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Suppose $\sigma(u,v)$ is a surface patch of surface S. A tangent vector at point p in the image of $\sigma$ can be expressed uniquely as a linear combination of $\sigma_{u}$ and $\sigma_{v}$. My surface S is in $\mathbb{R}^{3}$.

I want to understand maps $du : T_{p} (S) \to \mathbb{R}$. I mean how $du(v^-) = \lambda$ ?, where $v^{-} = \lambda \sigma_{u} + \mu \sigma_{v}$

Please help me to understand this map precisely. This is a part of the book on differential geometry by Andrew Pressley section -6.1.

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Ah, I think the confusion is here: If you let $u,v$ be coordinate functions on $S$ about $p \in S$ then;

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$$\left\{\frac{\partial}{\partial u}\Bigr|_p, \frac{\partial}{\partial v}\Bigr|_p\right\} \equiv \{\sigma_u, \sigma_v\}$$

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span $T_pS$ and the corresponding 1-covectors (or 1-forms); $\{du, dv\}$ span $T_p^* S$. Recall $du,dv$ have the property that;

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$$du(\sigma u) = 1 ; du(\sigma v) = 0 ; dv(\sigma v) = 1; dv(\sigma u) = 0$$

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Therefore, if you have $v = \lambda_1 \sigma_u + \lambda_2 \sigma_v$ then since $du,dv$ are also linear maps;

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$$du(v) = du(\lambda_1 \sigma u + \lambda_2 \sigma v) = \lambda_1 du(\sigma_u) + \lambda_2 du(\sigma_v) = \lambda_1$$

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  • $\begingroup$ Yes, but how $du(\sigma_{u}) =1$ and other values ? $\endgroup$
    – Arun Sharma
    Sep 15, 2016 at 14:47
  • $\begingroup$ Remember that if $\{v_1,...,v_n\}$ is a basis for a vector space $V$ and $\{v'_1,...,v'_n\}$ is the corresponding dual basis for $V$ i.e $V*$ then by definition $v'_i \cdot v_j = \delta^i_j$ i.e $\delta^i_j = 0$ if $i \not = j$ and $1$ if $i = j$. $\endgroup$
    – Faraad Armwood
    Sep 15, 2016 at 15:04
  • $\begingroup$ Does this answer your question? $\endgroup$
    – Faraad Armwood
    Sep 15, 2016 at 17:14

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