1
$\begingroup$

I am reading through, Pressley's elementary differential geometry, and in chapter 6 on first fundamental forms, there is a concept of symmetric bilinear form, which I misunderstand.

Here is the text:

Suppose that $\sigma(u, v)$ is a surface patch of $S$. Then, any tangent vector to $S$ at a point $p$ in the image of $\sigma$ can be expressed uniquely as a linear combination of $\sigma_u$ and $\sigma_v$. Define maps $du : T_pS \rightarrow R$ and $dv : T_pS \rightarrow R$ by $du(v) = \lambda, du(v) = \mu$ if $ v = \lambda \sigma_u + \mu \sigma_v $. for some $\lambda, \mu \in R$. It is easy to see that $du$ and $dv$ are linear maps. Then, using the fact that $< , >$ is a symmetric bilinear form, we have: $<v,v> = \lambda^2<\sigma_u,\sigma_u> + 2\lambda\mu<\sigma_u,\sigma_v> + \mu^2<\sigma_v, \sigma_v> $.

Could someone explain this concept? I found a definition of symmetric bilinear form on wikipedia, but did not find it enlightening. (i mainly do not understand where the last equality comes from but do see how dot product satisfies the definition :https://en.wikipedia.org/wiki/Symmetric_bilinear_form)

$\endgroup$

1 Answer 1

1
$\begingroup$

You are just giving a very nice way of describing length in terms of differential forms. This is simply using the fact that the dual basis vectors to the tangent space give are the projection maps on the coordinates for vectors in the tangent space. Let $(u,v)$ be the coordinates on $(U, \sigma)$ which is a chart on the surface $S$, then:

\begin{align*} w \in T_q(S) \Rightarrow w&= a_1 \sigma_u + a_2 \sigma_v \\&= du(w) \sigma_u + dv(w)\sigma_v \end{align*}

Therefore, in terms of the basis $\{\sigma_v,\sigma_u\}$, the vector $w$ has coordinates $(du(w),dv(w))$ and if the evaluation is clear from context, we simply write $w = (du, dv) = du \sigma_u + dv \sigma_v$. To see what's going on we will use the Euclidean inner product.

\begin{align*}\|w\| = \sqrt{\langle w,w\rangle} &= \sqrt{w \cdot w} \\&= \sqrt{(du \sigma_u + dv \sigma_v)\cdot(du \sigma_u + dv \sigma_v)} \\ &= \sqrt{\|\sigma_u\|^2 du^2 + 2 \sigma_u \cdot \sigma_v dudv + \|\sigma_v\|^2 dv^2} \end{align*}

$$ $$

$\textbf{What a symmetric bilinear form?}$ Let $V$ be a vector space. Then $B: V \times V \to \mathbb{K}$ is a symmetric bilinear form $\iff B(v,w)$ is linear in both components (where $\mathbb{K}$ is the scalar field) i.e:

\begin{align*} B_w(\lambda v_1+\lambda_2 v_2) &= \lambda_1B_w(v_1) + \lambda_2B_w(v_2) \\ &= \lambda_1B(v_1,w)+ \lambda_2B(v_2,w)\end{align*}

(similarly for the other component) and $B(v,w) = B(w,v)$. As an example you can take the dot product on $\mathbb{R}^n$.

$\endgroup$
1
  • $\begingroup$ makes sense! thx $\endgroup$ Commented Mar 10, 2017 at 5:06

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .