# Proving positive definiteness of the metric tensor.

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.

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

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.