I'm not a mathematician (I am a cartographer) and I want to know wether all equal area map projections are local symplectomorphisms or not. It is no doubt that the surfaces of a 2-sphere and Earth ellipsoid are symplectic manifolds and that all cartographic projections are local diffeomorphisms, but I'm not sure about retaining symplectic form by all equal area (azimuthal, conic, cylindrical, pseudoazimuthal, pseudoconical, pseudocylindrical) map projections. If not all of them are local symplectomorphisms, please, can you name those projections, which are not local symplectomorphisms.
HERE IS THE ANSWER TO A SIMILAR QUESTION
link to that question Why is an area preserving diffeomorphism a symplectomorphism (in $R^2$)
There are two different notions to define a volume, volume forms i.e. nowhere vanishing $n$ forms on an $n$ dimensional manifold, and measure from measure theory. If $\omega$ is a symplectic form on $M$ of dimension $2n$, then $\mathrm{Vol} = \frac{\omega^n}{n!}$ is a volume form on $M$. On $M = \mathbb{R}^{2n}$ endowed with its canonical symplectic form $\omega = \sum_{k = 1}^n dx_k \wedge dy_k$, we obtain the canonical volume form $\mathrm{Vol} = dx_1 \wedge dy_1 \wedge \cdots \wedge dx_n \wedge dy_n$ (this is why we divide by $n!$).
A symplectomorphism is a diffeomorphism $f$ such that $f^*\omega = \omega$ i.e.
$$
\forall x \in M, \forall (u,v) \in T_xM^2, \omega_{f(x)}(df(x)u,df(x)v) = \omega_x(u,v),
$$
and a volume preserving diffeomorphism is a diffeomorphism $f$ such that $f^*\mathrm{Vol} = \mathrm{Vol}$ i.e.
$$
\forall x \in M, \forall (u_1,\ldots,u_m) \in T_xM^m, \omega_{f(x)}(df(x)u_1,\ldots,df(x)u_m) = \omega_x(u_1,\ldots,u_m),
$$
where $m = \dim(M)$ ($= 2n$ in the symplectic case). In the case where $M$ is a symplectic surface, we simply have $\mathrm{Vol} = \omega$ hence the equivalence is trivial. In the general case, we only have symplectic $\Rightarrow$ volume preserving.
But if the question is "what is the link between $f^*\mathrm{Vol} = \mathrm{Vol}$ and $f$ preaserving volume in the measure theory sens ?" , then in that case, on any manifold $M$ equipped with a volume form $\mathrm{Vol}$ (whether or not it comes from a symplectic form), we can define a measure $\mu$ with,
$$
\mu(A) = \int_M \mathbf{1}_A \mathrm{Vol},
$$
on the open sets, extended to the Borel sets. In the case where $M = \mathbb{R}^n$ endowed with its canonical volume form, we obtain the Lebesgue measure. And we can prove that for all diffeomorphism $f$,
$$
f^*\mathrm{Vol} = \mathrm{Vol} \quad \Leftrightarrow \quad \forall A \subset M \textrm{ Borel}, \mu(f(A)) = \mu(A) \textrm{ and $f$ preserves orientation}.
$$
Indeed, assuming the left hand side $f$ preserves the orientation (trivial) and we have for all $A$ open (it is enough to show the right hand side for open sets),
\begin{align*}
\mu(f(A)) & = \int_M \mathbf{1}_{f(A)}\mathrm{Vol}\\
& = \int_M f^*\mathbf{1}_Af^*\mathrm{Vol} \textrm{ because $f^*\mathrm{Vol} = \mathrm{Vol}$,}\\
& = \int_M f^*(\mathbf{1}_A\mathrm{Vol})\\
& = \int_M \mathbf{1}_A\mathrm{Vol}\\
& = \mu(A).
\end{align*}
For the reciprocal, we may assume that $M = \mathbb{R}^m$ and the volume form is given by $\mathrm{Vol}_x = g(x)\det$ for some smooth $g : \mathbb{R}^m \rightarrow \mathbb{R}_+^*$ since we want to prove a local statement. Then, $\mu = g\lambda$ where $\lambda$ is the Lebesgue measure (in your case, $g = 1$). Let $x_0 \in \mathbb{R}^m$ and $B_\varepsilon$ the ball of center $x_0$ and radius $\varepsilon$ for all $\varepsilon > 0$. We have,
\begin{align*}
\mu(f^{-1}(B_\varepsilon)) & = \int_{f^{-1}(B_\varepsilon)} g(x) \, d\lambda\\
& = \int_{B_\varepsilon} g(f(x))\det(df(x)) \, d\lambda\\
& \textrm{ by change of variable ($\det(df(x)) > 0$ because $f$ preserves orientation),}\\
& \sim g(f(x_0))\det(df(x_0))\lambda(B_\varepsilon),
\end{align*}
when $\varepsilon \rightarrow 0$. Similarly, $\mu(B_\varepsilon) \sim g(x_0)\lambda(B_\varepsilon)$. We deduce that $g(f(x_0))\det(df(x_0)) = g(x_0)$, which implies $(f^*\mathrm{Vol})_{x_0} = \mathrm{Vol}_{x_0}$, and it is true for all $x_0$ thus $f^*\mathrm{Vol} = \mathrm{Vol}$. Notice that in the case where $f$ is volume preserving in the measure sens by $f$ reverses orientation, then $f^*\mathrm{Vol} = -\mathrm{Vol}$ so the two notion are not exactly identical.
SO MY QUESTION IS: Does it mean, that all equal area map projections are local symplectomorphisms?
Best Answer
The answer is YES. In the case of surfaces ($\mathbb{R}^2$) an area preserving local diffeomorphism $f: \mathbb{R}^2 \rightarrow \mathbb{R}^2$ defines a local symplectomorphism because $\mathrm{Vol} = \omega$ in the case of $ \mathbb{R}^2\ $. Therefore, all equal area map projections are local symplectomorphisms.