# Tensor Analysis On Manifolds: Question on an example manifold

Below is the excerpt from Tensor Analysis On Manifolds by Bishop & Goldberg. I do not understand the passage at all. I guess I have not understood some key fundamental concept properly. Below are my questions.

1. What exactly is the manifold we are trying to chart here? Is it a single parabola OR all parabolas in $$R^2$$ [in (+,+) & (+,-) quadrants]?

2. The example says "No neighborhood of a singular point is mapped 1-1 by $$\mu$$; such neighborhoods are 'folded' on themselves, and neighborhoods of (0,0) are folded twice". What does "folding" mean here?

3. When the determinant is zero, the matrix has vectors(basis?) which are linearly dependent. Here when the jacobian determinant is zero, does that mean at those singular points, the basis vectors become linearly dependent?

4. It says "For every point except those on these lines there is some neighborhood on which $$\mu$$ is an 'admissibile' coordinate map". Intuitively how is this 'admissibility' derived from the jacobian determinant being non-zero? I get that this line cannot be mapped continuously in some other chart, ie, $$\tau$$ $$\circ$$ $$\mu^-1$$ is not continuous. But how jacobian determinant being zero results in this?

5. It says "Each of the four connected regions of non-singular points is mapped by $$\mu$$ 1-1 onto V". In the figure, what are these four connected regions? Is it the line D's segments? (above line A, A to x, x to B and below B)?

I guess I have some serious gaps in my understanding. Probably I'm missing the big picture. Can somebody help me on this?

1. We are trying to chart $$\mathbb{R}^2$$ with the map $$\mu\colon\mathbb{R}^2\to\mathbb{R}^2$$ and analyse where and how it fails.
2. One way of interpreting "folding" here is a topological conjugate of $$(x,y)\mapsto(\lvert x\rvert, y)$$ that you can imagine as folding the plane in half with the singular locus $$x=0$$.
4. The inverse function theorem: if $$f\colon\mathbb{R}^n\to\mathbb{R}^n$$ is continuously differentiable with $$f'(0)$$ invertible, then there is a neighbourhood $$U\ni 0$$, $$V\ni f(0)$$ such that $$f^{-1}\colon V\to U$$ exists and is continuously differentiable with derivative $$(f^{-1})'(f(0))=f'(0)^{-1}$$.
5. The four regions of the $$xy$$-plane separated by the lines $$y=\pm x/\sqrt 2$$. Each is mapped bijectively, smoothly with smooth inverse, to $$V$$.
• What I mean is the following. Pick any ine through the origin, for example, the line $y=0$, going from $x=-1$ to $x=1$ for definiteness. The image under $\mu$ is the line segment $[(0,0),(1,0)]$, starting at $(1,0)$ going to $(0,0)$ and then back to $(1,0)$. So you can't parameterize the image with unit speed curve without having it suddenly changes direction from going left to going right. The same thing happen to the tangent line $A$ and $B$ (the $u\geq 0$ is the culpit). – user10354138 May 26 at 16:03