3
$\begingroup$

Let $M$ be a smooth closed manifold, and let $g_0$ be a Riemannian metric on $M$.

Let $U$ be a neighbourhood of $p \in M$, and suppose that we are given a metric $g$ on $U$, which satisfies $\| g-g_0|_U\|_{C^1} < \epsilon$ on $U$.

Can we extend $g$ to a metric $\tilde g$ on $M$ such that $\| \tilde g-g_0\|_{C^1} = \mathcal{o}(1)$ will hold everywhere on $M$? ( i.e. bounded by something which tends to zero when $\epsilon \to 0$).

I am fine with shrinking $U$ if that's necessary. (So the "extension" $\tilde g$ needs to coincide with $g$ on some small neighbourhood of $p$).

Edit:

I am actually interested in a more refined version of this problem- where the size of $U$ depends on $\epsilon$:

Specifically, $U_{\epsilon}=B_{\epsilon}(p)$ where $B_{\epsilon}(p)$ is the $\epsilon$-ball around $p$ w.r.t. the given metric $g$. So, now we are given the estimate $\| g-g_0|_{U_{\epsilon}}\|_{C^1} < \epsilon$ on $U$, and we ask if we can extend $g$ (perhaps after shrinking $U_{\epsilon}$) to a metric $\tilde g$ which is $o(1)$-close to $g_0$ in the $C^1$ sense.

Deane Yang gave a construction where the error $\| \tilde g-g_0\|_{C^1} \approx 1$, so it does not tend to zero with $\epsilon$.

His construction is to take $\tilde{g} = (1-\chi)g_0 + \chi g$, where $\chi$ is a smooth function that is identically $1$ on a neighbourhood of $p$ and compactly supported on $U_{\epsilon}$. This means that $|d\chi| \approx \frac{1}{\text{diam}(U_{\epsilon})}$ on average, so

$$ |d\tilde{g} - dg_0|\le |d\chi| |g-g_0| + |\chi||dg-dg_0| \le (\frac{1}{\text{diam}(U_{\epsilon})}+1)\epsilon. $$ Since $\text{diam}(U) \approx \epsilon$, we get an error of order $1$.

If that helps, I am interested in the special case where $M=\mathbb{T}^d=\mathbb{R}^d/\mathbb{Z}^d$ is the $d$-dimensional torus, and where $g_0$ is a metric on $\mathbb{T}^d$ which is isometric to one of the standard flat metrics induced on $\mathbb{T}^d$ by the Euclidean metric on $\mathbb{R}^d$.

Improve this question
$\endgroup$
3
  • 1
    $\begingroup$ A Riemannian metric is a section of a tensor bundle. For smooth maps between manifolds there are relative approximation theorems as e.g. in Chapter 2, section 3 in M. Hirsch's "Differential topology". I don't think the reference contains the exact statement you need but surely the techniques should extend to your setting. Basically, you extend locally and glue via a partition of unity. $\endgroup$
    – Igor Belegradek
    Aug 26, 2018 at 13:49
  • $\begingroup$ Thanks, I will take a look. One problem with applying naively partitions of unity arguments is that we want to control here derivatives (not just $C^0$ norms). However, perhaps in the specific case of the torus it's easier. $\endgroup$
    – Asaf Shachar
    Aug 26, 2018 at 13:52
  • $\begingroup$ Higher norms is a non-issue. The non-relative approximation theorems face the same difficulty (when extending from a chart to the manifold), and it is explained in the same section in Hirsch's book. $\endgroup$
    – Igor Belegradek
    Aug 26, 2018 at 15:27

1 Answer 1

Reset to default
3
$\begingroup$

Let $\chi$ be a smooth function that is identically $1$ on a neighborhood of $p$ and compactly supported on $U$. Let $\tilde{g} = (1-\chi)g_0 + \chi g$.

Improve this answer
$\endgroup$
6
  • 1
    $\begingroup$ Thanks, that is nice. However, in your construction the error depends also on the size of $U$, not only on $\epsilon$: $\tilde{g} - g_0= \chi (g-g_0)$, so $|d\tilde{g} - dg_0|\le |d\chi| |g-g_0| + |dg-dg_0| \le (|d\chi|+1)\epsilon$. My problem is that $|d\chi|$ should be roughly like $\frac{1}{\text{diam}(U)}$. So, if $U$ is small (say the size of $\epsilon$) we have a problem. I wonder whether there is a way to do something which does not depend on the size of $U$. $\endgroup$
    – Asaf Shachar
    Aug 26, 2018 at 15:43
  • $\begingroup$ Since I did not explicitly required something which is independent in the size of $U$, I think I will accept your answer. (It certainly clarified something for me). However, I will leave the question open for a while, to see if anyone has an idea regarding this refined version. $\endgroup$
    – Asaf Shachar
    Aug 26, 2018 at 15:47
  • $\begingroup$ Just to make sure, you want a way to do this such that the new $\epsilon$ does not depend on the size of $U$? $\endgroup$
    – Deane Yang
    Aug 26, 2018 at 22:05
  • $\begingroup$ Not really; To be more precise, In my "application", $U$ depends on $\epsilon$ (or if you want, then $\epsilon$ depends on $U$; they are "coupled"). Specifically, $U_{\epsilon}=B_{\epsilon}(p)$ where $B_{\epsilon}(p)$ is the $\epsilon$-ball around $p$, w.r.t. the given metric $g$. Thus, your construction gives an error which is on the order of $1$, instead of $\epsilon$. I am elaborating on the exact details in the question now. $\endgroup$
    – Asaf Shachar
    Aug 27, 2018 at 9:11
  • $\begingroup$ The reason why $U_{\epsilon}=B_{\epsilon}(p)$ is not random or mysterious by the way: I am trying to approximate a given metric with Euclidean metric in a $C^1$-sense. By looking at the metric expansion in normal coordinates, we see that every metric is $C^1$-close to being Euclidean- and the "closeness" (or error) depends linearly on the distance from the point of origin. $\endgroup$
    – Asaf Shachar
    Aug 27, 2018 at 9:32

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.