# Rank One Tensors .... Fortney Appendix A, Section A2 ....

• MHB
• Math Amateur
In summary, the conversation discusses the concept of tensors and their interpretation in the context of manifolds. The speaker is reading Jon Pierre Fortney's book and is seeking help understanding some statements and equations related to rank one tensors. The conversation delves into the concept of charts and how they represent different coordinates for points on a manifold. It also discusses how the coordinate functions are not necessarily a basis for the manifold, but the tangent space at a point can be considered a vector space. Overall, the conversation provides a deeper understanding of tensors and their role in calculus on manifolds.
Math Amateur
Gold Member
MHB
I am reading Jon Pierre Fortney's book: A Visual Introduction to Differential Forms and Calculus on Manifolds ... and am currently focused on Appendix A : Introduction to Tensors ...I need help to understand some statements/equations by Fortney concerning rank one tensors ...

Those remarks by Fortney read as follows:
View attachment 8786
View attachment 8787In the above text by Fortney we read the following:

" ... ... Suppose we change the coordinates from $$\displaystyle (x^1, x^2, \ ... \ ... \ , x^n )$$ to $$\displaystyle (u^1, u^2, \ ... \ ... \ , u^n )$$ using the $$\displaystyle n$$ functions

$$\displaystyle u^1 (x^1, x^2, \ ... \ ... \ , x^n ) = u_1$$

$$\displaystyle u^2 (x^1, x^2, \ ... \ ... \ , x^n ) = u_2$$

... ...

... ...

$$\displaystyle u^n (x^1, x^2, \ ... \ ... \ , x^n ) = u_n$$ ... ... "

My question is as follows:

What do the equations $$\displaystyle u^i (x^1, x^2, \ ... \ ... \ , x^n ) = u_i$$ mean ... ? ... how do we interpret them ...?

What would it mean for example if we wanted to form the differentials $$\displaystyle du^i$$ ... ?Help will be appreciated ...

Peter
EDIT ... Reflecting on the above ... a further question ... are the coordinate functions $$\displaystyle (x^1, x^2, \ ... \ ... \ , x^n )$$ essentially a basis for M ... (I am assuming the manifold is a vector space ... hmmm bt not sure it is ...?)

Hope someone can clarify ...

Peter

#### Attachments

• Fortney - 1 - Rank One Tensors ... PART 1 .png
4.8 KB · Views: 58
• Fortney - 2 - Rank One Tensors ... PART 2 .png
64.7 KB · Views: 69
Last edited:
Peter said:
What do the equations $$\displaystyle u^i (x^1, x^2, \ ... \ ... \ , x^n ) = u_i$$ mean ... ? ... how do we interpret them ...?

Hi Peter,

It means that $u_1$ is a function $\mathbb R^n \to \mathbb R$.
And $(u^1,\ldots,u^n)$ is an element of $\mathbb R^n$, just like $(x^1,\ldots,x^n)$.
The function $u_1$ maps $(x^1,\ldots,x^n)$ to $u^1$.

I'm not sure if you're already familiar with the concept of charts. Are you?
The concept is similar to the charts that we have in an atlas of the earth.
The surface of the Earth represents the manifold. And a chart represents a page with a map of a section of the earth.

In this case it means that we have two overlapping charts, let's call them $\phi$ and $\psi$, that map a subset of $M$ to $\mathbb R^n$.
The chart $\phi$ would map a point $p$ in $M$ to $(x^1,\ldots,x^n)$, which is one set of local coordinates.
And $\psi$ would map $p$ to $(u^1,\ldots,u^n)$, which is another set of coordinates for the same point, but in a different chart.

Peter said:
What would it mean for example if we wanted to form the differentials $$\displaystyle du^i$$ ... ?

$du^i$ is a function $T_pM\to \mathbb R$ just like $dx^i$, where $T_p M$ is the so called tangent space of $M$ at a point $p$.
However, since both $M$ and $T_p M$ are abstract spaces, we need more abstractions to map them to each other properly.

Peter said:
are the coordinate functions $$\displaystyle (x^1, x^2, \ ... \ ... \ , x^n )$$ essentially a basis for M ... (I am assuming the manifold is a vector space)

$M$ is a manifold, which means that it is locally isomorphic to Euclidean space $\mathbb R^n$.
It means that $M$ can be a vector space, but that this is not necessarily the case.
So we can not assume that.
If $M$ would be a vector space, it would be globally isomorphic to Euclidean space.

