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Let $D$ be an open subset of $\mathbb{R}^3$, and $f: D \to \mathbb{R}$ be a smooth function whose gradient $ \nabla f \neq 0$ on $D$. Consider the surface $M = \{(x_1,x_2,x_3) \in D \mid f(x_1,x_2,x_3)=0 \} $. Assign at each $p \in M$ the unit normal vector $N(p)$ from $\nabla f$, and the Riemannian metric $g$ induced on $M$ from the usual flat metric on $\mathbb{R}^3$.

I'm not sure what the induced metric means. Is it the same metric or something different? I tried to search for some answer but could not find any.

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Since $f$ is smooth and $\nabla f \neq 0$, $M$ is a smooth submanifold of $\Bbb R^3$, and at each point $p \in M$, we can identify $$T_p M = \ker (df)_p = \langle (\nabla f)_p \rangle^{\perp} = \langle N(p) \rangle^{\perp} \subset T_p \Bbb R^3$$ Then, the induced metric (or pullback metric) on $M$ is the pullback $\iota^* \bar g$, where $\bar g$ is the flat metric on $\Bbb R^3$: For any $p \in M$ and $X, Y \in T_p M$, $$(\iota^* \bar g)_p(X, Y) = \bar g_p(T_p \iota \cdot X, T_p \iota \cdot Y) .$$ Under the above identification of $T_p M$ with $T_p \Bbb R^3$, we're simply regarding vectors in $T_p M$ as tangent vectors in $T_p \Bbb R^3$, and so we may as well write this quantity as $$(\iota^* \bar g)_p(X, Y) = \bar g_p(X, Y).$$ Despite this formula, the metrics are emphatically not the same: $\bar g$ and the induced metric $\iota^* \bar g$ live on different manifolds, and $\iota^* \bar g$ contains strictly less information than $\bar g$.

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You have $f^{-1}(0) = M \subseteq \Bbb R^3$ and the inclusion $i: M \hookrightarrow \Bbb R^3$. The induced metric on $M$ is $i^\ast g$, where $g$ is the usual flat metric on $\Bbb R^3$ and $\ast$ denotes pull-back.

In general, if you have an abstract manifold $M$ and an immersion ${\bf x}: M \rightarrow \Bbb R^n$, $g$ being the usual flat metric in $\Bbb R^n$, then ${\bf x}^\ast g$ makes $(M,{\bf x}^\ast g)$ a Riemannian manifold, for which ${\bf x}$ will be an isometric immersion.

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  • $\begingroup$ Both answers are great, but since I can only accept one, I'm gonna accept Travis's answer. Thank you for your help. $\endgroup$
    – calm
    May 5 '16 at 20:18
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    $\begingroup$ He answered 40 seconds first and explained more things. I'd go for his answer too, don't worry. Glad to help! $\endgroup$
    – Ivo Terek
    May 5 '16 at 20:19

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