Remark on the Definition of Differentials .... Lafontaine page 5 ....

[Math Amateur]

I am reading the book "An Introduction to Differential Manifolds" (Springer) by Jacques Lafontaine ...

I am currently focused on Chapter 1: Differential Calculus ...

I need help in order to fully understand a remark by Lafontaine following his definition of differentials ...

Lafontaine's definition of differentials followed by the remark in question read as follows:View attachment 8514
View attachment 8515
At the start of the above remark, Lafontaine writes the following:"We can rewrite the definition in the form \(\displaystyle \overrightarrow{ f(a) f(x) } = L \cdot \vec{ax} + o( \vec{ax} )\) ... ... "
Can someone please explain (simply and in detail) how Lafontaine's definition can be rewritten in the form \(\displaystyle \overrightarrow{ f(a) f(x) } = L \cdot \vec{ax} + o( \vec{ax} )\)
Hope someone can help ...

Peter

 

[Evgeny.Makarov]

In any affine space, including \(\displaystyle \mathbb{R}^n\), there is an operation of plotting a vector \(\displaystyle \vec{v}\) from point \(\displaystyle P\), denoted by \(\displaystyle P+\vec{v}\). The result is another point, which is the end of the vector. This operation has the following properties:

1. \(\displaystyle P+\vec{0}=P\).
2. \(\displaystyle (P+\vec{u})+\vec{v}=P+(\vec{u}+\vec{v})\).
3. For any points \(\displaystyle P\) and \(\displaystyle Q\) there is a unique vector \(\displaystyle \vec{v}\) such as \(\displaystyle P+\vec{v}=Q\). This \(\displaystyle \vec{v}\) is denoted by \(\displaystyle \overrightarrow{PQ}\).

It follows from uniqueness that \(\displaystyle Q=P+\vec{v}\) implies \(\displaystyle \vec{v}=\overrightarrow{PQ}\). In the book $x$ is defined as $a+h$ (plotting vector $h$ from point $a$), so $h=\overrightarrow{ax}$.

 

[Math Amateur]

In any affine space, including \(\displaystyle \mathbb{R}^n\), there is an operation of plotting a vector \(\displaystyle \vec{v}\) from point \(\displaystyle P\), denoted by \(\displaystyle P+\vec{v}\). The result is another point, which is the end of the vector. This operation has the following properties:

1. \(\displaystyle P+\vec{0}=P\).
2. \(\displaystyle (P+\vec{u})+\vec{v}=P+(\vec{u}+\vec{v})\).
3. For any points \(\displaystyle P\) and \(\displaystyle Q\) there is a unique vector \(\displaystyle \vec{v}\) such as \(\displaystyle P+\vec{v}=Q\). This \(\displaystyle \vec{v}\) is denoted by \(\displaystyle \overrightarrow{PQ}\).

It follows from uniqueness that \(\displaystyle Q=P+\vec{v}\) implies \(\displaystyle \vec{v}=\overrightarrow{PQ}\). In the book $x$ is defined as $a+h$ (plotting vector $h$ from point $a$), so $h=\overrightarrow{ax}$.

Thanks for the help Evgeny ...

Still reflecting over this ... and still a bit puzzled ...

Can you show explicitly how to derive \(\displaystyle \overrightarrow{ f(a) f(x) } = L \cdot \vec{ax} + o( \vec{ax} )\)That, I think, would help a lot ...Thanks again,

Peter

 

[HOI]

Do you understand the distinction between a "vector space" and an "affine space"? In a vector space we have a defined "zero vector" while in an "affine space" we do not. Think of it as the distinction between a plane with a given coordinate system and just a plane. In an affine space we can designate a point "a" and then talk of the "vector" from "a" to any point, so that we have a vector space with "a" being the zero vector.

 

[Evgeny.Makarov]

Can you show explicitly how to derive

\(\displaystyle \overrightarrow{ f(a) f(x) } = L \cdot \vec{ax} + o( \vec{ax} )\)

The definition is
\[
f(a+h)=f(a)+Lh+o(h).
\]
Here $a$ and $f(a+h)$ are points, $h$ and $Lh$ are vectors and $+$ is the operation of drawing a vector from a given point, whose result is the endpoint of the vector. The function $o(h)$ also returns a vector whose magnitude divided by the magnitude of $h$ tends to $0$ as $|h|\to 0$.

Define $x=a+h$. Then $h=\overrightarrow{ax}$ from axiom 3 of affine spaces. So
\[
f(x)=f(a)+L\overrightarrow{ax}+o(\overrightarrow{ax}).
\]
Again by axiom 3
\[
\overrightarrow{f(a)f(x)}=L\overrightarrow{ax}+o(\overrightarrow{ax}).
\]

In a vector space we have a defined "zero vector" while in an "affine space" we do not.

I wouldn't put it quite like that. An affine space is a triple $(A,V,{+})$ where $A$ is a set of points, $V$ is a vector space, and $+$ satisfies the three axioms in post #2. So if we say that $A$ does not have a zero vector, this is true because $A$ does not contain any vector, just points. On the other hand, $V$ certainly contains a zero vector.

In an affine space we can designate a point "a" and then talk of the "vector" from "a" to any point, so that we have a vector space with "a" being the zero vector.

Yes, but again, I wouldn't mix points and vectors: they are objects of different nature. Any point can be designated as an origin, and then there is a 1-1 correspondence between points and vectors from the origin to those points.

One does not need to know the precise definition of affine spaces to understand how the remark in the book follows from the definition. It is sufficient to have high school understanding of vectors.