I have problems with the proof why every manifold admits a Riemannian metric. If the manifold $M$ has an atlas given by sets $U_i$, we can take a subordinate partition of unity $\{\phi_i\}_i$. Then $\sum \phi_i g_i$, where $g_i$ is the standard metric on $U_i \tilde = \mathbb R^n$, defines a riemannian metric.

I don't understand why this is well defined. Isn't there some transformation condition we need to to check when switching charts on their overlap?

  • $\begingroup$ Inner products (or positive definite bilinear forms) form a convex cone over the reals. Specifically, if $g_{1}$ and $g_{2}$ are inner products on a vector space $V$ and $\alpha, \beta \in \mathbb{R}$ with $\alpha, \beta > 0$, then $g = \alpha g_{1} + \beta g_{2}$ is also an inner product. On the overlaps, the inner product at a particular point is defined by the sum of inner products, but as you move out of the overlap $U_{i} \cap U_{j}$ to $U_{i}$ or $U_{j}$ (respectively), the contribution of $g_{i}$ or $g_{j}$ (respectively) tapers off to zero, leaving you with a well-defined metric. $\endgroup$ – THW Dec 12 '17 at 18:02
  • $\begingroup$ I understand what the definition does. I have the following problem: When for example defining a section by local functions, we still need to check if the sections glue to a global section ($s_i=g_{ij} s_j$). Why is that not the case here, resp how can I see that such a criterion this is fulfilled? $\endgroup$ – user509325 Dec 12 '17 at 18:29
  • 1
    $\begingroup$ We're not trying to check that locally-defined sections patch compatibly. We're just summing locally defined Riemannian metrics, and the sum of positive definite bilinear forms with nonnegative coefficients [given by the partition of unity] is again a positive definite bilinear form. $\endgroup$ – Ted Shifrin Dec 12 '17 at 18:34
  • 1
    $\begingroup$ Maybe the confusion is related to what is actually being summed. The local sections $g_{i}$ are local sections of the bundle of symmetric, positive definite (0,2) tensors. But $\phi_{i}g_{i}$ is a global section of the bundle of $(0, 2)$ tensors. So the summation is a sum of global sections of the bundle of $(0, 2)$ tensors (none of which are necessarily positive definite) that are pieced together in such a way as to form a global section of the bundle of symmetric, positive definite (0, 2) tensors. $\endgroup$ – THW Dec 13 '17 at 3:21
  • 1
    $\begingroup$ If you try to apply the same technique to construct a global section $v$ of an arbitrary (fiber subbundle of a) vector bundle, you run into the issue that $\phi_1 v_1 + \phi_2 v_2$ could vanish even when $\phi_i > 0$ and $v_i \ne 0$. It is only because we are working with a very particular bundle (the positive symmetric 2-tensors) that this construction works, thanks to the convexity property THW mentions. $\endgroup$ – Anthony Carapetis Dec 13 '17 at 3:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.