The terms "parabolic," "hyperbolic" and "elliptic" are used to classify certain differential equations. The terms "hyperbolic" and "elliptic" are also used to describe certain geometries. Is there a connection between these usages, and, if so, what is it?

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    $\begingroup$ Related $\endgroup$ Oct 19, 2017 at 13:44
  • $\begingroup$ It has to do with the characteristic curves of the equation, as an intuitive example, the wave equation has two families of nonparallel lines (degenerated hyperbola) and the heat equation has one family of lines (degenerated parabola). But I would like to read a more developed answer as well. $\endgroup$
    – Koto
    Oct 19, 2017 at 13:51
  • $\begingroup$ For “elliptic” in particular I had asked a similar question in the past, albeit a bit broader by taking other uses into account as well. $\endgroup$
    – MvG
    Oct 28, 2017 at 7:15
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    $\begingroup$ In my mind the elliptic/parabolic/hyperbolic language all stems from conic sections - the conic $a^{ij}x_i x_j + b^i x_i = 0$ and the PDE $a^{ij} \partial_i \partial_j u + b^i \partial_i u =0$ are classified near identically. Elliptic/hyperbolic geometries have models built from circles/hyperbolae, and volume growths given by circular/hyperbolic trigonometric functions. $\endgroup$ Oct 28, 2017 at 7:22
  • $\begingroup$ Related: math.stackexchange.com/q/4294133/169085 $\endgroup$
    – Alp Uzman
    May 7 at 10:02

1 Answer 1


I have just added an entry on my blog to explain this.

Let's consider the space $\mathbb{M}^4$, that is, the manifold $\mathbb{R}^4$ with coordinates $(x,y,z,t)$ but with the pseudo-Riemannian metric given in each tangent space by $$ ds^2=dx^2+dy^2+dz^2-dt^2. $$ This metric is left invariant by the group $O(3,1)$ when acting on $\mathbb{M}^4$. It also leaves invariant the cone $$ t^2-x^2-y^2-z^2=0 $$ so is logical that we pay attention to it.

Now, following TRTR from Penrose page 423, we can consider the family of hyperplanes inside $\mathbb{M}^4$ given by $$ z+t+\lambda(t-z)=2 $$ intersecting our cone in different 2-dimensional submanifolds $P_{\lambda}$ of $\mathbb{M}^4$.

Here we have a picture from that book, where we consider only $(x,z,t)$ for purposes of drawability.

enter image description here

If we analyse the case $\lambda=0$, we observe that we obtain the submanifold $P_0$, $E$ in the picture with a shape of a parabola, given by the embedding

\begin{align*} \phi \colon \mathbb{R}^2 &\longrightarrow \mathbb{M}^4\\ (u_1,u_2) &\mapsto (u_1, u_2, 1-\frac{u_1^2+u_2^2}{4},1+\frac{u_1^2+u_2^2}{4}) \end{align*}

We can look for the (possibly pseudo) metric inherited by $P_0$. Since $$ d\phi=\begin{pmatrix} 1 & 0 \\ 0 & 1\\ -\frac{u_1}{2} &-\frac{u_2}{2}\\ \frac{u_1}{2} &\frac{u_2}{2} \end{pmatrix} $$ we conclude that $P_0$ consists of $\mathbb{R}^2$ with the euclidean metric: $$ g=\begin{pmatrix} 1 & 0 \\ 0 & 1\\ \end{pmatrix} $$

That is, although we see it like a parabola, intrinsically is only the Euclidean plane, and that is the reason the latter is sometimes called parabolic geometry.

