local parametrisation $\Lambda=\{F_{\alpha}:U_{\alpha}\rightarrow O_{\alpha}\}$ which covers some smooth manifold $M,$ and let the partitions of unity subordinate to $\Lambda$ be denoted by $\{\rho_{\alpha}:U_{\alpha}\rightarrow [0,1]\}$. Now we can denote the $(2,0) -$ tensor as :
$$\delta_{\alpha}=\sum^n_{j=1}du^j_{\alpha}\otimes du^j_{\alpha}$$
Then it remains to show that the below expression is smooth and symmetric and positive definite:
$$\mathcal{M}=\sum_{\alpha}\rho_{\alpha}\delta_{\alpha}$$
Now it is not hard to see that it indeed is symmetric and smooth, so we just need to check the positive definite condition. Let us introduce some vectors for computations, notably $X=\sum_i X^i_{\beta}\frac{\partial}{\partial u^i_{\beta}}$, then we simply compute the metric using this vector which can be done in the following way:
$$\mathcal{M}(X,X)=\sum_{\alpha}\sum^n_{j=1}\rho_{\alpha}du^j_{\alpha}\otimes du^j_{\alpha}\big(\sum_i X^i_{\beta}\frac{\partial}{\partial u^i_{\beta}},\sum_k X^k_{\beta}\frac{\partial}{\partial u^k_{\beta}}\big)$$
$$=\sum_{\alpha}\sum_{j,p,q}\rho_{\alpha}\frac{\partial u^{j}_{\alpha}}{\partial u^{p}_{\beta}}\frac{\partial u^{j}_{\alpha}}{\partial u^q_{\beta}}du^p_{\alpha}\otimes du^q_{\alpha}\big(\sum_i X^i_{\beta}\frac{\partial}{\partial u^i_{\beta}},\sum_k X^k_{\beta}\frac{\partial}{\partial u^k_{\beta}}\big)$$
$$=\sum_{\alpha}\sum_{i,j,k}\rho_{\alpha}\frac{\partial u^{j}_{\alpha}}{\partial u^{p}_{\beta}}\frac{\partial u^{j}_{\alpha}}{\partial u^q_{\beta}}X^i_{\beta}X^k_{\beta}$$
$$=\sum_{\alpha}\rho_{\alpha}\sum_{j}\bigg(\sum_{i}X^i_{\beta}\frac{\partial u^{j}_{\alpha}}{\partial u^i_{\beta}}\bigg)^2\ge 0$$
Hence it is positive definite and we are done.
Is this how you do it?
Best Answer
In a local chart, there is an "obvious" choice. Namely set $g_p(u,v) = \langle u,v \rangle_{T_pM}$ for each $p \in U \subset M$. Now use a partition of unity, and check that convex combinations of inner products is an inner product.