MathNugget
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- Feb 1, 2024
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Given M a smooth manifold, [imath]M=f^{-1}(0)[/imath] for a differentiable function [imath]f: \mathbb{R}^n \rightarrow \mathbb{R}[/imath].
I have to prove that a vector field X on [imath]\mathbb{R}^n[/imath] is tangent in x on M if and only if [imath]X_x(f)=0[/imath].
Suppose [imath]X=\sum_{i=1}^nX_i\frac{\mathcal{d}}{dx_i}[/imath], and [imath]x=(x_1,...,x_n) \in M.[/imath]
I know that f(x)=0.
I think [imath]X_x(f)=\sum_{i=1}^n X_i(x)\frac{df}{dx_i}(x)[/imath].
I get stuck here, I am not creative enough to make up something. Do I say there's a neighbourhood around x (probably because [imath]f^{-1}(0)[/imath] is a smooth manifold), where f(x)=0, so the derivative would be 0?
I have to prove that a vector field X on [imath]\mathbb{R}^n[/imath] is tangent in x on M if and only if [imath]X_x(f)=0[/imath].
Suppose [imath]X=\sum_{i=1}^nX_i\frac{\mathcal{d}}{dx_i}[/imath], and [imath]x=(x_1,...,x_n) \in M.[/imath]
I know that f(x)=0.
I think [imath]X_x(f)=\sum_{i=1}^n X_i(x)\frac{df}{dx_i}(x)[/imath].
I get stuck here, I am not creative enough to make up something. Do I say there's a neighbourhood around x (probably because [imath]f^{-1}(0)[/imath] is a smooth manifold), where f(x)=0, so the derivative would be 0?