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Thread: (n - 1) dimensional submanifold of the manifold R^n

  1. #1
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    (n - 1) dimensional submanifold of the manifold R^n

    Let $\displaystyle A$ be a symmetric $\displaystyle n \times n$ matrix over $\displaystyle \mathbb{R}$.
    Let $\displaystyle 0 \neq b \in \mathbb{R}$.


    Show that the surface $\displaystyle M = \{x\in \mathbb{R}^n \mid x^T A x = b\}$ is an $\displaystyle (n - 1)$ dimensional submanifold of the manifold $\displaystyle \mathbb{R}^n$.


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  2. #2
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    Re: (n - 1) dimensional submanifold of the manifold R^n

    The differential of the smooth function $\displaystyle f: R^n \rightarrow R, f: x \mapsto x^TAx-b$ is
    $\displaystyle f_*: R^n \rightarrow R, f_*: v \mapsto 2x^TAv$
    Since $\displaystyle b \ne 0$, $\displaystyle x^tA$ never vanishes on the locus $\displaystyle f=0$ so $\displaystyle f_*$ has full rank.
    That is, all points of M are regular points. Hence M is a n-1 dimensional sub-manifold.
    Last edited by xxp9; Dec 19th 2012 at 06:22 PM.
    Thanks from vercammen
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    Re: (n - 1) dimensional submanifold of the manifold R^n

    Thank you!!!
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