May 19th: the Euclidean Bundle and its Projective Dual

May 20—2025

Recall, from the last post, we have shown that the potential inside a uniformly charged conducting sphere is constant, and so, it follows that the electric field inside must be 0. But, is there a better way for us to visualize this phenomenon? Arnold mentioned that Newton was able to find the electric field inside a sphere, an ellipsoid, and even two similar concentric ellipsoids (oppositely charged) is 0 by balancing forces. Interesting…

The goal of this post and probably the following one is to understand why the field between two oppositely charged similar concentric ellipsoids is 0.

We start by looking into the 2nd example that Arnold provided: the Euclidean bundle and its projective dual.

Given a symmetric matrix $A$ in $\mathbb{R}^n$ and $I$ the identity operator. We define the following 2 forms:

$((A - \lambda I)x, x)$ , a Euclidean bundle, and
$((A - \lambda I)^{-1}x, x)$ , its dual.

Notice, it suffices for us to consider diagonal matrices with eigenvalues $a_1, a_2, ..., a_n$ . This is because we can find an orthogonal eigenbasis corresponding to $A$ , in which the operator becomes diagonal. This simplifies things drastically:

$((A - \lambda I)x, x) = (a_1 - \lambda)x_1^2 + ... (a_n - \lambda)x_n^2$ and

$((A - \lambda I)^{-1}x, x) = \frac{x_1^2}{a_1-\lambda} + ... + \frac{x_n^2}{a_n - \lambda}$ .

For n = 2, I have created a graph showing the two dual curves given various eigenvalues and their change with respect to the changing of $\lambda$ :

https://www.desmos.com/calculator/tmfespq55k

Let’s focus on n = 2 for simplicity. What does it mean for the two curves to be dual? How do we see it?

Answer: The curves are dual in the sense that a point on one curve corresponds to the tangent line of the other curve. Let’s see what this actually means in terms of an example.

And some corresponding pictures of the (point, tangent line) pair I was talking about:

Correspondence in the other direction is also true:

We now use coordinates to describe this phenomenon. In the above example, the red ellipse is given by:

$x^2 + 4y^2 = 1$

and the blue ellipse is given by:

$x^2 + \frac{y^2}{4} = 1$ .

Let $(x_0, y_0)$ be an arbitrary point on the red ellipse. It follows that $x_0^2 + 4y_0^2 = 1$ and the associated tangent line at that point is $X_0x + Y_0y = 1$ , where $X_0^2 + \frac{Y_0^2}{4} = 1$ . 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-sidetrack-into-projective-duality/?preview_id=365&preview_nonce=1b6eba53cd&preview=true