1
$\begingroup$

Let $S$ be a surface embedded in 3 dimensional Euclidean space, $\mathbb{E}^3$. We define ${\bf a}_{\alpha}$ at some point $P \in S$ as $\frac{d{\bf p}}{dx^{\alpha}}$, where $\bf p$ is the position vector expressed in Cartesian coordinates: ${\bf p} = y^k {\bf E}_k$, ${\bf E}_k$ is the standard basis of $\mathbb{E}^3$ and $x^{\alpha}$ are coordinates used to locally chart out $S$.

Components of the metric tensor for $S$, $a_{\alpha\beta}$ are then given by an inner product ${\bf a}_{\alpha} \cdot {\bf a}_{\beta}$. We define a bilinear form $ds^2 : \mathbb{R}^2 \times\mathbb{R}^2 \to \mathbb{R}$ as $ds^2 = a_{\alpha\beta} dx^{\alpha} dx^{\beta}$.

It is clear that $ds^2$ is a symmetric form as $a_{\alpha\beta} = a_{\beta\alpha}$ but what makes it positive definite? Clearly, $ds^2 \geq 0$ so $a_{\alpha\beta} dx^{\alpha} dx^{\beta} \geq 0.$ Is the reasoning that this must be true $\forall$ $a_{\alpha\beta}$ and hence $ds^2 = 0$ only when $dx^{\alpha}dx^{\beta} = 0$? Or does the reasoning have something to do with the structure of $a_{\alpha\beta}$? We know that its diagonal elements must be positive. Does this imply positive definiteness?

Thanks.

$\endgroup$
1
  • $\begingroup$ The metric is the restriction to the surface of the positive definite metric on $\Bbb R^3$! $\endgroup$ – Ted Shifrin Oct 11 '15 at 7:06
0
$\begingroup$

The form $ds^2$ is actually postive definite.

Note that $(x^1, x^2)$ are called a local chart of $S$ is there is a smooth map $$ \phi : U \to S \subset \mathbb R^3$$ so that ${\bf a}_{1} \times {\bf a}_{2} \neq 0$ for all $(x^1, x^2) \in U$ (That is, it should be a regular parametrization). This condition would forces $ds^2$ to be positive definite as $$\|{\bf a}_{1} \times {\bf a}_{2}\|^2 = \det a_{\alpha\beta}$$ by the Lagrange's identity.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.