# How to draw a 2D space in 3D Euclidean space by metric tensor

• arpon
In summary, the conversation discusses the challenge of drawing a 2D space with a given metric tensor, specifically in the context of isometric embeddings into 3D Euclidean space. The metric tensor of a 2D space, such as a sphere, is not unique and does not determine the shape of the surface. Instead, the shape is determined by both the intrinsic and extrinsic curvatures, which can be described using a set of nonlinear differential equations. Finding solutions to these equations can be difficult and there is a wide range of possible shapes for a given metric tensor.
arpon
Suppose, I know the metric tensor of a 2D space. for example, the metric tensor of a sphere of radius R,
gij = ##\begin{pmatrix} R^2 & 0 \\ 0 & R^2\cdot sin^2\theta \end{pmatrix}##
,and I just know the metric tensor, but don't know that it is of a sphere.
Now I want to draw a 2D space(surface) which will have the metric tensor gij and will be embedded on 3D Euclidean space.

Last edited:
Your question does not always have an answer! Not every 2d manifold is isometrically embeddable into 3d Euclidean space.

When your question does have an answer, it might not be unique. For example, the metric

$$ds^2 = dx^2 + dy^2$$
could be an infinite plane, or it could be a cylinder, or it could be a non-circular cylinder, or it could be an infinite plane with a plane wave wiggle on it of arbitrary profile.

In general, you have a whole arbitrary function's worth of non-uniqueness for isometric embeddings into R^3. The reason for this is that surfaces embedded in R^3 have two curvatures: an intrinsic one (Gauss curvature) and an extrinsic one (mean curvature). As the terminology would imply, only the Gauss curvature is determined by the 2d metric tensor; the mean curvature is a property of the choice of embedding.

In general, you should write

$$g_{ij} = \partial_i \vec s(u, v) \cdot \partial_j \vec s(u,v)$$
where ##g_{ij}## is your 2d metric tensor, ##u, v## are your 2d coordinates, and ##\vec s(u,v)## is a vector (in R^3) describing your surface, as a function of ##u,v##. Your task is to solve these equations for ##\vec s##. You have a coupled set of nonlinear differential equations. It is not always easy to find solutions.

arpon

## 1. How do you define a 2D space in 3D Euclidean space?

To define a 2D space in 3D Euclidean space, we use a metric tensor. This tensor is a mathematical tool that allows us to measure distances and angles in a given space. By using a metric tensor, we can define a 2D space within the larger 3D space.

## 2. What is a metric tensor and how is it used to draw a 2D space in 3D Euclidean space?

A metric tensor is a mathematical object that describes the geometry of a given space. It is used to measure distances and angles between points in that space. In order to draw a 2D space in 3D Euclidean space, we use the metric tensor to define the relationships between points in the 2D space and the larger 3D space.

## 3. How is a metric tensor represented in 3D Euclidean space?

In 3D Euclidean space, a metric tensor is represented by a matrix with three rows and three columns. This matrix contains information about the distances and angles between points in the space. By manipulating this matrix, we can define a 2D space within the larger 3D space.

## 4. What are the steps involved in drawing a 2D space in 3D Euclidean space using a metric tensor?

The first step is to choose a basis for the 2D space, which will serve as the x and y axes. Then, we define a metric tensor using this basis. Next, we use the metric tensor to calculate the distances and angles between points in the 2D space. Finally, we use this information to plot the 2D space within the larger 3D space.

## 5. Can a 2D space be drawn in any 3D Euclidean space using a metric tensor?

Yes, a 2D space can be drawn in any 3D Euclidean space using a metric tensor, as long as the basis for the 2D space is chosen properly. The metric tensor allows us to define the relationships between points in the 2D space and the 3D space, regardless of the dimensions of the space.

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