Linear manifolds.

There are many similarities between manifolds and vector spaces. Choosing co-ordinates is much like choosing a basis. It is useful to develop this idea further.

Definition 2.7   Define linear co-ordinates $\psi$ on a set $V$ to be a bijection $\psi\colon V \to \mathbb{R}^n$.

Definition 2.8   Define two sets of linear co-ordinates $\psi$ and $\chi$ to be linearly equivalent if $\psi\circ \chi^{-1}$ is a linear isomorphism.

It is straightforward to prove that linear equivalence is an equivalence relation. We define

Definition 2.9   A linear atlas on a set $V$ is an equivalence class of linear co-ordinates.

Definition 2.10   We define a linear manifold to be a set $V$ with a choice of linear atlas.

We can define an addition and scalar multiplication on $V$ by choosing some linear co-ordinates $\psi$ from the linear atlas and defining

$$av + b w = \psi^{-1}(a \psi(v) + b \psi(w))$$

where $a$ and $b$ are real numbers and $v$ and $w$ are elements of $V$. We have to check that this is well-defined that is it is independent of the choice of $\psi$ from the equivalence class. If $\chi$ is another choice then we have

$$av + b w = \psi^{-1}(a \psi(v) + b \psi(w))$$

$$= \psi^{-1}(a \psi(\chi^{-1}\circ\chi(v)) + b \psi(\chi^{-1}\circ\chi(w)))$$

$$= \psi^{-1}(a (\psi\circ\chi^{-1})(\chi(v)) + b (\psi\circ\chi^{-1})(\chi (w)))$$

$$= \psi^{-1} (\psi\circ\chi^{-1})(a\chi(v) + b \chi (w))$$

$$= \chi^{-1}(a\chi(v) + b \chi (w))$$

where in moving from the third to the fourth lines we use the fact that $\psi\circ \chi^{-1}$ is linear. We have proved.

Proposition 2.2   A linear manifold has a natural vector space structure which makes all of the linear co-ordinates linear isomorphisms.

Because of Proposition 2.2 the theory of linear manifolds is really the theory of vector spaces. However it is an amusing exercise to translate everything in the theory of vector spaces into the linear manifold setting. For example a function $f \colon V \to \mathbb{R}$ is linear if $f \circ \psi^{-1}$ is linear some choice of linear co-ordinates $\psi$. It is then easy to prove that $f \circ \chi^{-1} \colon V \to \mathbb{R}$ for any choice of linear co-ordinates $\chi$. Indeed we just note that

$$f \circ \chi^{-1} = (f \circ \psi^{-1}) \circ (\psi \circ \chi^{-1}).$$