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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 $\mathbb{R}^d$.

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    $\begingroup$ 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
    Commented 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
    Commented 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
    Commented Aug 26, 2018 at 15:27

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

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    $\begingroup$ 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
    Commented 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
    Commented 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
    Commented 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
    Commented 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
    Commented Aug 27, 2018 at 9:32

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