But, what happens for others $\lambda$? Let's see. We have a far more complicated parametrization of our embedded manifold: $$ \phi(u_1,u_2)=(u_1,u_2, \frac{-1+{ \lambda }+\sqrt{-(1+{ \lambda })^{2}\left(-1+{\lambda}\left(u_1^{2}+u_2^{2}\right)\right)}}{2{ \lambda }}, $$ $$ \frac{2+{ \lambda }+\sqrt{-(1+{\lambda })^{2}\left(-1+{\lambda}\left(u_1^{2}+u_2^{2}\right)\right)+\frac{1-\sqrt{-(1+{\lambda})^{2}\left(-1+{\lambda}\left(u_1^{2}+u_2^{2}\right)\right)}}{{ \lambda }}}}{2(1+{ \lambda })}) $$ (computations made with Mathematica).

If we compute the (possibly pseudo) metric in this chart we obtain the inherited metric in $\mathbb{R}^2$

$$ g_{\lambda}=\begin{pmatrix} \frac{1-{ \lambda } u_2^{2}}{1-{ \lambda }\left(u_1^{2}+u_2^{2}\right)} & \frac{\lambda u_1 u_2}{1-{ \lambda }\left(u_1^{2}+u_2^{2}\right)} \\ \frac{\lambda u_1 u_2}{1-{ \lambda }\left(u_1^{2}+u_2^{2}\right)} & \frac{1-{ \lambda } u_1^{2}}{1-{ \lambda }\left(u_1^{2}+u_2^{2}\right)}\\ \end{pmatrix} $$

Next step is to show that for $\lambda>0$ we have an isometry into the usual sphere, and that for $\lambda<0$ we have an isometry into the hyperbolic space. This would justify the terms elliptic and hyperbolic for those geometries, as you can see at the picture above ($S$ and $H$ respectively). In order to keep it simple we can work with $\lambda=1$ and $\lambda=-1$.

For $\lambda=1$, we have $$ g=\begin{pmatrix} \frac{1- u_2^{2}}{1-\left(u_1^{2}+u_2^{2}\right)} & \frac{ u_1 u_2}{1-\left(u_1^{2}+u_2^{2}\right)} \\ \frac{u_1 u_2}{1-\left(u_1^{2}+u_2^{2}\right)} & \frac{1- u_1^{2}}{1-\left(u_1^{2}+u_2^{2}\right)}\\ \end{pmatrix} $$

But if we look at a typical chart of the usual sphere in $\mathbb{E}^3$ $$ \psi: (x,y)\longmapsto (x,y,\sqrt{1-x^2-y^2}) $$ and compute the metric we obtain just $$ g=\begin{pmatrix} \frac{1- y^{2}}{1-\left(x^{2}+y^{2}\right)} & \frac{ xy}{1-\left(x^{2}+y^{2}\right)} \\ \frac{xy}{1-\left(x^{2}+y^{2}\right)} & \frac{1- x^{2}}{1-\left(x^{2}+y^{2}\right)}\\ \end{pmatrix} $$ so they are the same.

And for $\lambda=-1$ we have $$ g=\begin{pmatrix} \frac{1+ u_2^{2}}{1+u_1^{2}+u_2^{2}} & \frac{ -u_1 u_2}{1+u_1^{2}+u_2^{2}} \\ \frac{-u_1 u_2}{1+u_1^{2}+u_2^{2}} & \frac{1+ u_1^{2}}{1+u_1^{2}+u_2^{2}}\\ \end{pmatrix} $$

But if we look at the typical model for hyperbolic plane that consists of the pseudosphere embedded in $\mathbb{M}^3$: $$ x^2+y^2-z^2=-1 $$ with a chart given by $$ \psi: (x,y)\longmapsto (x,y,\sqrt{1+x^2+y^2}) $$ we obtain the inherited metric in $\mathbb{R}^2$ (keep an eye: inherited from the Minkowski metric, not the Euclidean one): $$ g=\begin{pmatrix} \frac{1+ y^{2}}{1+x^{2}+y^{2}} & \frac{ -xy}{1+x^{2}+y^{2}} \\ \frac{-xy}{1+x^{2}+y^{2}} & \frac{1+ x^{2}}{1+x^{2}+y^{2}}\\ \end{pmatrix} $$ so we are done.


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