The Geometry of the Legendre Transformation (sidetrack into projective duality)

Jan 25—2025

We will study equations of the form $x = t\dot x + f(\dot x)$ (1). And the geometry hidden behind them.

In the (t, x, p)-space, equation (1) defines a surface M. Note that intersection of M with the planes p = C (where C is a constant). are straight lines: $x = Ct + f(C)$ .

we will find the integral curves on M (and then, find the integral curves on the (t, x)-plane).

First, we find the direction field (i.e. the field of traces of the contact planes) on M through the relations:

$x = pt + f(p)$

$dx = pdt + tdp + f'dp$ (condition of belonging to the tangent plane of the surface of M)

$dx = pdt$ (condition of belonging to the contact plane)

We obtain the 1-form $(t + f')dp = 0$ . On the (t, p)-plane, this form describes the structure of a field of lines. Note that a unique line (line parallel to the t-axis) is assigned by the 1-form to each point EXCEPT for the points t = -f’ (these points are assigned a space of lines, i.e. dp can be of any value).

Why does this occur? Looking back at the relations, we see that when t = -f’, the tangent plane coincides exactly with the contact plane. These points are exceptional and the set of these points is the discriminant curve t = -f'(p).

On the surface M, the discriminant curve is described by the relation t = – f'(p) and x = pt + f(p). On the (t, x)-plane, this curve can be described by the parameterization $\gamma(C) = (t = -f'(C), x = Ct + f(C))$ .

On the (x, p)-plane, the integral curves are given by p = C (where C is constant). On the surface M, the integral curves are given by (t, Ct + f(C), C). Projecting the curves onto the (t, x)-plane gives us the integral curves $x = Ct + f(C)$ .

So, what is the relation between the discriminant curve and the integral curves on the (t, x)-plane?

Claim: the integral curves $x = Ct+ f(C)$ are the tangent lines of $\gamma(C)$ .

The velocity vector at each point on $\gamma(C)$ is given by:

$\dot {\gamma}(C) = (-f''(C), \frac{d(-Cf'(C) + f(C))}{dC}$ = (-f”(C), -Cf”(C))$.

Note that the 1-form $dx = Cdt$ vanishes on these vectors. Hence, the family of tangent lines along the curve $\gamma(C)$ can be described by this 1-form. We find at each point (-f'(C), Cf'(C) + f(C)), the tangent line is $x = Ct + f(C)$ .

A Curve and its Family of Tangent Lines

Identifying the plane as the space of all lines (except for the vertical lines) and the dual plane of points:

And also

A shift in perspective – how they all connect together.

The duality between a point and the space of lines centered at the point:

Taking this to the 3-dim. space. Duality between a hyperplane (passing through the origin) and its normal vector (up to multiplication by a non-zero constant):

Arnold’s ODE

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Exploring Legendre Transformations Using Examples – Menagerie of Mathematics

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[…] More details here: https://menagerieofmathematics.wordpress.com/2025/01/25/the-geometry-of-the-legendre-transformation-… […]

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May 19th: the Euclidean Bundle and its Projective Dual – Menagerie of Mathematics

May 20, 2025

[…] Let be an arbitrary point on the red ellipse. It follows that and the associated tangent line at that point is , where . This is true for any point on the red ellipse. By circling around the red ellipse once, the corresponding tangent line also induces a circling around the blue ellipse. The other way applies as well. I discussed more about this in my previous post on the Legendre transformation: https://menagerieofmathematics.wordpress.com/2025/01/25/the-geometry-of-the-legendre-transformation-… […]

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The Geometry of the Legendre Transformation (sidetrack into projective duality)

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