We can say however that $T_p M$, the tangent space of $M$ at point $p$, is a vector space.

The point $(x^1,\ldots,x^n) \in \mathbb R^n$ is a representation of a point $p \in M$ according to a local chart.
If $M$ would be the vector space $\mathbb R^n$, then we might pick the identity function as a local chart.
Then $(x^1,\ldots,x^n)$ would simply be a point in $M$.
It's just that a manifold is an abstraction of this.

Klaas van Aarsen said:
Hi Peter,

It means that $u_1$ is a function $\mathbb R^n \to \mathbb R$.
And $(u^1,\ldots,u^n)$ is an element of $\mathbb R^n$, just like $(x^1,\ldots,x^n)$.
The function $u_1$ maps $(x^1,\ldots,x^n)$ to $u^1$.

I'm not sure if you're already familiar with the concept of charts. Are you?
The concept is similar to the charts that we have in an atlas of the earth.
The surface of the Earth represents the manifold. And a chart represents a page with a map of a section of the earth.

In this case it means that we have two overlapping charts, let's call them $\phi$ and $\psi$, that map a subset of $M$ to $\mathbb R^n$.
The chart $\phi$ would map a point $p$ in $M$ to $(x^1,\ldots,x^n)$, which is one set of local coordinates.
And $\psi$ would map $p$ to $(u^1,\ldots,u^n)$, which is another set of coordinates for the same point, but in a different chart.
$du^i$ is a function $T_pM\to \mathbb R$ just like $dx^i$, where $T_p M$ is the so called tangent space of $M$ at a point $p$.
However, since both $M$ and $T_p M$ are abstract spaces, we need more abstractions to map them to each other properly.
$M$ is a manifold, which means that it is locally isomorphic to Euclidean space $\mathbb R^n$.
It means that $M$ can be a vector space, but that this is not necessarily the case.
So we can not assume that.
If $M$ would be a vector space, it would be globally isomorphic to Euclidean space.

We can say however that $T_p M$, the tangent space of $M$ at point $p$, is a vector space.

The point $(x^1,\ldots,x^n) \in \mathbb R^n$ is a representation of a point $p \in M$ according to a local chart.
If $M$ would be the vector space $\mathbb R^n$, then we might pick the identity function as a local chart.
Then $(x^1,\ldots,x^n)$ would simply be a point in $M$.
It's just that a manifold is an abstraction of this.
Thanks Klaas ... for an extremely helpful post ...

... and yes ... familiar (but not confident...) with concept of charts ...

Peter

## What is a Rank One Tensor?

A Rank One Tensor is a mathematical object that represents a linear transformation from one vector space to another. It is a type of tensor that can be represented by a single vector and is also known as a first-order tensor.

## What are some examples of Rank One Tensors?

Some examples of Rank One Tensors include force, velocity, and electric field. These are all physical quantities that can be represented by a single vector and have a magnitude and direction.

## How is a Rank One Tensor different from a Scalar or a Vector?

A Scalar is a single number that has magnitude but no direction, while a Vector is a quantity that has both magnitude and direction. A Rank One Tensor is a type of vector that can represent a linear transformation from one vector space to another.

## What are the properties of Rank One Tensors?

One of the main properties of Rank One Tensors is that they are invariant under a change of basis. This means that their components will change, but their overall meaning and physical significance will remain the same.

## How are Rank One Tensors used in scientific research?

Rank One Tensors are used in a variety of scientific fields, including physics, engineering, and mathematics. They are particularly useful in describing physical quantities that have both magnitude and direction, such as forces, velocities, and electric fields. They are also used in tensor calculus to solve complex equations and model physical systems.

• Topology and Analysis
Replies
2
Views
1K
• Topology and Analysis
Replies
2
Views
997
• Topology and Analysis
Replies
26
Views
3K
• Topology and Analysis
Replies
3
Views
2K
• Topology and Analysis
Replies
1
Views
1K
• Topology and Analysis
Replies
6
Views
3K
• Topology and Analysis
Replies
2
Views
2K
• Topology and Analysis
Replies
9
Views
2K
• Topology and Analysis
Replies
2
Views
1K
• Topology and Analysis
Replies
2
Views
2K